A008459 -id:A008459 - cp's OEIS frontend

A102221 Column 0 of triangular matrix A102220, which equals [2*I - A008459]^(-1).

1, 1, 5, 55, 1077, 32951, 1451723, 87054773, 6818444405, 675900963271, 82717196780955, 12248810651651333, 2158585005685222491, 446445657799551807541, 107087164031952038620481, 29487141797206760561836055, 9238158011747884080353808245

Offset: 0

Paul D. Hanna, Dec 31 2004

nonn

a(n) is the number of ways to form an ordered pair of n-permutations and then choose a subset of its common descent set. Cf. A192721. - Geoffrey Critzer, Apr 29 2023

Alois P. Heinz, Table of n, a(n) for n = 0..248

Cf. A000275, A008459, A102220, A102222, A192721.
Row sums of A192722.
Column k=2 of A326322.

b:= proc(n) option remember; `if`(n=0, 1,
      add(b(n-i)*binomial(n, i)/i!, i=1..n))
    end:
a:= n-> b(n)*n!:
seq(a(n), n=0..20);  # Alois P. Heinz, May 11 2016

Rest[CoefficientList[Series[1/(2-BesselJ[0, 2*I*Sqrt[x]]), {x, 0, 20}], x] * Range[0, 20]!^2] (* Vaclav Kotesovec, Mar 02 2014 *)
m = 20; CoefficientList[1/(2 - BesselI[0, 2 Sqrt[x]]) + O[x]^m, x] Range[0, m - 1]!^2 (* Jean-François Alcover, Jun 11 2019, after Vladeta Jovovic *)
b[n_] := b[n] = If[n==0, 1, Sum[b[n-i] Binomial[n, i]/i!, {i, 1, n}]];
a[n_] := b[n] n!;
a /@ Range[0, 20] (* Jean-François Alcover, Dec 03 2020, after Alois P. Heinz *)

a(n)=if(n==0,1,sum(k=0,n-1,binomial(n,k)^2*a(k)))

L = taylor(1/(1-x*hypergeometric((1,),(2,2),x)),x,0,14).list()
[factorial(i)^2*c for (i,c) in enumerate(L)] # Peter Luschny, Jul 28 2015

a(n) = Sum_{k=0..n-1} C(n, k)^2*a(k) for n>0, with a(0)=1.
a(n) = A102220(n+k, k)/C(n+k, k)^2 for k>=0.
Sum_{n>=0} a(n)*x^n/n!^2 = 1/(2-BesselI(0,2*sqrt(x))). - Vladeta Jovovic, Jul 17 2006
a(n) ~ c * (n!)^2 / r^n, where r = 0.81712266563155429332453954757369795... is the root of the equation BesselJ(0, 2*I*sqrt(x))=2, and c = 0.833570458821600548332410448635741072476086046022299770387... = 1/(sqrt(r) * BesselI(1, 2*sqrt(r))). - Vaclav Kotesovec, Mar 02 2014, updated Apr 01 2018
From Geoffrey Critzer, Apr 29 2023: (Start)
Sum_{n>=0} a(n)*z^n/(n!)^2 = 1/(2-E(z)) where E(z) = Sum_{n>=0} z^n/(n!)^2.
a(n) = Sum_{k=0..n-1} A192721(n,k)*2^k. (End)

Content moved from A192723 to this sequence by Alois P. Heinz, Sep 11 2019

A102223 Column 0 of triangular matrix A102222, which equals -log[2*I - A008459].

Original entry on oeis.org

0, 1, 3, 22, 323, 7906, 290262, 14919430, 1022475715, 90094491994, 9923239949978, 1335853771297750, 215797095378591542, 41198645313603207990, 9176288655853717238830, 2358300288047799986966722

Offset: 0

Paul D. Hanna, Dec 31 2004

nonn

Triangle A008459 consists of squared binomial coefficients.

			a(2) = 3 = 1 + (1*0*0 + 4*1*1)/2,
a(3) = 22 = 1 + (1*0*0 + 9*1*1 + 9*2*3)/3,
a(4) = 323 = 1 + (1*0*0 + 16*1*1 + 36*2*3 + 16*3*22)/4,
a(5) = 7906 = 1 + (1*0*0 + 25*1*1 + 100*2*3 + 100*3*22 + 25*4*323)/5.

