A131295 a(n)=ds_4(a(n-1))+ds_4(a(n-2)), a(0)=0, a(1)=1; where ds_4=digital sum base 4.

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

Offset

0,4

Comments

The digital sum analog (in base 4) of the Fibonacci recurrence.

When starting from index n=3, periodic with Pisano period A001175(3)=8.

Also a(n)==A004090(n) modulo 3 (A004090(n)=digital sum of Fib(n)).

For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=5=A131319(4) for the base p=4.

a(n) and Fib(n)=A000045(n) are congruent modulo 3 which implies that (a(n) mod 3) is equal to (Fib(n) mod 3)=A082115(n-1) (for n>0). Thus (a(n) mod 3) is periodic with the Pisano period = A001175(3)=8 too. - Hieronymus Fischer

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)-3*(floor(a(n-1)/4)+floor(a(n-2)/4)).

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

a(n)=A002265(a(n-1))+A002265(a(n-2))+A010873(a(n-1))+A010873(a(n-2)).

a(n)=Fib(n)-3*sum{1<k<n, Fib(n-k+1)*floor(a(k)/4)}, where Fib(n)=A000045(n).

Example

a(8)=3, since a(6)=5=11(base 4), ds_4(5)=2,
a(7)=4=10(base 4), ds_4(4)=1 and so a(8)=2+1.

Mathematica

nxt[{a_, b_}]:={b, Total[IntegerDigits[a, 4]]+Total[IntegerDigits[b, 4]]}; NestList[ nxt, {0, 1}, 110][[All, 1]] (* Harvey P. Dale, Jul 30 2018 *)

Crossrefs

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131294, A131296, A131297, A131318, A131319, A131320.

Sequence in context: A262565 A096289 A367848 * A102642 A183228 A133304

Adjacent sequences: A131292 A131293 A131294 * A131296 A131297 A131298

Keyword

nonn,base

Author

Hieronymus Fischer, Jun 27 2007

Status

approved