Displaying 1-10 of 45 results found.
Pisano period of the 5-Fibonacci numbers, A052918 preceded by 0.
+20
9
1, 3, 8, 6, 2, 24, 6, 12, 8, 6, 24, 24, 12, 6, 8, 24, 36, 24, 40, 6, 24, 24, 22, 24, 10, 12, 8, 6, 116, 24, 64, 48, 24, 36, 6, 24, 76, 120, 24, 12, 28, 24, 88, 24, 8, 66, 96, 24, 42, 30, 72, 12, 52, 24, 24, 12, 40, 348, 58, 24, 124, 192, 24, 96, 12, 24, 66, 36, 88, 6, 70, 24, 148
COMMENTS
Period of the sequence defined by reading A052918 modulo n.
MAPLE
F := proc(k, n) option remember; if n <= 1 then n; else k*procname(k, n-1)+procname(k, n-2) ; end if; end proc:
Pper := proc(k, m) local cha, zer, n, fmodm ; cha := [] ; zer := [] ; for n from 0 do fmodm := F(k, n) mod m ; cha := [op(cha), fmodm] ; if fmodm = 0 then zer := [op(zer), n] ; end if; if nops(zer) = 5 then break; end if; end do ; if [op(1..zer[2], cha) ] = [ op(zer[2]+1..zer[3], cha) ] and [op(1..zer[2], cha)] = [ op(zer[3]+1..zer[4], cha) ] and [op(1..zer[2], cha)] = [ op(zer[4]+1..zer[5], cha) ] then return zer[2] ; elif [op(1..zer[3], cha) ] = [ op(zer[3]+1..zer[5], cha) ] then return zer[3] ; else return zer[5] ; end if; end proc:
k := 5 ; seq( Pper(k, m), m=1..80) ;
MATHEMATICA
Table[s = t = Mod[{0, 1}, n]; cnt = 1; While[tmp = Mod[5*t[[2]] + t[[1]], n]; t[[1]] = t[[2]]; t[[2]] = tmp; s!= t, cnt++]; cnt, {n, 100}] (* Vincenzo Librandi, Dec 20 2012, after T. D. Noe *)
Squares of A052918(n-1) (generalized Fibonacci).
+20
8
0, 1, 25, 676, 18225, 491401, 13249600, 357247801, 9632441025, 259718659876, 7002771375625, 188815108482001, 5091005157638400, 137268324147754801, 3701153746831741225, 99793882840309258276, 2690733682941518232225
COMMENTS
See the comment in A099279. This is example a=5. a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using half-squares (1/2 X 1 pieces, always placed so that the shorter sides are horizontal) and (1/2,1/2)-fences if there are 5 kinds of half-squares available. A (w,g)-fence is a tile composed of two w X 1 pieces separated horizontally by a gap of width g. a(n+1) also equals the number of tilings of an n-board using (1/4,1/4)-fences and (1/4,3/4)-fences if there are 5 kinds of (1/4,1/4)-fences available. - Michael A. Allen, Mar 30 2023
FORMULA
a(n) = A052918(n-1)^2, n >= 1, a(0) = 0. a(n) = 26*a(n-1) + 26*a(n-2) - a(n-3), n >= 3; a(0)=0, a(1)=1, a(2)=25.
a(n) = 27*a(n-1) - a(n-2) - 2*(-1)^n, n >= 2; a(0)=0, a(1)=1.
a(n) = 2*(T(n, 27/2) - (-1)^n)/29 with twice the Chebyshev's T(n, x) polynomials of the first kind. 2*T(n, 27/2) = A090248(n). G.f.: x*(1-x)/((1-27*x+x^2)*(1+x)) = x*(1-x)/(1-26*x-26*x^2+x^3).
a(n) = (1 - (-1)^n)/2 + 25*Sum_{r=1..n-1} r*a(n-r). - Michael A. Allen, Mar 30 2023
MAPLE
with (combinat):seq(fibonacci(n, 5)^2, n=0..16); # Zerinvary Lajos, Apr 09 2008
MATHEMATICA
LinearRecurrence[{26, 26, -1}, {0, 1, 25}, 30] (* Harvey P. Dale, Sep 25 2019 *)
PROG
(Magma) [(2/29)*(Evaluate(ChebyshevFirst(n), 27/2) -(-1)^n): n in [0..30]]; // G. C. Greubel, Aug 21 2022 (SageMath)
def A099365(n): return (2/29)*(chebyshev_T(n, 27/2) - (-1)^n)
Odd composite integers m such that A052918(m-J(m,29)) == 0 (mod m) and gcd(m,29)=1, where J(m,29) is the Jacobi symbol.
+20
6
9, 15, 27, 45, 91, 121, 135, 143, 1547, 1573, 1935, 2015, 6543, 6721, 8099, 10403, 10877, 10905, 13319, 13741, 13747, 14399, 14705, 16109, 16471, 18901, 19043, 19109, 19601, 19951, 20591, 22753, 24639, 26599, 26937, 27593
COMMENTS
The generalized Lucas sequences of integer parameters (a,b) defined by U(m+2)=a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfy the identity
U(p-J(p,D)) == 0 (mod p) when p is prime, b=-1 and D=a^2+4.
This sequence contains the odd composite integers with U(m-J(m,D)) == 0 (mod m).
For a=5 and b=-1, we have D=29 and U(m) recovers A052918(m). If even numbers greater than 2 that are coprime to 29 are allowed, then 26, 442, 6994, ... would also be terms. - Jianing Song, Jan 09 2021
REFERENCES
D. Andrica and O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
MATHEMATICA
Select[Range[3, 28000, 2], CoprimeQ[#, 29] && CompositeQ[#] && Divisible[Fibonacci[#-JacobiSymbol[#, 29], 5], #] &]
CROSSREFS
Cf. A052918, A071904, A081264 (a=1, b=-1), A327653 (a=3, b=-1), A340096 (a=7, b=-1), A340097 (a=3, b=1), A340098 (a=5, b=1), A340099 (a=7, b=1).
EXTENSIONS
Coprime condition added to definition by Georg Fischer, Jul 20 2022
Odd composite integers such that A052918(m-1)^2 == 1 (mod m).
+20
4
9, 15, 25, 27, 35, 45, 65, 75, 91, 121, 135, 143, 175, 225, 275, 325, 385, 455, 533, 595, 615, 675, 935, 1035, 1107, 1325, 1359, 1431, 1495, 1547, 1573, 1935, 2015, 2255, 2275, 2775, 3025, 3059, 3575, 3605, 4025, 4081, 4235, 4355, 5005, 5089, 5475, 5525, 5719, 5993, 6165
COMMENTS
If p is a prime, then A052918(p-1)^2 == 1 (mod p). This sequence contains the odd composite integers for which the congruence holds.
The generalized Lucas sequence of integer parameters (a,b) defined by U(n+2) = a*U(n+1)-b*U(n) and U(0)=0, U(1)=1, satisfies the identity U^2(p) == 1 (mod p) whenever p is prime and b=-1,1 (this property is a form of pseudoprimality).
For a=5, b=-1, U(n) recovers A052918(n-1), for n=1,2,....
REFERENCES
D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (to appear, 2020).
MATHEMATICA
Select[Range[3, 20000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 5]*Fibonacci[#, 5] - 1, #] &]
Odd composite integers m such that A052918(2*m-J(m,29)) == 1 (mod m), where J(m,29) is the Jacobi symbol.