Cf. A008459, A102220, A102222.

a(n)=if(n<1,0,1+sum(k=0,n-1,binomial(n,k)^2*k*a(k))/n)

a(n) = 1 + (1/n)*Sum_{k=0..n-1} C(n, k)^2*k*a(k) for n>0, with a(0)=0.

Sum_{n>=0} a(n)*x^n/n!^2 = -log(2-BesselI(0,2*sqrt(x))). - Vladeta Jovovic, Jul 16 2006

A288876 a(n) = binomial(n+4, n)^2. Square of the fifth diagonal sequence of A007318 (Pascal). Fifth diagonal sequence of A008459.

Original entry on oeis.org

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

Offset: 0

Wolfdieter Lang, Jul 27 2017

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/(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.
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)).
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.

Cf. A007318, A008459.

The squares of the first diagonals are in A000012, A000290(n+1), A000537, A001249 (for d = 0..3).

[Binomial(n+4, n)^2: n in [0..30]]; // Vincenzo Librandi, Aug 02 2017

Table[Binomial[n + 4, n]^2, {n, 0, 30}] (* Michael De Vlieger, Jul 30 2017 *)

a(n) = binomial(n+4, n)^2 \\ Felix Fröhlich, Jul 31 2017

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.
From Amiram Eldar, Sep 20 2022: (Start)
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)

A055133 Matrix inverse of A008459 (squares of entries of Pascal's triangle).

Original entry on oeis.org

1, -1, 1, 3, -4, 1, -19, 27, -9, 1, 211, -304, 108, -16, 1, -3651, 5275, -1900, 300, -25, 1, 90921, -131436, 47475, -7600, 675, -36, 1, -3081513, 4455129, -1610091, 258475, -23275, 1323, -49, 1, 136407699, -197216832, 71282064, -11449536, 1033900, -59584, 2352, -64, 1

Offset: 0

Christian G. Bower, Apr 25 2000

Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0,2*sqrt(-y)). Then this triangle is the generalized Riordan array (1/E(y), y) with respect to the sequence n!^2 as defined in Wang and Wang. - Peter Bala, Jul 24 2013

			Table T(n,k) (with rows n >= 0 and columns k >= 0) begins as follows:
      1;
     -1,       1;
      3,      -4,     1;
    -19,      27,    -9,     1;
    211,    -304,   108,   -16,   1;
  -3651,    5275, -1900,   300, -25,   1;
  90921, -131436, 47475, -7600, 675, -36, 1;
  ... [edited by _Petros Hadjicostas_, Aug 24 2019]
From _Peter Bala_, Jul 24 2013: (Start)
Function   |        Real zeros to 5 decimal places
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
R(5,x)     | 1, 5.40649,  7.23983
R(10,x)    | 1, 5.26894, 12.97405, 18.53109
R(15,x)    | 1, 5.26894, 12.94909, 24.04769, 33.87883
R(20,x)    | 1, 5.26894, 12.94909, 24.04216, 38.54959, 53.32419
= = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = =
E(alpha*x) | 1, 5.26894, 12.94909, 24.04216, 38.54835, 56.46772, ...
where alpha = -1.44579 64907 ... ( = -(A115365/2)^2).
Note: The n-th zero of E(alpha*x) may be calculated in Maple 17 using the instruction evalf( (BesselJZeros(0,n)/BesselJZeros(0,1))^2 ). (End)

Alois P. Heinz, Rows n = 0..99, flattened
J. Riordan, Inverse relations and combinatorial identities, Amer. Math. Monthly, 71 (1964), 485-498; see p. 493 with beta_{n,k} = |T(n,k)|.
W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, 308(24) (2008), 6466-6500.

Cf. A000275, A008459 (matrix inverse), A115365.

