cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 22 results. Next

A002432 Denominators of zeta(2*n)/Pi^(2*n).

Original entry on oeis.org

2, 6, 90, 945, 9450, 93555, 638512875, 18243225, 325641566250, 38979295480125, 1531329465290625, 13447856940643125, 201919571963756521875, 11094481976030578125, 564653660170076273671875, 5660878804669082674070015625, 62490220571022341207266406250
Offset: 0

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Author

Keywords

Comments

Also denominators in expansion of Psi(x).
For n >= 1 a(n) is always divisible by 3 (by the von Staudt-Clausen theorem, see A002445).
A comment due to G. Campbell: The original approach taken by Euler was to derive the infinite product for sin(Pi*x)/(Pi*x) equal to (1 - x^2/1^2) (1 - x^2/2^2)(1 - x^2/3^2) ... treating sin(Pi*x)/(Pi*x) as if it were a polynomial. Differentiating the logarithm of both sides and equating coefficients gives all of the zeta function values for 2, 4, 6, 8, .... - M. F. Hasler, Mar 29 2015
Note that 2n+1 divides a(n) for every n. If 2n+1 > 9 is composite, then (2n+1)^2 divides a(n). If 2n+1 is prime, then (2n+1)^2 does not divide a(n). My theorem: for n > 4, (2n+1)^2 divides a(n) if and only if the number 2n+1 is composite. - Thomas Ordowski, Nov 07 2022

Examples

			(zeta(2n)/Pi^(2n), n = 0, 1, 2, 3, ...) = (-1/2, 1/6, 1/90, 1/945, 1/9450, 1/93555, 691/638512875, 2/18243225, 3617/325641566250, ...), i.e.: zeta(0) = -1/2, zeta(2) = Pi^2/6, zeta(4) = Pi^4/90, zeta(6) = Pi^6/945, zeta(8) = Pi^8/9450, zeta(10) = Pi^10/93555, zeta(12) = Pi^12*691/638512875, ...
In Maple, series(Psi(x),x,20) gives -1*x^(-1) + (-gamma) + 1/6*Pi^2*x + (-Zeta(3))*x^2 + 1/90*Pi^4*x^3 + (-Zeta(5))*x^4 + 1/945*Pi^6*x^5 + (-Zeta(7))*x^6 + 1/9450*Pi^8*x^7 + (-Zeta(9))*x^8 + 1/93555*Pi^10*x^9 + ...
a(5) = 93555 = 10!/(2^9 * B(10)) = 3628800/(512*5/66). - _Frank Ellermann_, Apr 03 2020
		

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 88.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 84.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A046988 (numerators), A006003.

Programs

  • Maple
    seq(denom(Zeta(2*n)/Pi^(2*n)),n=0..24); # Martin Renner, Sep 07 2016
    A002432List := proc(len) series(-x*cot(x)/2, x, 2*len+1):
    seq(denom(coeff(%, x, n)), n=0..2*len-1, 2) end:
    A002432List(17); # Peter Luschny, Jun 07 2020
  • Mathematica
    Table[Denominator[Zeta[2 n]/Pi^(2 n)], {n, 0, 30}] (* Artur Jasinski, Mar 11 2010 *)
    Denominator[Zeta[2*Range[0, 20]]] (* Harvey P. Dale, Sep 05 2013 *)
  • PARI
    a(n)=numerator(bestappr(Pi^(2*n)/zeta(2*n))) \\ Requires sufficient realprecision. The standard value of 38 digits yields erroneous values for n>9. \p99 is more than enough to get the 3 lines of displayed data. - M. F. Hasler, Mar 29 2015
    
  • PARI
    a002432(n) = denominator(polcoeff((1-x*cotan(x))/2,n*2))
    default(seriesprecision, 33); for(i=0,16,print1(a002432(i),",")) \\ Chris Boyd, Dec 21 2015

