A131294 a(n)=ds_3(a(n-1))+ds_3(a(n-2)), a(0)=0, a(1)=1; where ds_3=digital sum base 3.

0, 1, 1, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3, 3, 2, 3

Offset

0,4

Comments

The digital sum analog (in base 3) of the Fibonacci recurrence.

When starting from index n=3, periodic with Pisano period A001175(2)=3.

a(n) and Fib(n)=A000045(n) are congruent modulo 2 which implies that (a(n) mod 2) is equal to (Fib(n) mod 2)=A011655(n). Thus (a(n) mod 2) is periodic with the Pisano period A001175(2)=3 too.

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

Links

Table of n, a(n) for n=0..100.

Index entries for Colombian or self numbers and related sequences

Index entries for linear recurrences with constant coefficients, signature (0, 0, 1).

Formula

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

a(n) = floor(a(n-1)/3)+floor(a(n-2)/3)+(a(n-1)mod 3)+(a(n-2)mod 3).

a(n) = A002264(a(n-1))+A002264(a(n-2))+A010872(a(n-1))+A010872(a(n-2)).

a(n) = Fib(n)-2*sum{1<k<n, Fib(n-k+1)*floor(a(k)/3)}, where Fib(n)=A000045(n).

Example

a(5)=3, since a(3)=2, ds_3(2)=2, a(4)=3=10(base 3),
ds_3(3)=1 and so a(5)=2+1.

Mathematica

nxt[{a_, b_}]:={b, Total[IntegerDigits[a, 3]]+Total[IntegerDigits[b, 3]]}; Transpose[NestList[nxt, {0, 1}, 100]][[1]] (* Harvey P. Dale, Aug 02 2016 *)

Crossrefs

Cf. A000045, A010073, A010074, A010075, A010076, A010077, A131295, A131296, A131297, A131318, A131319, A131320.

Sequence in context: A048198 A096006 A182128 * A276869 A102313 A262955

Adjacent sequences: A131291 A131292 A131293 * A131295 A131296 A131297

Keyword

nonn,base

Author

Hieronymus Fischer, Jun 27 2007

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

Status

approved