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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). 13
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
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)
MATHEMATICA
nxt[{a_, b_}]:={b, Total[IntegerDigits[a, 8]]+Total[IntegerDigits[b, 8]]}; NestList[ nxt, {0, 1}, 100][[All, 1]] (* or *) PadRight[{0, 1, 1}, 100, {7, 8, 8, 2, 3, 5, 8, 6, 7, 13, 13, 12, 11, 9, 6, 8}] (* Harvey P. Dale, Apr 19 2020 *)
CROSSREFS
Sequence in context: A065115 A262263 A072987 * A182445 A010074 A355702
KEYWORD
nonn,base
AUTHOR
EXTENSIONS
Incorrect comment removed by Michel Marcus, Apr 29 2018
STATUS
approved

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Last modified April 23 07:57 EDT 2024. Contains 371905 sequences. (Running on oeis4.)