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A254748
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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.
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4
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2, 4, 6, 8, 12, 14, 16, 18, 20, 24, 26, 28, 30, 32, 38, 40, 42, 44, 48, 50, 52, 54, 56, 60, 62, 64, 66, 68, 72, 74, 80, 84, 86, 88, 90, 92, 96, 98, 100, 102, 104, 108, 110, 112, 114, 120, 122, 124, 126, 128, 132, 134, 138, 140, 144, 146, 148, 150, 152, 158, 160, 164, 168, 170, 172, 174
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OFFSET
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1,1
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COMMENTS
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Zerosuperdivisor numbers. Numbers n such that A247477(n) = 0.
A000027 = zerosuperdivisor numbers U onesuperdivisor numbers U twosuperdivisor numbers U threesuperdivisor numbers U ...
Conjecture: Perfect numbers (A000396) are zerosuperdivisor numbers.
Conjecture: Average of twin prime pairs (A014574) are zerosuperdivisor numbers.
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
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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