1,4

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 (benoit7848c(AT)orange.fr), Apr 27 2002

Odd part of tangent numbers A000182 (even part is 2^A101921(n)). - Ralf Stephan, Dec 21 2004

(-1)^n*a(n) is the numerator of Euler(2n+1,1). - njas, Nov 10 2009

H. Cohn, Bull. Am. Math. Soc., Sept. 1965, 681ff, esp. p. 688.

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

Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")

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 (benoit7848c(AT)orange.fr), Dec 30 2003

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

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009: (Start)

a(n) = denominator(4*n/((2^(2*n)-1)*bernoulli(2*n)))

(End)

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

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

Numerator given by A037239.

Different from A089171.

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009: (Start)

Equals A160469(n)/A048896(n-1)

Equals A089171(n)*A089170(n-1)

(End)

Sequence in context: A146667 A146462 A089171 this_sequence A046990 A059212 A058899

Adjacent sequences: A002422 A002423 A002424 this_sequence A002426 A002427 A002428

nonn,frac,easy,new

N. J. A. Sloane (njas(AT)research.att.com).

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

Formula and cross-references edited by Johannes W. Meijer (meijgia(AT)hotmail.com), May 21 2009