A251653 5-step Fibonacci sequence starting with 0,0,1,0,0.

0, 0, 1, 0, 0, 1, 2, 4, 7, 14, 28, 55, 108, 212, 417, 820, 1612, 3169, 6230, 12248, 24079, 47338, 93064, 182959, 359688, 707128, 1390177, 2733016, 5372968, 10562977, 20766266, 40825404, 80260631, 157788246, 310203524, 609844071, 1198921876, 2357018348, 4633776065, 9109763884, 17909324244, 35208804417

Offset: 0

Author

Arie Bos, Dec 06 2014

Keywords

• nonn
• easy

Comments

Doubling the entries > 1 as 1, 2, 2, 4, 4, 7, 7, 14, 14, 28, 28, 55, 55,... (offset 0) gives Nyblom's palindromic binary strings having no 5-runs of 1's. - R. J. Mathar, Mar 28 2025

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• M. A. Nyblom, Counting Palindromic Binary Strings Without r-Runs of Ones, J. Int. Seq. 16 (2013) #13.8.7, P_5(n)
• H. Prodinger, Counting Palindromes According to r-Runs of Ones Using Generating Functions, J. Int. Seq. 17 (2014) # 14.6.2, even length, r=4.
• Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1).

Crossrefs

Other 5-step sequences are A000322, A001591, A023424, A074048, A124312, A124313, A122997, A135055, A135056, A145029

Programs

J

(see www.jsoftware.com) First construct the generating matrix
1  1  1  1  1
1  2  2  2  2
2  3  4  4  4
4  6  7  8  8
8 12 14 15 16
Given that matrix one can produce the first 5*200 numbers by
, M(+/ . *)^:(i.250) 0 0 1 0 0x

Mathematica

LinearRecurrence[{1, 1, 1, 1, 1}, {0, 0, 1, 0, 0}, 100] (* G. C. Greubel, May 27 2016 *)

Formula

a(n+5) = a(n) + a(n+1) + a(n+2) + a(n+3) + a(n+4).

G.f.: x^2*(x^2 + x - 1)/(x^5 + x^4 + x^3 + x^2 + x - 1). - Chai Wah Wu, May 27 2016