A010074 a(n) = sum of base-7 digits of a(n-1) + sum of base-7 digits of a(n-2).

0, 1, 1, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7, 2, 3, 5, 8, 7, 3, 4, 7, 5, 6, 11, 11, 10, 9, 7, 4, 5, 9, 8, 5, 7, 6, 7, 7

Offset

0,4

Comments

The digital sum analog (in base 7) of the Fibonacci recurrence. - Hieronymus Fischer, Jun 27 2007

a(n) and Fib(n)=A000045(n) are congruent modulo 6 which implies that (a(n) mod 6) is equal to (Fib(n) mod 6) = A082117(n-1) (for n>0). Thus (a(n) mod 6) is periodic with the Pisano period A001175(6)=24. - 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(7) for the base p=7. - Hieronymus Fischer, Jun 27 2007

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for Colombian or self numbers and related sequences

Formula

Periodic from n=3 with period 24. - Franklin T. Adams-Watters, Mar 13 2006

From Hieronymus Fischer, Jun 27 2007: (Start)

a(n) = a(n-1)+a(n-2)-6*(floor(a(n-1)/7)+floor(a(n-2)/7)).

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

a(n) = (a(n-1)+a(n-2)+6*(A010876(a(n-1))+A010876(a(n-2))))/7.

a(n) = Fib(n)-6*sum{1<k<n, Fib(n-k+1)*floor(a(k)/7)} where Fib(n)=A000045(n). (End)

Mathematica

nxt[{a_, b_}]:={b, Total[IntegerDigits[a, 7]]+Total[IntegerDigits[b, 7]]}; Transpose[NestList[nxt, {0, 1}, 80]][[1]] (* Harvey P. Dale, Oct 12 2013 *)

Crossrefs

Cf. A000045, A010073, A010075, A010076, A010077, A131294, A131295, A131296, A131297, A131318, A131319, A131320.

Sequence in context: A072987 A010075 A182445 * A355702 A116918 A271621

Adjacent sequences: A010071 A010072 A010073 * A010075 A010076 A010077

Keyword

nonn,base

Author

N. J. A. Sloane, Leonid Broukhis

Extensions

Incorrect comment removed by Michel Marcus, Apr 29 2018

Status

approved