OFFSET
0,4
COMMENTS
The digital sum analog (in base 11) of the Fibonacci recurrence.
When starting from index n=3, periodic with Pisano period A001175(10)=60.
a(n) and Fib(n)=A000045(n) are congruent modulo 10 which implies that (a(n) mod 10) is equal to (Fib(n) mod 10)=A003893(n). Thus (a(n) mod 10) is periodic with the Pisano period A001175(10)=60 too.
For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=19=A131319(11) for the base p=11.
LINKS
FORMULA
a(n) = a(n-1)+a(n-2)-10*(floor(a(n-1)/11)+floor(a(n-2)/11)).
a(n) = floor(a(n-1)/11)+floor(a(n-2)/11)+(a(n-1)mod 11)+(a(n-2)mod 11).
a(n) = Fib(n)-10*sum{1<k<n, Fib(n-k+1)*floor(a(k)/11)}, where Fib(n)=A000045(n).
EXAMPLE
a(10)=5, since a(8)=11=10(base 11), ds_11(11)=1,
a(9)=4, ds_11(4)=4 and so a(10)=1+4.
MATHEMATICA
nxt[{a_, b_}]:={b, Total[IntegerDigits[a, 11]]+Total[IntegerDigits[b, 11]]}; NestList[nxt, {0, 1}, 80][[All, 1]] (* or *) PadRight[{0, 1, 1}, 80, {10, 11, 11, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 17, 17, 14, 11, 5, 6, 11, 7, 8, 15, 13, 8, 11, 9, 10, 19, 19, 18, 17, 15, 12, 7, 9, 16, 15, 11, 6, 7, 13, 10, 13, 13, 6, 9, 15, 14, 9, 13, 12, 5, 7, 12, 9, 11}] (* Harvey P. Dale, Jul 24 2017 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Hieronymus Fischer, Jun 27 2007
EXTENSIONS
Incorrect comment removed by Michel Marcus, Apr 29 2018
STATUS
approved