Formula

Sum_{n>=1} 2/(n^2 + 1) = Pi*coth(Pi)-1. 2*Sum_{k>=1} (-1)^(k + 1)/n^(2*k) = 2/(n^2+1). - Shmuel Spiegel (shmualm(AT)hotmail.com), Aug 13 2001
zeta(2n)/(2i * ( log(1-i)-log(1+i) ))^(2n) = zeta(2n)/(-i*log(-1))^(2n). - Eric Desbiaux, Dec 12 2008
zeta(2n) = Sum_{k >= 1} k^(-2n) = (-1)^(n-1)*B_{2n}*2^(2n-1)*Pi^(2n)/(2n)!.
a(n) = -A046988(n)*A010050(n)*A002445(n)/(A009117(n)*A000367(n))
a(n) = sqrt(denominator(Sum_{i>=1} A000005(i)/i^2n)). - Enrique Pérez Herrero, Jan 19 2012
Sum_{k >= 1} zeta(2k)*x^(2k) = (1-Pi*x*cot(Pi*x))/2. - Chris Boyd, Dec 21 2015
a(n) = denominator([x^(2*n)] -x*cot(x)/2). - Peter Luschny, Jun 07 2020

Extensions

Formula and link from Henry Bottomley, Mar 10 2000
Formula corrected by Bjoern Boettcher, May 15 2003
Corrected and edited by M. F. Hasler, Mar 29 2015
a(0) = 2 prepended by Peter Luschny, Jun 07 2020

A073747 Decimal expansion of coth(1).

Original entry on oeis.org

1, 3, 1, 3, 0, 3, 5, 2, 8, 5, 4, 9, 9, 3, 3, 1, 3, 0, 3, 6, 3, 6, 1, 6, 1, 2, 4, 6, 9, 3, 0, 8, 4, 7, 8, 3, 2, 9, 1, 2, 0, 1, 3, 9, 4, 1, 2, 4, 0, 4, 5, 2, 6, 5, 5, 5, 4, 3, 1, 5, 2, 9, 6, 7, 5, 6, 7, 0, 8, 4, 2, 7, 0, 4, 6, 1, 8, 7, 4, 3, 8, 2, 6, 7, 4, 6, 7, 9, 2, 4, 1, 4, 8, 0, 8, 5, 6, 3, 0, 2, 9, 4, 6, 7, 9
Offset: 1

Views

Author

Rick L. Shepherd, Aug 07 2002

Keywords

Comments

coth(x) = (e^x + e^(-x))/(e^x - e^(-x)).
Because the continued fraction for coth(1) is all positive odd numbers in sequence, the second Mathematica program below also generates the sequence. - Harvey P. Dale, Oct 15 2011
By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 14 2019

Examples

			1.31303528549933130363616124693...
		

References

  • Samuel M. Selby, editor, CRC Basic Mathematical Tables, CRC Press, 1970, p. 218.

Crossrefs

Cf. A005408 (continued fraction: odd numbers), A073821 (continued fraction exp. is even numbers), A073744 (tanh(1)=1/A073747), A073742 (sinh(1)), A073743 (cosh(1)), A073745 (csch(1)), A073746 (sech(1)), A349004.

Programs

  • Mathematica
    RealDigits[Coth[1],10,120][[1]] (* or *) RealDigits[ FromContinuedFraction[ Range[1,1001,2]],10,120][[1]] (* Harvey P. Dale, Oct 15 2011 *) (* see Comments, above, for the second program *)
  • PARI
    1/tanh(1)

Formula

Equals 1 + Sum_{n>=1} (2^(2*n)*B(2*n))/(2*n)! = 1 + Sum_{n>=1} (-1)^(n+1)*2*(A046988(n+1) / A002432(n+1)). - Terry D. Grant, May 30 2017
Equals 1 + BesselI(3/2, 1)/BesselI(1/2, 1). - Terry D. Grant, Jun 18 2018
Equals 1 + Sum_{k>=1} csch(2^k) (Ohtsuka, 2015; Stenger, 2017). - Amiram Eldar, Oct 04 2021

A013676 Decimal expansion of zeta(18).

