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%I #26 Feb 10 2018 13:04:33
%S 1,1,3,2,1,8,19,18,6,1,15,69,147,162,90,20,1,24,176,624,1251,1500,
%T 1070,420,70,1,35,370,1920,5835,11253,14240,11830,6230,1890,252,1,48,
%U 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
%N 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.
%C The length of row n of this irregular triangle is 2*n+1.
%C A008459 gives the squares of the entries of Pascal's triangle A007318.
%C 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!.
%C 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.
%H M. Dukes, C. D. White, <a href="http://arxiv.org/abs/1603.01589">Web Matrices: Structural Properties and Generating Combinatorial Identities</a>, arXiv:1603.01589 [math.CO], 2016.
%F T(n, k) = Sum_{i = 0..k} binomial(2*n - i, k-i)*binomial(n, i)^2.
%F From _Peter Bala_, Feb 06 2018: (Start)
%F T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*binomial(n+i,i)^2.
%F T(n,k) = Sum_{i = 0..k} binomial(n,i)*binomial(n,k-i)*binomial(n+k-i,k-i).
%F 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).
%F 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).
%F 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.
%F R(n,x) = Sum_{k >= 0} binomial(n+k,k)^2 * x^k/(1 + x)^(k+1).
%F (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.
%F 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.
%F 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)
%F T(n,k) = binomial(2*n, k)*hypergeom([-k, -n, -n],[1, -2*n], 1). - _Peter Luschny_, Feb 10 2018
%e The irregular triangle T(n, k) begins:
%e n\k 0 1 2 3 4 5 6 7 8 9 10 11 12 ..,
%e 0: 1
%e 1: 1 3 2
%e 2: 1 8 19 18 6
%e 3: 1 15 69 147 162 90 20
%e 4: 1 24 176 624 1251 1500 1070 420 70
%e 5: 1 35 370 1920 5835 11253 14240 11830 6230 1890 252
%e 6: 1 48 687 4850 20385 55908 104959 137886 127050 80640 33642 8316 924
%e ...
%e 7: 1 63 1169 10703 58821 214123 545629 1004307 1356194 1347318 974862 500346 172788 36036 3432,
%e 8: 1 80 1864 21392 147952 683648 2240560 5401952 9793891 13507416 14210532 11337480 6749358 2906904 856284 154440 12870,
%e 9: 1 99 2826 39624 335376 1906128 7747152 23429808 54099027 96918753 135916002 149594148 128769102 85783698 43366752 16087500 4131270 656370 48620.
%e ...
%e 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.
%p T := (n,k) -> binomial(2*n, k)*hypergeom([-k, -n, -n], [1, -2*n], 1):
%p seq(seq(simplify(T(n,k)),k=0..2*n),n=0..7); # _Peter Luschny_, Feb 10 2018
%t 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 *)
%Y 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).
%Y Cf. A007318, A008459, A063007, A000984, A002457.
%K nonn,tabf,easy
%O 0,3
%A _Wolfdieter Lang_, Jul 27 2017
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