T:= proc(n) local M;
       M:= Matrix(n+1, (i, j)-> binomial(i-1, j-1)^2)^(-1);
       seq(M[n+1, i], i=1..n+1)
    end:
seq(T(n), n=0..10);  # Alois P. Heinz, Mar 14 2013

T[n_] := Module[{M}, M = Table[Binomial[i-1, j-1]^2, {i, 1, n+1}, {j, 1, n+1}] // Inverse; Table[M[[n+1, i]], {i, 1, n+1}]]; Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Nov 28 2015, after Alois P. Heinz *)

T(n, k) = (-1)^(n+k)*A000275(n-k)*C(n, k)^2.
From Peter Bala, Jul 24 2013: (Start)
Let E(y) = Sum_{n >= 0} y^n/n!^2 = BesselJ(0,2*sqrt(-y)). Generating function: E(x*y)/E(y) = 1 + (-1 + x)*y + (3 - 4*x + x^2)*y^2/2!^2 + (-19 + 27*x - 9*x^2 + x^3)*y^3/3!^2 + ....
The n-th power of this array has a generating function E(x*y)/E(y)^n. In particular, the matrix inverse A008459 has a generating function E(y)*E(x*y).
Recurrence equation for the row polynomials: R(n,x) = x^n - Sum_{k = 0..n-1} binomial(n,k)^2*R(k,x) with initial value R(0,x) = 1.
There appears to be a connection between the zeros of the Bessel function E(x) and the real zeros of the row polynomials R(n,x). Let alpha denote the root of E(x) = 0 that is smallest in absolute magnitude. Numerically, alpha = -1.44579 64907 ... ( = -(A115365/2)^2). It appears that the real zeros of R(n,x) approach zeros of E(alpha*x) as n increases. A numerical example is given below. Indeed, it may be the case that lim_{n -> inf} R(n,x)/R(n,0) = E(alpha*x) for arbitrary complex x. (End)

A101981 Column 0 of triangle A101980, which is the matrix logarithm of A008459 (squared entries of Pascal's triangle).

Original entry on oeis.org

0, 1, -1, 4, -33, 456, -9460, 274800, -10643745, 530052880, -32995478376, 2510382661920, -229195817258100, 24730000147369440, -3113066087894608560, 452168671458789789504, -75059305956331837485345, 14121026957032156557396000, -2988687741694684876495689040

Offset: 0

Paul D. Hanna, Dec 23 2004

sign

This sequence is a signed version of A002190 and is related to Bessel functions.

Alois P. Heinz, Table of n, a(n) for n = 0..100
O. Arizmendi, T. Hasebe, F. Lehner, C. Vargas, Relations between cumulants in noncommutative probability, arXiv preprint arXiv:1408.2977 [math.CO], 2014.
Christian Günther, Kai-Uwe Schmidt, Lq norms of Fekete and related polynomials, arXiv:1602.01750 [math.NT], 2016.

Cf. A008459, A002190, A101980, A101982.

a:= n-> (-1)^(n+1)*coeff (series (-ln(BesselJ(0, 2*sqrt(x))), x, n+1), x, n)*(n!)^2: seq (a(n), n=0..30); # Alois P. Heinz, Oct 27 2012

a[n_] := (-1)^(n+1) n!^2 SeriesCoefficient[-Log[BesselJ[0, 2 Sqrt[x]]], {x, 0, n}];
Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Jul 26 2018 *)

{a(n)=sum(m=1,n,(-1)^(m-1)* (matrix(n+1,n+1,i,j,if(i>j,binomial(i-1,j-1)^2))^m/m)[n+1,1])}

a(n) = (-1)^(n+1)*A002190(n) for n>=0.

a(n) = 1 - Sum_{j=1..k-1} binomial(k, j)*binomial(k-1, j-1)*a(j) for n >= 1. See Günther & Schmidt link p.5. - Michel Marcus, Jun 17 2017

A101514 Shifts one place left under the square binomial transform (A008459): a(0) = 1, a(n) = Sum_{k=0..n-1} C(n-1,k)^2*a(k).

