A131296 a(n) = ds_5(a(n-1))+ds_5(a(n-2)), a(0)=0, a(1)=1; where ds_5=digital sum base 5.

0, 1, 1, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3, 5, 4, 5, 5, 2, 3

Offset

0,4

Comments

The digital sum analog (in base 5) of the Fibonacci recurrence.

When starting from index n=3, periodic with Pisano period A001175(4)=6.

a(n) and Fib(n)=A000045(n) are congruent modulo 4 which implies that (a(n) mod 4) is equal to (Fib(n) mod 4)=A079343(n). Thus (a(n) mod 4) is periodic with the Pisano period A001175(4)=6 too.

For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(5) for the base p=5.

Links

Harvey P. Dale, Table of n, a(n) for n = 0..1000

Index entries for Colombian or self numbers and related sequences

Formula

a(n) = a(n-1)+a(n-2)-4*(floor(a(n-1)/5)+floor(a(n-2)/5)).

a(n) = floor(a(n-1)/5)+floor(a(n-2)/5)+(a(n-1)mod 5)+(a(n-2)mod 5).

a(n) = A002266(a(n-1))+A002266(a(n-2))+A010874(a(n-1))+A010874(a(n-2)).

a(n) = Fib(n)-4*sum{1<k<n, Fib(n-k+1)*floor(a(k)/5)}, where Fib(n)=A000045(n).

Example

a(10)=3, since a(8)=5=10(base 5), ds_5(5)=1,
a(9)=2, ds_5(2)=2 and so a(10)=1+2.

Mathematica

nxt[{a_, b_}]:={b, Total[IntegerDigits[a, 5]]+Total[IntegerDigits[b, 5]]}; NestList[nxt, {0, 1}, 100][[;; , 1]] (* Harvey P. Dale, Sep 01 2024 *)

Crossrefs

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131295, A131297, A131318, A131319, A131320.

Sequence in context: A102149 A321782 A104204 * A371216 A267808 A239852

Adjacent sequences: A131293 A131294 A131295 * A131297 A131298 A131299

Keyword

nonn,base

Author

Hieronymus Fischer, Jun 27 2007

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

Status

approved