login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

a(n) = Sum_{k=0..n} binomial(n,k)^(2*k+1).
4

%I #7 Mar 03 2014 14:06:34

%S 1,2,10,272,24226,12053252,40086916024,429254371605824,

%T 23527609330364490754,10714627376371224032350052,

%U 16964729291782419425708732425300,109783535843179466164398767001178968704,6782057095273243388704415924996348722446049600

%N a(n) = Sum_{k=0..n} binomial(n,k)^(2*k+1).

%C Ignoring initial term a(0), equals the logarithmic derivative of A206157.

%F Limit n->infinity a(n)^(1/n^2) = r^(2*r^2/(1-2*r)) = 2.3520150420944489879258119..., where r = 0.70350607643066243... (see A220359) is the root of the equation (1-r)^(2*r-1) = r^(2*r). - _Vaclav Kotesovec_, Mar 03 2014

%e L.g.f.: L(x) = 2*x + 10*x^2/2 + 272*x^3/3 + 24226*x^4/4 + 12053252*x^5/5 +...

%e where exponentiation yields A206157:

%e exp(L(x)) = 1 + 2*x + 7*x^2 + 102*x^3 + 6261*x^4 + 2423430*x^5 + 6686021554*x^6 +...

%e Illustration of initial terms:

%e a(1) = 1^1 + 1^3 = 2;

%e a(2) = 1^1 + 2^3 + 1^5 = 10;

%e a(3) = 1^1 + 3^3 + 3^5 + 1^7 = 272;

%e a(4) = 1^1 + 4^3 + 6^5 + 4^7 + 1^9 = 24226;

%e a(5) = 1^1 + 5^3 + 10^5 + 10^7 + 5^9 + 1^11 = 12053252; ...

%t Table[Sum[Binomial[n,k]^(2*k+1), {k, 0, n}], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 03 2014 *)

%o (PARI) {a(n)=sum(k=0,n,binomial(n,k)^(2*k+1))}

%o for(n=0,16,print1(a(n),", "))

%Y Cf. A206157 (exp), A184731, A206154, A206156, A206152, A220359.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 04 2012