Original entry on oeis.org

1, 1, 2, 7, 35, 236, 2037, 21695, 277966, 4198635, 73558135, 1475177880, 33495959399, 853167955357, 24182881926558, 757554068775721, 26068954296880361, 980202973852646786, 40079727064364154465, 1774594774575753650941, 84756211791797266285252

Offset: 0

Paul D. Hanna, Dec 06 2004

nonn

Equals the main diagonal of symmetric square array A101515 shift right.

Empirical: a(n) = sum((number of standard immaculate tableaux of shape m)^2, m|=n), where this sum is over all compositions m of n > 0. - John M. Campbell, Jul 07 2017

			The binomial transform of the rows of the term-wise product of this sequence with the rows of Pascal's triangle produces the symmetric square array A101515, in which the main diagonal equals this sequence shift left:
BINOMIAL[1*1] = [(1),1,1,1,1,1,1,1,1,...],
BINOMIAL[1*1,1*1] = [1,(2),3,4,5,6,7,8,9,...],
BINOMIAL[1*1,1*2,2*1] = [1,3,(7),13,21,31,43,57,73,...],
BINOMIAL[1*1,1*3,2*3,7*1] = [1,4,13,(35),77,146,249,393,...],
BINOMIAL[1*1,1*4,2*6,7*4,35*1] = [1,5,21,77,(236),596,1290,...],
BINOMIAL[1*1,1*5,2*10,7*10,35*5,236*1] = [1,6,31,146,596,(2037),...],...
Thus the square binomial transform shifts this sequence one place left:
a(5) = 236 = 1^2*(1) + 4^2*(1) + 6^2*(2) + 4^2*(7) + 1^2*(35),
a(6) = 2037 = 1^2*(1) + 5^2*(1) + 10^2*(2) + 10^2*(7) + 5^2*(35) + 1^2*(236).

Vaclav Kotesovec, Table of n, a(n) for n = 0..330
John M. Campbell, Semisimple algebras related to immaculate tableaux, arXiv:2507.02539 [math.CO], 2025. See p. 7.

Cf. A008459, A101515, A101516.

a:= proc(n) option remember; if n<=0 then 1 else
      add(binomial(n-1,k)^2 *a(k), k=0..n-1) fi
    end:
seq(a(n), n=0..25);  # Alois P. Heinz, Sep 05 2008

a[0] = 1;
a[n_] := Sum[Binomial[n - 1, k]^2 a[k], {k, 0, n - 1}];
Table[a[i], {i, 0, 20}] (* Philip B. Zhang, Oct 10 2014 *)

{a(n)=if(n==0,1,sum(k=0,n-1,binomial(n-1,k)^2*a(k)))}
for(n=0,20,print1(a(n),", "))

E.g.f. satisfies: B(x)/A(x) = Sum_{n>=0} x^n/n!^2 where A(x) = Sum_{n>=0} a(n)*x^n/n!^2 and B(x) = Sum_{n>=0} a(n+1)*x^n/n!^2. - Paul D. Hanna, Oct 10 2014

Typo in definition corrected by Philip B. Zhang, Oct 10 2014

A102220 Triangular matrix, read by rows, equal to [2*I - A008459]^(-1), i.e., the matrix inverse of the difference of twice the identity matrix and the triangular matrix of squared binomial coefficients.

Original entry on oeis.org

1, 1, 1, 5, 4, 1, 55, 45, 9, 1, 1077, 880, 180, 16, 1, 32951, 26925, 5500, 500, 25, 1, 1451723, 1186236, 242325, 22000, 1125, 36, 1, 87054773, 71134427, 14531391, 1319325, 67375, 2205, 49, 1, 6818444405, 5571505472, 1138150832, 103334336, 5277300, 172480, 3920, 64, 1

Offset: 0

Paul D. Hanna, Dec 31 2004

Column 0 forms A102221. Row sums form twice column 0 for n>0. Matrix logarithm is A102222.

