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%I #34 Apr 02 2015 05:26:41
%S 2,4,6,8,12,14,16,18,20,24,26,28,30,32,38,40,42,44,48,50,52,54,56,60,
%T 62,64,66,68,72,74,80,84,86,88,90,92,96,98,100,102,104,108,110,112,
%U 114,120,122,124,126,128,132,134,138,140,144,146,148,150,152,158,160,164,168,170,172,174
%N Numbers without superdivisors: numbers n such that n/k + n fails to divide at least one of (n/k)^(n/k) + n, (n/k)^n + n/k or n^(n/k) + n/k for any divisor k of n.
%C Zerosuperdivisor numbers. Numbers n such that A247477(n) = 0.
%C A000027 = zerosuperdivisor numbers U onesuperdivisor numbers U twosuperdivisor numbers U threesuperdivisor numbers U ...
%C Conjecture: Perfect numbers (A000396) are zerosuperdivisor numbers.
%C Conjecture: Average of twin prime pairs (A014574) are zerosuperdivisor numbers.
%C None of these numbers are odd or 10 mod 12 or 36 mod 40 or 78 mod 84 or 136 mod 144 or ... - _Charles R Greathouse IV_, Feb 19 2015
%H Michael De Vlieger and Charles R Greathouse IV, <a href="/A254748/b254748.txt">Table of n, a(n) for n = 1..10000</a> (first 3592 terms from Michael De Vlieger)
%e 2 is in this sequence because 2/1 + 2 does not divide (2/1)^(2/1) + 2, (2/1)^2 + 2/1, 2^(2/1) + 2/1 and 2/2 + 2 does not divide (2/2)^(2/2) + 2, (2/2)^2 + 2/2, 2^(2/2) + 2/2: 4 does not divide 6, 6, 6 and 3 does not divide 3, 2, 3.
%t superdivisors[n_] := Select[Range@ n, And[Mod[(n/#)^(n/#) + n, n/# + n] == 0, Mod[(n/#)^n + n/#, n/# + n] == 0, Mod[n^(n/#) + n/#, n/# + n] == 0] &] /. {} -> 0; Position[Array[superdivisors, 174], 0] // Flatten (* _Michael De Vlieger_, Feb 09 2015 *)
%o (PARI) is(n)=fordiv(n,d,my(m=n/d,k=d+n); if(Mod(d,k)^d==-n && Mod(d,k)^n==-d && Mod(n,k)^d==-d, return(0))); 1 \\ _Charles R Greathouse IV_, Feb 19 2015
%Y Cf. A000027, A000396, A014574, A247477 (definition of superdivisor).
%K nonn,easy
%O 1,1
%A _Juri-Stepan Gerasimov_, Feb 07 2015
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