A290310 Irregular triangle read by rows. Row n gives the coefficients of the polynomial multiplying the exponential function in the e.g.f. of the (n+1)-th diagonal sequences of triangle A008459 (Pascal squares). T(n,k) for n >= 0 and k = 0..2*n.

1, 1, 3, 2, 1, 8, 19, 18, 6, 1, 15, 69, 147, 162, 90, 20, 1, 24, 176, 624, 1251, 1500, 1070, 420, 70, 1, 35, 370, 1920, 5835, 11253, 14240, 11830, 6230, 1890, 252, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 63, 1169, 10703, 58821, 214123, 545629, 1004307, 1356194, 1347318, 974862, 500346, 172788, 36036, 3432

Offset: 0

Author

Wolfdieter Lang, Jul 27 2017

Keywords

• nonn
• tabf
• easy

Comments

The length of row n of this irregular triangle is 2*n+1.

A008459 gives the squares of the entries of Pascal's triangle A007318.

The e.g.f. of the (n+1)-th diagonal sequence of the squares of Pascal's triangle (A008459) is EDP2(n, t) = Sum_{m=0..n} binomial(n+m, m)^2*t^m/m!, for n >= 0. It turns out to be EDP2(n, t) = exp(t)*Sum_{k=0..2*n} T(n, k)*t^k/k!.

This has been computed from the corresponding o.g.f.: GDP2(n, x) = Sum_{m=0..n} binomial(n+m, m)^2*x^m, which is GDP2(n, x) = Sum_{m=0..n} binomial(n,m)^2*x^m / (1 - x)^(2*n+1) (see the triangle A008459, and comments in A288876 on how to compute these o.g.f.s). To obtain the e.g.f.s from the o.g.f.s the formulas (23) - (25) of the W. Lang link given in A060187 have been used.

Examples

			The irregular triangle T(n, k) begins:
  n\k 0  1   2    3     4     5      6      7      8     9    10   11  12 ..,
  0:  1
  1:  1  3   2
  2:  1  8  19   18     6
  3:  1 15  69  147   162    90     20
  4:  1 24 176  624  1251  1500   1070    420     70
  5:  1 35 370 1920  5835 11253  14240  11830   6230  1890   252
  6:  1 48 687 4850 20385 55908 104959 137886 127050 80640 33642 8316 924
  ...
  7: 1 63 1169 10703 58821 214123 545629 1004307 1356194 1347318 974862 500346 172788 36036 3432
  ...
n = 3: The e.g.f. of the fourth diagonal sequence of A008459 is A001249 = [1, 16, 100, ...] is exp(t)*(1 + 15*t + 69*t^2/2! + 147*t^3/3! + 162*t^4/4! + 90*t^5/5! + 20*t^6/6!). The corresponding o.g.f. from which the e.g.f. has been computed is (1 + x)*(1 + 8*x + x^2)/(1 - x)^7 = (1 + 9*x + 9*x^2 + x^3)/(1 - x)^7.
		

Links

• M. Dukes, C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Crossrefs

T(n, n) = A005258(n). The squares of the first diagonals are in A000012, A000290(n+1), A000537(n+1), A001249, A288876 (for d = 0..4).

Cf. A007318, A008459, A063007, A000984, A002457.

Programs

Maple

T := (n,k) -> binomial(2*n, k)*hypergeom([-k, -n, -n], [1, -2*n], 1):
seq(seq(simplify(T(n,k)),k=0..2*n),n=0..7); # Peter Luschny, Feb 10 2018

Mathematica

Table[Sum[Binomial[2 n - i, k - i] Binomial[n, i]^2, {i, 0, k}], {n, 0, 7}, {k, 0, 2 n}] // Flatten (* Michael De Vlieger, Jul 30 2017 *)

Formula

T(n, k) = Sum_{i = 0..k} binomial(2*n - i, k-i)*binomial(n, i)^2.

From Peter Bala, Feb 06 2018: (Start)

T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*binomial(n+i,i)^2.

T(n,k) = Sum_{i = 0..k} binomial(n,i)*binomial(n,k-i)*binomial(n+k-i,k-i).

T(n,2*n) = binomial(2*n,n) = A000984(n); T(n+1,2*n+1) = 3*(2*n+1)!/n!^2 = 3*A002457(n).

Recurrence: (2*n-k)*(2*n-k-1)T(n,k) = (5*n^2+2*n*k+1-4*n-k)*T(n-1,k) - (n-1)^2*T(n-2,k).

n-th row polynomial R(n,x) = (1 + x)^n * P(n,2*x + 1) = (1 + x)^n * the n-th row polynomial of A063007, where P(n,x) is the n-th Legendre polynomial.

R(n,x) = Sum_{k >= 0} binomial(n+k,k)^2 * x^k/(1 + x)^(k+1).

(x - 1)^(2*n)/x^n * R(n,1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)^2*x^k, the n-th row polynomial of A008459.

R(n,x) = (1 + x)^n o (1 + x)^n where o denotes the black diamond product of power series as defined in Dukes and White.

R(n,x) = coefficient of u^n*v^n in the expansion of the rational function 1/((1+x)*(1+u)(1+v) - x). (End)

T(n,k) = binomial(2*n, k)*hypergeom([-k, -n, -n],[1, -2*n], 1). - Peter Luschny, Feb 10 2018