			Rows begin:
[1],
[1,1],
[5,4,1],
[55,45,9,1],
[1077,880,180,16,1],
[32951,26925,5500,500,25,1],
[1451723,1186236,242325,22000,1125,36,1],...
and equal the term-by-term product of column 0
with the squared binomial coefficients (A008459):
[(1)1^2],
[(1)1^2,(1)1^2],
[(5)1^2,(1)2^2,(1)1^2],
[(55)1^2,(5)3^2,(1)3^2,(1)1^2],
[(1077)1^2,(55)4^2,(5)6^2,(1)4^2,(1)1^2],...
The matrix inverse is [2*I - A008459]:
[1],
[ -1,1],
[ -1,-4,1],
[ -1,-9,-9,1],
[ -1,-16,-36,-16,1],...

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A008459, A102221, A102222.

b:= proc(n) option remember; `if`(n=0, 1,
      add(b(n-i)*binomial(n, i)/i!, i=1..n))
    end:
T:= (n, k)-> binomial(n, k)^2*b(n-k)*(n-k)!:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Sep 10 2019

nmax = 10;
M = Inverse[2 IdentityMatrix[nmax+1] - Table[Binomial[n, k]^2, {n, 0, nmax}, {k, 0, nmax}]];
T[n_, k_] := M[[n+1, k+1]];
Table[T[n, k], {n, 0, nmax}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 06 2019 *)

{T(n,k)=(matrix(n+1,n+1,i,j,if(i==j,2,0)-binomial(i-1,j-1)^2)^-1)[n+1,k+1]}

T(n,k) = C(n,k)^2*A102221(n-k). T(n,0) = A102221(n). 2*A102221(n) = Sum_{k=0..n} T(n,k) for n>0.

A102222 Logarithm of triangular matrix A102220, which equals [2*I - A008459]^(-1).

Original entry on oeis.org

0, 1, 0, 3, 4, 0, 22, 27, 9, 0, 323, 352, 108, 16, 0, 7906, 8075, 2200, 300, 25, 0, 290262, 284616, 72675, 8800, 675, 36, 0, 14919430, 14222838, 3486546, 395675, 26950, 1323, 49, 0, 1022475715, 954843520, 227565408, 24793216, 1582700, 68992, 2352, 64, 0

Offset: 0

Paul D. Hanna, Dec 31 2004

Column 0 forms A102223.

			Rows begin:
[0],
[1,0],
[3,4,0],
[22,27,9,0],
[323,352,108,16,0],
[7906,8075,2200,300,25,0],
[290262,284616,72675,8800,675,36,0],...
which equals the term-by-term product of column 0
with the squared binomial coefficients (A008459):
[(0)1^2],
[(1)1^2,(0)1^2],
[(3)1^2,(1)2^2,(0)1^2],
[(22)1^2,(3)3^2,(1)3^2,(0)1^2],
[(323)1^2,(22)4^2,(3)6^2,(1)4^2,(0)1^2],...

Cf. A008459, A102220, A102223.

PARI
```
{T(n,k)=if(n
				
```

T(n, k) = C(n, k)^2*A102223(n-k). T(n, 0) = A102223(n). T(n, n) = 0 for n>=0. [A102222] = Sum_{m=1..inf} [A008459 - I]^m/m.

A097085 Row sums of triangle A097084, in which the n-th diagonal equals the n-th row transformed by triangle A008459 (squared binomial coefficients).

Original entry on oeis.org

1, 2, 4, 10, 26, 70, 204, 618, 1908, 6010, 19316, 63034, 208210, 695594, 2346748, 7983450, 27364842, 94439262, 327922692, 1145029314, 4018618374, 14169874350, 50179643628, 178410716622, 636679332588, 2279906714610, 8190512723940

Offset: 0

Paul D. Hanna, Jul 23 2004

nonn

a(n) is also the number of anagram compositions of 2n or of 2n+1. A composition of n is an ordered sequence of positive integers whose sum is n. An anagram composition of n can be divided into two consecutive subsequences with exactly the same parts, with a central part between the subsequences permitted. - Gregory L. Simay, Oct 30 2015

			[1,2,3,4][3,2,1,4] is an anagram composition of 20 enumerated by a(10), [3,2,1] 5 [2,1,3] is an anagram composition of 17 enumerated by a(8), [3467] 8 [7643] is an anagram composition of 48 enumerated by a(24). - _Gregory L. Simay_, Oct 30 2015

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A097084.