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 3, 8, 1, 7, 2, 9, 3, 2, 6, 4, 9, 9, 9, 8, 3, 9, 8, 5, 6, 4, 6, 1, 6, 4, 4, 6, 2, 1, 9, 3, 9, 7, 3, 0, 4, 5, 4, 6, 9, 7, 2, 1, 8, 9, 5, 3, 3, 3, 1, 1, 4, 3, 1, 7, 4, 4, 2, 9, 9, 8, 7, 6, 3, 0, 0, 3, 9, 5, 4, 2, 6, 5, 0, 0, 4, 5, 6, 3, 8, 0, 0, 1, 9, 6, 8, 6, 6, 8, 9, 8, 9, 6, 4
Offset: 1

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Author

Keywords

Examples

			1.0000038172932649998398564616446219397...
		

References

  • Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 811.

Programs

Formula

zeta(18) = Sum_{n >= 1} (A010052(n)/n^9) = Sum_{n >= 1} ( (floor(sqrt(n))-floor(sqrt(n-1)))/n^9 ). - Mikael Aaltonen, Mar 06 2015
zeta(18) = 43867*Pi^18/38979295480125 = A046988(9)*Pi^18/A002432(9). - Alonso del Arte, Feb 12 2016
zeta(18) = Product_{k>=1} 1/(1 - 1/prime(k)^18). - Vaclav Kotesovec, May 02 2020

A117972 Numerator of zeta'(-2n), n >= 0.

Original entry on oeis.org

1, -1, 3, -45, 315, -14175, 467775, -42567525, 638512875, -97692469875, 9280784638125, -2143861251406875, 147926426347074375, -48076088562799171875, 9086380738369043484375, -3952575621190533915703125
Offset: 0

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Author

Eric W. Weisstein, Apr 06 2006

Keywords

Comments

In A160464 the coefficients of the ES1 matrix are defined. This matrix led to the discovery that the successive differences of the ES1[1-2*m,n] coefficients for m = 1, 2, 3, ..., are equal to the values of zeta'(-2n), see also A094665 and A160468. - Johannes W. Meijer, May 24 2009
A048896(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2,
a(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011
From Andrey Zabolotskiy, Sep 23 2021: (Start)
zeta'(-2n), which is mentioned in the Name, is irrational. For n > 0, a(n) is the numerator of the rational fraction g(n) = Pi^(2n)*zeta'(-2n)/zeta(2n+1). The denominator is 4*A048896(n-1). g(n) = f(n) for n > 0, where f(n) is given in the Formula section. Also, f(n) = Bernoulli(2n)/z(n)/4 (see Formula section) for all n.
For n = 0, zeta'(0) = -log(2Pi)/2, g(0) can be set to 0 because of the infinite denominator. However, a(0) is set to 1 because it is the numerator of f(0).
It seems that -4*f(n)*alpha_n = A000182(n), where alpha_n = A191657(n, p(n)) / A191658(n, p(n)) [where p(n) = A000041(n)] is the n-th "elementary coefficient" from the paper by Izaurieta et al. (End)

Examples

			-1/4, 3/4, -45/8, 315/4, -14175/8, 467775/8, -42567525/16, ...
-zeta(3)/(4*Pi^2), (3*zeta(5))/(4*Pi^4), (-45*zeta(7))/(8*Pi^6), (315*zeta(9))/(4*Pi^8), (-14175*zeta(11))/(8*Pi^10), ...
		

