login
A099376
An inverse Chebyshev transform of x^3.
4
0, 1, 4, 14, 48, 165, 572, 2002, 7072, 25194, 90440, 326876, 1188640, 4345965, 15967980, 58929450, 218349120, 811985790, 3029594040, 11338026180, 42550029600, 160094486370, 603784920024, 2282138106804, 8643460269248
OFFSET
0,3
COMMENTS
The sequence is 0,0,0,1,0,4,0,14,0,...with zeros restored. Second binomial transform of (-1)^n*A003518(n). Second binomial transform of expansion of x^3*c(-x)^8, where c(x) is g.f. of A000108. The g.f. is transformed to x^3 under the Chebyshev transformation A(x) -> (1/(1+x^2))*A(x/(1+x^2)). For a sequence b(n), this corresponds to taking Sum_{k=0..floor(n/2)} C(n-k,k) * (-1)^k * b(n-2k), or Sum_{k=0..n} C((n+k)/2,k) * b(k) * (-1)^((n-k)/2) * (1+(-1)^(n-k))/2.
Let X_n be the set of all noncrossing set partitions of an n-element set which either do not contain {n-1,n} as a block, or which do not contain the block {n} whenever 1 and n-1 are in the same block. For n>0, (-1)^n*a(n) gives the value of the Möbius function of X_{n+2} ordered by dual refinement between the discrete and the full partition. For example, X_3 is a chain consisting of 3 elements and its Möbius function between least and greatest element therefore takes the value a(1)=0. - Henri Mühle, Jan 10 2017
LINKS
FORMULA
G.f.: (1-2*x)^4*( sqrt((1+2*x)/(1-2*x)) - 1)^8/(256*x^5).
a(n) = Sum_{k=0..n} (k+1)*C(n, (n-k)/2)*(-1)^k*( C(3, k) -3*C(2, k) +3*C(1, k) -C(0, k) )*(1+(-1)^(n-k))/(n+k+2).
a(n) = A002057(n-1). - Michael Somos, Jul 31 2005
Given an ellipse with eccentricity e and major and minor axis a and b respectively, then ((a-b)/ (a+b))^2 = 1*(e/2)^4 +4*(e/2)^6 +14*(e/2)^8 +48*(e/2)^10 + ... - Michael Somos, Apr 11 2007
E.g.f.: exp(2x)*(Bessel_I(1,2x) - Bessel_I(3,2x)). - Paul Barry, Jun 04 2007
D-finite with recurrence (n+3)*(n-1)*a(n) -2*n*(2*n+1)*a(n-1)=0. - R. J. Mathar, Sep 26 2012
a(n) = A000108(n+2) - 2*A000108(n+1) for n>0. - Henri Mühle, Jan 10 2017, corrected Sep 25 2021
G.f.: ( (1-2*x)*c(x) - (1-x) )/x^2, where c(x) is the gf of A000108. - G. C. Greubel, May 05 2021
From Peter Bala, Aug 30 2023: (Start)
a(n) = 2*n/((n+2)*(n+3)) * binomial(2*n+2, n+1).
a(n) = 2*Sum_{k = 0..n-1} 1/(n+1)*binomial(n+1, k)*binomial(n+1, k+2). (End)
MATHEMATICA
Table[CatalanNumber[n+2] -2CatalanNumber[n+1], {n, 0, 30}] (* or *)
Table[4 Binomial[2#+3, #]/(#+4) &[n-1], {n, 0, 30}] (* Michael De Vlieger, Jan 10 2017, latter after Harvey P. Dale at A002057 *)
PROG
(PARI) {a(n)= if(n<1, 0, n++; 2* binomial(2*n, n-2)/n)} /* Michael Somos, Apr 11 2007 */
(Magma) [Catalan(n+2) -2*Catalan(n+1): n in [0..30]]; // G. C. Greubel, May 05 2021
(Sage) [catalan_number(n+2) -2*catalan_number(n+1) for n in (0..30)] # G. C. Greubel, May 05 2021
CROSSREFS
Partial sums of A026016.
Sequence in context: A094827 A094667 A370051 * A002057 A047048 A071745
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Oct 13 2004
STATUS
approved