A089170 -id:A089170 - cp's OEIS frontend

A002425 Denominator of Pi^(2n)/(Gamma(2n)(1-2^(-2n))zeta(2n)).

1, 1, 1, 17, 31, 691, 5461, 929569, 3202291, 221930581, 4722116521, 968383680827, 14717667114151, 2093660879252671, 86125672563201181, 129848163681107301953, 868320396104950823611, 209390615747646519456961

Offset: 1

N. J. A. Sloane

Differs from the absolute values of A275994 the first time at index 60.
Consider the C(k)-summation process for divergent series: the series Sum((-1)^n*(n+1)^k) == 1 - 2^k + 3^k - 4^k + ..., summable C(1) to the value 1/2 for k = 0, is for each k >= 1 exactly summable C(k+1) to the sum s(k+1) = (2^(k+1)-1)*B(k+1)/ (k+1) and so a(n) = abs(numerator(s(2n))). - Benoit Cloitre, Apr 27 2002
Odd part of tangent numbers A000182 (even part is 2^A101921(n)). - Ralf Stephan, Dec 21 2004
(-1)^n*a(n+1) is the numerator of Euler(2n+1,1). - N. J. A. Sloane, Nov 10 2009 (a misprint corrected by Vladimir Shevelev, Sep 18 2017)
a(n) is the absolute value of the constant term of the Euler polynomial E_{2n-1} times the even part of 2n. - Peter Luschny, Nov 26 2010
From Vladimir Shevelev, Aug 31 2017: (Start)
Let E_m(x) = x^m + Sum_{odd k=1..m} e_k(m)*x^(m-k) be the Euler polynomial, let 2*n-1 <= m. Show that the expression c(m,n) = |e_(2*n-1)(m)|/binomial(m,2*n-1) does not depend on m and c(m,n) = a(n)/A006519(2*n). Indeed, by the formula in the Shevelev link |e_(2*n-1)(m)| = binomial(m,2*n-1)*(4^n-1)*B_(2*n)/n. On the other hand, by Cloitre's formula, we have a(n) = (4^n-1)*|B_(2*n)|*2^A001511(n) /n. Taking into account that 2^A001511 = A006519(2*n) we obtain the claimed equality. Since sign(e_k(n)) = (-1)^((k+1)/2), we have the following application of the sequence: e_k(n) = (-1)^((k+1)/2))*a((k+1)/2)*binomial(n,k)/A006519(k+1). (End)

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. 73.
S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51.
Konrad Knopp, Theory and application of infinite series, Divergent series, Dover, p. 479
L. Oettinger, Archiv. Math. Phys., 26 (1856), see esp. p. 5.
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).

N. J. A. Sloane, Table of n, a(n) for n = 1..300
H. Cohn, Some elementary aspects of modular functions in several variables, Bull. Am. Math. Soc., Sept. 1965, 681ff, esp. p. 688.
Ren Guan, K_0 groups of noncommutative R^2n, arXiv:2208.06253 [math.RA], 2022. See p. 22.
S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51. [Annotated scanned copy of pages 38-51 only, plus notes]
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
Vladimir Shevelev, On a Luschny question, arXiv:1708.08096 [math.NT], 2017.

Numerator given by A037239.
Different from A089171, A275994.
Cf. A048896, A089170, A089171, A160469, A335956.

[Denominator(4*n/((4^n-1)*Bernoulli(2*n))): n in [1..20]]; // G. C. Greubel, Jul 03 2019

A002425 := n -> (-1)^n*euler(2*n-1,0)*2^padic[ordp](2*n,2); # Peter Luschny, Nov 26 2010
A002425_list := proc(n) 1/(1+1/exp(z)); series(%,z,2*n+4);
seq(numer((-1)^i*(2*i+1)!*coeff(%,z,2*i+1)),i=0..n) end;
A002425_list(17); # Peter Luschny, Jul 12 2012

a[n_]:= (-1)^(n-1) * Numerator[EulerE[2n-1, 1]]; Table[a[n], {n, 1, 20}] (* Jean-François Alcover, Sep 20 2011, after N. J. A. Sloane's comment *)
a[n_]:= If[n<1, 0, With[{m = 2n-1}, Numerator[ m! SeriesCoefficient[ Tan[x/2], {x, 0, m}]]]] (* Michael Somos, Sep 14 2013 *)
Table[2*(4^n-1)*Zeta[1-2n] // Abs // Numerator, {n, 1, 20}] (* Jean-François Alcover, Oct 16 2013 *)

