|
%I #19 Jan 02 2023 12:30:52
%S 14,22,26,28,30,38,42,44,46,50,52,54,56,58,60,62,70,74,76,78,82,84,86,
%T 88,90,92,94,98,100,102,104,106,108,110,112,114,116,118,120,122,124,
%U 126,134,138,140,142,146,148,150,152,154,156,158,162,164,166,168,170
%N Numbers n that are not coprime to the numerator of zeta(2*n)/(Pi^(2*n)).
%C Equivalently, n not coprime to the numerator of 2^(2n-1)*Bernoulli(2n)/(2n)! (see _Lekraj Beedassy_ comment in A046988).
%C Conjecture 1: for n>=1, a(n) is identical to 2*A072823(n+1).
%C Conjecture 2: The corresponding GCDs are powers of 2.
%C Verified for n <= 10000, e.g.,
%C GCD = 2 for 14, 22, 26, 28, 30, 38, 42, 44, 46, 50, 52, 54, 56, 58, ...
%C GCD = 4 for 60, 92, 108, 116, 120, 124, 156, 172, 180, 184, 188, ...
%C GCD = 8 for 248, 376, 440, 472, 488, 496, 504, 632, 696, 728, 744, ...
%C GCD = 16 for 1008, 1520, 1776, 1904, 1968, 2000, 2016, 2032, 2544, ...
%C GCD = 32 for 4064, 6112, 7136, 7648, 7904, 8032, 8096, 8128, 8160
%C 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 From _Chris Boyd_, Jan 25 2016: (Start)
%C 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 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 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 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 Chris Boyd, <a href="/A266214/b266214.txt">Table of n, a(n) for n = 1..10000</a>
%H C. Boyd in reply to R. Israel, <a href="http://list.seqfan.eu/oldermail/seqfan/2015-December/015906.html">A046988 query</a>, SeqFan list, Dec 22 2015.
%p select(n -> igcd(n,numer(2^(2*n-1)*bernoulli(2*n)/(2*n)!)) > 1), [$1..1000]);
%t Select[Range@ 170, ! CoprimeQ[#, Numerator[Zeta[2 #]/Pi^(2 #)]] &] (* _Michael De Vlieger_, Dec 24 2015 *)
%o (PARI) test(n) = if(gcd(numerator(2^(2*n-1)*bernfrac(2*n)/(2*n)!),n)!=1,1,0)
%o for(i=1,200,if(test(i),print1(i,", ")))
%Y Cf. A010036, A072823, A131262, A171499.
%K nonn,easy
%O 1,1
%A _Chris Boyd_, _Robert Israel_, Dec 24 2015
|
