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 A002425 Denominator of Pi^(2n)/(Gamma(2n)*(1-2^(-2n))*Zeta(2n)). (Formerly M5036 N2174) 26

%I M5036 N2174

%S 1,1,1,17,31,691,5461,929569,3202291,221930581,4722116521,

%T 968383680827,14717667114151,2093660879252671,86125672563201181,

%U 129848163681107301953,868320396104950823611,209390615747646519456961

%N Denominator of Pi^(2n)/(Gamma(2n)*(1-2^(-2n))*Zeta(2n)).

%C 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

%C Odd part of tangent numbers A000182 (even part is 2^A101921(n)). - _Ralf Stephan_, Dec 21 2004

%C (-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)

%C 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

%C From _Vladimir Shevelev_, Aug 31 2017: (Start)

%C 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)

%D 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.

%D S. A. Joffe, Sums of like powers of natural numbers, Quart. J. Pure Appl. Math. 46 (1914), 33-51.

%D Konrad Knopp, Theory and application of infinite series, Divergent series, Dover, p. 479

%D L. Oettinger, Archiv. Math. Phys., 26 (1856), see esp. p. 5.

%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H N. J. A. Sloane, <a href="/A002425/b002425.txt">Table of n, a(n) for n = 1..300</a>

%H H. Cohn, <a href="https://doi.org/10.1090/S0002-9904-1965-11343-X">Some elementary aspects of modular functions in several variables</a>, Bull. Am. Math. Soc., Sept. 1965, 681ff, esp. p. 688.

%H S. A. Joffe, <a href="/A001896/a001896.pdf">Sums of like powers of natural numbers</a>, Quart. J. Pure Appl. Math. 46 (1914), 33-51. [Annotated scanned copy of pages 38-51 only, plus notes]

%H Konrad Knopp, <a href="http://www.hti.umich.edu/cgi/t/text/text-idx?sid=b88432273f115fb346725f1a42422e19;c=umhistmath;idno=ACM1954.0001.001">Theorie und Anwendung der unendlichen Reihen</a>, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")

%H Vladimir Shevelev, <a href="https://arxiv.org/abs/1708.08096">On a Luschny question</a>, arXiv:1708.08096 [math.NT], 2017.

%F 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

%F This is different from the sequence of numerators of the expansion of cosec(x)-cot(x) - see A089171.

%F a(n) = denominator(4*n/((2^(2*n)-1)*bernoulli(2*n))). - _Johannes W. Meijer_, May 24 2009

%F E.g.f.: a(n) = numerator((2*n+1)!*[x^(2*n+1)](1/(1+1/exp(x)))). - _Peter Luschny_, Jul 12 2012

%F 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

%p A002425 := n -> (-1)^n*euler(2*n-1,0)*2^padic[ordp](2*n,2); # _Peter Luschny_, Nov 26 2010

%p A002425_list := proc(n) 1/(1+1/exp(z)); series(%,z,2*n+4);

%p seq(numer((-1)^i*(2*i+1)!*coeff(%,z,2*i+1)),i=0..n) end;

%p A002425_list(17); # _Peter Luschny_, Jul 12 2012

%t a[n_] := (-1)^(n-1) * Numerator[EulerE[2n - 1, 1]]; Table[a[n], {n, 1, 17}] (* _Jean-François Alcover_, Sep 20 2011, after _N. J. A. Sloane_'s comment *)

%t a[ n_] := If[ n < 1, 0, With[{m = 2 n - 1}, Numerator[ m! SeriesCoefficient[ Tan[x/2], {x, 0, m}]]]] (* _Michael Somos_, Sep 14 2013 *)

%t Table[2*(4^n-1)*Zeta[1-2n] // Abs // Numerator, {n, 1, 18}] (* _Jean-François Alcover_, Oct 16 2013 *)

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

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

%o (PARI) a(n)=(-1)^n*4*bitand(n,-n)*polylog(1-2*n,-1); \\ _Peter Luschny_, Nov 22 2012

%o (Sage)

%o def A002425_list(n):

%o T = [0]*n; T[0] = 1; S = [0]*n; k2 = 0

%o for k in (1..n-1): T[k] = k*T[k-1]

%o for k in (1..n):

%o if is_odd(k): S[k-1] = 4*k2; k2 += 1

%o else: S[k-1] = S[k2-1]+2*k2-1

%o for j in (k..n-1): T[j] = (j-k)*T[j-1]+(j-k+2)*T[j]

%o return [T[j]>>S[j] for j in (0..n-1)]

%o A002425_list(18) # _Peter Luschny_, Nov 17 2012

%Y Numerator given by A037239.

%Y Different from A089171.

%Y Contribution from _Johannes W. Meijer_, May 24 2009: (Start)

%Y Equals A160469(n)/A048896(n-1);

%Y equals A089171(n)*A089170(n-1). (End)

%K nonn,frac,easy

%O 1,4

%A _N. J. A. Sloane_

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

%E Formula and cross-references edited by _Johannes W. Meijer_, May 21 2009

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Last modified October 20 15:22 EDT 2018. Contains 316388 sequences. (Running on oeis4.)