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A131297
a(n) = ds_11(a(n-1))+ds_11(a(n-2)), a(0)=0, a(1)=1; where ds_11=digital sum base 11.
13
0, 1, 1, 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, 10, 11, 11, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 17, 17, 14, 11
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.
a(n)==A074867(n) modulo 10 (A074867(n)=digital product analog base 10 of the Fibonacci recurrence).
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.
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 *)
KEYWORD
nonn,base
AUTHOR
Hieronymus Fischer, Jun 27 2007
EXTENSIONS
Incorrect comment removed by Michel Marcus, Apr 29 2018
STATUS
approved