

A266214


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


3



14, 22, 26, 28, 30, 38, 42, 44, 46, 50, 52, 54, 56, 58, 60, 62, 70, 74, 76, 78, 82, 84, 86, 88, 90, 92, 94, 98, 100, 102, 104, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126, 134, 138, 140, 142, 146, 148, 150, 152, 154, 156, 158, 162, 164, 166, 168, 170
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OFFSET

1,1


COMMENTS

Equivalently, n not coprime to the numerator of 2^(2n1)*Bernoulli(2n)/(2n)! (see Lekraj Beedassy comment in A046988).
Conjecture 1: for n>=1, a(n) is identical to 2*A072823(n+1).
Conjecture 2: The corresponding GCDs are powers of 2.
Verified for n <= 10000, e.g.,
GCD = 2 for 14, 22, 26, 28, 30, 38, 42, 44, 46, 50, 52, 54, 56, 58, ...
GCD = 4 for 60, 92, 108, 116, 120, 124, 156, 172, 180, 184, 188, ...
GCD = 8 for 248, 376, 440, 472, 488, 496, 504, 632, 696, 728, 744, ...
GCD = 16 for 1008, 1520, 1776, 1904, 1968, 2000, 2016, 2032, 2544, ...
GCD = 32 for 4064, 6112, 7136, 7648, 7904, 8032, 8096, 8128, 8160
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.
From Chris Boyd, Jan 25 2016: (Start)
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^(2n2)*N_2n/(D_2n*(2n)!)), where N_2n and D_2n are odd coprime integers such that Bernoulli(2n) = N_2n/2D_2n.
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.
Case 2: even factors. An even n is a term if 2 divides 2^(2n2) at a higher multiplicity than it divides (2n)!. The multiplicity of 2 in 2^(2n2) is 2n2, 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 1bits in their binary expansion. It follows that for m > 1, 2*A072823(m) values belong to this sequence.
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)


LINKS

Chris Boyd, Table of n, a(n) for n = 1..10000
C. Boyd in reply to R. Israel, A046988 query, SeqFan list, Dec 22 2015.


MAPLE

select(n > igcd(n, numer(2^(2*n1)*bernoulli(2*n)/(2*n)!)) > 1), [$1..1000]);


MATHEMATICA

Select[Range@ 170, ! CoprimeQ[#, Numerator[Zeta[2 #]/Pi^(2 #)]] &] (* Michael De Vlieger, Dec 24 2015 *)


PROG

(PARI) test(n) = if(gcd(numerator(2^(2*n1)*bernfrac(2*n)/(2*n)!), n)!=1, 1, 0)
for(i=1, 200, if(test(i), print1(i, ", ")))


CROSSREFS

Cf. A010036, A072823, A131262, A171499.
Sequence in context: A166628 A199250 A199195 * A113190 A169804 A092112
Adjacent sequences: A266211 A266212 A266213 * A266215 A266216 A266217


KEYWORD

nonn,easy


AUTHOR

Chris Boyd, Robert Israel, Dec 24 2015


STATUS

approved



