A001249 -id:A001249 - cp's OEIS frontend

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

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)

A286724 Triangle read by rows. A generalization of unsigned Lah numbers, called L[2,1].

Original entry on oeis.org

1, 2, 1, 8, 8, 1, 48, 72, 18, 1, 384, 768, 288, 32, 1, 3840, 9600, 4800, 800, 50, 1, 46080, 138240, 86400, 19200, 1800, 72, 1, 645120, 2257920, 1693440, 470400, 58800, 3528, 98, 1, 10321920, 41287680, 36126720, 12042240, 1881600, 150528, 6272, 128, 1, 185794560, 836075520, 836075520, 325140480, 60963840, 6096384, 338688, 10368, 162, 1, 3715891200, 18579456000, 20901888000, 9289728000, 2032128000, 243855360, 16934400, 691200, 16200, 200, 1

Offset: 0

Wolfdieter Lang, Jun 16 2017

These generalized unsigned Lah numbers are the instance L[2,1] of the Sheffer triangles called L[d,a], with integers d >= 1 and integers 0 <= a < d with gcd(d,a) = 1. The standard unsigned Lah numbers are L[1,0] = A271703.
The Sheffer structure of L[d,a] is ((1 - d*t)^(-2*a/d), t/(1 - d*t)). This follows from the defining property
  risefac[d,a](x, n) = Sum_{m=0..n} L[d,a](n, m)*fallfac[d,a](x, m), where risefac[d,a](x, n):= Product_{0..n-1} (x  + (a+d*j)) for n >= 1 and risefac[d,a](x, 0) := 1, and fallfac[d,a](x, n):= Product_{0..n-1} (x - (a+d*j)) = for n >= 1 and fallfac[d,a](x, 0) := 1. Such rising and falling factorials arise in the generalization of Stirling numbers of both kinds S2[d,a] and S1[d,a]. See the Peter Bala link under A143395 for these falling factorials called there [t;a,b,c]_n with t=x, a=d, b=0, c=a.
In matrix notation: L[d,a] = S1phat[d,a].S2hat[d,a] with the unsigned scaled Stirling1 and the scaled Stirling2 generalizations with Sheffer structures S1phat[d,a] = ((1 - d*t)^(-a/d), -(1/d)*(log(1 - d*t))) and S2hat[d,a] = (exp(a*t), (1/d)*(exp(d*t) - 1). See, e.g., S1phat[2,1] = A028338 and S2hat[2,1] = A039755.
The a- and z-sequences for these Sheffer matrices have e.g.f.s 1 + d*t and ((1 + d*t)/t)*(1 - (1 + d*t)^(-2*a/d)), respectively. See a W. Lang link under A006232 for these types of sequences.
E.g.f. of row polynomials R[d,a](n, x) := Sum_{m=0..n} L[d,a](n, m)*x^m
  (1 - d*x)^(-2*a/d)*exp(t*x/(1 - d*x)) (this is the e.g.f. for the triangle).
E.g.f. of column m: (1 - d*t)^(-2*a/d)*(t/(1 - d*t))^m/m, m >= 0.
Meixner type identity for (monic) row polynomials: (D_x/(1 + d*D_x)) * R[d,a](n, x) = n*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1. The series in the differentiations D_x = d/dx terminates.
General Sheffer recurrence for row polynomials (see the Roman reference, p. 50, Corollary 3.7.2, rewritten for the present Sheffer notation):
  R[d,a](n, x) = [(2*a+x)*1 + 2*d*(a + x)*D_x + d^2*x*(D_x)^2]*R[d,a](n-1, x), n >= 1, with R[d,a](0, x) = 1.
The inverse matrix L^(-1)[d,a] is Sheffer (g[d,a](-t), -f[d,a](-t)) with L[d,a] Sheffer (g[d,a](t), f[d,a](t)) from above. This means (see the column e.g.f. of Sheffer matrices) that L^(-1)[d,a](n, m) = (-1)^(n-m)*L[d,a](n, m). Therefore, the recurrence relations can easily be rewritten for L^(-1)[d,a] by replacing a -> -a and d -> -d.
fallfac[d,a](x, n) = Sum_{m=0..n} L^(-1)[d,a](n, m)*risefac[d,a](x, m), n >= 0.
From Wolfdieter Lang, Aug 12 2017: (Start)
The Sheffer row polynomials R[d,a](n, x) belong to the Boas-Buck class and satisfy therefore the Boas-Buck identity (see the reference, and we use the notation of Rainville, Theorem 50, p. 141, adapted to an exponential generating function) (E_x - n*1)*R[d,a](n, x) = - n*(2*a*1 + d*E_x) * Sum_{k=0..n-1} d^k*R(d,a;n-1-k,x)/(n-1-k)!, with E_x = x*d/dx (Euler operator).
This implies a recurrence for the sequence of column m: L[d,a](n, m) = (n!*(2*a + d*m)/(n-m))*Sum_{p=0..n-1-m} d^p*L[d,a](n-1-p, m)/(n-1-p)!, for n > m>=0, and input L[d,a](m, m) = 1. For the present [d,a] = [2,1] instance see the formula and example sections. (End)
From Wolfdieter Lang, Sep 14 2017: (Start)
The diagonal sequences are 2^D*D!*(binomial(m+D, m))^2, m >= 0, for D >= 0 (main diagonal D = 0). From the o.g.f.s obtained via Lagrange's theorem. See the second W. Lang link below for the general Sheffer case.
The o.g.f. of the diagonal D sequence is 2^D*D!*Sum_{m=0..D} A008459(D, m)*x^m /(1- x)^(2*D + 1),  D >= 0. (End)
It appears that this is also the matrix square of unsigned triangle of coefficients of Laguerre polynomials n!*L_n(x), abs(A021009(n, k)). - Ali Pourzand, Mar 10 2025 [This observation is correct. - Peter Luschny, Mar 10 2025]

