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A091853
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a(n) is the number of numbers k such that k is obtained by permuting the digits of n and gcd(n,k) > 1.
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2
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0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 6, 1, 4, 6
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OFFSET
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1,12
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COMMENTS
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For a number like 12345 which is a multiple of three and does not contain zero, all 5! = 120 permutations yield a distinct number divisible by 3, thus a(12345) = 120. However, 120 occurs for the first time at n = 10236, which is also a multiple of three. - Antti Karttunen, Dec 16 2017
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LINKS
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EXAMPLE
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1 is the only number obtained by permuting the digits of 1, gcd(1,1) = 1, hence a(1) = 0.
001, 010, 100 are the numbers obtained by permuting the digits of 100, gcd(100,1) = 1, gcd(100,10) = 10, gcd(100,100) = 100, hence a(100) = 2.
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MATHEMATICA
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Table[Count[Union@ Map[# Boole[! CoprimeQ[#, n]] &@ FromDigits@ # &, Permutations@ IntegerDigits@ n], _?(# > 0 &)], {n, 105}] (* Michael De Vlieger, Dec 16 2017 *)
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PROG
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(PARI) A091853(n) = { my(digs=digits(n), nd=length(digs), k, p, s = Set([])); for(j=0, nd!-1, p=numtoperm(nd, j); if(1<gcd(n, k=fromdigits(vector(nd, i, digs[p[i]]))), s = setunion(Set([k]), s))); length(s); }; \\ Antti Karttunen, Dec 16 2017
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CROSSREFS
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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