+20
4
9, 15, 25, 27, 45, 75, 91, 121, 125, 135, 143, 147, 175, 225, 275, 325, 375, 441, 483, 625, 675, 735, 755, 1125, 1323, 1547, 1573, 1875, 1935, 2015, 2205, 2275, 2485, 2943, 3025, 3125, 3375, 3575, 3675, 3775, 3843, 4375, 5525, 5625, 5819, 6543, 6615, 6721
COMMENTS
The generalized Lucas sequences of integer parameters (a,b) defined by U(m+2)=a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfy U(2*p-J(p,D)) == 1 (mod p) whenever p is prime, k is a positive integer, b=-1 and D=a^2+4.
The composite integers m with the property U(k*m-J(m,D)) == U(k-1) (mod m) are called generalized Lucas pseudoprimes of level k- and parameter a. Here b=-1, a=5, D=29 and k=2, while U(m) is A052918(m).
REFERENCES
D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021).
D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted).
MATHEMATICA
Select[Range[3, 15000, 2], CoprimeQ[#, 29] && CompositeQ[#] && Divisible[Fibonacci[2*#-JacobiSymbol[#, 29], 5] - 1, #] &]
Odd composite integers m such that A052918(3*m-J(m,29)) == 5 (mod m), where J(m,29) is the Jacobi symbol.
+20
3
9, 27, 33, 35, 65, 81, 99, 121, 221, 243, 297, 363, 513, 585, 627, 705, 729, 891, 1089, 1539, 1541, 1881, 2145, 2187, 2299, 2673, 3267, 3605, 4181, 4573, 4579, 5265, 5633, 6721, 6993, 7865, 8019, 8979, 9131, 9801, 10307, 10877, 10881, 13333, 13741, 14001, 14705, 14989
COMMENTS
The generalized Lucas sequences of integer parameters (a,b) defined by U(m+2)=a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfy U(3*p-J(p,D)) == a (mod p) whenever p is prime, k is a positive integer, b=-1 and D=a^2+4.
The composite integers m with the property U(k*m-J(m,D)) == U(k-1) (mod m) are called generalized Lucas pseudoprimes of level k- and parameter a.
Here b=-1, a=5, D=29 and k=3, while U(m) is A052918(m).
REFERENCES
D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer, 2020.
D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021).
D. Andrica, O. Bagdasar, On generalized pseudoprimality of level k (submitted).
MATHEMATICA
Select[Range[3, 15000, 2], CoprimeQ[#, 29] && CompositeQ[#] && Divisible[Fibonacci[3*#-JacobiSymbol[#, 29], 5] - 5, #] &]
Even composite positive integers m such that A052918(m-1)^2 == 1 (mod m).
+20
0
4, 8, 16, 32, 68, 248, 268, 544, 1328, 4216, 4768, 9112, 9376, 12664, 20128, 22112, 24536, 25544, 30488, 43262, 61574, 125792, 148004, 304792, 398248, 493646, 648848, 913456, 1036664, 1975784, 2350792, 3672454, 4248488, 5422688, 6318188, 6768928, 7079656, 8560724
COMMENTS
If p is a prime, then A052918(p-1)^2 == 1 (mod p). This sequence contains the even composite integers for which the congruence holds.
The generalized Lucas sequence of integer parameters (a,b) defined by U(m+2) = a*U(m+1)-b*U(m) and U(0)=0, U(1)=1, satisfies the identity U^2(p) == 1 (mod p) whenever p is prime and b=-1,1.
For a=5, b=-1, U(m) recovers A052918(m-1), for m=1,2,....
REFERENCES
D. Andrica, O. Bagdasar, Recurrent Sequences: Key Results, Applications and Problems. Springer (2020)
D. Andrica, O. Bagdasar, On some new arithmetic properties of the generalized Lucas sequences, Mediterr. J. Math. (to appear, 2021)
MATHEMATICA
Select[Range[2, 20000, 2], CompositeQ[#] && Divisible[Fibonacci[#, 5]*Fibonacci[#, 5] - 1, #] &]
Pell numbers: a(0) = 0, a(1) = 1; for n > 1, a(n) = 2*a(n-1) + a(n-2).
(Formerly M1413 N0552)
+10
762
0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741, 13860, 33461, 80782, 195025, 470832, 1136689, 2744210, 6625109, 15994428, 38613965, 93222358, 225058681, 543339720, 1311738121, 3166815962, 7645370045, 18457556052, 44560482149, 107578520350, 259717522849
COMMENTS
Sometimes also called lambda numbers.
Also denominators of continued fraction convergents to sqrt(2): 1, 3/2, 7/5, 17/12, 41/29, 99/70, 239/169, 577/408, 1393/985, 3363/2378, 8119/5741, 19601/13860, 47321/33461, 114243/80782, ... = A001333/ A000129. Number of lattice paths from (0,0) to the line x=n-1 consisting of U=(1,1), D=(1,-1) and H=(2,0) steps (i.e., left factors of Grand Schroeder paths); for example, a(3)=5, counting the paths H, UD, UU, DU and DD. - Emeric Deutsch, Oct 27 2002 a(2*n) with b(2*n) := A001333(2*n), n >= 1, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = +1 (see Emerson reference). a(2*n+1) with b(2*n+1) := A001333(2*n+1), n >= 0, give all (positive integer) solutions to Pell equation b^2 - 2*a^2 = -1. Bisection: a(2*n+1) = T(2*n+1, sqrt(2))/sqrt(2) = A001653(n), n >= 0 and a(2*n) = 2*S(n-1,6) = 2* A001109(n), n >= 0, with T(n,x), resp. S(n,x), Chebyshev's polynomials of the first, resp. second kind. S(-1,x)=0. See A053120, resp. A049310. - Wolfdieter Lang, Jan 10 2003 Consider the mapping f(a/b) = (a + 2b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 3/2, 7/5, 17/12, 41/29, ... converging to 2^(1/2). Sequence contains the denominators. - Amarnath Murthy, Mar 22 2003 This is also the Horadam sequence (0,1,1,2). Limit_{n->oo} a(n)/a(n-1) = sqrt(2) + 1 = A014176. - Ross La Haye, Aug 18 2003 Number of 132-avoiding two-stack sortable permutations.
For n > 0, the number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 2, s(n) = 3.