Crossrefs

From Johannes W. Meijer, May 24 2009: (Start)
Absolute values equal row sums of A160468. (End)

Programs

  • Maple
    # Without rational arithmetic
    a := n -> (-1)^n*(2*n)!*2^(add(i,i=convert(n,base,2))-2*n);
    # Peter Luschny, May 02 2009
  • Mathematica
    Table[Numerator[(2 n)!/2^(2 n + 1) (-1)^n], {n, 0, 30}]
  • Maxima
    L:taylor(1/x*sin(sqrt(x))^2,x,0,15); makelist(denom(coeff(L,x,n))*(-1)^(n+1),n,0,15); /* Vladimir Kruchinin, May 30 2011 */

Formula

a(n) = numerator(f(n)) where f(n) = (2*n)!/2^(2*n + 1)(-1)^n, from the Mathematica code.
From Terry D. Grant, May 28 2017: (Start)
|a(n)| = A049606(2n).
a(n) = -numerator(Bernoulli(2n)/z(n)) where Bernoulli(2n) = A000367(n) / A002445(n) and z(n) = A046988(n) / A002432(n) for n > 0. (End) [Corrected by Andrey Zabolotskiy, Sep 23 2021]

Extensions

First term added, offset changed and edited by Johannes W. Meijer, May 15 2009

A276592 Numerator of the rational part of the sum of reciprocals of even powers of odd numbers, i.e., Sum_{k>=1} 1/(2*k-1)^(2*n).

Original entry on oeis.org

1, 1, 1, 17, 31, 691, 5461, 929569, 3202291, 221930581, 4722116521, 56963745931, 14717667114151, 2093660879252671, 86125672563201181, 129848163681107301953, 868320396104950823611, 209390615747646519456961, 14129659550745551130667441, 16103843159579478297227731
Offset: 1

Views

Author

Martin Renner, Sep 07 2016

Keywords

Comments

Apart from signs, same as A089171 and A279370. - Peter Bala, Feb 07 2019

Crossrefs

Programs

  • Maple
    seq(numer(sum(1/(2*k-1)^(2*n),k=1..infinity)/Pi^(2*n)),n=1..22);
  • Mathematica
    a[n_]:=Numerator[Pi^(-2 n) (1-2^(-2 n)) Zeta[2 n]]  (* Steven Foster Clark, Mar 10 2023 *)
    a[n_]:=Numerator[(-1)^n SeriesCoefficient[1/(E^x+1),{x,0,2 n-1}]] (* Steven Foster Clark, Mar 10 2023 *)
    a[n_]:=Numerator[(-1)^n Residue[Zeta[s] Gamma[s] (1-2^(1-s)),{s,1-2 n}]] (* Steven Foster Clark, Mar 11 2023 *)

Formula

a(n)/A276593(n) + A276594(n)/A276595(n) = A046988(n)/A002432(n).
a(n)/A276593(n) = (-1)^(n+1) * B_{2*n} * (2^(2*n) - 1) / (2 * (2*n)!), where B_n is the Bernoulli number. - Seiichi Manyama, Sep 03 2018

A276593 Denominator of the rational part of the sum of reciprocals of even powers of odd numbers, i.e., Sum_{k>=1} 1/(2*k-1)^(2*n).

Original entry on oeis.org

8, 96, 960, 161280, 2903040, 638668800, 49816166400, 83691159552000, 2845499424768000, 1946321606541312000, 408727537373675520000, 48662619743783485440000, 124089680346647887872000000, 174221911206693634572288000000, 70734095949917615636348928000000
Offset: 1

Views

Author

Martin Renner, Sep 07 2016

Keywords

Comments

A276592(n)/a(n) * Pi^(2*n) = Sum_{k>=1} 1/(2*k-1)^(2*n) > 1. So Pi^(2*n) > a(n)/A276592(n). - Seiichi Manyama, Sep 03 2018

Examples

			From _Seiichi Manyama_, Sep 03 2018: (Start)
n |    Pi^(2*n)   |   a(n)/A276592(n)
--+---------------+------------------------------------
1 |        9.8... |           8
2 |       97.4... |          96
3 |      961.3... |         960
4 |     9488.5... |      161280/17     =     9487.0...
5 |    93648.0... |     2903040/31     =    93646.4...
6 |   924269.1... |   638668800/691    =   924267.4...
7 |  9122171.1... | 49816166400/5461   =  9122169.2... (End)
		

