A211781
Numbers n such that Sum_(d_A000005(x).
2, 3, 4, 5, 7, 11, 13, 17, 19, 23, 27, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, 199, 211, 223, 225, 227, 229, 233, 239, 241, 251, 252
Offset: 1
Keywords
Examples
Number 225 is in sequence because 1/1 + 2/3 + 2/5 + 3/9 + 4/15 + 3/25 + 6/45 + 6/75 = 3 (integer).
Programs
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Mathematica
t = {}; Do[d2 = Sum[DivisorSigma[0,d]/d, {d, Most[Divisors[n]]}]; If[IntegerQ[d2], AppendTo[t, n]], {n, 2, 252}]; t (* T. D. Noe, Apr 26 2012 *)
Comments