A247477 -id:A247477 - cp's OEIS frontend

A254748 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.

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

Offset: 1

Juri-Stepan Gerasimov, Feb 07 2015

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

			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.

Michael De Vlieger and Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 3592 terms from Michael De Vlieger)

Cf. A000027, A000396, A014574, A247477 (definition of superdivisor).

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 *)

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

A272538 Call n/m a superdivisor of n if n/m + n divides (n/m)^(n/m) + n, (n/m)^n + n/m and n^(n/m) + n/m. This sequence gives the smallest number with n superdivisors.

Original entry on oeis.org

2, 1, 406, 2926, 81810, 521626, 4276350, 1590715126, 266048146

Offset: 0

Juri-Stepan Gerasimov, Jan 19 2015

			a(0) = 2 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.
a(0) = 2 with superdivisors {}.
a(1) = 1 with superdivisor {1}.
a(2) = 406 with superdivisors {29, 203}.
a(3) = 2926 with superdivisors {77, 209, 1463}.
a(4) = 81810 with superdivisors {405, 909, 2727, 13635}.
a(5) = 521626 with superdivisors {4921, 7049, 13727, 37259, 260813}.
a(6) = 4276350 with superdivisors {2925, 16575, 54825, 237575, 427635, 712725}.
a(7) = 1590715126 with superdivisors {607607, 939029, 4253249, 10329319, 61181351, 113622509, 795357563}.
a(8) = 266048146 with superdivisors {538559, 655291, 1000181, 1461803, 2714777, 10232621, 19003439, 133024073}.

Cf. A247477, A254748.

superdivisors(n)=select(d->Mod(d, d+n)^d==-n && Mod(d, d+n)^n==-d && Mod(n, d+n)^d==-d, divisors(n))
a(n)=my(k); while(#superdivisors(k++)!=n, ); k \\ Charles R Greathouse IV, May 06 2016

a(4)-a(6) from Charles R Greathouse IV, Feb 19 2015.
Edited by Charles R Greathouse IV, May 06 2016
a(7)-a(8) from Charles R Greathouse IV, May 23 2016

cp's OEIS Frontend

A254748 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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