A272858 Numbers m such that Product(1 + p_i) = Product(1 + e_i), where m = Product((p_i)^e_i).
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
Offset: 1
Keywords
Examples
92160 is included because 92160 = 2^11 * 3^2 * 5 and (2+1)*(3+1)*(5+1) = (11+1)*(2+1)*(1+1).
Links
- Giuseppe Coppoletta and Giovanni Resta, Table of n, a(n) for n = 1..6058 (terms < 10^18, first 100 terms from G. Coppoletta)
Programs
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Mathematica
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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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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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)]
Formula
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.
Comments