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A272858
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Numbers m such that Product(1 + p_i) = Product(1 + e_i), where m = Product((p_i)^e_i).
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3
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1, 4, 27, 72, 96, 108, 486, 800, 1280, 3125, 6272, 10976, 12500, 14336, 21600, 28800, 30375, 34560, 36000, 38880, 48600, 54675, 84375, 92160, 96000, 121500, 134456, 153600, 169344, 217728, 218700, 225000, 247808, 262440, 296352, 300000, 337500, 340736, 387072, 395136, 489888, 666792, 703125, 750141, 781250, 823543, 857304, 885735
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OFFSET
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1,2
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COMMENTS
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A048102 is clearly a subsequence, as for any prime p, p^p satisfy the herein condition. Similarly, A122406 is also a subsequence. More generally, if a number is a term, then any permutation of the exponents in its prime factorization (i.e., any permutation of its prime signature) gives also a term.
The condition defining this sequence coincides with the condition in A272859 at least for the terms of A114129.
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LINKS
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FORMULA
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If N is a positive integer and N = Product_{i=1..k} (p_i)^e_i is its prime factorization, then N is in A272858 iff Product_{i=1..k} (1 + p_i) = Product_{i=1..k} (1 + e_i).
For a number with three different prime factors N = p1^e1 * p2^e2 * p3^e3, the defining condition can be expressed as: p1 + p2 + p3 + p1*p2 + p1*p3 + p2*p3 + p1*p2*p3 = e1 + e2 + e3 + e1*e2 + e1*e3 + e2*e3 + e1*e2*e3.
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EXAMPLE
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92160 is included because 92160 = 2^11 * 3^2 * 5 and (2+1)*(3+1)*(5+1) = (11+1)*(2+1)*(1+1).
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MATHEMATICA
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ok[n_] := Block[{p, e}, {p, e} = Transpose@ FactorInteger@ n; Times @@ (1+p) == Times @@ (1+e)]; Select[Range[10^6], ok] (* Giovanni Resta, May 08 2016 *)
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PROG
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(Sage)
def d(n):
v = factor(n)
d1 = prod(1 + w[0] for w in v)
d2 = prod(1 + w[1] for w in v)
return d1 == d2
[k for k in (1..10000) if d(k)]
(PARI) is(n)=my(f=factor(n)); prod(i=1, #f~, f[i, 1]+1)==prod(i=1, #f~, f[i, 2]) \\ Charles R Greathouse IV, Sep 08 2016
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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