1,10

Call 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. Then a(n) is the largest superdivisor of n, or 0 if n has no superdivisors.

Conjecture: smallest k such that k/m = n and k/m + k divides (k/m)^(k/m) + k, (k/m)^k + k/m, k^(k/m) + k/m, or 0 if no such k exists: 2, 1, 10, 0, 36, 0, 78, 0, 136, 0, 210, 0, 312, 0, 406, 0, ...

Conjecture:

1 = odd superdivisor of 2n + 1 (or A005408(n));

m = even superdivisor of m*(2m + 2)*n + m*(2m + 1).

That is,

2 = even superdivisor of 12n + 10 (or A017641(n)),

4 = even superdivisor of 40n + 36,

...

Smallest n with more than 1 superdivisor is n = 406 with superdivisors {2, 14}. - Michael De Vlieger, Feb 09 2015

Smallest k such that number of superdivisors of k is equal to n: 2, 1, 406, 2926, ... - Juri-Stepan Gerasimov, Feb 12 2015

Conjecture: the superdivisor constant is equal to 1/2 + sum_{n >= 1} 1/(4*A000217(2n)) - Sum_{n >= 1} 1/b(n) - Sum_{n >= 1} 1/c(n) - Sum_{n >= 1} 1/d(n), ... = 0.64.., where b(n) = numbers with 2 superdivisors {or 406, 430, 646, 666, 826, 1090, 1236, 1246, 1378, 1596, 1666, 1750, 2002, 2028, 2346, 2410, 2506, 2782, 2796, 2850, ...), c(n) = numbers with 3 superdivisors {or 2926, ...), d(n) = numbers with 4 superdivisors, ... - Juri-Stepan Gerasimov, Feb 18 2015

A000027 = A254748 U 1-superdivisor numbers U 2-superdivisor numbers U 3-superdivisor numbers U 4-superdivisor numbers U ... - Juri-Stepan Gerasimov, Feb 19 2015

Let n = k*d with d odd. Then, k is a superdivisor of n iff d^(d-1) == 1 (mod k+1) and d^(k-1) == -1 (mod k+1). (Sometimes the numbers d are called the superdivisors of n, as in A272538 and possibly A254748.) - Charlie Neder, Jun 02 2019

Michael De Vlieger, Table of n, a(n) for n = 1..10000

a(10) = 2 because 10/2 + 10 = 15 divides (10/2)^(10/2) + 10 = 3135, (10/2)^10 + 10/2 = 9765630, 10^(10/2) + 10/2 = 100005, i.e., 3135/15 = 209, 9765630/15 = 651042, 100005/15 = 6667.

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; Min /@ Array[superdivisors, 94] (* Michael De Vlieger, Feb 09 2015 *)

(PARI) a(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(m))); 0 \\ Charles R Greathouse IV, Feb 19 2015

Cf. A000217, A005408, A017641, A254748.

Sequence in context: A307196 A123737 A083037 * A072927 A307194 A331218

Adjacent sequences: A247474 A247475 A247476 * A247478 A247479 A247480

Juri-Stepan Gerasimov, Jan 19 2015

approved