

A288876


a(n) = binomial(n+4, n)^2. Square of the fifth diagonal sequence of A007318 (Pascal). Fifth diagonal sequence of A008459.


5



1, 25, 225, 1225, 4900, 15876, 44100, 108900, 245025, 511225, 1002001, 1863225, 3312400, 5664400, 9363600, 15023376, 23474025, 35820225, 53509225, 78411025, 112911876, 160022500, 223502500, 308002500, 419225625, 564110001, 751034025, 990046225, 1293121600, 1674446400, 2150733376
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OFFSET

0,2


COMMENTS

This is also the square of the fifth (k = 4) column sequence (without leading zeros) of the Pascal triangle A007318. For the triangle with the squares of the entries of Pascal's triangle see A008459.
For the square of the (d+1)th diagonal sequence of A007318, PD2(d,n) = binomial(d + n, n)^2, d >= 0, one finds the o.g.f. GPD2(d, x) = Sum_{n>=0} PD2(d,n)*x^n in the following way. Compute the compositional inverse (Lagrange inversion formula) of y(t,x) = x*(1  t/(1x)) w.r.t. x, that is x = x(t,y). Then log(1  x(t,y)) = Sum_{d=0} y^(d+1)/(d+1)*GPD2(d, x). The r.h.s. can be called the logarithmic generating function (l.g.f.) of the o.g.f.s of the square of the diagonals of Pascal's triangle.
This computation was inspired by an article by P. Bala (see a link in A112007) on the diagonal sequences of special Sheffer triangles (1, f(t)) (Sheffer triangles are there called exponential Riordan triangles, and f is called F). This can be generalized to Sheffer (g, f). For general Riordan triangles R = (G(x, F(x)) a similar analysis can be done. The present entry is then obtained for example of the Pascal triangle P = (1/(1x), x/(1x)).
The o.g.f.s for the square of the diagonals of Pascal's triangle turn out to be GPD2(d, x) = P(d,x)/(1  x)^(2*d+1), with the numerator polynomials given by row n of triangle A008459 (squares of the entries of Pascal's triangle): P(d, x) = Sum_{k=0..d} A008459(d, k)*x^k.


LINKS



FORMULA

a(n) = binomial(n+4, n)^2, n >= 0.
O.g.f.: (1 + 16*x + 36*x^2 + 16*x^3 + x^4)/(1  x)^9. (See a comment above and row n=4 of A008459.)
E.g.f: exp(x)*(1 + 24*x + 176*x^2/2! + 624*x^3/3! + 1251*x^4/4!+ 1500*x^5/5!+ 1070*x^6/6! + 420*x^7/7! + 70*x^8/8!), computed from the o.g.f with the formulas (23)  (25) of the W. Lang link given in A060187.
Sum_{n>=0} 1/a(n) = 160*Pi^2/3  1576/3.
Sum_{n>=0} (1)^n/a(n) = 512*log(2)/3  352/3. (End)


MATHEMATICA



PROG



CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



