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 A074867 a(n)=M[a(n-1)]+M[a(n-2)] where a(0)=a(1)=1 and M(n) is the product of the digits of n in base 10. 3
 1, 1, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 7, 7, 14, 11, 5, 6, 11, 7, 8, 15, 13, 8, 11, 9, 10, 9, 9, 18, 17, 15, 12, 7, 9, 16, 15, 11, 6, 7, 13, 10, 3, 3, 6, 9, 15, 14, 9, 13, 12, 5, 7, 12, 9, 11, 10, 1, 1, 2, 3, 5, 8, 13, 11, 4, 5, 9, 14, 13, 7, 10, 7, 7, 14, 11, 5, 6, 11, 7, 8, 15, 13 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS Periodic with least period 60. - Christopher N. Swanson (cswanson(AT)ashland.edu), Jul 22 2003 The digital product analog (in base 10) of the Fibonacci recurrence. - Hieronymus Fischer, Jul 01 2007 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. - Hieronymus Fischer, Jul 01 2007 a(n)==A131297(n) modulo 10 (A131297(n)=digital sum analog base 11 of the Fibonacci recurrence). - Hieronymus Fischer, Jul 01 2007 For general bases p>1, we have the inequality 1<=a(n)<=2p-2 (for n>0). Actually, a(n)<=18. - Hieronymus Fischer, Jul 01 2007 LINKS Harvey P. Dale, Table of n, a(n) for n = 1..1000 FORMULA a(n)=a(n-1)+a(n-2)-10*(floor(a(n-1)/10)+floor(a(n-2)/10)). This is valid, since a(n)<100. - Hieronymus Fischer, Jul 01 2007 a(n)=ds_10(a(n-1))+ds_10(a(n-2))-(floor(a(n-1)/10)+floor(a(n-2)/10)) where ds_10(x) is the digital sum of x in base 10. - Hieronymus Fischer, Jul 01 2007 a(n)=(a(n-1)mod 10)+(a(n-2)mod 10)=A010879(a(n-1))+A010879(a(n-2)). - Hieronymus Fischer, Jul 01 2007 a(n)=A131297(n) if A131297(n)<=10. - Hieronymus Fischer, Jul 01 2007 a(n)=Fib(n)-10*sum{1

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Last modified April 25 12:19 EDT 2019. Contains 322456 sequences. (Running on oeis4.)