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A010075 a(n) = sum of base-8 digits of a(n-1) + sum of base-8 digits of a(n-2). 12
0, 1, 1, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8, 7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

The digital sum analog (in base 8) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007

a(n) and Fib(n)=A000045(n) are congruent modulo 7 which implies that (a(n) mod 7) is equal to (Fib(n) mod 7). Thus (a(n) mod 7) is periodic with the Pisano period A001175(7)=16. - Hieronymus Fischer, Jun 27 2007

For general bases p>2, the inequality 2<=a(n)<=2p-3 holds for n>2. Actually, a(n)<=11=A131319(8) for the base p=8. - Hieronymus Fischer, Jun 27 2007

LINKS

Table of n, a(n) for n=0..95.

Index entries for Colombian or self numbers and related sequences

FORMULA

Periodic from n=3 with period 16. - Franklin T. Adams-Watters, Mar 13 2006

From Hieronymus Fischer, Jun 27 2007: (Start)

a(n) = a(n-1)+a(n-2)-7*(floor(a(n-1)/8)+floor(a(n-2)/8)).

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

a(n) = (a(n-1)+a(n-2)+7*(A010877(a(n-1))+A010877(a(n-2))))/8.

a(n) = Fib(n)-7*sum{1<k<n, Fib(n-k+1)*floor(a(k)/8)} where Fib(n)=A000045(n). (End)

CROSSREFS

Cf. A000045, A010073, A010074, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Sequence in context: A065115 A262263 A072987 * A182445 A010074 A116918

Adjacent sequences:  A010072 A010073 A010074 * A010076 A010077 A010078

KEYWORD

nonn,base

AUTHOR

N. J. A. Sloane, Leonid Broukhis

EXTENSIONS

Incorrect comment removed by Michel Marcus, Apr 29 2018

STATUS

approved

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Last modified September 18 18:29 EDT 2019. Contains 327180 sequences. (Running on oeis4.)