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A206849 a(n) = Sum_{k=0..n} binomial(n^2, k^2). 11

%I #23 Jan 19 2019 06:42:36

%S 1,2,6,137,13278,4098627,8002879629,66818063663192,

%T 1520456935214867934,167021181249536494996841,

%U 102867734705055054467692090431,179314863425920182637610314008444247,1094998941099523423274757578750950802034789

%N a(n) = Sum_{k=0..n} binomial(n^2, k^2).

%H Seiichi Manyama, <a href="/A206849/b206849.txt">Table of n, a(n) for n = 0..57</a>

%H Vaclav Kotesovec, <a href="/A206849/a206849.jpg">Limits, graph for 500 terms</a>

%F Ignoring the initial term a(0), equals the logarithmic derivative of A206848.

%F Equals the row sums of triangle A226234.

%F From _Vaclav Kotesovec_, Mar 03 2014: (Start)

%F Limit n->infinity a(n)^(1/n^2) = 2

%F Lim sup n->infinity a(n)/(2^(n^2)/n) = sqrt(2/Pi) * JacobiTheta3(0,exp(-4)) = Sqrt[2/Pi] * EllipticTheta[3, 0, 1/E^4] = 0.827112271364145742...

%F Lim inf n->infinity a(n)/(2^(n^2)/n) = sqrt(2/Pi) * JacobiTheta2(0,exp(-4)) = Sqrt[2/Pi] * EllipticTheta[2, 0, 1/E^4] = 0.587247586271786487...

%F (End)

%e L.g.f.: L(x) = 2*x + 6*x^2/2 + 137*x^3/3 + 13278*x^4/4 + 4098627*x^5/5 +...

%e where exponentiation yields the g.f. of A206848:

%e exp(L(x)) = 1 + 2*x + 5*x^2 + 53*x^3 + 3422*x^4 + 826606*x^5 + 1335470713*x^6 +...

%e Illustration of terms: by definition,

%e a(1) = C(1,0) + C(1,1);

%e a(2) = C(4,0) + C(4,1) + C(4,4);

%e a(3) = C(9,0) + C(9,1) + C(9,4) + C(9,9);

%e a(4) = C(16,0) + C(16,1) + C(16,4) + C(16,9) + C(16,16);

%e a(5) = C(25,0) + C(25,1) + C(25,4) + C(25,9) + C(25,16) + C(25,25);

%e a(6) = C(36,0) + C(36,1) + C(36,4) + C(36,9) + C(36,16) + C(36,25) + C(36,36); ...

%e Numerically, the above evaluates to be:

%e a(1) = 1 + 1 = 2;

%e a(2) = 1 + 4 + 1 = 6;

%e a(3) = 1 + 9 + 126 + 1 = 137;

%e a(4) = 1 + 16 + 1820 + 11440 + 1 = 13278;

%e a(5) = 1 + 25 + 12650 + 2042975 + 2042975 + 1 = 4098627;

%e a(6) = 1 + 36 + 58905 + 94143280 + 7307872110 + 600805296 + 1 = 8002879629; ...

%t Table[Sum[Binomial[n^2, k^2],{k,0,n}],{n,0,20}] (* _Vaclav Kotesovec_, Mar 03 2014 *)

%o (PARI) {a(n)=sum(k=0, n,binomial(n^2,k^2))}

%o for(n=0, 20, print1(a(n), ", "))

%Y Cf. A206848 (exp), A226234, A206847, A206850, A206851.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 15 2012

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