cp's OEIS Frontend

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.

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A077047 Doubly restricted composition numbers: number of compositions of floor(n(n+2)/2) into exactly n positive integers each no more than n+1.

Original entry on oeis.org

1, 1, 3, 12, 85, 780, 9331, 134512, 2306025, 45433800, 1018872811, 25506741084, 707972099627, 21518492021208, 712601187601395, 25491847538274240, 981272544393935569, 40392787067756440272, 1772592132899627652691
Offset: 0

Views

Author

Henry Bottomley, Oct 22 2002

Keywords

Comments

a(n) is the maximum number of ordered partitions when using n numbers each ranging from 0 to n. This maximum occurs when partitioning n^2/2 for n even, or (n^2 - 1)/2 or (n^2 + 1)/2 for n odd. Example for a(3)=12: the partitions of 4 are (1,1,2) and (0,2,2), each having 3 ordered arrangements, and (0,1,3) having 6 arrangements; hence 3+3+6=12. For 5 the partitions are (1,2,2) and (1,1,3), with 3 ordered arrangements each, and (0,2,3) having 6 arrangements. - J. M. Bergot, Jul 11 2015
Largest coefficient of (1 + x + x^2 + ... + x^n)^n. - Vaclav Kotesovec, Mar 26 2016

Examples

			a(3)=12 since the compositions of [3*5/2]=7 into exactly 3 positive integers each no more than 4 are 1+2+4, 1+3+3, 1+4+2, 2+1+4, 2+2+3, 2+3+2, 2+4+1, 3+1+3, 3+2+2, 3+3+2, 4+1+2, 4+2+1.
		

Crossrefs

Programs

  • Maple
    f:= proc(n) if n::odd then coeff(add(x^i,i=0..n)^n,x,(n^2-1)/2)
    else coeff(add(x^i,i=0..n)^n,x,n^2/2) fi end proc:
    map(f, [$0..40]); # Robert Israel, Nov 16 2016
  • Mathematica
    Table[Max[CoefficientList[Expand[Sum[x^k, {k, 0, n}]^n], x]], {n, 0, 20}] (* Vaclav Kotesovec, Mar 26 2016 *)
    Table[Max[CoefficientList[((x^(n+1)-1)/(x-1))^n, x]], {n, 0, 20}] (* Vladimir Reshetnikov, Nov 16 2016 *)

Formula

a(n) = A077042(n+1, n).
a(n) ~ exp(1) * sqrt(6/Pi) * n^(n-3/2). - Vaclav Kotesovec, Mar 26 2016

A273975 Three-dimensional array written by antidiagonals in k,n: T(k,n,h) with k >= 1, n >= 0, 0 <= h <= n*(k-1) is the coefficient of x^h in the polynomial (1 + x + ... + x^(k-1))^n = ((x^k-1)/(x-1))^n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 1, 1, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 3, 2, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 6, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 4, 3, 2, 1, 1, 3, 6, 10, 12, 12, 10, 6, 3, 1, 1, 4, 10
Offset: 1

Views

Author

Andrey Zabolotskiy, Nov 10 2016

Keywords

Comments

Equivalently, T(k,n,h) is the number of ordered sets of n nonnegative integers < k with the sum equal to h.
From Juan Pablo Herrera P., Nov 21 2016: (Start)
T(k,n,h) is the number of possible ways of randomly selecting h cards from k-1 sets, each with n different playing cards. It is also the number of lattice paths from (0,0) to (n,h) using steps (1,0), (1,1), (1,2), ..., (1,k-1).
Shallow diagonal sums of each triangle with fixed k give the k-bonacci numbers. (End)
T(k,n,h) is the number of n-dimensional grid points of a k X k X ... X k grid, which are lying in the (n-1)-dimensional hyperplane which is at an L1 distance of h from one of the grid's corners, and normal to the corresponding main diagonal of the grid. - Eitan Y. Levine, Apr 23 2023