for(n=1,20,print1(abs(numerator(2*bernfrac(2*n)*(4^n-1)/(2*n))),","))

a(n)=if(n<1,0,(-1)^n/n*(1-4^n)*bernfrac(2*n)*2^valuation(2*n,2))

a(n)=(-1)^n*4*bitand(n,-n)*polylog(1-2*n,-1); \\ Peter Luschny, Nov 22 2012

def A002425_list(n):
    T = [0]*n; T[0] = 1; S = [0]*n; k2 = 0
    for k in (1..n-1): T[k] = k*T[k-1]
    for k in (1..n):
        if is_odd(k): S[k-1] = 4*k2; k2 += 1
        else: S[k-1] = S[k2-1]+2*k2-1
        for j in (k..n-1): T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]
    return [T[j]>>S[j] for j in (0..n-1)]
A002425_list(20)  # Peter Luschny, Nov 17 2012

[denominator(4*n/((4^n-1)*bernoulli(2*n))) for n in (1..20)] # G. C. Greubel, Jul 03 2019

a(n) = (-1)^n/n*(1 - 4^n)*B(2*n)*2^A001511(n) where B(k) denotes the k-th Bernoulli number. - Benoit Cloitre, Dec 30 2003
This is different from the sequence of numerators of the expansion of cosec(x) - cot(x) - see A089171.
From Johannes W. Meijer, May 24 2009: (Start)
a(n) = denominator(4*n/((2^(2*n)-1)*bernoulli(2*n))).
Equals A160469(n)/A048896(n-1).
Equals A089171(n)*A089170(n-1). (End)
E.g.f.: a(n) = numerator((2*n+1)!*[x^(2*n+1)](1/(1+1/exp(x)))). - Peter Luschny, Jul 12 2012
a(n) = numerator(abs(2*(4^n-1)*zeta(1-2*n))). - Jean-François Alcover, Oct 16 2013
For every positive integers n,k we have a(n) = (-1)^(n+k)*N(2*n-1,k) + 2*(-1)^(n-1)*A006519(2*n)*(1^(2*n-1)-2^(2*n-1)+..+(-1)^k*(k-1)^(2*n-1)), where N(n,k) is the numerator of Euler(n,k). So, the right hand side is an invariant of k. - Vladimir Shevelev, Sep 19 2017
a(n) = numerator(r(n)) where r(n) = (-1)^binomial(2*n, 2)*Sum_{k=1..2*n}(-1)^k*Stirling2(2*n, k)*2^(-k)*(k-1)!. - Peter Luschny, May 24 2020
a(n) = 2*(-1)^n*A335956(2*n)*zeta(1-2*n). - Peter Luschny, Aug 30 2020

The n=15 term was formerly incorrectly given as 86125672563301143.

Formula and cross-references edited by Johannes W. Meijer, May 21 2009

A156769 a(n) = denominator(2^(2n-2)/factorial(2n-1)).

Original entry on oeis.org

1, 3, 15, 315, 2835, 155925, 6081075, 638512875, 10854718875, 1856156927625, 194896477400625, 49308808782358125, 3698160658676859375, 1298054391195577640625, 263505041412702261046875, 122529844256906551386796875, 4043484860477916195764296875

Offset: 1

Johannes W. Meijer, Feb 15 2009

Resembles A036279, the denominators in the Taylor series for tan(x). The first difference occurs at a(12).
The numerators of the two formulas for this sequence lead to A001316, Gould's sequence.
Stephen Crowley indicated on Aug 25 2008 that a(n) = denominator(Zeta(2*n)/Zeta(1-2*n)) and here numerator((Zeta(2*n)/Zeta(1-2*n))/(2*(-1)^(n)*(Pi)^(2*n))) leads to Gould's sequence.
This sequence appears in the Eta and Zeta triangles A160464 and A160474. Its resemblance to the sequence of the denominators of the Taylor series for tan(x) led to the conjecture A156769(n) = A036279(n)*A089170(n-1). - Johannes W. Meijer, May 24 2009

G. C. Greubel, Table of n, a(n) for n = 1..250

Cf. A036279 Denominators in Taylor series for tan(x).
Cf. A001316 Gould's sequence appears in the numerators.
Cf. A000265, A036279, A089170, A117972, A160464, A160469 (which resembles the numerators of the Taylor series for tan(x)), A160474. - Johannes W. Meijer, May 24 2009

