OFFSET
0,2
COMMENTS
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,...,14} with no two adjacent letters identical. -Milan Janjic, Jan 31 2015
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..800
M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7.
Index entries for linear recurrences with constant coefficients, signature (14).
FORMULA
a(n) = Sum_{k=0..n} A097805(n,k)*(-1)^(n-k)*15^k. - Philippe Deléham, Dec 04 2009
a(0) = 1; for n>0, a(n) = 15*14^(n-1). - Vincenzo Librandi, Dec 05 2009
a(0) = 1, a(1) = 15, a(n) = 14*a(n-1). - Vincenzo Librandi, Dec 10 2012
E.g.f.: (15*exp(14*x) -1)/14. - G. C. Greubel, Sep 24 2019
MAPLE
k:=15; seq(`if`(n=0, 1, k*(k-1)^(n-1)), n = 0..25); # G. C. Greubel, Sep 24 2019
MATHEMATICA
Join[{1}, 15*14^Range[0, 25]] (* Vladimir Joseph Stephan Orlovsky, Jul 11 2011 *)
CoefficientList[Series[(1+x)/(1-14x), {x, 0, 30}], x] (* Vincenzo Librandi, Dec 10 2012 *)
PROG
(PARI) vector(26, n, k=15; if(n==1, 1, k*(k-1)^(n-2))) \\ G. C. Greubel, Sep 24 2019
(Magma) k:=15; [1] cat [k*(k-1)^(n-1): n in [1..25]]; // G. C. Greubel, Sep 24 2019
(Sage) k=15; [1]+[k*(k-1)^(n-1) for n in (1..25)] # G. C. Greubel, Sep 24 2019
(GAP) k:=15;; Concatenation([1], List([1..25], n-> k*(k-1)^(n-1) )); # G. C. Greubel, Sep 24 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Dec 04 2009
STATUS
approved