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

Author

Hieronymus Fischer, Jun 27 2007

Keywords

• nonn
• base

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.

Examples

			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.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Index entries for Colombian or self numbers and related sequences

Crossrefs

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

Programs

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 *)

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{1A000045(n).

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018