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A339012
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Written in factorial base, n ends in a(n) consecutive non-0 digits.
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3
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0, 1, 0, 2, 0, 2, 0, 1, 0, 3, 0, 3, 0, 1, 0, 3, 0, 3, 0, 1, 0, 3, 0, 3, 0, 1, 0, 2, 0, 2, 0, 1, 0, 4, 0, 4, 0, 1, 0, 4, 0, 4, 0, 1, 0, 4, 0, 4, 0, 1, 0, 2, 0, 2, 0, 1, 0, 4, 0, 4, 0, 1, 0, 4, 0, 4, 0, 1, 0, 4, 0, 4, 0, 1, 0, 2, 0, 2, 0, 1, 0, 4, 0, 4, 0, 1, 0
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OFFSET
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0,4
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COMMENTS
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Also, a(n) is the least p for which n mod (p+2)! < (p+1)!. A small remainder like this means a 0 digit at position p in the factorial base representation of n, where the least significant digit is position p=0. The least such p means only nonzero digits below.
Those n with a(n)=p are characterized by remainders n mod (p+2)!, per the formula below. These remainders are terms of A227157 which is factorial base digits all nonzero. A227157 can be taken by rows where row p lists the terms having p digits in factorial base. Each digit ranges from 1 up to 1,2,3,... respectively so there are p! values in a row, and so the asymptotic density of terms p here is p!/(p+2)! = 1/((p+2)*(p+1)) = 1/A002378(p+1) = 1/2, 1/6, etc.
The smallest n with a(n)=p is the factorial base repunit n = 11..11 with p 1's = A007489(p).
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LINKS
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FORMULA
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a(n)=p iff n mod (p+2)! is a term in row p of A227157 (row p terms having p digits), including p=0 by reckoning an initial A227157(0) = 0 as no digits.
a(n)=0 iff n mod 2 = 0.
a(n)=1 iff n mod 6 = 1, which is A016921.
a(n)=2 iff n mod 24 = 3 or 5.
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EXAMPLE
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n = 10571 written in factorial base is 2,0,4,0,1,2,1. It ends in 3 consecutive nonzero digits (1,2,1) so a(10571) = 3. Remainder 10571 mod (3+2)! = 11 is in A227157 row 3.
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MATHEMATICA
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a[n_] := Module[{k = n, m = 2, r}, While[{k, r} = QuotientRemainder[k, m]; r != 0, m++]; m - 2]; Array[a, 30, 0] (* Amiram Eldar, Feb 15 2021 after Kevin Ryde's PARI code *)
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PROG
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(PARI) a(n) = my(b=2, r); while([n, r]=divrem(n, b); r!=0, b++); b-2;
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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