A272538 -id:A272538 - 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

A371513 a(n) is the smallest number m with n divisors d such that d^m mod m = d.

Original entry on oeis.org

1, 2, 6, 42, 30, 105, 910, 561, 1365, 5005, 5565, 11305, 36465, 140505, 239785, 41041, 682465, 873145, 185185, 418285, 1683969, 2113665, 5503785, 1242241, 6697405, 8549905, 31932901, 11996985, 31260405, 30534805, 47031061, 825265, 27265161, 32306365, 55336645, 21662641

Offset: 0

Juri-Stepan Gerasimov, Apr 10 2024

nonn

			a(0) = 1 with divisors {};
a(1) = 2 with divisor {1};
a(2) = 6 with divisors {1, 3};
a(3) = 42 with divisors {1, 7, 21};
a(4) = 30 with divisors {1, 6, 10, 15};
a(5) = 105 with divisors {1, 7, 15, 21, 35};
a(6) = 910 with divisors {1, 35, 65, 91, 130, 455};
a(7) = 561 with divisors {1, 3, 11, 17, 33, 51, 187};
a(8) = 1365 with divisors {1, 13, 21, 91, 105, 195, 273, 455};
a(9) = 5005 with divisors {1, 11, 55, 65, 77, 143, 385, 715, 1001};
a(10) = 5565 with divisors {1, 7, 15, 21, 35, 105, 265, 371, 1113, 1855};
a(11) = 11305 with divisors {1, 17, 19, 35, 85, 119, 323, 595, 665, 1615, 2261}.

Cf. A182816, A272538, A279024, A371883, A371884.

f[n_] := DivisorSum[n, 1 &, PowerMod[#, n, n] == # &]; seq[max_] := Module[{t = Table[0, {max}], c = 0, n = 1, i}, While[c < max, i = f[n] + 1; If[i <= max && t[[i]] == 0, c++; t[[i]] = n]; n++]; t]; seq[18] (* Amiram Eldar, Apr 11 2024 *)

from sympy import divisors
from itertools import count, islice
def f(n, divs): return sum(1 for d in divs if pow(d, n, n) == d%n)
def agen(verbose=False): # generator of terms
    adict, n = dict(), 0
    for k in count(1):
        divs = divisors(k)[1:]
        if len(divs) < n: continue
        v = f(k, divs)
        if v not in adict:
            adict[v] = k
            if verbose: print("FOUND", v, k)
        while n in adict: yield adict[n]; n += 1
print(list(islice(agen(), 15))) # Michael S. Branicky, Apr 10 2024, updated Apr 17 2024 after Jon E. Schoenfield

a(12)-a(25) from Michael S. Branicky, Apr 10 2024

a(26)-a(35) from Jon E. Schoenfield, Apr 10 2024

A279680 Definition: m < n is an extradivisor of n if for some positive k < n, m | n | k^(n+1) + m and n | (n-k)^(n+1) + m. This sequence gives the smallest number with n extradivisors.

Original entry on oeis.org

1, 2, 45, 105, 1365, 1305, 4305, 11445

Offset: 0

Juri-Stepan Gerasimov, Dec 16 2016

			a(0) = 1 with extradivisors {};
a(1) = 2 with extradivisor {1};
a(2) = 45 with extradivisors {5, 9};
a(3) = 105 with extradivisors {5, 21, 35};
a(4) = 1365 with extradivisors {35, 105, 195, 455};
a(5) = 1305 with extradivisors {5, 9, 29, 45, 261}.

Cf. A272538, A279024.

First /@ Values@ KeySort@ PositionIndex@ Table[Count[DeleteCases[Most@ Divisors@ n, d_ /; EvenQ@ d], m_ /; Total@ Boole@ Map[Function[k, And[Mod[PowerMod[k, (n + 1), n] + m, n] == 0, Mod[PowerMod[(n - k), (n + 1), n] + m, n] == 0]], Range[n - 1]] > 0], {n, 1500}] (* Michael De Vlieger, Dec 17 2016, Version 10 *)

a(3)-a(7) from Michael De Vlieger, Dec 07 2016

Definition edited by N. J. A. Sloane, Jun 19 2020

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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A371513 a(n) is the smallest number m with n divisors d such that d^m mod m = d.

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A279680 Definition: m < n is an extradivisor of n if for some positive k < n, m | n | k^(n+1) + m and n | (n-k)^(n+1) + m. This sequence gives the smallest number with n extradivisors.

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