			The triangle T(n, m) begins:
  n\m        0         1         2         3        4       5      6     7   8 9
  0:         1
  1:         2         1
  2:         8         8         1
  3:        48        72        18         1
  4:       384       768       288        32        1
  5:      3840      9600      4800       800       50       1
  6:     46080    138240     86400     19200     1800      72      1
  7:    645120   2257920   1693440    470400    58800    3528     98     1
  8:  10321920  41287680  36126720  12042240  1881600  150528   6272   128   1
  9: 185794560 836075520 836075520 325140480 60963840 6096384 338688 10368 162 1
  ...
From _Wolfdieter Lang_, Aug 12 2017: (Start)
Recurrence for column elements with m >= 1, and input column m = 0: T(3, 2) = (3/2)*T(2, 1) + 2*3*T(2, 2) = (3/2)*8 + 6 = 18.
Four term recurrence: T(3, 2) = T(2, 1) + 2*5*T(2, 2) - 4*2^2*T(1, 2) = 8 + 10 + 0 = 18.
Meixner type identity, n=2: 2*R(1, x) = (D_x - 2*(D_x)^2)*R(2, x), 2*(2 + x) = (8 + 2*x) - 2*2.
Sheffer recurrence: R(2, x) = (2 + x)*(2 + x) + 4*(1 + x)*1 + 0 = 8 + 8*x + x^2.
Boas-Buck recurrence for column m = 2 and n = 4: T(4, 2) = (2*4!*3/2)*(1*T(3, 2)/3! + 2*T(2, 2)/2!) = 4!*3*(18/3! + 1) = 288. (End)
Diagonal sequence D = 1: o.g.f. 2*1!*(1 + 1*x)/(1- x)^3 generating
{2*(binomial(m+1, m))^2}_{m >= 0} = {2, 8, 18, 32, ...}. - _Wolfdieter Lang_, Sep 14 2017

Ralph P. Boas, jr. and R. Creighton Buck, Polynomial Expansions of analytic functions, Springer, 1958, pp. 17 - 21, (last sign in eq. (6.11) should be -).
Earl D. Rainville, Special Functions, The Macmillan Company, New York, 1960, ch. 8, sect. 76, 140 - 146.
Steven Roman, The Umbral Calculus, Academic press, Orlando, London, 1984, p. 50.

Peter Bala, The white diamond product of power series
Wolfdieter Lang, On Sums of Powers of Arithmetic Progressions, and Generalized Stirling, Eulerian and Bernoulli Numbers, arXiv:math/1707.04451 [math.NT], July 2017.
Wolfdieter Lang, On Generating functions of Diagonal Sequences of Sheffer and Riordan Number Triangles, arXiv:1708.01421 [math.NT], August 2017.
Emanuele Munarini, Combinatorial identities involving the central coefficients of a Sheffer matrix, Applicable Analysis and Discrete Mathematics (2019) Vol. 13, 495-517.

