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A074867
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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.
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3
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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
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OFFSET
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1,3
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COMMENTS
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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.
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.
a(n)==A131297(n) modulo 10 (A131297(n)=digital sum analog base 11 of the Fibonacci recurrence).
For general bases p>1, we have the inequality 1<=a(n)<=2p-2 (for n>0). Actually, a(n)<=18.
(End)
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LINKS
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FORMULA
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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.
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.
a(n) = (a(n-1)mod 10)+(a(n-2)mod 10) = A010879(a(n-1))+A010879(a(n-2)).
a(n) = Fib(n)-10*sum{1<k<n, Fib(n-k+1)*floor(a(k)/10)} where Fib(n)=A000045(n).
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MATHEMATICA
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nxt[{a_, b_}]:={b, Times@@IntegerDigits[a]+Times@@IntegerDigits[b]}; Transpose[ NestList[nxt, {1, 1}, 90]][[1]] (* Harvey P. Dale, Feb 01 2015 *)
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CROSSREFS
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Cf. A000045, A010073, A010074, A010075, A010076, A131294, A131295, A131296, A131297, A131318, A131319, A131320.
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KEYWORD
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base,easy,nonn
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AUTHOR
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EXTENSIONS
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More terms from Christopher N. Swanson (cswanson(AT)ashland.edu), Jul 22 2003
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STATUS
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approved
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