Number of (s(0), s(1), ..., s(n)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n, s(0) = 1, s(n) = 2. (End)
Counts walks of length n from a vertex of a triangle to another vertex to which a loop has been added. - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, Pisot sequence P(2,5). See A008776 for definition of Pisot sequences. - David W. WilsonThe Pell primality test is "If N is an odd prime, then P(N)-Kronecker(2,N) is divisible by N". "Most" composite numbers fail this test, so it makes a useful pseudoprimality test. The odd composite numbers which are Pell pseudoprimes (i.e., that pass the above test) are in A099011. - Jack Brennen, Nov 13 2004 (0!a(1), 1!a(2), 2!a(3), 3!a(4), ...) and (1,-2,-2,0,0,0,...) form a reciprocal pair under the list partition transform and associated operations described in A133314. - Tom Copeland, Oct 29 2007 Let C = (sqrt(2)+1) = 2.414213562..., then for n > 1, C^n = a(n)*(1/C) + a(n+1). Example: C^3 = 14.0710678... = 5*(0.414213562...) + 12. Let X = the 2 X 2 matrix [0, 1; 1, 2]; then X^n * [1, 0] = [a(n-1), a(n); a(n), a(n+1)]. a(n) = numerator of n-th convergent to (sqrt(2)-1) = 0.414213562... = [2, 2, 2, ...], the convergents being [1/2, 2/5, 5/12, ...]. - Gary W. Adamson, Dec 21 2007 A = sqrt(2) = 2/2 + 2/5 + 2/(5*29) + 2/(29*169) + 2/(169*985) + ...; B = ((5/2) - sqrt(2)) = 2/2 + 2/(2*12) + 2/(12*70) + 2/(70*408) + 2/(408*2378) + ...; A+B = 5/2. C = 1/2 = 2/(1*5) + 2/(2*12) + 2/(5*29) + 2/(12*70) + 2/(29*169) + ... - Gary W. Adamson, Mar 16 2008 Related convergents (numerator/denominator):
Binomial transform of the sequence:= 0,1,0,2,0,4,0,8,0,16,..., powers of 2 alternating with zeros. - Philippe Deléham, Oct 28 2008 a(n) is also the sum of the n-th row of the triangle formed by starting with the top two rows of Pascal's triangle and then each next row has a 1 at both ends and the interior values are the sum of the three numbers in the triangle above that position. - Patrick Costello (pat.costello(AT)eku.edu), Dec 07 2008
Starting with offset 1 = eigensequence of triangle A135387 (an infinite lower triangular matrix with (2,2,2,...) in the main diagonal and (1,1,1,...) in the subdiagonal). - Gary W. Adamson, Dec 29 2008 In general, denominators, a(k,n) and numerators, b(k,n), of continued fraction convergents to sqrt((k+1)/k) may be found as follows:
let a(k,0) = 1, a(k,1) = 2k; for n > 0, a(k,2n) = 2*a(k,2n-1) + a(k,2n-2)
and a(k,2n+1) = (2k)*a(k,2n) + a(k,2n-1);
let b(k,0) = 1, b(k,1) = 2k+1; for n > 0, b(k,2n) = 2*b(k,2n-1) + b(k,2n-2)
and b(k,2n+1) = (2k)*b(k,2n) + b(k,2n-1).
For example, the convergents to sqrt(2/1) start 1/1, 3/2, 7/5, 17/12, 41/29.
In general, if a(k,n) and b(k,n) are the denominators and numerators, respectively, of continued fraction convergents to sqrt((k+1)/k) as defined above, then
k*a(k,2n)^2 - a(k,2n-1)*a(k,2n+1) = k = k*a(k,2n-2)*a(k,2n) - a(k,2n-1)^2 and
b(k,2n-1)*b(k,2n+1) - k*b(k,2n)^2 = k+1 = b(k,2n-1)^2 - k*b(k,2n-2)*b(k,2n);
for example, if k=1 and n=3, then a(1,n) = a(n+1) and
1*a(1,6)^2 - a(1,5)*a(1,7) = 1*169^2 - 70*408 = 1;
1*a(1,4)*a(1,6) - a(1,5)^2 = 1*29*169 - 70^2 = 1;
b(1,5)*b(1,7) - 1*b(1,6)^2 = 99*577 - 1*239^2 = 2;
b(1,5)^2 - 1*b(1,4)*b(1,6) = 99^2 - 1*41*239 = 2.
(End)
Starting with offset 1 = row sums of triangle A155002, equivalent to the statement that the Fibonacci sequence convolved with the Pell sequence prefaced with a "1": (1, 1, 2, 5, 12, 29, ...) = (1, 2, 5, 12, 29, ...). - Gary W. Adamson, Jan 18 2009 It appears that P(p) == 8^((p-1)/2) (mod p), p = prime; analogous to [Schroeder, p. 90]: Fp == 5^((p-1)/2) (mod p). Example: Given P(11) = 5741, == 8^5 (mod 11). Given P(17) = 11336689, == 8^8 (mod 17) since 17 divides (8^8 - P(17)). - Gary W. Adamson, Feb 21 2009 Another combinatorial interpretation of a(n-1) arises from a simple tiling scenario. Namely, a(n-1) gives the number of ways of tiling a 1 X n rectangle with indistinguishable 1 X 2 rectangles and 1 X 1 squares that come in two varieties, say, A and B. For example, with C representing the 1 X 2 rectangle, we obtain a(4)=12 from AAA, AAB, ABA, BAA, ABB, BAB, BBA, BBB, AC, BC, CA and CB. - Martin Griffiths, Apr 25 2009 a(n+1) = 2*a(n) + a(n-1), a(1)=1, a(2)=2 was used by Theon from Smyrna. - Sture Sjöstedt, May 29 2009 The n-th Pell number counts the perfect matchings of the edge-labeled graph C_2 x P_(n-1), or equivalently, the number of domino tilings of a 2 X (n-1) cylindrical grid. - Sarah-Marie Belcastro, Jul 04 2009 As a fraction: 1/79 = 0.0126582278481... or 1/9799 = 0.000102051229...(1/119 and 1/10199 for sequence in reverse). - Mark Dols, May 18 2010 Limit_{n->oo} (a(n)/a(n-1) - a(n-1)/a(n)) tends to 2.0. Example: a(7)/a(6) - a(6)/a(7) = 169/70 - 70/169 = 2.0000845... - Gary W. Adamson, Jul 16 2010 Starting (1, 2, 5, ...) = INVERTi transform of A006190: (1, 3, 10, 33, 109, ...). - Gary W. Adamson, Aug 06 2010 [u,v] = [a(n), a(n-1)] generates all Pythagorean triples [u^2-v^2, 2uv, u^2+v^2] whose legs differ by 1. - James R. Buddenhagen, Aug 14 2010 An elephant sequence, see A175654. For the corner squares six A[5] vectors, with decimal values between 21 and 336, lead to this sequence (without the leading 0). For the central square these vectors lead to the companion sequence A078057. - Johannes W. Meijer, Aug 15 2010 Let the 2 X 2 square matrix A=[2, 1; 1, 0] then a(n) = the (1,1) element of A^(n-1). - Carmine Suriano, Jan 14 2011 Define a t-circle to be a first-quadrant circle tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the t-circle with radius 1. Then for n > 0, define C(n) to be the next larger t-circle which is tangent to C(n - 1). C(n) has radius A001333(2n) + a(2n)*sqrt(2) and each of the coordinates of its point of intersection with C(n + 1) is a(2n + 1) + ( A001333(2n + 1)*sqrt(2))/2. See similar Comments for A001109 and A001653, Sep 14 2005. - Charlie Marion, Jan 18 2012 Pell numbers could also be called "silver Fibonacci numbers", since, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio, while a(n+1) = ceiling(delta*a(n)), if n is even and a(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1+sqrt(2) is the silver ratio. - Vladimir Shevelev, Feb 22 2013 a(n) is the number of compositions (ordered partitions) of n-1 into two sorts of 1's and one sort of 2's. Example: the a(3)=5 compositions of 3-1=2 are 1+1, 1+1', 1'+1, 1'+1', and 2. - Bob Selcoe, Jun 21 2013 Between every two consecutive squares of a 1 X n array there is a flap that can be folded over one of the two squares. Two flaps can be lowered over the same square in 2 ways, depending on which one is on top. The n-th Pell number counts the ways n-1 flaps can be lowered. For example, a sideway representation for the case n = 3 squares and 2 flaps is \\., .