Cf. A006232, A025167, A028338, A039755, A143395, A271703, A021009.
Column sequences (no leading zeros): A000165, A014479, A286725.
Diagonal sequences: A000012, 2*A000290(m+1), 8*A000537(n+1), 48*A001249, 384*A288876. - Wolfdieter Lang, Sep 14 2017
Row sums are A025167. - Michael Somos, Sep 27 2017

T := (n, k) -> ifelse(n < k, 0, ifelse(k = 0, n!*2^n, (n/k)*T(n-1, k-1) + 2*n*T(n-1, k))): seq(seq(T(n, k), k = 0..n), n = 0..10);  # Peter Luschny, Mar 10 2025

T[ n_, k_] := Coefficient[ Integrate[ Exp[-x^2 - y x] HermiteH[n, x]^2, {x, -Infinity, Infinity}] / (Sqrt[Pi] Exp[y^2 / 4]), y, 2 k]; (* Michael Somos, Sep 27 2017 *)

# Using the function A021009_triangle, displays as a matrix. Following the observation of Ali Pourzand.
print(A021009_triangle(9)^2)  # Peter Luschny, Mar 10 2025

T(n, m) = L[2,1](n, m) = Sum_{k=m..n} A028338(n, k)*A039755(k, m).
Three term recurrence for column elements with m >= 1: T(n, m) = (n/m)*T(n-1, m-1) + 2*n*T(n-1, m) with T(n, m) = 0 for n < m and the column m = 0 is T(n, 0) = (2*n)!! = n*2^n = A000165(n). (From the a- and z-sequences {1, 2, repeat(0)} and {2, repeat(0)}, respectively.)
Four term recurrence: T(n, m) = T(n-1, m-1) + 2*(2*n-1)*T(n-1, m) - 4*(n-1)^2*T(n-2, m), n >= m >= 0, with T(0, 0) = 1, T(-1, m) = 0, T(n, -1) = 0 and T(n, m) = 0 if n < m.
E.g.f. of row polynomials R(n, x) = R[2,1](n, x) (i.e., e.g.f. of the triangle): (1/(1-2*t))*exp(x*t/(1-2*t)).
E.g.f. of column m sequences: (t^m/(1-2*t)^(m+1))/m!, m >= 0.
Meixner type identity: Sum_{k=0..n-1} (-1)^k*2^k*(D_x)^(k+1)*R(n, x) = n*R(n-1, x), n >= 1, with R(0, x) = 1 and D_x = d/dx.
Sheffer recurrence: R(n, x) = [(2 + x)*1 + 4*(1 + x)*D_x + 4*x*(D_x)^2]*R(n-1, x), n >= 1, and R(0, x) = 1.
Boas-Buck recurrence for column m (see a comment above): T(n, m) = (2*n!*(1 + m)/(n-1))*Sum_{p=0..n-1-m} 2^p*T(n-1-p, m)/(n-1-p)!, for n > m >= 0, and input T(m, m) = 1. - Wolfdieter Lang, Aug 12 2017
Explicit form (from the diagonal sequences with the o.g.f.s given as a comment above): T(n, m) = 2^(n-m)*(n-m)!*(binomial(n, n-m))^2 for n >= m >= 0. - Wolfdieter Lang, Sep 23 2017
Let R(n,x) denote the n-th row polynomial. Then x^n*R(n,x) = x^n o x^n, where o denotes the deformed Hadamard product of power series defined in Bala, Section 3.1. - Peter Bala, Jan 18 2018

A128629 A triangular array generated by moving Pascal sequences to prime positions and embedding new sequences at the nonprime locations. (cf. A007318 and A000040).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, 1, 4, 9, 10, 5, 1, 1, 6, 10, 16, 15, 6, 1, 1, 5, 18, 20, 25, 21, 7, 1, 1, 8, 15, 40, 35, 36, 28, 8, 1, 1, 9, 27, 35, 75, 56, 49, 36, 9, 1

Offset: 1

Alford Arnold, Mar 29 2007

The array can be constructed by beginning with A007318 (Pascal's triangle) placing each diagonal on a prime row. The other rows are filled in by mapping the prime factorization of the row number to the known sequences on the prime rows and multiplying term by term.