[Denominator(4^(n-1)/Factorial(2*n-1)): n in [1..25]]; // G. C. Greubel, Jun 19 2021

a := n ->(2*n-1)!*2^(add(i,i=convert(n-1,base,2))-2*n+2); # Peter Luschny, May 02 2009

a[n_] := Denominator[4^(n-1)/(2n-1)!];
Array[a, 15] (* Jean-François Alcover, Jun 20 2018 *)

[denominator(4^(n-1)/factorial(2*n-1)) for n in (1..25)] # G. C. Greubel, Jun 19 2021

a(n) = denominator( Product_{k=1..n-1} 2/(k*(2*k+1)) ).
G.f.: (1/2)*z^(1/2)*sinh(2*z^(1/2)).
From Johannes W. Meijer, May 24 2009: (Start)
a(n) = abs(A117972(n))/A000265(n).
a(n) = A036279(n)*A089170(n-1). (End)
a(n) = A049606(2*n-1). - Zhujun Zhang, May 29 2019

A160469 The left hand column of the triangle A160468.

Original entry on oeis.org

1, 1, 2, 17, 62, 1382, 21844, 929569, 6404582, 443861162, 18888466084, 1936767361654, 58870668456604, 8374643517010684, 689005380505609448, 129848163681107301953, 1736640792209901647222, 418781231495293038913922

Offset: 1

Johannes W. Meijer, May 24 2009

Resembles A002430, the numerators of the Taylor series for tan(x). The first difference occurs at a(12). (Its resemblance to this sequence led to the conjecture A160469(n) = A002430(n)*A089170(n-1).)

Equals the first left hand column of A160468.
Equals A002430(n)*A089170(n-1).
Equals (A002430(n)/A036279(n))*(A117972(n)/A000265(n)).
Equals A048896(n-1)*A002425(n).
Cf. A156769 (which resembles the denominators of the Taylor series for tan(x)).

a(n) = A002430(n)*A089170(n-1) with A002430 (n) = numer((-1)^(n-1)*2^(2*n)*(2^(2*n)-1)* bernoulli(2*n)/(2*n)!) and A089170 (n-1) = numer(2*bernoulli(2*n)* (4^n-1)/(2*n))/ numer((4^n-1)*bernoulli(2*n)/(2*n)!) for n = 1, 2, 3, ....

A110841 a(n) is the number of prime factors, with multiplicity, of abs(A014509(n)).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 1, 4, 2, 7, 7, 2, 2, 4, 3, 3, 7, 1, 6, 4, 5, 14, 4, 9, 5, 10, 3, 11, 2, 5, 3, 7, 11, 5, 3, 4, 15, 6, 5, 19, 10, 6, 13, 15, 5, 10, 5, 5, 6, 7, 5, 15, 7, 5, 2, 13, 4, 3, 10, 5, 9, 7, 5, 4, 9, 5, 4, 1, 7, 4, 4, 5, 3, 11, 13, 10, 5, 5, 7, 6

Offset: 0

Jonathan Vos Post, Sep 16 2005

			a(10) = 2 because A014509(10) = 529 = 23^2.
a(8) = a(19) = 1 since A014509(8) and A014509(19) are prime.

Sean A. Irvine, Table of n, a(n) for n = 0..91

Cf. A001222, A073276, A075178, A076547, A076549, A079294, A083687, A084217, A085092, A085737, A089170, A089644, A089655.

a(n) = my(b=bernfrac(2*n), c=floor(abs(b))*sign(b)); if (c==0, 0, bigomega(c)); \\ Michel Marcus, Mar 29 2020

a(n) = A001222(abs(A014509(n))).

More terms from Michel Marcus, Mar 29 2020
a(51)-a(65) from Jinyuan Wang, Apr 02 2020
More terms from Sean A. Irvine, Jul 29 2024

cp's OEIS Frontend

A002425 Denominator of Pi^(2n)/(Gamma(2n)*(1-2^(-2n))*zeta(2n)).

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A156769 a(n) = denominator(2^(2*n-2)/factorial(2*n-1)).

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A160469 The left hand column of the triangle A160468.

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A110841 a(n) is the number of prime factors, with multiplicity, of abs(A014509(n)).

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A002425 Denominator of Pi^(2n)/(Gamma(2n)(1-2^(-2n))zeta(2n)).

A156769 a(n) = denominator(2^(2n-2)/factorial(2n-1)).