cp's OEIS Frontend

A247477 Largest m such that n/m + n divides (n/m)^(n/m) + n, (n/m)^n + n/m and n^(n/m) + n/m, or 0 if no such m exists.

%I A247477 #67 Jun 14 2025 23:48:30
%S A247477 1,0,1,0,1,0,1,0,1,2,1,0,1,0,1,0,1,0,1,0,1,2,1,0,1,0,1,0,1,0,1,0,1,2,
%T A247477 1,4,1,0,1,0,1,0,1,0,1,2,1,0,1,0,1,0,1,0,1,0,1,2,1,0,1,0,1,0,1,0,1,0,
%U A247477 1,2,1,0,1,0,1,4,1,6,1,0,1,2,1,0,1,0,1,0,1,0,1,0,1,2
%N A247477 Largest m such that n/m + n divides (n/m)^(n/m) + n, (n/m)^n + n/m and n^(n/m) + n/m, or 0 if no such m exists.
%C A247477 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.
%C A247477 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, ...
%C A247477 Conjecture:
%C A247477 1 = odd superdivisor of 2n + 1 (or A005408(n));
%C A247477 m = even superdivisor of m*(2m + 2)*n + m*(2m + 1).
%C A247477 That is,
%C A247477 2 = even superdivisor of 12n + 10 (or A017641(n)),
%C A247477 4 = even superdivisor of 40n + 36,
%C A247477 ...
%C A247477 Smallest n with more than 1 superdivisor is n = 406 with superdivisors {2, 14}. - _Michael De Vlieger_, Feb 09 2015
%C A247477 Smallest k such that number of superdivisors of k is equal to n: 2, 1, 406, 2926, ... - _Juri-Stepan Gerasimov_, Feb 12 2015
%C A247477 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
%C A247477 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
%C A247477 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
%H A247477 Michael De Vlieger, <a href="/A247477/b247477.txt">Table of n, a(n) for n = 1..10000</a>
%e A247477 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.
%t A247477 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 *)
%o A247477 (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
%Y A247477 Cf. A000217, A005408, A017641, A254748.
%K A247477 nonn
%O A247477 1,10
%A A247477 _Juri-Stepan Gerasimov_, Jan 19 2015