A010076 a(n) = sum of base-9 digits of a(n-1) + sum of base-9 digits of a(n-2).

0, 1, 1, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9, 8, 9, 9

Offset

0,4

Comments

The digital sum analog (in base 9) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007

a(n) and Fib(n)=A000045(n) are congruent modulo 8 which implies that (a(n) mod 8) is equal to (Fib(n) mod 8) = A079344(n). Thus (a(n) mod 8) is periodic with the Pisano period A001175(8)=12. - Hieronymus Fischer, Jun 27 2007

For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=13=A131319(9) for the base p=9. - Hieronymus Fischer, Jun 27 2007

Links

Table of n, a(n) for n=0..74.

Index entries for Colombian or self numbers and related sequences

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

Formula

Periodic from n=3 with period 12. - Franklin T. Adams-Watters, Mar 13 2006

From Hieronymus Fischer, Jun 27 2007: (Start)

a(n) = a(n-1)+a(n-2)-8*(floor(a(n-1)/9)+floor(a(n-2)/9)).

a(n) = floor(a(n-1)/9)+floor(a(n-2)/9)+(a(n-1)mod 9)+(a(n-2)mod 9).

a(n) = (a(n-1)+a(n-2)+8*(A010878(a(n-1))+A010878(a(n-2))))/9.

a(n) = Fib(n)-8*sum{1<k<n, Fib(n-k+1)*floor(a(k)/9)} where Fib(n)=A000045(n). (End)

Mathematica

PadRight[{0, 1, 1}, 100, {8, 9, 9, 2, 3, 5, 8, 13, 13, 10, 7, 9}] (* Paolo Xausa, Aug 25 2024 *)

Crossrefs

Cf. A000045, A001175, A010073, A010074, A010075, A010077, A010878, A079344, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Sequence in context: A069638 A237568 A272918 * A182444 A138183 A076591

Adjacent sequences: A010073 A010074 A010075 * A010077 A010078 A010079

Keyword

nonn,base

Author

N. J. A. Sloane, Leonid Broukhis

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

Status

approved