//, \./, ./_., ._\., where . is an empty square. - Jean M. Morales, Sep 18 2013 Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A005319(k)*(a(n-2k+1) - a(n-2k)) + a(n-4k) = A075870(k)*(a(n-2k+2) - a(n-2k+1)) - a(n-4k+2). - Charlie Marion, Nov 26 2013 An alternative formulation of the combinatorial tiling interpretation listed above: Except for n=0, a(n-1) is the number of ways of partial tiling a 1 X n board with 1 X 1 squares and 1 X 2 dominoes. - Matthew Lehman, Dec 25 2013 Define a(-n) to be a(n) for n odd and -a(n) for n even. Then a(n) = A077444(k)*a(n-2k+1) + a(n-4k+2). This formula generalizes the formula used to define this sequence. - Charlie Marion, Jan 30 2014 a(n-1) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 1; 1, 1, 1; 0, 1, 1], [0, 1, 1; 0, 1, 1; 1, 1, 1], [0, 1, 0; 1, 1, 1; 1, 1, 1] or [0, 0, 1; 1, 1, 1; 1, 1, 1]. - R. J. Mathar, Feb 03 2014 a(n+1) counts closed walks on K2 containing two loops on the other vertex. Equivalently the (1,1) entry of A^(n+1) where the adjacency matrix of digraph is A=(0,1;1,2). - David Neil McGrath, Oct 28 2014 For n >= 1, a(n) equals the number of ternary words of length n-1 avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015 This is a divisibility sequence (i.e., if n|m then a(n)|a(m)). - Tom Edgar, Jan 28 2015 A strong divisibility sequence, that is, gcd(a(n), a(m)) = a(gcd(n, m)) for all positive integers n and m. - Michael Somos, Jan 03 2017 a(n) is the number of compositions (ordered partitions) of n-1 into two kinds of parts, n and n', when the order of the 1 does not matter, or equivalently, when the order of the 1' does not matter. Example: When the order of the 1 does not matter, the a(3)=5 compositions of 3-1=2 are 1+1, 1+1'=1+1, 1'+1', 2 and 2'. (Contrast with entry from Bob Selcoe dated Jun 21 2013.) - Gregory L. Simay, Sep 07 2017 Number of weak orderings R on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n and for which {1,...,n} has exactly one minimal element for the weak ordering R. - J. Devillet, Sep 28 2017 Also the number of matchings in the (n-1)-centipede graph. - Eric W. Weisstein, Sep 30 2017 Let A(r,n) be the total number of ordered arrangements of an n+r tiling of r red squares and white tiles of total length n, where the individual tile lengths can range from 1 to n. A(r,0) corresponds to a tiling of r red squares only, and so A(r,0)=1. Let A_1(r,n) = Sum_{j=0..n} A(r,j) and let A_s(r,n) = Sum_{j=0..n} A_(s-1)(r,j). Then A_0(1,n) + A_2(3,n-4) + A_4(5,n-8) + ... + A_(2j) (2j+1, n-4j) = a(n) without the initial 0. - Gregory L. Simay, May 25 2018 (1, 2, 5, 12, 29, ...) is the fourth INVERT transform of (1, -2, 5, -12, 29, ...), as shown in A073133. - Gary W. Adamson, Jul 17 2019 Number of 2-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020 Also called the 2-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence. - Michael A. Allen, Jan 23 2023 Named by Lucas (1878) after the English mathematician John Pell (1611-1685). - Amiram Eldar, Oct 02 2023 a(n) is the number of compositions of n when there are F(i) parts of size i, with i,n >= 1, F(n) the Fibonacci numbers, A000045(n) (see example below). - Enrique Navarrete, Dec 15 2023
REFERENCES
J. Austin and L. Schneider, Generalized Fibonacci sequences in Pythagorean triple preserving sequences, Fib. Q., 58:1 (2020), 340-350.
P. Bachmann, Niedere Zahlentheorie (1902, 1910), reprinted Chelsea, NY, 1968, vol. 2, p. 76.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 941.
J. M. Borwein, D. H. Bailey, and R. Girgensohn, Experimentation in Mathematics, A K Peters, Ltd., Natick, MA, 2004. x+357 pp. See p. 53.
John Derbyshire, Prime Obsession, Joseph Henry Press, 2004, see p. 16.
S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.1.
Shaun Giberson and Thomas J. Osler, Extending Theon's Ladder to Any Square Root, Problem 3858, Elementa, No. 4 1996.
R. P. Grimaldi, Ternary strings with no consecutive 0's and no consecutive 1's, Congressus Numerantium, 205 (2011), 129-149.
Thomas Koshy, Pell and Pell-Lucas Numbers with Applications, Springer, New York, 2014.
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 43.
J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 224.
Manfred R. Schroeder, "Number Theory in Science and Communication", 5th ed., Springer-Verlag, 2009, p. 90.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 34.
D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 62.
LINKS
Elena Barcucci, Antonio Bernini, and Renzo Pinzani, A Gray code for a regular language, Semantic Sensor Networks Workshop 2018, CEUR Workshop Proceedings (2018) Vol. 2113. M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28. A. F. Horadam, Pell Identities, Fib. Quart., Vol. 9, No. 3, 1971, pp. 245-252, 263. Shirley Law, Hopf Algebra of Sashes, in FPSAC 2014, Chicago, USA; Discrete Mathematics and Theoretical Computer Science (DMTCS) Proceedings, 2014, 621-632. Eric Weisstein's World of Mathematics, Matching.
FORMULA
G.f.: x/(1 - 2*x - x^2). - Simon Plouffe in his 1992 dissertation. G.f.: Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (2*k + x)/(1 + 2*k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 1 + k)/(1 + k*x) ) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (x + 3 - k)/(1 - k*x) ) may all be proved using telescoping series. - Peter Bala, Jan 04 2015 a(n) = 2*a(n-1) + a(n-2), a(0)=0, a(1)=1.
a(n) = ((1 + sqrt(2))^n - (1 - sqrt(2))^n)/(2*sqrt(2)).
For initial values a(0) and a(1), a(n) = ((a(0)*sqrt(2)+a(1)-a(0))*(1+sqrt(2))^n + (a(0)*sqrt(2)-a(1)+a(0))*(1-sqrt(2))^n)/(2*sqrt(2)). - Shahreer Al Hossain, Aug 18 2019 a(n) = Sum_{i, j, k >= 0: i+j+2k = n} (i+j+k)!/(i!*j!*k!).
a(n)^2 + a(n+1)^2 = a(2n+1) (1999 Putnam examination).
a(n) = ((-i)^(n-1))*S(n-1, 2*i), with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x)=0, S(-2, x)= -1. Binomial transform of expansion of sinh(sqrt(2)x)/sqrt(2). E.g.f.: exp(x)sinh(sqrt(2)x)/sqrt(2). - Paul Barry, May 09 2003 a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2k+1)*2^k. - Paul Barry, May 13 2003 Unreduced g.f.: x(1+x)/(1 - x - 3x^2 - x^3); a(n) = a(n-1) + 3*a(n-2) + a(n-2). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*2^(n-2k). - Mario Catalani (mario.catalani(AT)unito.it), Jul 23 2004
Apart from initial terms, inverse binomial transform of A052955. - Paul Barry, May 23 2004 a(n+1) = Sum_{k=0..n} binomial((n+k)/2, (n-k)/2)*(1+(-1)^(n-k))*2^k/2. - Paul Barry, Aug 28 2005 a(n) = F(n, 2), the n-th Fibonacci polynomial evaluated at x=2. - T. D. Noe, Jan 19 2006 Let F(n) = a(n) = Pell numbers, L(n) = A002203 = companion Pell numbers ( A002203): For a >= b and odd b, F(a+b) + F(a-b) = L(a)*F(b).
