A242035 -id:A242035 - cp's OEIS frontend

A242050 Denominator of A242035(n) / A246053(floor(n/2)).

2, 2, 2, 2, 1, 7, 1, 2, 1, 1, 1, 5, 1, 1, 7, 10, 1, 7, 1, 1, 1, 1, 3, 7, 1, 5, 1, 1, 1, 1, 1, 2, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 5, 1, 1, 3, 1, 1, 1, 1, 62, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 0

Peter Luschny, Aug 17 2014

nonn

Let h(x) = zeta(2*x)*(4^x-2) and r(n) = h([(n+1)/2])*h([n/2])/ h(n). Then the denominator of r([n/2]) almost divides the numerator of r(n), where 'a almost divides b' means that a divides b or the denominator of the quotient b/a (reduced to lowest terms) is very small compared to the size of b.

			a(34) = 1 because A246053(17) divides A242035(34).
a(35) = 7 because A242035(35)/A246053(17) = 8416843291197378595121525029762095062482279733/7.

Peter Luschny, Table of n, a(n) for n = 0..10000

Cf. A242035, A246053.

h = lambda x: zeta(2*x)*(4^x-2)
r = lambda n: h((n+1)//2)*h(n//2)/h(n)
A242050 = lambda n: (r(n).numerator()/r(n//2).denominator()).denominator()
[A242050(n) for n in range(1000)]

A246053 The denominator of h(n-k)h(k)/h(n) where h(x) = zeta(2x)*(4^x-2) and k = floor(n/2).

Original entry on oeis.org

2, 2, 7, 62, 381, 365, 1414477, 573370, 118518239, 5749691557, 274638832071, 768018339627, 29741482024668555, 860983513348179, 65053034220152267, 1850237821952082716222, 16555640865486520478399, 962703047799452264039, 29167285342563717499865628061

Offset: 0

Peter Luschny, Aug 12 2014

There are terms that are not squarefree. For example, a(113) is divisible by 631^2 and a(114) is divisible by 103^2. Most terms appear to be divisible by numerator(bernoulli(2*n)/factorial(2*n)) but not all. The first two exceptions are a(1437) and a(23766). - Hans Havermann, Aug 16 2014

			a( 0) = 2
a( 1) = 2
a( 2) = 7
a( 3) = 2 * 31
a( 4) = 3 * 127
a( 5) = 5 * 73
a( 6) = 23 * 89 * 691
a( 7) = 2 * 5 * 7 * 8191
a( 8) = 7 * 31 * 151 * 3617
a( 9) = 43867 * 131071
a(10) = 3 * 283 * 617 * 524287
a(11) = 3 * 7 * 11 * 127 * 131 * 337 * 593
a(12) = 3 * 5 * 47 * 103 * 178481 * 2294797
a(13) = 3 * 13 * 31 * 601 * 1801 * 657931

Hans Havermann, Table of n, a(n) for n = 0..200
Hans Havermann, Factorization table of n, a(n) for n = 0..150
Dinesh S. Thakur, A note on numerators of Bernoulli numbers, Proc. Amer. Math. Soc. 140 (2012), 3673-3676.

Cf. A242035 (numerator), A240978 (largest prime divisor), A242050, A246051, A246052.

h = lambda x: zeta(2*x)*(4^x-2)
A246053 = lambda n: Integer((h((n+1)//2)*h(n//2)/h(n)).denominator())
[A246053(n) for n in range(19)]

a(n) = A246052(n, floor(n/2)).

cp's OEIS Frontend

A242050 Denominator of A242035(n) / A246053(floor(n/2)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Sage

A246053 The denominator of h(n-k)h(k)/h(n) where h(x) = zeta(2x)*(4^x-2) and k = floor(n/2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Sage

Formula

cp's OEIS Frontend

A242050 Denominator of A242035(n) / A246053(floor(n/2)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Sage

A246053 The denominator of h(n-k)*h(k)/h(n) where h(x) = zeta(2*x)*(4^x-2) and k = floor(n/2).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Sage

Formula

A246053 The denominator of h(n-k)h(k)/h(n) where h(x) = zeta(2x)*(4^x-2) and k = floor(n/2).