Examples

			For first few k and for first few n, the rows with h = 0..n*(k-1) are given:
k=1:  1;  1;  1;  1;  1; ...
k=2:  1;  1, 1;  1, 2, 1;  1, 3, 3, 1;  1, 4, 6, 4, 1; ...
k=3:  1;  1, 1, 1;  1, 2, 3, 2, 1;  1, 3, 6, 7, 6, 3, 1; ...
k=4:  1;  1, 1, 1, 1;  1, 2, 3, 4, 3, 2, 1; ...
For example, (1 + x + x^2)^3 = 1 + 3*x + 6*x^2 + 7*x^3 + 6*x^4 + 3*x^5 + x^6, hence T(3,3,2) = T(3,3,4) = 6.
From _Eitan Y. Levine_, Apr 23 2023: (Start)
Example for the repeated cumulative sum formula, for (k,n)=(3,3) (each line is the cumulative sum of the previous line, and the first line is the padded, alternating 3rd row from Pascal's triangle):
  1  0  0 -3  0  0  3  0  0 -1
  1  1  1 -2 -2 -2  1  1  1
  1  2  3  1 -1 -3 -2 -1
  1  3  6  7  6  3  1
which is T(3,3,h). (End)
		

Crossrefs

k-nomial arrays for fixed k=1..10: A000012, A007318, A027907, A008287, A035343, A063260, A063265, A171890, A213652, A213651.
Arrays for fixed n=0..6: A000012, A000012, A004737, A109439, A277949, A277950, A277951.
Central n-nomial coefficients for n=1..9, i.e., sequences with h=floor(n*(k-1)/2) and fixed n: A000012, A000984 (A001405), A002426, A005721 (A005190), A005191, A063419 (A018901), A025012, (A025013), A025014, A174061 (A025015), A201549, (A225779), A201550. Arrays: A201552, A077042, see also cfs. therein.
Triangle n=k-1: A181567. Triangle n=k: A163181.

Programs

  • Mathematica
    a = Table[CoefficientList[Sum[x^(h-1),{h,k}]^n,x],{k,10},{n,0,9}];
    Flatten@Table[a[[s-n,n+1]],{s,10},{n,0,s-1}]
    (* alternate program *)
    row[k_, n_] := Nest[Accumulate,Upsample[Table[((-1)^j)*Binomial[n,j],{j,0,n}],k],n][[;;n*(k-1)+1]] (* Eitan Y. Levine, Apr 23 2023 *)

Formula

T(k,n,h) = Sum_{i = 0..floor(h/k)} (-1)^i*binomial(n,i)*binomial(n+h-1-k*i,n-1). [Corrected by Eitan Y. Levine, Apr 23 2023]
From Eitan Y. Levine, Apr 23 2023: (Start)
(T(k,n,h))_{h=0..n*(k-1)} = f(f(...f(g(P))...)), where:
(x_i)_{i=0..m} denotes a tuple (in particular, the LHS contains the values for 0 <= h <= n*(k-1)),
f repeats n times,
f((x_i){i=0..m}) = (Sum{j=0..i} x_j)_{i=0..m} is the cumulative sum function,
g((x_i){i=0..m}) = (x(i/k) if k|i, otherwise 0)_{i=0..m*k} is adding k-1 zeros between adjacent elements,
and P=((-1)^i*binomial(n,i))_{i=0..n} is the n-th row of Pascal's triangle, with alternating signs. (End)
From Eitan Y. Levine, Jul 27 2023: (Start)
Recurrence relations, the first follows from the sequence's defining polynomial as mentioned in the Smarandache link:
T(k,n+1,h) = Sum_{i = 0..s-1} T(k,n,h-i)
T(k+1,n,h) = Sum_{i = 0..n} binomial(n,i)*T(k,n-i,h-i*k) (End)

A077044 Largest coefficient in expansion of (1 + x + x^2 + ... + x^(n-1))^5 = ((1-x^n)/(1-x))^5, i.e., the coefficient of x^floor(5*(n-1)/2) and of x^ceiling(5*(n-1)/2); also number of compositions of floor(5*(n+1)/2) into exactly 5 positive integers each no more than n.

Original entry on oeis.org

0, 1, 10, 51, 155, 381, 780, 1451, 2460, 3951, 6000, 8801, 12435, 17151, 23030, 30381, 39280, 50101, 62910, 78151, 95875, 116601, 140360, 167751, 198780, 234131, 273780, 318501, 368235, 423851, 485250, 553401, 628160, 710601, 800530
Offset: 0

Views

Author

Henry Bottomley, Oct 22 2002

Keywords

Examples

			a(2)=10 since the compositions of floor(5*(2+1)/2) = 7 into exactly 5 positive integers each no more than 2 are: 1+1+1+2+2, 1+1+2+1+2, 1+1+2+2+1, 1+2+1+1+2, 1+2+1+2+1, 1+2+2+1+1, 2+1+1+1+2, 2+1+1+2+1, 2+1+2+1+1, 2+2+1+1+1.
		