For a >= b and even b, F(a+b) + F(a-b) = F(a)*L(b).
For a >= b and odd b, F(a+b) - F(a-b) = F(a)*L(b).
For a >= b and even b, F(a+b) - F(a-b) = L(a)*F(b).
F(n+m) + (-1)^m*F(n-m) = F(n)*L(m).
F(n+m) - (-1)^m*F(n-m) = L(n)*F(m).
F(n+m+k) + (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = F(n)*L(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) + (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = L(n)*L(m)*F(k).
F(n+m+k) + (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) + (-1)^k*F(n-m-k)) = L(n)*F(m)*L(k).
F(n+m+k) - (-1)^k*F(n+m-k) - (-1)^m*(F(n-m+k) - (-1)^k*F(n-m-k)) = 8*F(n)*F(m)*F(k). (End)
a(n+1) = Sum_{k=0..n} binomial(n+1,2k+1) * 2^k = Sum_{k=0..n} A034867(n,k) * 2^k = (1/n!) * Sum_{k=0..n} A131980(n,k) * 2^k. - Tom Copeland, Nov 30 2007 a(n) (n >= 3) is the determinant of the (n-1) X (n-1) tridiagonal matrix with diagonal entries 2, superdiagonal entries 1 and subdiagonal entries -1. - Emeric Deutsch, Aug 29 2008 fract((1+sqrt(2))^n) = (1/2)*(1 + (-1)^n) - (-1)^n*(1+sqrt(2))^(-n) = (1/2)*(1 + (-1)^n) - (1-sqrt(2))^n.
See A001622 for a general formula concerning the fractional parts of powers of numbers x > 1, which satisfy x - x^(-1) = floor(x). a(n) = round((1+sqrt(2))^n/(2*sqrt(2))) for n > 0. (End) [last formula corrected by Josh Inman, Mar 05 2024] a(n) = ((4+sqrt(18))*(1+sqrt(2))^n + (4-sqrt(18))*(1-sqrt(2))^n)/4 offset 0. - Al Hakanson (hawkuu(AT)gmail.com), Aug 08 2009
If p[i] = Fibonacci(i) and if A is the Hessenberg matrix of order n defined by A[i,j] = p[j-i+1] when i<=j, A[i,j]=-1 when i=j+1, and A[i,j]=0 otherwise, then, for n >= 1, a(n) = det A. - Milan Janjic, May 08 2010 a(n) = 3*a(n-1) - a(n-2) - a(n-3), n > 2. - Gary Detlefs, Sep 09 2010 a(n) = 2*(a(2k-1) + a(2k))*a(n-2k) - a(n-4k).
a(n) = 2*(a(2k) + a(2k+1))*a(n-2k-1) + a(n-4k-2). (End)
G.f.: x/(1 - 2*x - x^2) = sqrt(2)*G(0)/4; G(k) = ((-1)^k) - 1/(((sqrt(2) + 1)^(2*k)) - x*((sqrt(2) + 1)^(2*k))/(x + ((sqrt(2) - 1)^(2*k + 1))/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Dec 02 2011 In general, for n > k, a(n) = a(k+1)*a(n-k) + a(k)*a(n-k-1). See definition of Pell numbers and the formula for Sep 04 2008. - Charlie Marion, Jan 17 2012 (1) Expression a(n+1) via a(n): a(n+1) = a(n) + sqrt(2*a^2(n) + (-1)^n);
(2) a(n+1)^2 - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = sqrt(2) - 1 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
G.f.: G(0)/(2+2*x) - 1/(1+x), where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 10 2013 G.f.: Q(0)*x/2, where Q(k) = 1 + 1/(1 - x*(4*k+2 + x)/( x*(4*k+4 + x) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013 a(n) = Sum_{r=0..n-1} Sum_{k=0..n-r-1} binomial(r+k,k)*binomial(k,n-k-r-1). - Peter Luschny, Nov 16 2013 a(k*n) = a(k)*a(k*n-k+1) + a(k-1)*a(k*n-k). - Charlie Marion, Mar 27 2014 a(k*n) = 2*a(k)*(a(k*n-k)+a(k*n-k-1)) + (-1)^k*a(k*n-2k). - Charlie Marion, Mar 30 2014 a(n+1) = (1+sqrt(2))*a(n) + (1-sqrt(2))^n. - Art DuPre, Apr 04 2014 a(n+1) = (1-sqrt(2))*a(n) + (1+sqrt(2))^n. - Art DuPre, Apr 04 2014 a(n) = F(n) + Sum_{k=1..n} F(k)*a(n-k), n >= 0 where F(n) the Fibonacci numbers A000045. - Ralf Stephan, May 23 2014 a(n) = 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1) for n >= 2. - Peter Luschny, Dec 17 2015 a(n+1) = Sum_{k=0..n} binomial(n,k)*2^floor(k/2). - Tony Foster III, May 07 2017 a(n) = exp((i*Pi*n)/2)*sinh(n*arccosh(-i))/sqrt(2). - Peter Luschny, Mar 07 2018 Some properties:
(1) a(n)^2 - a(n-2)^2 = 2*a(n-1)*(a(n) + a(n-2)) (see A005319); (2) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(3) Sum_{k=2..n+1} a(k)*a(k-1) = a(n+1)^2 if n is odd, else a(n+1)^2 - 1 if n is even;
(4) a(n) - a(n-2*k+1) = ( A077444(k) - 1)*a(n-2*k+1) + a(n-4*k+2); (5) Sum_{k=n..n+9} a(k) = 41* A001333(n+5). (End) a(m+r)*a(n+s) - a(m+s)*a(n+r) = -(-1)^(n+s)*a(m-n)*a(r-s).
a(n)^2 - a(n+1)*a(n-1) = (-1)^(n-1).
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2.
a(m)*a(n+1) - a(m+1)*a(n) = (-1)^n*a(m-n).
Sum_{m>=1} arctan(2/a(2*m+1)) = arctan(1/2).
Sum_{m>=2} arctan(2/a(2*m+1)) = arctan(1/12).
In general, for n > 0,
Sum_{m>=n} arctan(2/a(2*m+1)) = arctan(1/a(2*n)). (End)
Sum_{i=0..n} a(i)*J(n-i) = (a(n+1) + a(n) - J(n+2))/2 for J(n) = A001045(n). - Greg Dresden, Jan 05 2022 Sum_{n >= 1} 1/(a(2*n) + 1/a(2*n)) = 1/2.
Sum_{n >= 1} 1/(a(2*n+1) - 1/a(2*n+1)) = 1/4. Both series telescope - see A075870 and A005319. Product_{n >= 1} ( 1 + 2/a(2*n) ) = 1 + sqrt(2).
Product_{n >= 2} ( 1 - 2/a(2*n) ) = (1/3)*(1 + sqrt(2)). (End)
Sum_{n >=1} 1/a(n) = 1.84220304982752858079237158327980838... - R. J. Mathar, Feb 05 2024 a(n) = ((3^(n+1) + 1)^(n-1) mod (9^(n+1) - 2)) mod (3^(n+1) - 1). - Joseph M. Shunia, Jun 06 2024
EXAMPLE
G.f. = x + 2*x^2 + 5*x^3 + 12*x^4 + 29*x^5 + 70*x^6 + 169*x^7 + 408*x^8 + 985*x^9 + ...
From the comment on compositions with Fibonacci number of parts, F(n), there are F(1)=1 type of 1, F(2)=1 type of 2, F(3)=2 types of 3, F(4)=3 types of 4, F(5)=5 types of 5 and F(6)=8 types of 6.
