A220493 Fibonacci 15-step numbers, a(n) = a(n-1) + a(n-2) + ... + a(n-15).
1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32767, 65533, 131064, 262124, 524240, 1048464, 2096896, 4193728, 8387328, 16774400, 33548288, 67095552, 134189056, 268374016, 536739840, 1073463296, 2146893825, 4293722117, 8587313170
Offset: 1
Links
- Robert Israel, Table of n, a(n) for n = 1..3320
- M. Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, J. Int. Seq. 18 (2015) # 15.4.7.
- Tony D. Noe and Jonathan Vos Post, Primes in Fibonacci n-step and Lucas n-step Sequences, J. of Integer Sequences, Vol. 8 (2005), Article 05.4.4.
- Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1,1,1,1,1,1,1,1,1).
Crossrefs
Programs
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Maple
f:= gfun:-rectoproc({a(n) = add(a(n-i),i=1..15), seq(a(n)=0,n=-14..0),a(1)=1},a(n),remember): map(f, [$1..100]); # Robert Israel, Feb 19 2019
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Mathematica
FibonacciSequence[n_, kMax_] := Module[{a, s}, a = Join[{1}, Table[0, {n - 1}]]; lst = {}; Table[s = Plus @@ a; a = RotateLeft[a]; a[[n]] = s, {k, 1, kMax}]]; FibonacciSequence[15, 50] (* T. D. Noe, Feb 20 2013 *)
Formula
G.f.: x/(1-Sum_{k=1..15} x^k). - Robert Israel, Feb 19 2019
Comments