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A038148
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Number of 3-infinitary divisors of n: if n = Product p(i)^r(i) and d = Product p(i)^s(i), each s(i) has a digit a <= b in its ternary expansion everywhere that the corresponding r(i) has a digit b, then d is a 3-infinitary-divisor of n.
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8
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1, 2, 2, 3, 2, 4, 2, 2, 3, 4, 2, 6, 2, 4, 4, 4, 2, 6, 2, 6, 4, 4, 2, 4, 3, 4, 2, 6, 2, 8, 2, 6, 4, 4, 4, 9, 2, 4, 4, 4, 2, 8, 2, 6, 6, 4, 2, 8, 3, 6, 4, 6, 2, 4, 4, 4, 4, 4, 2, 12, 2, 4, 6, 3, 4, 8, 2, 6, 4, 8, 2, 6, 2, 4, 6, 6, 4, 8, 2, 8, 4, 4, 2, 12, 4, 4, 4, 4, 2, 12, 4, 6, 4, 4, 4, 12, 2, 6, 6, 9, 2, 8, 2, 4, 8
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OFFSET
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1,2
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COMMENTS
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Multiplicative: If e = sum d_k 3^k, then a(p^e) = prod (d_k+1). - Christian G. Bower, May 19 2005
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LINKS
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FORMULA
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EXAMPLE
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2^3*3 is a 3-infinitary-divisor of 2^5*3^2 because 2^3*3 = 2^10*3^1 and 2^5*3^2 = 2^12*3^2 in ternary expanded power. All corresponding digits satisfy the condition. 1 <= 1, 0 <= 2, 1 <= 2.
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MAPLE
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A038148 := proc(n) if n= 1 then 1; else ifa := ifactors(n)[2] ;
a := 1; for f in ifa do e := convert(op(2, f), base, 3) ; a := a*mul(d+1, d=e) ; end do: end if; end proc:
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MATHEMATICA
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a[1] = 1; a[n_] := (k = 1; Do[k = k * Times @@ (IntegerDigits[f, 3] + 1), {f, FactorInteger[n][[All, 2]]}]; k); Table[a[n], {n, 1, 102}](* Jean-François Alcover, Feb 03 2012, after R. J. Mathar *)
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PROG
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(PARI)
A006047(n) = { my(m=1, d); while(n, d = (n%3); m *= (1+d); n = (n-d)/3); m; };
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CROSSREFS
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KEYWORD
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nonn,nice,easy,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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