A045979 -id:A045979 - cp's OEIS frontend

A295770 Numbers k such that Bernoulli number B_{k} has denominator 4686.

70, 350, 4970, 5110, 7070, 8890, 9590, 9730, 13790, 15610, 15890, 16030, 17990, 18410, 19810, 21770, 22190, 23170, 24290, 25550, 26530, 26810, 27230, 28070, 30310, 32270, 32690, 33530, 34930, 36470, 38990, 39830, 40390, 43190, 44450, 45010, 48650, 49070, 49630, 51730

Offset: 1

Paolo P. Lava, Nov 27 2017

4686 = 2*3*11*71.
All terms are multiples of a(1) = 70.
For these numbers numerator(B_{k}) mod denominator(B_{k}) = 289.

			Bernoulli B_{70} is 1505381347333367003803076567377857208511438160235/4686, hence 70 is in the sequence.

Seiichi Manyama, Table of n, a(n) for n = 1..1000

Cf. A045979, A051222, A051225, A051226, A051227, A051228, A051229, A051230, A119456, A119480, A249134, A255684, A271634, A271635, A272138, A272139, A272140, A272183, A272184, A272185, A272186, A272369.

with(numtheory): P:=proc(q, h) local n;  for n from 2 by 2 to q do
if denom(bernoulli(n))=h then print(n); fi; od; end: P(10^6,4686);
# Alternative: # according to Robert Israel code in A282773
with(numtheory): filter:= n ->
select(isprime, map(`+`, divisors(n), 1)) = {2, 3, 11, 71}:
select(filter, [seq(i, i=1..10^5)]);

70 Position[Array[Denominator@ BernoulliB[70 #] &, 10^3], 4686][[All, 1]] (* Michael De Vlieger, Nov 27 2017 *)
Select[70*Range[750],Denominator[BernoulliB[#]]==4686&] (* Harvey P. Dale, Nov 23 2023 *)

isok(n) = denominator(bernfrac(n)) == 4686; \\ Michel Marcus, Nov 27 2017

lista(nn) = forstep(n=70, nn, 70, if(denominator(bernfrac(n)) == 4686, print1(n, ", "))) \\ Iain Fox, Nov 27 2017

A337596 Largest m such that k^n (mod m) is always either 0, +1, or -1.

Original entry on oeis.org

3, 5, 9, 16, 11, 13, 4, 32, 27, 25, 23, 16, 4, 29, 31, 64, 4, 37, 4, 41, 49, 23, 47, 32, 11, 53, 81, 29, 59, 61, 4, 128, 67, 8, 71, 73, 4, 8, 79, 41, 83, 49, 4, 89, 31, 47, 4, 97, 4, 125, 103, 53, 107, 109, 121, 113, 9, 59, 4, 61, 4, 8, 127, 256, 131, 67, 4, 137

Offset: 1

Elliott Line, Sep 02 2020

nonn

For a given n, for all k, k^n mod a(n) will always be either 0, 1 or a(n)-1. This will not be true for numbers larger than a(n).

It appears that a(m) = 4 for m in A045979. - Michel Marcus, Sep 04 2020

			For n = 5 all fifth powers of natural numbers: 1,32,243,1024, etc. are either a multiple of 11, or 1 greater or 1 less than a multiple of 11. There is no greater number than 11 for which all fifth powers are at most 1 different from a multiple. So a(5) = 11.

Cf. A010872, A070430, A167176, A000035, A070596, A070636, A109718.

Cf. residues: A096008 (for n=2), A096087 (for n=3).

More terms from Michel Marcus, Sep 04 2020

cp's OEIS Frontend

A295770 Numbers k such that Bernoulli number B_{k} has denominator 4686.

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