			Row six begins 1 6 18 40 75 126 ... because rows two and three are
1 2 3 4 5 6 ...
1 3 6 10 15 21 ...
The array begins
1 1 1 1 1 1 1 1 1 A000012
1 2 3 4 5 6 7 8 9 A000027
1 3 6 10 15 21 28 36 45 A000217
1 4 9 16 25 36 49 64 81 A000290
1 4 10 20 35 56 84 120 165 A000292
1 6 18 40 75 126 196 288 405 A002411
1 5 15 35 70 126 210 330 495 A000332
1 8 27 64 125 216 343 512 729 A000578
1 9 36 100 225 441 784 1296 2025 A000537
1 8 30 80 175 336 588 960 1485 A002417
1 6 21 56 126 252 462 792 1287 A000389
1 12 54 160 375 756 1372 2304 3645 A019582
1 7 28 84 210 462 924 1716 3003 A000579
1 10 45 140 350 756 1470 2640 4455 A027800
1 12 60 200 525 1176 2352 4320 7425 A004302
1 16 81 256 625 1296 2401 4096 6561 A000583
1 8 36 120 330 792 1716 3432 6435 A000580
1 18 108 400 1125 2646 5488 10368 18225 A019584
1 9 45 165 495 1287 3003 6435 12870 A000581
1 16 90 320 875 2016 4116 7680 13365 A119771
1 15 90 350 1050 2646 5880 11880 22275 A001297
1 12 63 224 630 1512 3234 6336 11583 A027810
1 10 55 220 715 2002 5005 11440 24310 A000582
1 24 162 640 1875 4536 9604 18432 32805 A019583
1 16 100 400 1225 3136 7056 14400 27225 A001249
1 14 84 336 1050 2772 6468 13728 27027 A027818
1 27 216 1000 3375 9261 21952 46656 91125 A059827
1 20 135 560 1750 4536 10290 21120 40095 A085284

Cf. A000040 A007318.

Cf. A064553 (second diagonal), A080688 (second diagonal resorted).

A128629 := proc(n,m) if n = 1 then 1; elif isprime(n) then p := numtheory[pi](n) ; binomial(p+m-1,p) ; else a := 1 ; for p in ifactors(n)[2] do a := a* procname(op(1,p),m)^ op(2,p) ; od: fi; end: # R. J. Mathar, Sep 09 2009

A-number added to each row of the examples by R. J. Mathar, Sep 09 2009

A373449 Number A(n,k) of (binary) heaps of length n whose element set is a subset of [k]; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 3, 1, 0, 1, 4, 6, 5, 1, 0, 1, 5, 10, 14, 7, 1, 0, 1, 6, 15, 30, 25, 11, 1, 0, 1, 7, 21, 55, 65, 53, 16, 1, 0, 1, 8, 28, 91, 140, 173, 100, 26, 1, 0, 1, 9, 36, 140, 266, 448, 400, 222, 36, 1, 0, 1, 10, 45, 204, 462, 994, 1225, 1122, 386, 56, 1, 0

Offset: 0

Alois P. Heinz, Jun 05 2024

These heaps may contain repeated elements.

			A(3,1) = 1: 111.
A(3,2) = 5: 111, 211, 212, 221, 222.
A(3,3) = 14: 111, 211, 212, 221, 222, 311, 312, 313, 321, 322, 323, 331, 332, 333.
(The examples use max-heaps.)
Square array A(n,k) begins:
  1, 1,  1,   1,    1,     1,     1,     1,      1, ...
  0, 1,  2,   3,    4,     5,     6,     7,      8, ...
  0, 1,  3,   6,   10,    15,    21,    28,     36, ...
  0, 1,  5,  14,   30,    55,    91,   140,    204, ...
  0, 1,  7,  25,   65,   140,   266,   462,    750, ...
  0, 1, 11,  53,  173,   448,   994,  1974,   3606, ...
  0, 1, 16, 100,  400,  1225,  3136,  7056,  14400, ...
  0, 1, 26, 222, 1122,  4147, 12428, 32028,  73644, ...
  0, 1, 36, 386, 2336, 10036, 34242, 98922, 251922, ...

Alois P. Heinz, Antidiagonals n = 0..200, flattened
Eric Weisstein's World of Mathematics, Heap
Wikipedia, Binary heap

Columns k=0-2 give: A000007, A000012, A091980(n+1).
Rows n=0-6 give: A000012, A001477, A000217, A000330, A001296, A207361, A001249(k-1).
Main diagonal gives A373450.
Cf. A373451.

