A251655 4-step Fibonacci sequence starting with 0, 1, 1, 1.

0, 1, 1, 1, 3, 6, 11, 21, 41, 79, 152, 293, 565, 1089, 2099, 4046, 7799, 15033, 28977, 55855, 107664, 207529, 400025, 771073, 1486291, 2864918, 5522307, 10644589, 20518105, 39549919, 76234920, 146947533, 283250477, 545982849, 1052415779, 2028596638

Offset: 0

Author

Arie Bos, Dec 06 2014

Keywords

• nonn
• easy

Links

• Tamás Lengyel and Diego Marques, The 2-adic Order of Some Generalized Fibonacci Numbers, INTEGERS, 17, 2017, A5.
• Index entries for linear recurrences with constant coefficients, signature (1,1,1,1).

Crossrefs

Other 4-step Fibonacci sequences are A000078, A000288, A001630, A001631, A001648, A073817, A100532, A251654, A251656, A251672, A251703, A251704, A251705.

Programs

J

(see www.jsoftware.com) First construct the generating matrix
   [M=: (#.@}: + {:)\"1&.|: <:/~i.4
1 1 1 1
1 2 2 2
2 3 4 4
4 6 7 8
Given that matrix, one can produce the first 4*250 numbers with
, M(+/ . *)^:(i.250) 0 1 1 1x

Mathematica

LinearRecurrence[Table[1, {4}], {0, 1, 1, 1}, 36] (* Michael De Vlieger, Dec 09 2014 *)

Formula

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

G.f.: x*(x-1)*(1+x)/(-1+x+x^2+x^3+x^4) . - R. J. Mathar, Mar 28 2025

a(n) = A000078(n+2)-A000078(n). - R. J. Mathar, Mar 28 2025