A254748 -id:A254748 - 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.

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, 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, 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

Offset: 1

Juri-Stepan Gerasimov, Jan 19 2015

nonn

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

			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.

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

Cf. A000217, A005408, A017641, A254748.

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 *)

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

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

A272887 Number of ways to write prime(n) as (4x + 2)y + 4*x + 1 where x and y are nonnegative integers.

Original entry on oeis.org

0, 1, 2, 1, 2, 2, 3, 2, 2, 4, 1, 2, 4, 2, 2, 4, 4, 2, 2, 3, 2, 2, 4, 6, 3, 4, 2, 4, 4, 4, 1, 4, 4, 4, 6, 2, 2, 2, 4, 4, 6, 4, 2, 2, 6, 3, 2, 2, 4, 4, 6, 4, 3, 6, 4, 4, 8, 2, 2, 4, 2, 6, 4, 4, 2, 4, 2, 3, 4, 6, 4, 6, 2, 4, 4, 2, 8, 2, 4, 4, 8, 2, 4, 4, 4, 4, 9, 2, 8, 2, 6, 4, 2, 4, 4, 6, 8, 6, 2, 2, 2, 6, 4, 8, 4

Offset: 1

Juri-Stepan Gerasimov, May 16 2016

nonn

Number of distinct values of k such that k/p_n + k divides (k/p_n)^(k/p_n) + k, (k/p_n)^k + k/p_n and k^(k/p_n) + k/p_n where p_n = prime(n) is n-th prime.
a(1) = 0, a(n+1) = number of odd divisors of 1+prime(n+1).
Conjectures:
1) a(Fermat prime(n)) >= n, i.e. a(A019434(1)=3) = 1, a(A019434(2)=5) = 2, a(A019434(3)=17) = 3, a(A019434(4)=257) = 4, a(A019434(5)=65537) = 12 > 5, ...
2) a(2^(2^n)+1) > n;
3) a(2^(2^n)+1) < a(2^(2^(n+1))+1).

			a(3) = 2 because (4*0+2)*2+4*0+1 = 5 for (x=0, y=2) and (4*1+2)*0+4*1+1 = 5 for (x=1, y=0) where 5 is the 3rd prime.

Cf. A000215 (Fermat numbers), A001227, A000668 (Mersenne primes n such that a(n)=1), A019434 (Fermat primes), A069283, A192869 (primes n such that a(n) = 1 or 2), A206581 (primes n such that a(n)=2), A254748.

{0}~Join~Table[Count[Divisors[Prime[n + 1] + 1], ?OddQ], {n, 120}] (* _Michael De Vlieger, Jun 24 2016 *)

a(n+1) = A001227(A000040(n+1) + 1).

More terms from Alois P. Heinz, May 17 2016

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.

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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.

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A272887 Number of ways to write prime(n) as (4x + 2)y + 4*x + 1 where x and y are nonnegative integers.

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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.

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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.

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A272887 Number of ways to write prime(n) as (4*x + 2)*y + 4*x + 1 where x and y are nonnegative integers.

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Formula

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A272887 Number of ways to write prime(n) as (4x + 2)y + 4*x + 1 where x and y are nonnegative integers.