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 A002802 a(n) = (2*n+3)!/(6*n!*(n+1)!). (Formerly M4724 N2019) 35
 1, 10, 70, 420, 2310, 12012, 60060, 291720, 1385670, 6466460, 29745716, 135207800, 608435100, 2714556600, 12021607800, 52895074320, 231415950150, 1007340018300, 4365140079300, 18839025605400, 81007810103220, 347176329013800, 1483389769422600 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS For n >= 1 a(n) is also the number of rooted bicolored unicellular maps of genus 1 on n+2 edges. - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 20 2001 a(n) = A051133(n+1)/3 = A000911(n)/6. - Zerinvary Lajos, Jun 02 2007 REFERENCES C. Jordan, Calculus of Finite Differences. Budapest, 1939, p. 449. 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 T. D. Noe, Table of n, a(n) for n = 0..200 R. Cori, G. Hetyei, Counting genus one partitions and permutations, arXiv preprint arXiv:1306.4628 [math.CO], 2013. R. Cori, G. Hetyei, How to count genus one partitions, FPSAC 2014, Chicago, Discrete Mathematics and Theoretical Computer Science (DMTCS), Nancy, France, 2014, 333-344; http://www.dmtcs.org/dmtcs-ojs/index.php/proceedings/article/viewFile/dmAT0130/4488 Alain Goupil and Gilles Schaeffer, Factoring N-Cycles and Counting Maps of Given Genus, Europ. J. Combinatorics (1998) 19 819-834. T. R. S. Walsh and A. B. Lehman, Counting rooted maps by genus. I, J. Comb. Theory B 13 (1972), 192-218 (Tab.1). Liang Zhao and Fengyao Yan, Note on Total Positivity for a Class of Recursive Matrices, Journal of Integer Sequences, Vol. 19 (2016), Article 16.6.5. FORMULA G.f.: (1 - 4*x)^(-5/2) = 1F0(5/2;;4x). Asymptotic expression for a(n) is a(n) ~ (n+2)^(3/2) * 4^(n+2) / (sqrt(Pi) * 48). a(n) = Sum_{a+b+c+d+e=n} f(a)*f(b)*f(c)*f(d)*f(e) with f(n) = binomial(2n, n) = A000984(n). - Philippe Deléham, Jan 22 2004 a(n-1) = (1/4)*Sum_{=1..n} k*(k+1)*binomial(2*k, k). - Benoit Cloitre, Mar 20 2004 Also convolution of A000984 with A002697, also convolution of A000302 with A002457. a(n) = ((2n+3)(2n+1)/(3*1)) * binomial(2n, n) a(n) = binomial(2n+4, 4) * binomial(2n, n) / binomial(n+2, 2) a(n) = binomial(n+2, 2) * binomial(2n+4, n+2) / binomial(4, 2) = binomial(2n+4, n+2) * (n+2)*(n+1) / 12. - Rui Duarte, Oct 08 2011 n*a(n) + 2*(-2*n-3)*a(n-1) = 0. - R. J. Mathar, Jan 31 2014 a(n) = 4^n*hypergeom([-n,-3/2], [1], 1). - Peter Luschny, Apr 26 2016 Boas-Buck recurrence: a(n) = (10/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+2, 2). See a comment there. - Wolfdieter Lang, Aug 10 2017 a(n) = (-4)^n*binomial(-5/2, n). - Peter Luschny, Oct 23 2018 EXAMPLE G.f. = 1 + 10*x + 70*x^2 + 420*x^3 + 2310*x^4 + 12012*x^5 + 60060*x^6 + ... MAPLE seq(simplify(4^n*hypergeom([-n, -3/2], [1], 1)), n=0..22); # Peter Luschny, Apr 26 2016 MATHEMATICA Table[(2*n+3)!/(6*n!*(n+1)!), {n, 0, 20}] (* Vladimir Joseph Stephan Orlovsky, Dec 13 2008 *) PROG (PARI) {a(n) = if( n<0, 0, (2*n + 3)! / (6 * n! * (n+1)!))}; /* Michael Somos, Sep 16 2013 */ (PARI) {a(n) = 2^(n+3) * polcoeff( pollegendre(n+4), n) / 3}; /* Michael Somos, Sep 16 2013 */ CROSSREFS Cf. A035309, A000108 (for genus 0 maps), A046521 (third column). Sequence in context: A005567 A174434 A073391 * A101029 A122892 A125347 Adjacent sequences:  A002799 A002800 A002801 * A002803 A002804 A002805 KEYWORD nonn,easy AUTHOR STATUS approved

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Last modified December 9 19:14 EST 2018. Contains 318023 sequences. (Running on oeis4.)