This site is supported by donations to The OEIS Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 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 (list; graph; refs; listen; history; text; internal format)
 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. 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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified October 16 23:46 EDT 2019. Contains 328103 sequences. (Running on oeis4.)