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%I #19 Jun 18 2021 14:47:05
%S 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,
%T 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,
%U 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
%N a(n) = M(a(n-1)) + M(a(n-2)) where a(1)=a(2)=1 and M(k) is the product of the digits of k in base 10.
%C Periodic with least period 60. - Christopher N. Swanson (cswanson(AT)ashland.edu), Jul 22 2003
%C From _Hieronymus Fischer_, Jul 01 2007: (Start)
%C The digital product analog (in base 10) of the Fibonacci recurrence.
%C 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.
%C a(n)==A131297(n) modulo 10 (A131297(n)=digital sum analog base 11 of the Fibonacci recurrence).
%C For general bases p>1, we have the inequality 1<=a(n)<=2p-2 (for n>0). Actually, a(n)<=18.
%C (End)
%H Harvey P. Dale, <a href="/A074867/b074867.txt">Table of n, a(n) for n = 1..1000</a>
%F From _Hieronymus Fischer_, Jul 01 2007: (Start)
%F 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.
%F 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.
%F a(n) = (a(n-1)mod 10)+(a(n-2)mod 10) = A010879(a(n-1))+A010879(a(n-2)).
%F a(n) = A131297(n) if A131297(n)<=10.
%F a(n) = Fib(n)-10*sum{1<k<n, Fib(n-k+1)*floor(a(k)/10)} where Fib(n)=A000045(n).
%F a(n) = A000045(n)-10*sum{1<k<n, A000045(n-k+1)*A059995(a(k))}. (End)
%t nxt[{a_,b_}]:={b,Times@@IntegerDigits[a]+Times@@IntegerDigits[b]}; Transpose[ NestList[nxt,{1,1},90]][[1]] (* _Harvey P. Dale_, Feb 01 2015 *)
%Y Cf. A000045, A010073, A010074, A010075, A010076, A131294, A131295, A131296, A131297, A131318, A131319, A131320.
%K base,easy,nonn
%O 1,3
%A _Felice Russo_, Sep 11 2002
%E More terms from Christopher N. Swanson (cswanson(AT)ashland.edu), Jul 22 2003
%E Definition adapted to offset by _Georg Fischer_, Jun 18 2021
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