The following table gives the number of compositions of n=6 with Fibonacci number of parts:
Composition, number of such compositions, number of compositions of this type:
6, 1, 8;
5+1, 2, 10;
4+2, 2, 6;
3+3, 1, 4;
4+1+1, 3, 9;
3+2+1, 6, 12;
2+2+2, 1, 1;
3+1+1+1, 4, 8;
2+2+1+1, 6, 6;
2+1+1+1+1, 5, 5;
1+1+1+1+1+1, 1, 1;
for a total of a(6)=70 compositions of n=6. (End).
MAPLE
A000129 := proc(n) option remember; if n <=1 then n; else 2*procname(n-1)+procname(n-2); fi; end;
a:= n-> (<<2|1>, <1|0>>^n)[1, 2]: seq(a(n), n=0..40); # Alois P. Heinz, Aug 01 2008 A000129 := n -> `if`(n<2, n, 2^(n-1)*hypergeom([1-n/2, (1-n)/2], [1-n], -1)):
MATHEMATICA
Expand[Table[((1 + Sqrt[2])^n - (1 - Sqrt[2])^n)/(2Sqrt[2]), {n, 0, 30}]] (* Artur Jasinski, Dec 10 2006 *) LinearRecurrence[{2, 1}, {0, 1}, 60] (* Harvey P. Dale, Jan 04 2012 *) a[ n_] := With[ {s = Sqrt@2}, ((1 + s)^n - (1 - s)^n) / (2 s)] // Simplify; (* Michael Somos, Jun 01 2013 *) a[ n_] := ChebyshevU[n - 1, I] / I^(n - 1); (* Michael Somos, Oct 30 2021 *)
PROG
(PARI) default(realprecision, 2000); for (n=0, 4000, a=contfracpnqn(vector(n, i, 1+(i>1)))[2, 1]; if (a > 10^(10^3 - 6), break); write("b000129.txt", n, " ", a)); \\ Harry J. Smith, Jun 12 2009 (PARI) {a(n) = imag( (1 + quadgen( 8))^n )}; /* Michael Somos, Jun 01 2013 */ (PARI) {a(n) = if( n<0, -(-1)^n, 1) * contfracpnqn( vector( abs(n), i, 1 + (i>1))) [2, 1]}; /* Michael Somos, Jun 01 2013 */ (PARI) {a(n) = polchebyshev(n-1, 2, I) / I^(n-1)}; /* Michael Somos, Oct 30 2021 */ (Sage) [lucas_number1(n, 2, -1) for n in range(30)] # Zerinvary Lajos, Apr 22 2009 (Haskell)
a000129 n = a000129_list !! n
a000129_list = 0 : 1 : zipWith (+) a000129_list (map (2 *) $ tail a000129_list)
(Maxima)
a[0]:0$
a[1]:1$
a[n]:=2*a[n-1]+a[n-2]$
(Maxima) makelist((%i)^(n-1)*ultraspherical(n-1, 1, -%i), n, 0, 24), expand; /* Emanuele Munarini, Mar 07 2018 */ (Magma) [0] cat [n le 2 select n else 2*Self(n-1) + Self(n-2): n in [1..35]]; // Vincenzo Librandi, Aug 08 2015 (GAP) a := [0, 1];; for n in [3..10^3] do a[n] := 2 * a[n-1] + a[n-2]; od; A000129 := a; # Muniru A Asiru, Oct 16 2017 (Python)
from itertools import islice
def A000129_gen(): # generator of terms a, b = 0, 1
yield from [a, b]
while True:
a, b = b, a+2*b
yield b
CROSSREFS
a(n) = A054456(n-1, 0), n>=1 (first column of triangle). INVERT transform of Fibonacci numbers ( A000045). The following sequences (and others) belong to the same family: A001333, A000129, A026150, A002605, A046717, A015518, A084057, A063727, A002533, A002532, A083098, A083099, A083100, A015519. Cf. A034867, A131980, A133156, A143808, A135387, A153346, A001622, A006497, A014176 (growth power), A098316, A154325, A021083, A243399, A008555. Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), this sequence (k=2), A006190 (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).
KEYWORD
nonn,easy,core,cofr,nice,frac
a(n) = 3*a(n-1) + a(n-2), with a(0)=0, a(1)=1.
(Formerly M2844)
+10
147
0, 1, 3, 10, 33, 109, 360, 1189, 3927, 12970, 42837, 141481, 467280, 1543321, 5097243, 16835050, 55602393, 183642229, 606529080, 2003229469, 6616217487, 21851881930, 72171863277, 238367471761, 787274278560, 2600190307441
COMMENTS
Denominators of continued fraction convergents to (3+sqrt(13))/2. - Benoit Cloitre, Jun 14 2003 a(n) and A006497(n) occur in pairs: (a,b): (1,3), (3,11), (10,36), (33,119), (109,393), ... such that b^2 - 13a^2 = 4(-1)^n. - Gary W. Adamson, Jun 15 2003 Form the 4-node graph with matrix A = [1,1,1,1; 1,1,1,0; 1,1,0,1; 1,0,1,1]. Then this sequence counts the walks of length n from the vertex with degree 5 to one (any) of the other vertices. - Paul Barry, Oct 02 2004 a(n+1) is the diagonal sum of the exponential Riordan array (exp(3x),x). - Paul Barry, Jun 03 2006 Number of paths in the right half-plane from (0,0) to the line x=n-1, consisting of steps U=(1,1), D=(1,-1), h=(1,0) and H=(2,0). Example: a(3)=10 because we have hh, H, UD, DU, hU, Uh, UU, hD, Dh and DD. - Emeric Deutsch, Sep 03 2007 Equals INVERT transform of A000129. Example: a(5) = 109 = (29, 12, 5, 2, 1) dot (1, 1, 3, 10, 33) = (29 + 12 + 15 + 20 + 33). - Gary W. Adamson, Aug 06 2010 For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the main diagonal, 1's along the superdiagonal and subdiagonal, and 0's everywhere else. - John M. Campbell, Jul 08 2011 These numbers could also be called "bronze Fibonacci numbers". Indeed, for n >= 1, F(n+1) = ceiling(phi*F(n)), if n is even and F(n+1) = floor(phi*F(n)), if n is odd, where phi is the golden ratio; analogously, for Pell numbers ( A000129), or "silver Fibonacci numbers", P(n+1) = ceiling(delta*a(n)), if n is even and P(n+1) = floor(delta*a(n)), if n is odd, where delta = delta_S = 1 + sqrt(2) is the silver ratio. Here, for n >= 1, we have a(n+1) = ceiling(c*a(n)), if n is even and a(n+1) = floor(c*a(n)), if n is odd, where c = (3 + sqrt(13))/2 is the bronze ratio (cf. comment in A098316). - Vladimir Shevelev, Feb 23 2013 Let p(n,x) denote the Fibonacci polynomial, defined by p(1,x) = 1, p(2,x) = x, p(n,x) = x*p(n-1,x) + p(n-2,x). Let q(n,x) be the numerator polynomial of the rational function p(n, x + 1 + 1/x). Then q(n,1) = a(n). - Clark Kimberling, Nov 04 2013 The (1,1)-entry of the matrix A^n where A = [0,1,0; 1,2,1; 1,1,2]. - David Neil McGrath, Jul 18 2014 a(n+1) counts closed walks on K2, containing three loops on the other vertex. Equivalently the (1,1)-entry of A^(n+1) where the adjacency matrix of digraph is A = (0,1; 1,3). - David Neil McGrath, Oct 29 2014 For n >= 1, a(n) equals the number of words of length n-1 on alphabet {0,1,2,3} avoiding runs of zeros of odd lengths. - Milan Janjic, Jan 28 2015 Apart from the initial 0, this is the p-INVERT transform of (1,0,1,0,1,0,...) for p(S) = 1 - 3 S. See A291219. - Clark Kimberling, Sep 02 2017 This is a divisibility sequence (i.e., if n|m then a(n)|a(m)).