Crossrefs

Programs

  • Maple
    seq(denom(sum(1/(2*k-1)^(2*n),k=1..infinity)/Pi^(2*n)),n=1..22);
  • Mathematica
    a[n_]:=Denominator[(1-2^(-2 n)) Zeta[2 n]] (* Steven Foster Clark, Mar 10 2023 *)
    a[n_]:=Denominator[1/2 SeriesCoefficient[1/(E^x+1),{x,0,2 n-1}]] (* Steven Foster Clark, Mar 10 2023 *)
    a[n_]:=Denominator[1/2 Residue[Zeta[s] Gamma[s] (1-2^(1-s)) x^(-s),{s,1-2 n}]] (* Steven Foster Clark, Mar 11 2023 *)

Formula

A276592(n)/a(n) + A276594(n)/A276595(n) = A046988(n)/A002432(n).
A276592(n)/a(n) = (-1)^(n+1) * B_{2*n} * (2^(2*n) - 1) / (2 * (2*n)!), where B_n is the Bernoulli number. - Seiichi Manyama, Sep 03 2018

A276595 Denominator of the rational part of the sum of reciprocals of even powers of even numbers, i.e., Sum_{k>=1} 1/(2*k)^(2*n).

Original entry on oeis.org

24, 1440, 60480, 2419200, 95800320, 2615348736000, 149448499200, 21341245685760000, 10218188434341888000, 1605715325396582400000, 28202200078783610880000, 3387648273463487338905600000, 372269041039943663616000000, 75786531374911731038945280000000
Offset: 1

Views

Author

Martin Renner, Sep 07 2016

Keywords

Comments

Denominator of Bernoulli(2*n)/(2*(2*n)!). - Robert Israel, Sep 18 2016

Crossrefs

Programs

  • Maple
    seq(denom(sum(1/(2*k)^(2*n),k=1..infinity)/Pi^(2*n)),n=1..24);
    seq(denom(bernoulli(2*n)/2/(2*n)!),n=1..24); # Robert Israel, Sep 18 2016
  • Mathematica
    Table[Denominator[Zeta[2*n]/(2*Pi)^(2*n)], {n, 1, 30}] (* Terry D. Grant, Jun 19 2018 *)
  • PARI
    a(n) = denominator(bernfrac(2*n)/(2*(2*n)!)); \\ Michel Marcus, Jul 05 2018

Formula

A276592(n)/A276593(n) + A276594(n)/a(n) = A046988(n)/A002432(n).
Zeta(2n) = (-1)^(n-1)*(A276594(n)/a(n))*((2*Pi)^(2n)), according to Euler. - Terry D. Grant, Jun 19 2018

A100594 Floor of Pi^(2*n)/Zeta(2*n).

Original entry on oeis.org

6, 90, 945, 9450, 93555, 924041, 9121612, 90030844, 888579011, 8769948429, 86555983552, 854273468992, 8431341566236, 83214006759229, 821289329637860, 8105800788023426, 80001047145799660, 789578687036411293
Offset: 1

Views

Author

Joseph Biberstine (jrbibers(AT)indiana.edu), Nov 30 2004

Keywords

Examples

			a(1)=6 because Zeta(2*1)=Pi^2/6 implies Pi^2/Zeta(2)=6 and floor(6)=6.
a(6)=924041 because Zeta(2*6)=691/638512875*Pi^12 implies Pi^12/Zeta(12)=638512875/691 and floor(638512875/691)=924041.
		

