cp's OEIS Frontend

A266214 Numbers n that are not coprime to the numerator of zeta(2*n)/(Pi^(2*n)).

%I A266214 #20 Dec 23 2024 14:53:44
%S A266214 14,22,26,28,30,38,42,44,46,50,52,54,56,58,60,62,70,74,76,78,82,84,86,
%T A266214 88,90,92,94,98,100,102,104,106,108,110,112,114,116,118,120,122,124,
%U A266214 126,134,138,140,142,146,148,150,152,154,156,158,162,164,166,168,170
%N A266214 Numbers n that are not coprime to the numerator of zeta(2*n)/(Pi^(2*n)).
%C A266214 Equivalently, n not coprime to the numerator of 2^(2n-1)*Bernoulli(2n)/(2n)! (see _Lekraj Beedassy_ comment in A046988).
%C A266214 Conjecture 1: for n>=1, a(n) is identical to 2*A072823(n+1).
%C A266214 Conjecture 2: The corresponding GCDs are powers of 2.
%C A266214 Verified for n <= 10000, e.g.,
%C A266214 GCD = 2 for 14, 22, 26, 28, 30, 38, 42, 44, 46, 50, 52, 54, 56, 58, ...
%C A266214 GCD = 4 for 60, 92, 108, 116, 120, 124, 156, 172, 180, 184, 188, ...
%C A266214 GCD = 8 for 248, 376, 440, 472, 488, 496, 504, 632, 696, 728, 744, ...
%C A266214 GCD = 16 for 1008, 1520, 1776, 1904, 1968, 2000, 2016, 2032, 2544, ...
%C A266214 GCD = 32 for 4064, 6112, 7136, 7648, 7904, 8032, 8096, 8128, 8160
%C A266214 Taking GCDs vertically, column 1 = "14, 60, 248, 1008, 4064, ..." appears to be essentially the same as A171499 and A131262; (ii) column 2 = "22, 92, 376, 1520, 6112, ..." appears to be essentially the same as A010036.
%C A266214 From _Chris Boyd_, Jan 25 2016: (Start)
%C A266214 Determining whether n is a term of this sequence can be approached by considering odd and even factors separately, and exploiting the fact that numerator(zeta(2n)/(Pi^(2n))) = numerator(2^(2n-2)*N_2n/(D_2n*(2n)!)), where N_2n and D_2n are odd coprime integers such that Bernoulli(2n) = N_2n/2D_2n.
%C A266214 Case 1: odd factors. n is a term if it has an odd prime factor p (necessarily irregular) that divides N_2n at a higher multiplicity than it divides (2n)!. No such factor p of N_2n up to 2n = 10000 is of sufficient multiplicity, and the apparent scarcity of squared and higher power factors of N_2n values (see A090997) suggests that no such p is likely to exist.
%C A266214 Case 2: even factors. An even n is a term if 2 divides 2^(2n-2) at a higher multiplicity than it divides (2n)!. The multiplicity of 2 in 2^(2n-2) is 2n-2, and in (2n)! is 2n minus the number of 1's in the binary expansion of 2n (see A005187). Qualifying n values are therefore those where the number of 1's in the binary expansion of 2n is greater than 2. Except for its first term, A072823 comprises integers with three or more 1-bits in their binary expansion. It follows that for m > 1, 2*A072823(m) values belong to this sequence.
%C A266214 In summary, this sequence is essentially a supersequence of 2*A072823. Conjectures 1 and 2 are true if there are no irregular odd primes p that divide n and the numerator of Bernoulli(2n)/(2n)!. (End)
%H A266214 Chris Boyd, <a href="/A266214/b266214.txt">Table of n, a(n) for n = 1..10000</a>
%H A266214 C. Boyd in reply to R. Israel, <a href="https://web.archive.org/web/*/http://list.seqfan.eu/oldermail/seqfan/2015-December/015906.html">A046988 query</a>, SeqFan list, Dec 22 2015.
%p A266214 select(n -> igcd(n,numer(2^(2*n-1)*bernoulli(2*n)/(2*n)!)) > 1), [$1..1000]);
%t A266214 Select[Range@ 170, ! CoprimeQ[#, Numerator[Zeta[2 #]/Pi^(2 #)]] &] (* _Michael De Vlieger_, Dec 24 2015 *)
%o A266214 (PARI) test(n) = if(gcd(numerator(2^(2*n-1)*bernfrac(2*n)/(2*n)!),n)!=1,1,0)
%o A266214 for(i=1,200,if(test(i),print1(i,", ")))
%Y A266214 Cf. A010036, A072823, A131262, A171499.
%K A266214 nonn,easy
%O A266214 1,1
%A A266214 _Chris Boyd_, _Robert Israel_, Dec 24 2015