gcd(a(n),a(n+k)) = a(gcd(n, k)) for all positive integers n and k. (End)
Numbers of straight-chain fatty acids involving oxo and/or hydroxy groups, if cis-/trans isomerism and stereoisomerism are neglected. - Stefan Schuster, Apr 04 2018 Number of 3-compositions of n restricted to odd parts (and allowed zeros); see Hopkins & Ouvry reference. - Brian Hopkins, Aug 17 2020 Also called the 3-metallonacci sequence; the g.f. 1/(1-k*x-x^2) gives the k-metallonacci sequence.
a(n+1) is the number of tilings of an n-board (a board with dimensions n X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are 3 kinds of squares available. (End)
a(n) is the number of compositions of n when there are P(k) sorts of parts k, with k,n >= 1, P(k) = A000129(k) is the k-th Pell number (see example below). - Enrique Navarrete, Dec 15 2023
REFERENCES
H. L. Abbott and D. Hanson, A lattice path problem, Ars Combin., 6 (1978), 163-178.
A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 128.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
L.-N. Machaut, Query 3436, L'Intermédiaire des Mathématiciens, 16 (1909), 62-63. - N. J. A. Sloane, Mar 08 2022
LINKS
W. F. Klostermeyer, M. E. Mays, L. Soltes and G. Trapp, A Pascal rhombus, Fibonacci Quarterly, 35 (1976), 318-328.
FORMULA
G.f.: x/(1 - 3*x - x^2).
A006497(n)^2 - 13*a(n)^2 = 4(-1)^n. (End)
a(n) = U(n-1, (3/2)i)(-i)^(n-1), i^2 = -1. - Paul Barry, Nov 19 2003 a(n) = Sum_{k=0..n-1} binomial(n-k-1,k)*3^(n-2*k-1). - Paul Barry, Oct 02 2004 a(n) = F(n, 3), the n-th Fibonacci polynomial evaluated at x=3.
Let M = {{0, 1}, {1, 3}}, v[1] = {0, 1}, v[n] = M.v[n - 1]; then a(n) = Abs[v[n][[1]]]. - Roger L. Bagula, May 29 2005 [Or a(n) = [M^(n+1)]_{1,1}. - L. Edson Jeffery, Aug 27 2013 a(n+1) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(k-j).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(k,j)*C(n-j,k)*2^(n-j-k).
a(n+1) = Sum_{k=0..floor(n/2)} C(n-k,k)*3^(n-2*k).
a(n) = Sum_{k=0..n} C(k,n-k)*3^(2*k-n). (End)
E.g.f.: exp(3*x/2)*sinh(sqrt(13)*x/2)/(sqrt(13)/2). - Paul Barry, Jun 03 2006 a(n) = (ap^n - am^n)/(ap - am), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2.
Let C = (3 + sqrt(13))/2 = exp arcsinh(3/2) = 3.3027756377... Then C^n, n > 0 = a(n)*(1/C) + a(n+1). Let X = the 2 X 2 matrix [0, 1; 1, 3]. Then X^n = [a(n-1), a(n); a(n), a(n+1)]. - Gary W. Adamson, Dec 21 2007 1/3 = 3/(1*10) + 3/(3*33) + 3/(10*109) + 3/(33*360) + 3/(109*1189) + ... . - Gary W. Adamson, Mar 16 2008 a(n) = ((3 + sqrt(13))^n - (3 - sqrt(13))^n)/(2^n*sqrt(13)). - Al Hakanson (hawkuu(AT)gmail.com), Jan 12 2009
a(p) == 13^((p-1)/2) mod p, for odd primes p. - Gary W. Adamson, Feb 22 2009 Limit_{k->oo} a(n+k)/a(k) = ( A006497(n) + a(n)*sqrt(13))/2. Limit_{n->oo} A006497(n)/a(n) = sqrt(13). (End) Sum_{k>=1} (-1)^(k-1)/(a(k)*a(k+1)) = (sqrt(13)-3)/2. - Vladimir Shevelev, Feb 23 2013 (1) Expression a(n+1) via a(n): a(n+1) = (3*a(n) + sqrt(13*a^2(n) + 4*(-1)^n)/2;
(2) a^2(n+1) - a(n)*a(n+2) = (-1)^n;
(3) Sum_{k=1..n} (-1)^(k-1)/(a(k)*a(k+1)) = a(n)/a(n+1);
(4) a(n)/a(n+1) = (sqrt(13)-3)/2 + r(n), where |r(n)| < 1/(a(n+1)*a(n+2)). (End)
Sum_{n >= 1} 1/( a(2*n) + 1/a(2*n) ) = 1/3; Sum_{n >= 1} 1/( a(2*n + 1) - 1/a(2*n + 1) ) = 1/9. - Peter Bala, Mar 26 2015 Some properties:
(1) a(n)*a(n+1) = 3*Sum_{k=1..n} a(k)^2;
(2) a(n)^2 + a(n+1)^2 = a(2*n+1);
(3) a(n)^2 - a(n-2)^2 = 3*a(n-1)*(a(n) + a(n-2));
(4) a(m*(p+1)) = a(m*p)*a(m+1) + a(m*p-1)*a(m);
(5) a(n-k)*a(n+k) = a(n)^2 + (-1)^(n+k+1)*a(k)^2;
(6) a(2*n) = a(n)*(3*a(n) + 2*a(n-1));
(7) 3*Sum_{k=2..n+1} a(k)*a(k-1) is equal to a(n+1)^2 if n odd, and is equal to a(n+1)^2 - 1 if n is even;
(8) a(n) - a(n-2*k+1) = alpha(k)*a(n-2*k+1) + a(n-4*k+2), where alpha(k) = (ap^(2*k-1) + am^(2*k-1), with ap = (3 + sqrt(13))/2, am = (3 - sqrt(13))/2;
(9) 131|Sum_{k=n..n+9} a(k), for all positive n. (End)
a(n)^2 - a(n+r)*a(n-r) = (-1)^(n-r)*a(r)^2 - Catalan's identity.
arctan(1/a(2n)) - arctan(1/a(2n+2)) = arctan(a(2)/a(2n+1)).
arctan(1/a(2n)) = Sum_{m>=n} arctan(a(2)/a(2m+1)).
The same formula holds for Fibonacci numbers and Pell numbers. (End)
a(n+2) = 3^(n+1) + Sum_{k=0..n} a(k)*3^(n-k). - Greg Dresden and Gavron Campbell, Feb 22 2022 G.f.: x/(1 - 3*x - x^2) = Sum_{n >= 0} x^(n+1) *( Product_{k = 1..n} (m*k + 3 - m + x)/(1 + m*k*x) ) for arbitrary m (a telescoping series). - Peter Bala, May 08 2024
EXAMPLE
The following table gives the number of compositions of n=6:
Composition, number of such compositions, number of compositions of this type:
6, 1, 70;
5+1, 2, 58;
4+2, 2, 48;
3+3, 1, 25;
4+1+1, 3, 36;
3+2+1, 6, 60;
2+2+2, 1, 8;
3+1+1+1, 4, 20;
2+2+1+1, 6, 24;
2+1+1+1+1, 5, 10;
1+1+1+1+1+1, 1, 1;
for a total of a(6)=360 compositions of n=6. (End).