A:= proc(n, k) option remember; `if`(n=0, 1,
     (g-> (f-> add(A(f, j)*A(n-1-f, j), j=1..k)
             )(min(g-1, n-g/2)))(2^ilog2(n)))
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);

A[n_, k_] := A[n, k] = If[n == 0, 1,
   Function[g, Function[f, Sum[A[f, j]*A[n-1-f, j], {j, 1, k}]][
   Min[g-1, n-g/2]]][2^(Length[IntegerDigits[n, 2]]-1)]];
Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Jun 08 2024, after Alois P. Heinz *)

A(n,k) = Sum_{j=0..k} binomial(k,j) * A373451(n,k-j).

A265010 Numbers which are the product of two tetrahedral numbers.

Original entry on oeis.org

0, 1, 4, 10, 16, 20, 35, 40, 56, 80, 84, 100, 120, 140, 165, 200, 220, 224, 286, 336, 350, 364, 400, 455, 480, 560, 660, 680, 700, 816, 840, 880, 969, 1120, 1140, 1144, 1200, 1225, 1330, 1456, 1540, 1650, 1680, 1771, 1820, 1960, 2024, 2200, 2240

Offset: 1

R. J. Mathar, Nov 30 2015

nonn

This is for the tetrahedral numbers A000292 what A085780 is for the triangular numbers.

The subsequence of numbers with more than one factorization starts 0, 560 (= 65*10 = 1*560), 19600 (= 560*15 = 19600*1), 28560 (=816 *35 = 7140*4), 43680, 292600, 416640, ...

			Contains 480=4*120, 560=1*560, 660=4*165, 680=1*680, 700=20*35, ....

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A085780. Subsequences: A000292, A001249, A027790, A105939, A104474, A104677.

# reuses code of A000292
isA265010 := proc(n)
    if n = 0 then
        return true;
    end if;
    for d in numtheory[divisors](n) do
        if isA000292(d) and isA000292(n/d) then
            return true;
        end if;
    end do:
    false;
end proc:
for n from 0 to 4000 do
    if isA265010(n) then
        printf("%d, ",n);
    end if;
end do:

lim = 2240; t = Table[Binomial[n + 2, 3], {n, 0, 10^3}]; f[n_] := Select[{#, n/#} & /@ Select[Divisors[n], # <= Sqrt@ n && MemberQ[t, #] &], MemberQ[t, Last@ #] &]; Select[Range@ lim, Length@ f@ # > 0 &] (* Michael De Vlieger, Nov 30 2015 *)

{n: n = A000292(i)*A000292(j) for some i,j>=0}.

A270272 a(n) = binomial(n+3,n)^3.

Original entry on oeis.org

1, 64, 1000, 8000, 42875, 175616, 592704, 1728000, 4492125, 10648000, 23393656, 48228544, 94196375, 175616000, 314432000, 543338496, 909853209, 1481544000, 2352637000, 3652264000, 5554637011, 8291469824, 12167000000, 17576000000, 25025203125, 35158608576, 48787170264

Offset: 0

Benedict W. J. Irwin, Mar 14 2016

Index entries for linear recurrences with constant coefficients, signature (10,-45,120,-210,252,-210,120,-45,10,-1).

Cf. A000578, A000292, A001249.

A270272:=n->binomial(n+3,n)^3: seq(A270272(n), n=0..50); # Wesley Ivan Hurt, Jul 17 2017

Mathematica
```
Table[Binomial[n+3,n]^3,{n,0,30}]
```

a(n) = binomial(n+3,n)^3; \\ Michel Marcus, Jul 17 2017

G.f.: 3F2(4,4,4;1,1;z).
G.f.: (1 + 54x + 405x^2 + 760x^3 + 405x^4 + 54x^5 + x^6)/(x-1)^10.
a(n) = (6 + 11n + 6n^2 + n^3)^3/216.
a(n) = A000292(n+1)^3.
Sum_{n>=0} 1/a(n) = 783/4 - 162*zeta(3). - Jaume Oliver Lafont, Jul 17 2017
Sum_{n>=0} (-1)^n/a(n) = 1296*log(2) + 405*zeta(3)/2 - 4563/4. - Amiram Eldar, Sep 20 2022

A169801 a(n) = ((n-1)^2n^2(n+1)^2)/6 - 2Sum_{l=2..n}Sum_{k=2..n}(n-k+1)(n-l+1)(k-1)(l-1).

