OFFSET
1,2
COMMENTS
Related to the formula Sum_{k>0} sin(kx)/k^(2n+1)=(-1)^(n+1)/2*x^(2n+1)/(2n+1)! * Sum_{i=0..2n} (2Pi/x)^i*B(i)*C(2n+1,i). - Benoit Cloitre, May 01 2002
Named after the English mathematician and astronomer James Whitbread Lee Glaisher (1848-1928). - Amiram Eldar, Jun 16 2021
REFERENCES
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. 76.
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).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 1..254 (first 50 terms from T. D. Noe)
Shaun Cooper, Cubic elliptic functions, Res. Lett. Inf. Math. Sci., Vol. 5 (2003), pp. 23-59, see page 30.
Ira M. Gessel, On the Almkvist-Meurman Theorem for Bernoulli Polynomials, Integers (2023) Vol. 23, #A14.
J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., Vol. 31 (1899), pp. 216-235.
René Gy, Bernoulli-Stirling Numbers, Integers, Vol. 20, (2020), #A67.
J. C. P. Miller, Letter to N. J. A. Sloane, Mar 26 1971.
N. J. A. Sloane, Transforms.
Wikipedia, Bernoulli Polynomials.
FORMULA
To get these numbers, expand the e.g.f. (3/2)/(1+exp(x)+exp(-x)), multiply coefficient of x^n by (n+1)! and take absolute values.
Or expand the e.g.f. (3/2)/(1+2*cos(x)) and multiply coefficient of x^n by (n+1)!. - Herb Conn, Feb 25 2002
a(n) = Sum_{i=0, 2n} B(i)*C(2n+1, i)*3^i where B(i) are the Bernoulli numbers, C(2n, i) the binomial numbers. - Benoit Cloitre, May 01 2002
a(n) = (-1)^n * (6*n + 3) * s(2*n), if n>0, where s(n) are the cubic Bernoulli numbers. - Michael Somos, Feb 26 2004
E.g.f.: 3*x / (2 + 4*cos(x)) = Sum_{n>=0} a(n) * x^(2*n+1) / (2*n+1)!. - Michael Somos, Feb 26 2004
E.g.f.: E(x) = (3/2)/(1+2*cos(x)) - 1/2 = x^2/(3*G(0)+x^2); G(k) = 2*(2*k+1)*(k+1) - x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction Euler's kind, 1-step). Let f[n]:=coeftayl(E(x), x=0, n) then: A002111[n]=f[2*n+2]*((2*n+3)!). - Sergei N. Gladkovskii, Jan 14 2012
a(n) = Sum_{k=0..2n+1} Sum_{j=0..k} Sum_{v=0..j} ((-1)^(n-v+1)/(j+1))* binomial(2*n+1,k)*binomial(j,v)*(3*v)^k. - Peter Luschny, Jun 03 2013
a(n) ~ (2*n+1)! * sqrt(3) * (3/(2*Pi))^(2*n+1). - Vaclav Kotesovec, Jul 30 2013
From Peter Bala, Mar 02 2015: (Start)
a(n) = (-1)^(n+1)*3^(2*n+1)*B(2*n+1,1/3), where B(n,x) denotes the n-th Bernoulli polynomial. Cf. A009843, A069852, A069994.
Conjecturally, a(n) = the unsigned numerator of B(2*n+1,1/3). Cf. A033470.
Essentially a bisection of |A083007|.
G.f. for signed version of sequence: 1/2 + 1/2*Sum_{n >= 0} { 1/(n+1) * Sum_{k = 0..n} (-1)^(k+1)*binomial(n,k)/( (1 - (3*k + 1)*x)*(1 - (3*k + 2)*x) ) } = x^2 - 5*x^4 + 49*x^6 - .... (End)
EXAMPLE
G.f. = x + 5*x^2 + 49*x^3 + 809*x^4 + 20317*x^5 + 722813*x^6 + 34607305*x^7 + ...
MAPLE
read transforms; t1 := (3/2)/(1+exp(x)+exp(-x)); series(t1, x, 50): t2 := SERIESTOLISTMULT(t1); [seq(n*t2[n], n=1..nops(t5))];
MATHEMATICA
s[n_] := CoefficientList[Series[(1/2)*(Sin[t/2]/Sin[3*(t/2)]), {t, 0, 32}], t][[n + 1]]*n!*(-1)^Floor[n/2]; a[n_] := (-1)^n*(6*n + 3)*s[2*n]; Table[a[n], {n, 1, 16}] (* Jean-François Alcover, Mar 22 2011, after Michael Somos' formula *)
a[ n_] := If[ n < 1, 0, (2 n + 1)! SeriesCoefficient[ 3 / (2 + 4 Cos[x]), {x, 0, 2 n}]]; (* Michael Somos, Jun 01 2012 *)
PROG
(PARI) {a(n) = if( n<1, 0, n*=2; (n+1)! * polcoeff( 3 / (2 + 4 * cos( x + O(x^n))), n))}; /* Michael Somos, Feb 26 2004 */
(PARI) a(n)=if(n<1, 0, -(-1)^n*sum(i=0, 2*n, binomial(2*n+1, i)*bernfrac(i)*3^i)) \\ Benoit Cloitre, May 01 2002
(Sage)
def A002111(n):
return add(add(add(((-1)^(n+1-v)/(j+1))*binomial(2*n+1, k)*binomial(j, v)*(3*v)^k for v in (0..j)) for j in (0..k)) for k in (0..2*n+1))
[A002111(n) for n in (1..16)] # Peter Luschny, Jun 03 2013
CROSSREFS
KEYWORD
nonn,nice,easy
AUTHOR
STATUS
approved