MAPLE
a[0]:=0: a[1]:=1: for n from 2 to 35 do a[n]:= 3*a[n-1]+a[n-2] end do: seq(a[n], n=0..30); # Emeric Deutsch, Sep 03 2007 seq(combinat[fibonacci](n, 3), n=0..30); # R. J. Mathar, Dec 07 2011
MATHEMATICA
a[n_] := (MatrixPower[{{1, 3}, {1, 2}}, n].{{1}, {1}})[[2, 1]]; Table[ a[n], {n, -1, 24}] (* Robert G. Wilson v, Jan 13 2005 *) LinearRecurrence[{3, 1}, {0, 1}, 30] (* or *) CoefficientList[Series[x/ (1-3x-x^2), {x, 0, 30}], x] (* Harvey P. Dale, Apr 20 2011 *) Table[If[n==0, a1=1; a0=0, a2=a1; a1=a0; a0=3*a1+a2], {n, 0, 30}] (* Jean-François Alcover, Apr 30 2013 *)
PROG
(PARI) a(n)=if(n<1, 0, contfracpnqn(vector(n, i, 2+(i>1)))[2, 1])
(Sage) [lucas_number1(n, 3, -1) for n in range(0, 30)] # Zerinvary Lajos, Apr 22 2009 (Magma)[ n eq 1 select 0 else n eq 2 select 1 else 3*Self(n-1)+Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 19 2011 (Haskell)
a006190 n = a006190_list !! n
a006190_list = 0 : 1 : zipWith (+) (map (* 3) $ tail a006190_list) a006190_list
(PARI) concat([0], Vec(x/(1-3*x-x^2)+O(x^30))) \\ Joerg Arndt, Apr 30 2013 (GAP) a:=[0, 1];; for n in [3..30] do a[n]:=3*a[n-1]+a[n-2]; od; a; # Muniru A Asiru, Mar 31 2018
CROSSREFS
Row sums of Pascal's rhombus ( A059317). Also row sums of triangle A054456(n, m). Sequences with g.f. 1/(1-k*x-x^2) or x/(1-k*x-x^2): A000045 (k=1), A000129 (k=2), this sequence (k=3), A001076 (k=4), A052918 (k=5), A005668 (k=6), A054413 (k=7), A041025 (k=8), A099371 (k=9), A041041 (k=10), A049666 (k=11), A041061 (k=12), A140455 (k=13), A041085 (k=14), A154597 (k=15), A041113 (k=16), A178765 (k=17), A041145 (k=18), A243399 (k=19), A041181 (k=20).
Table by antidiagonals of T(n,k) = n*T(n,k-1) + T(n,k-2) starting with T(n,1) = 1.
+10
57
1, 1, 1, 1, 2, 2, 1, 3, 5, 3, 1, 4, 10, 12, 5, 1, 5, 17, 33, 29, 8, 1, 6, 26, 72, 109, 70, 13, 1, 7, 37, 135, 305, 360, 169, 21, 1, 8, 50, 228, 701, 1292, 1189, 408, 34, 1, 9, 65, 357, 1405, 3640, 5473, 3927, 985, 55, 1, 10, 82, 528, 2549, 8658, 18901, 23184, 12970, 2378, 89
COMMENTS
Columns of the array are generated from Fibonacci polynomials f(x). They are: (1), (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 +1), ... If column headings start 0, 1, 2, ... then the terms in the n-th column are generated from the n-th degree Fibonacci polynomial. For example, column 5 (8, 70, 360, ...) is generated from f(x), x = 1,2,3,...; fifth-degree polynomial x^5 + 4x^3 + 3x; e.g., f(2) = 70 = 2^5 + 4*8 + 3*2. - Gary W. Adamson, Apr 02 2006 The ratio of two consecutive entries of the sequence in the n-th row approaches (n + sqrt(n^2 + 4))/2. Example: The sequence beginning (1, 3, 10, 33, ...) tends to 3.302775... = (3 + sqrt(13))/2. - Gary W. Adamson, Aug 12 2013 As to the array sequences, (n+1)-th sequence is the INVERT transform of the n-th sequence. - Gary W. Adamson, Aug 20 2013 The array can be extended infinitely above the Fibonacci row by taking successive INVERTi transforms, resulting in:
...
1, -2, 5, -12, 29, -70, ...
1, -1, 2, -3, 5, -8, ...
l, 0, 1, 0, 1, 0, ...
1, 1, 2, 3, 5, 8, ...
1, 2, 5, 12, 29, 70, ...
...
This results in an infinite array in which sequences above the (1, 0, 1, 0, ...) are reflections of the sequences below, except for the alternate signs. Any sequence in the (+ sign) row starting (1, n, ...) is the (2*n-th) INVERT transform of the same sequence but with alternate signs. Example: (1, 2, 5, 12, ...) is the (2*2) = fourth INVERT transform of (1, -2, 5, -12, ...) by inspection. Conjecture: This "reflection" principle will result from taking successive INVERT transforms of any aerated sequence starting 1, ... and with positive signs. Likewise, the rows above the aerated sequence are successive INVERTi transforms of the aerated sequence. - Gary W. Adamson, Jul 14 2019 Row n is the n-metallonacci sequence.
T(n,k) is the number of tilings of a (k-1)-board (a board with dimensions (k-1) X 1) using unit squares and dominoes (with dimensions 2 X 1) if there are n kinds of squares available. (End)
FORMULA
T(n,k) = [[0,1; 1,n]^{k+1}]_{1,1}, n,k in {1,2,...}. - L. Edson Jeffery, Sep 23 2012
EXAMPLE
Table begins:
1, 1, 2, 3, 5, 8, 13, ...
1, 2, 5, 12, 29, 70, 169, ...
1, 3, 10, 33, 109, 360, 1189, ...
1, 4, 17, 72, 305, 1292, 5473, ... etc.
MAPLE
option remember;
if k <= 1 then
k;
else
n*procname(n, k-1)+procname(n, k-2) ;
end if;
end proc:
MATHEMATICA
T[n_, 1]:= 1; T[n_, k_]:= T[n, k] = If[k<0, 0, n*T[n, k-1] + T[n, k-2]]; Table[T[n-k+1, k], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Aug 12 2019 *)
PROG
(PARI) T(n, k) = if(k==1, 1, k<0, 0, n*T(n, k-1)+T(n, k-2));
for(n=1, 15, for(k=1, n, print1(T(n-k+1, k), ", "))) \\ G. C. Greubel, Aug 12 2019 (Sage)
def T(n, k):
if (k<0): return 0
elif (k==1): return 1
else: return n*T(n, k-1) + T(n, k-2)
[[T(n-k+1, k) for k in (1..n)] for n in (1..15)] # G. C. Greubel, Aug 12 2019 (GAP)
T:= function(n, k)
if k<0 then return 0;
elif k=1 then return 1;
else return n*T(n, k-1) + T(n, k-2);
fi;
end;
Flat(List([1..15], n-> List([1..n], k-> T(n-k+1, k) ))); # G. C. Greubel, Aug 12 2019
CROSSREFS
Rows are A000045, A000129, A006190, A001076, A052918, A005668, A054413, A041025, A099371, A041041, A049666, A041061, A140455, A041085, etc.
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