Crossrefs

Programs

  • Maple
    seq(simplify(floor(Pi^(2*k)/Zeta(2*k))),k=1..24);
  • Mathematica
    Table[Floor[Pi^(2*n)/Zeta[2*n]],{n,20}] (* Terry D. Grant, May 28 2017 *)
  • PARI
    {a(n)=if(n<1, 0, floor(-2*(2*n)!/(-4)^n/bernfrac(2*n)))} /* Michael Somos, Feb 18 2007 */

A231273 Numerator of zeta(4n)/(zeta(2n) * Pi^(2n)).

Original entry on oeis.org

1, 1, 1, 691, 3617, 174611, 236364091, 3392780147, 7709321041217, 26315271553053477373, 261082718496449122051, 2530297234481911294093, 5609403368997817686249127547, 61628132164268458257532691681, 354198989901889536240773677094747
Offset: 0

Views

Author

Leo Depuydt, Nov 07 2013

Keywords

Comments

Integer component of the numerator of a close variant of Euler's infinite prime product zeta(2n) = Product_{k>=1} (prime(k)^(2n))/(prime(k)^(2n)-1), namely with all minus signs changed into plus signs, as follows: zeta(4n)/zeta(2n) = Product_{k>=1} (prime(k)^(2n))/(prime(k)^(2n)+1). The transcendental component is Pi^(2n).
For a detailed account of the results, including proof and relation to the zeta function, see Links for the PDF file submitted as supporting material.
The reference to Apostol is to a discussion of the equivalence of 1) zeta(2s)/zeta(s) and 2) a related infinite prime product, that is, Product_{sigma>1} prime(n)^s/(prime(n)^s + 1), with s being a complex variable such that s = sigma + i*t where sigma and t are real (following Riemann), using a type of proof different from the one posted below involving zeta(4n)/zeta(2n). On this, see also Hardy and Wright cited below. - Leo Depuydt, Nov 22 2013, Nov 27 2013
The background of the sequence is now described in the link below to L. Depuydt, The Prime Sequence ... . - Leo Depuydt, Aug 22 2014
From Robert Israel, Aug 22 2014: (Start)
Numerator of (-1)^n*B(4*n)*4^n*(2*n)!/(B(2*n)*(4*n)!), where B(n) are the Bernoulli numbers (see A027641 and A027642).
Not the same as abs(A001067(2*n)): they differ first at n=17.
(End)

References

  • T. M. Apostol, Introduction to Analytic Number Theory, Springer, 1976, p. 231.
  • G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Fourth Edition, Clarendon Press, 1960, p. 255.

Crossrefs

Cf. A231327 (corresponding denominator).
Cf. A114362 and A114363 (closely related results).
Cf. A001067, A046968, A046988, A098087, A141590, A156036 (same number sequence, though in various transformations (alternation of signs, intervening numbers, and so on)).

Programs

  • Maple
    seq(numer((-1)^n*bernoulli(4*n)*4^n*(2*n)!/(bernoulli(2*n)*(4*n)!)),n=0..100); # Robert Israel, Aug 22 2014
  • Mathematica
    Numerator[Table[Zeta[4n]/(Zeta[2n] * Pi^(2n)), {n, 0, 15}]] (* T. D. Noe, Nov 18 2013 *)

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

Original entry on oeis.org

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
Offset: 1

Views

Author

Chris Boyd, Robert Israel, Dec 24 2015

Keywords

Comments

Equivalently, n not coprime to the numerator of 2^(2n-1)*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^(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.
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^(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.
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)

Crossrefs

Programs

  • Maple
    select(n -> igcd(n,numer(2^(2*n-1)*bernoulli(2*n)/(2*n)!)) > 1), [$1..1000]);
  • Mathematica
    Select[Range@ 170, ! CoprimeQ[#, Numerator[Zeta[2 #]/Pi^(2 #)]] &] (* Michael De Vlieger, Dec 24 2015 *)
  • PARI
    test(n) = if(gcd(numerator(2^(2*n-1)*bernfrac(2*n)/(2*n)!),n)!=1,1,0)
    for(i=1,200,if(test(i),print1(i,", ")))
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