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)

Crossrefs

Cf. A000045, A010073, A010074, A010075, A010077, 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