Original entry on oeis.org

0, 0, 4, 64, 400, 1600, 4900, 12544, 28224, 57600, 108900, 193600, 327184, 529984, 828100, 1254400, 1849600, 2663424, 3755844, 5198400, 7075600, 9486400, 12545764, 16386304, 21160000, 27040000, 34222500, 42928704, 53406864, 65934400, 80820100, 98406400

Offset: 0

N. J. A. Sloane, May 19 2010, May 22 2010

Created in an attempt to repair a formula in A045996, which however turned out to be correct after all.

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

f[n_] := ((n - 1)^2*n^2*(n + 1)^2)/6 - 2*Sum[(n - k + 1)*(n - l + 1)*(k - 1) (l - 1), {k, 2, n}, {l, 2, n}]; Array[f, 31] (* Robert G. Wilson v, May 23 2010 *)

a(n) = A099764(n-1) - 2*Sum_{l=2..n}Sum_{k=2..n}(n-k+1)*(n-l+1)*(k-1)*(l-1) = A099764(n-1)/9 = 4*A001249(n-2). - R. J. Mathar, May 23 2010

G.f.: 4*x^2*(1+x)*(x^2+8*x+1)/(1-x)^7. - R. J. Mathar, May 23 2010

A288959 a(n) = n^2*(n^2 - 1)^2/2.

Original entry on oeis.org

0, 18, 288, 1800, 7200, 22050, 56448, 127008, 259200, 490050, 871200, 1472328, 2384928, 3726450, 5644800, 8323200, 11985408, 16901298, 23392800, 31840200, 42688800, 56455938, 73738368, 95220000, 121680000, 154001250, 193179168, 240330888, 296704800, 363690450

Offset: 1

Eric W. Weisstein, Jun 20 2017

Except for n = 2, gives the detour index of the n X n rook and rook complement graph.

Also the detour index of the n X n king and n X n queen graphs. - Eric W. Weisstein, Dec 16 2017

Eric Weisstein's World of Mathematics, Detour Index
Eric Weisstein's World of Mathematics, King Graph
Eric Weisstein's World of Mathematics, Queen Graph
Eric Weisstein's World of Mathematics, Rook Complement Graph
Eric Weisstein's World of Mathematics, Rook Graph
Index entries for linear recurrences with constant coefficients, signature (7, -21, 35, -35, 21, -7, 1).

Table[n^2 (n^2 - 1)^2/2, {n, 20}]
LinearRecurrence[{7, -21, 35, -35, 21, -7, 1}, {0, 18, 288, 1800, 7200, 22050, 56448}, 20]
CoefficientList[Series[-((18 x (1 + x) (1 + 8 x + x^2))/(-1 + x)^7), {x, 0, 20}], x]
18 Binomial[Range[0, 20] + 2, 3]^2 (* Eric W. Weisstein, Dec 20 2017 *)

a(n) = n^2*(n^2-1)^2/2; \\ Altug Alkan, Dec 20 2017

a(n) = n^2*(n^2 - 1)^2/2.
a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7).
G.f.: (-18*x^2*(1+x)*(1+8*x+x^2))/(-1+x)^7.
a(n) = 18 *A001249(n-2). - R. J. Mathar, Dec 17 2017

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.

Original entry on oeis.org

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

Wolfdieter Lang, Jul 27 2017

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.

			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.

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

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.

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

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 *)

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

cp's OEIS Frontend

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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A286724 Triangle read by rows. A generalization of unsigned Lah numbers, called L[2,1].

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A128629 A triangular array generated by moving Pascal sequences to prime positions and embedding new sequences at the nonprime locations. (cf. A007318 and A000040).

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A373449 Number A(n,k) of (binary) heaps of length n whose element set is a subset of [k]; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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A265010 Numbers which are the product of two tetrahedral numbers.

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A270272 a(n) = binomial(n+3,n)^3.

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A169801 a(n) = ((n-1)^2*n^2*(n+1)^2)/6 - 2*Sum_{l=2..n}Sum_{k=2..n}(n-k+1)*(n-l+1)*(k-1)*(l-1).

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A169801 a(n) = ((n-1)^2n^2(n+1)^2)/6 - 2Sum_{l=2..n}Sum_{k=2..n}(n-k+1)(n-l+1)(k-1)(l-1).