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%I #42 May 25 2020 09:20:31
%S 1,10,115,1360,16265,195660,2361925,28577440,346316645,4201744870,
%T 51023399190,620022989200,7538489480075,91696845873760,
%U 1115794688036920,13581508654978560,165357977228808925,2013721466517360650
%N a(n) = a(n,m) = sum(k=0,n,binomial(m*k,k)*binomial(m*(n-k),n-k)) for m=5.
%C More generally : a(n,m)=sum(k=0,n,binomial(m*k,k)*binomial(m*(n-k),n-k)) is asymptotic to 1/2*m/(m-1)*(m^m/(m-1)^(m-1))^n. See A000302, A006256, A078995 for cases m=2,3 and 4.
%H Robert Israel, <a href="/A079678/b079678.txt">Table of n, a(n) for n = 0..828</a>
%H D. Merlini, R. Sprugnoli and M. C. Verri, <a href="http://dx.doi.org/10.1006/jcta.2002.3273">The tennis ball problem</a>, J. Combin. Theory, A 99 (2002), 307-344.
%H Rui Duarte and António Guedes de Oliveira, <a href="http://arxiv.org/abs/1302.2100">Short note on the convolution of binomial coefficients</a>, arXiv:1302.2100 [math.CO], 2013.
%F a(n) = 5/8*(3125/256)^n*(1+c/sqrt(n)+o(n^-1/2)) where c=0.356...
%F c = sqrt(2)/sqrt(5*Pi) = 0.3568248232305542229... - _Vaclav Kotesovec_, May 25 2020
%F a(n) = sum(k=0,n,binomial(5*k+l,k)*binomial(5*(n-k)-l,n-k)) for every real number l. - _Rui Duarte_ and António Guedes de Oliveira, Feb 16 2013
%F a(n) = sum(k=0,n,4^(n-k)*binomial(5n+1,k)). - _Rui Duarte_ and António Guedes de Oliveira, Feb 17 2013
%F a(n) = sum(k=0,n,5^(n-k)*binomial(4n+k,k)). - _Rui Duarte_ and António Guedes de Oliveira, Feb 17 2013
%F G.f.: hypergeom([1/5, 2/5, 3/5, 4/5], [1/4, 1/2, 3/4], (3125/256)*x)^2 satisfies
%F ((3125/2)*g^3*x^4-128*g^3*x^3)*g''''+((-3125*g^2*x^4+256*g^2*x^3)*g'+12500*g^3*x^3-576*g^3*x^2)*g'''+(-(9375/4)*g^2*x^4+192*g^2*x^3)*g''^2+(((28125/4)*g*x^4-576*g*x^3)*(g')^2+(-18750*g^2*x^3+864*g^2*x^2)*g'+22500*g^3*x^2-408*g^3*x)*g''+(-(46875/16)*x^4+240*x^3)*(g')^4+(9375*g*x^3-432*g*x^2)*(g')^3+(-11250*g^2*x^2+204*g^2*x)*(g')^2+(7500*g^3*x-12*g^3)*g'+120*g^4 = 0. - _Robert Israel_, Jul 16 2015
%p seq(add(binomial(5*k,k)*binomial(5*(n-k),n-k),k=0..n), n=0..30); # _Robert Israel_, Jul 16 2015
%t m = 5; Table[Sum[Binomial[m k, k] Binomial[m (n - k), n - k], {k, 0, n}], {n, 0, 17}] (* _Michael De Vlieger_, Sep 30 2015 *)
%o (PARI) main(size)=my(k,n,m=5); concat(1,vector(size,n, sum(k=0,n, binomial(m*k,k)*binomial(m*(n-k),n-k)))) \\ _Anders Hellström_, Jul 16 2015
%o (PARI) a(n) = sum(k=0,n,4^(n-k)*binomial(5*n+1,k));
%o vector(30, n, a(n-1)) \\ _Altug Alkan_, Sep 30 2015
%K nonn
%O 0,2
%A _Benoit Cloitre_, Jan 26 2003
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