This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A078995 #64 Aug 15 2025 09:10:39
%S A078995 1,8,72,664,6184,57888,543544,5113872,48180456,454396000,4288773152,
%T A078995 40503496536,382701222296,3617396099936,34203591636048,
%U A078995 323492394385824,3060238763412072,28955508198895584,274018698082833760,2593539713410178528,24550565251665845664
%N A078995 a(n) = Sum_{k=0..n} C(4*k,k)*C(4*(n-k),n-k).
%H A078995 Vincenzo Librandi, <a href="/A078995/b078995.txt">Table of n, a(n) for n = 0..200</a>
%H A078995 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 (Y_n for s=4).
%H A078995 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 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL16/Duarte/duarte3.html">J. Int. Seq. 16 (2013) #13.7.6</a>.
%F A078995 a(n) = 2/3*(256/27)^n*(1+c/sqrt(n)+o(n^-1/2)) where c = 2/3*sqrt(2/(3*Pi)) = 0.307105910641187... 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 for cases m=2 and 3. - _Benoit Cloitre_, Jan 26 2003, corrected and extended by _Vaclav Kotesovec_, Nov 06 2012
%F A078995 243*n*(8*n - 17)*(3*n - 1)*(3*n - 4)*(3*n - 2)*(3*n - 5)*a(n) = 72*(3*n - 5)*(3*n - 4)*(6912*n^4 - 33120*n^3 + 58256*n^2 - 47798*n + 15309)*a(n - 1) - 3072*(2*n - 3)*(6912*n^5 - 55008*n^4 + 175696*n^3 - 282180*n^2 + 227825*n - 73710)*a(n - 2) + 262144*(n - 2)*(4*n - 7)*(2*n - 3)*(2*n - 5)*(4*n - 9)*(8*n - 9)*a(n - 3). - _Vladeta Jovovic_, Jul 16 2004
%F A078995 Shorter recurrence: 81*n*(3*n-2)*(3*n-1)*(8*n-11)*a(n) = 24*(4608*n^4-14400*n^3+15776*n^2-7346*n+1215)*a(n-1) - 2048*(2*n-3)*(4*n-5)*(4*n-3)*(8*n-3)*a(n-2). - _Vaclav Kotesovec_, Nov 06 2012
%F A078995 a(n) = Sum_{k=0..n} binomial(4*k+l,k) * binomial(4*(n-k)-l,n-k) for every real number l. - _Rui Duarte_ and António Guedes de Oliveira, Feb 16 2013
%F A078995 From Rui Duarte and António Guedes de Oliveira, Feb 17 2013: (Start)
%F A078995 a(n) = Sum_{k=0..n} 3^(n-k) * binomial(4*n+1,k).
%F A078995 a(n) = Sum_{k=0..n} 4^(n-k) * binomial(3*n+k,k). (End)
%F A078995 G.f.: g^2/(3*g-4)^2 where g=ogf(A002293) satisfies g = 1+x*g^4. - _Mark van Hoeij_, May 06 2013
%F A078995 a(n) = [x^n] 1/((1-4*x) * (1-x)^(3*n+1)). - _Seiichi Manyama_, Aug 03 2025
%F A078995 a(n) = Sum_{k=0..n} 4^k * (-3)^(n-k) * binomial(4*n+1,k) * binomial(4*n-k,n-k). - _Seiichi Manyama_, Aug 15 2025
%p A078995 series(eval(g/(3*g-4), g=RootOf(g = 1+x*g^4,g))^2, x=0, 30); # _Mark van Hoeij_, May 06 2013
%t A078995 Table[Sum[Binomial[4*k, k]*Binomial[4*(n - k), n - k], {k, 0, n}], {n, 0, 20}] (* _Vaclav Kotesovec_, Nov 06 2012 *)
%o A078995 (PARI) a(n) = sum(k=0, n, binomial(4*k, k)*binomial(4*(n-k), n-k)); \\ _Michel Marcus_, May 09 2020
%Y A078995 See A049235 for more information.
%Y A078995 Cf. A006256, A079563, A079678, A079679.
%Y A078995 Cf. A005810, A147855, A226733, A226761, A385605, A386811.
%K A078995 nonn
%O A078995 0,2
%A A078995 _N. J. A. Sloane_, Jan 19 2003