Cf. A263897. - Gregory L. Simay, Oct 30 2015

b:= proc(n, i, p) option remember; `if`(n=0, p!^2,
      `if`(i<1, 0, add(b(n-i*j, i-1, p+j)/j!^2, j=0..n/i)))
    end:
a:= proc(n) option remember; b(n$2, 0)+`if`(n>0, a(n-1), 0) end:
seq(a(n), n=0..30);  # Alois P. Heinz, Oct 30 2015

b[n_, i_, p_] := b[n, i, p] = If[n == 0, p!^2, If[i < 1, 0, Sum[b[n - i*j, i - 1, p + j]/j!^2, {j, 0, n/i}]]];
a[n_] := a[n] = b[n, n, 0] + If[n > 0, a[n - 1], 0];
Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jun 04 2018, after Alois P. Heinz *)

a(n) = Sum_{j=0..n} A263897(j). - Gregory L. Simay, Oct 30 2015

A097084 Triangle, read by rows, where the n-th diagonal equals the n-th row transformed by triangle A008459 (squared binomial coefficients).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 10, 10, 1, 1, 5, 18, 28, 17, 1, 1, 6, 27, 74, 69, 26, 1, 1, 7, 39, 137, 245, 151, 37, 1, 1, 8, 52, 236, 586, 676, 298, 50, 1, 1, 9, 68, 372, 1194, 2126, 1634, 540, 65, 1, 1, 10, 85, 552, 2322, 5152, 6620, 3578, 913, 82, 1, 1, 11, 105, 777, 3954, 12002, 19292, 18082, 7249, 1459, 101, 1

Offset: 0

Paul D. Hanna, Jul 23 2004

Row sums form A097085.

			T(8,3) = 236 = (1)*1^2 + (5)*3^2 + (18)*3^2 + (28)*1^2
= Sum_{j=0..3} T(5,j)*C(3,j)^2.
Rows begin:
[1],
[1,1],
[1,2,1],
[1,3,5,1],
[1,4,10,10,1],
[1,5,18,28,17,1],
[1,6,27,74,69,26,1],
[1,7,39,137,245,151,37,1],
[1,8,52,236,586,676,298,50,1],...

Alois P. Heinz, Rows n = 0..140, flattened

Cf. A097085, A008459.

T:= proc(n, k) option remember;
      `if`(n=k or k=0, 1, `if`(k<0 or k>n, 0,
       add(T(n-k, j)*binomial(k, j)^2, j=0..k)))
    end:
seq(seq(T(n,k), k=0..n), n=0..12);  # Alois P. Heinz, Oct 30 2015

T[, 0] = 1; T[n, n_] = 1; T[n_, k_] /; 0 < k < n := T[n, k] = Sum[T[n - k, j]*Binomial[k, j]^2, {j, 0, k}]; T[, ] = 0;
Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 24 2016 *)

PARI
```
T(n,k)=if(n
				
```

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

cp's OEIS Frontend

A102221 Column 0 of triangular matrix A102220, which equals [2*I - A008459]^(-1).

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A102223 Column 0 of triangular matrix A102222, which equals -log[2*I - A008459].

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A288876 a(n) = binomial(n+4, n)^2. Square of the fifth diagonal sequence of A007318 (Pascal). Fifth diagonal sequence of A008459.

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A055133 Matrix inverse of A008459 (squares of entries of Pascal's triangle).

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A101981 Column 0 of triangle A101980, which is the matrix logarithm of A008459 (squared entries of Pascal's triangle).

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A101514 Shifts one place left under the square binomial transform (A008459): a(0) = 1, a(n) = Sum_{k=0..n-1} C(n-1,k)^2*a(k).

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A102220 Triangular matrix, read by rows, equal to [2*I - A008459]^(-1), i.e., the matrix inverse of the difference of twice the identity matrix and the triangular matrix of squared binomial coefficients.

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