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

%I A295770 #22 Nov 23 2023 13:23:53
%S A295770 70,350,4970,5110,7070,8890,9590,9730,13790,15610,15890,16030,17990,
%T A295770 18410,19810,21770,22190,23170,24290,25550,26530,26810,27230,28070,
%U A295770 30310,32270,32690,33530,34930,36470,38990,39830,40390,43190,44450,45010,48650,49070,49630,51730
%N A295770 Numbers k such that Bernoulli number B_{k} has denominator 4686.
%C A295770 4686 = 2*3*11*71.
%C A295770 All terms are multiples of a(1) = 70.
%C A295770 For these numbers numerator(B_{k}) mod denominator(B_{k}) = 289.
%H A295770 Seiichi Manyama, <a href="/A295770/b295770.txt">Table of n, a(n) for n = 1..1000</a>
%e A295770 Bernoulli B_{70} is 1505381347333367003803076567377857208511438160235/4686, hence 70 is in the sequence.
%p A295770 with(numtheory): P:=proc(q, h) local n;  for n from 2 by 2 to q do
%p A295770 if denom(bernoulli(n))=h then print(n); fi; od; end: P(10^6,4686);
%p A295770 # Alternative: # according to Robert Israel code in A282773
%p A295770 with(numtheory): filter:= n ->
%p A295770 select(isprime, map(`+`, divisors(n), 1)) = {2, 3, 11, 71}:
%p A295770 select(filter, [seq(i, i=1..10^5)]);
%t A295770 70 Position[Array[Denominator@ BernoulliB[70 #] &, 10^3], 4686][[All, 1]] (* _Michael De Vlieger_, Nov 27 2017 *)
%t A295770 Select[70*Range[750],Denominator[BernoulliB[#]]==4686&] (* _Harvey P. Dale_, Nov 23 2023 *)
%o A295770 (PARI) isok(n) = denominator(bernfrac(n)) == 4686; \\ _Michel Marcus_, Nov 27 2017
%o A295770 (PARI) lista(nn) = forstep(n=70, nn, 70, if(denominator(bernfrac(n)) == 4686, print1(n, ", "))) \\ _Iain Fox_, Nov 27 2017
%Y A295770 Cf. A045979, A051222, A051225, A051226, A051227, A051228, A051229, A051230, A119456, A119480, A249134, A255684, A271634, A271635, A272138, A272139, A272140, A272183, A272184, A272185, A272186, A272369.
%K A295770 nonn,easy
%O A295770 1,1
%A A295770 _Paolo P. Lava_, Nov 27 2017