Search: id:A010077 Results 1-1 of 1 results found. %I A010077 %S A010077 0,1,1,2,3,5,8,13,12,7,10,8,9,17,17,16,15,13,10,5,6,11,8,10,9,10,10, %T A010077 2,3,5,8,13,12,7,10,8,9,17,17,16,15,13,10,5,6,11,8,10,9,10,10, %U A010077 2,3,5,8,13,12,7,10,8,9,17,17,16,15,13,10,5,6,11,8,10,9,10,10 %N A010077 a(n) = sum of digits of a(n-1) + sum of digits of a(n-2); a(0) = 0, a(1) = 1. %C A010077 The digital sum analogue (in base 10) of the Fibonacci recurrence. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007 %C A010077 a(n) and Fib(n)=A000045(n) are congruent modulo 9 which implies that (a(n) mod 9) is equal to (Fib(n) mod 9) = A007887(n). Thus (a(n) mod 9) is periodic with the Pisano period A001175(9)=24. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007 %C A010077 a(n)==A004090(n) modulo 9 (A004090(n)=digital sum of Fib(n)). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007 %C A010077 For general bases p>2, we have the inequality 2<=a(n)<=2p-3 (for n>2). Actually, a(n)<=17=A131319(10) for the base p=10. - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007 %F A010077 Periodic from n=3 with period 24. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Mar 13 2006 %F A010077 a(n) = A030132(n-4) + A030132(n-3) for n>3. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jul 04 2007 %F A010077 a(n)=a(n-1)+a(n-2)-9*(floor(a(n-1)/10)+floor(a(n-2)/10)). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007 %F A010077 a(n)=floor(a(n-1)/10)+floor(a(n-2)/10)+(a(n-1)mod 10)+(a(n-2)mod 10). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007 %F A010077 a(n)=A059995(a(n-1))+A059995(a(n-2))+A010879(a(n-1))+A010879(a(n-2)). - Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Jun 27 2007 %F A010077 a(n)=Fib(n)-9*sum{1