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 A288876 #23 Mar 13 2025 22:02:31
%S A288876 1,25,225,1225,4900,15876,44100,108900,245025,511225,1002001,1863225,
%T A288876 3312400,5664400,9363600,15023376,23474025,35820225,53509225,78411025,
%U A288876 112911876,160022500,223502500,308002500,419225625,564110001,751034025,990046225,1293121600,1674446400,2150733376
%N A288876 a(n) = binomial(n+4, n)^2. Square of the fifth diagonal sequence of A007318 (Pascal). Fifth diagonal sequence of A008459.
%C A288876 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.
%C A288876 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/(1-x)) 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.
%C A288876 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/(1-x), x/(1-x)).
%C A288876 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.
%F A288876 a(n) = binomial(n+4, n)^2, n >= 0.
%F A288876 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.)
%F A288876 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.
%F A288876 From _Amiram Eldar_, Sep 20 2022: (Start)
%F A288876 Sum_{n>=0} 1/a(n) = 160*Pi^2/3 - 1576/3.
%F A288876 Sum_{n>=0} (-1)^n/a(n) = 512*log(2)/3 - 352/3. (End)
%t A288876 Table[Binomial[n + 4, n]^2, {n, 0, 30}] (* _Michael De Vlieger_, Jul 30 2017 *)
%o A288876 (PARI) a(n) = binomial(n+4, n)^2 \\ _Felix Fröhlich_, Jul 31 2017
%o A288876 (Magma) [Binomial(n+4, n)^2: n in [0..30]]; // _Vincenzo Librandi_, Aug 02 2017
%Y A288876 Cf. A007318, A008459.
%Y A288876 The squares of the first diagonals are in A000012, A000290(n+1), A000537, A001249 (for d = 0..3).
%K A288876 nonn,easy
%O A288876 0,2
%A A288876 _Wolfdieter Lang_, Jul 27 2017