Programs

  • Magma
    [(230*n^4+70*n^2+27-(30*n^2+27)*(-1)^n)/384: n in [0..40]]; // Vincenzo Librandi, Sep 05 2011
    
  • Mathematica
    LinearRecurrence[{2,2,-6,0,6,-2,-2,1},{0,1,10,51,155,381,780,1451},40] (* Harvey P. Dale, Mar 05 2015 *)
  • PARI
    a(n)=(230*n^4+70*n^2-30*n^2*(-1)^n)\/384 \\ Charles R Greathouse IV, Sep 25 2012

Formula

a(n) = (230*n^4 + 70*n^2 + 27 - (30*n^2 + 27)*(-1)^n)/384 = A077042(n, 5).
a(n) = 2*a(n-1) + 2*a(n-2) - 6*a(n-3) + 6*a(n-5) - 2*a(n-6) - 2*a(n-7) + a(n-8).
G.f.: -x*(1 + 8*x + 29*x^2 + 39*x^3 + 29*x^4 + 8*x^5 + x^6) / ( (1+x)^3*(x-1)^5 ). - R. J. Mathar, Sep 04 2011

A077046 Doubly restricted composition numbers: number of compositions of [n^2/2] into exactly n positive integers each strictly less than n.

Original entry on oeis.org

0, 1, 3, 19, 155, 1751, 24017, 398567, 7635987, 167729959, 4123838279, 112835748609, 3386455204288, 110976634957761, 3932912125462725, 150186639579545295, 6137695417757646851, 267654541150392845543, 12391407810082341898091, 607584722159224093306229
Offset: 1

Views

Author

Henry Bottomley, Oct 22 2002

Keywords

Examples

			a(3) = 3 since the compositions of [3^2/2]=4 into exactly 3 positive integers each strictly less than 3 are: 1+1+2, 1+2+1 and 2+1+1.
		

Crossrefs

Formula

a(n) = A077042(n-1, n).
a(n) ~ exp(-1)*sqrt(6/Pi)*n^(n-3/2). - Vaclav Kotesovec, Mar 26 2016

A201549 Number of arrays of n integers in -5..5 with sum zero.

Original entry on oeis.org

1, 1, 11, 91, 891, 8801, 88913, 908755, 9377467, 97464799, 1018872811, 10701243741, 112835748609, 1193692544825, 12663809507129, 134678108144591, 1435345208419771, 15326122342137035, 163920458145421109
Offset: 0

Views

Author

R. H. Hardin, Dec 02 2011

Keywords

Comments

Also largest coefficient of (1+x+...+x^10)^n. - Vaclav Kotesovec, Aug 09 2013

Examples

			Some solutions for n=6
.-5...-5...-1...-2...-1....3....4....1....1...-1....3....4....5....0...-5....5
.-2....4...-1....1....0...-4....2....1....1...-5...-4...-4....1...-3....5...-5
..0...-3....1....3...-1...-4....0...-1....2...-4...-4...-4...-5...-3...-2....5
..4....3....3...-3....4....5....1....2....2....5....3....5...-3....2....2...-5
..5...-4...-1...-4...-4....1...-2...-4...-5....2....0....4....1....4...-4....3
.-2....5...-1....5....2...-1...-5....1...-1....3....2...-5....1....0....4...-3
		

Crossrefs

Column 5 of A201552. Row 11 of A077042.

Programs

  • Maple
    seq(add((-1)^k * binomial(n, k)*binomial(6*n-11*k-1, n-1), k = 0..floor(n/2)), n = 0..20); # Peter Bala, Oct 16 2024
  • Mathematica
    Table[Coefficient[Expand[Sum[x^j,{j,0,10}]^n],x^(5*n)],{n,1,20}] (* Vaclav Kotesovec, Aug 09 2013 *)
  • PARI
    {a(n) = polcoeff((sum(k=0, 10, x^k))^n, 5*n, x)} \\ Seiichi Manyama, Dec 14 2018

Formula

From Vaclav Kotesovec, Aug 09 2013: (Start)
Recurrence: 726*(n-2) * (n-1) * (2*n-1) * (3*n-7) * (3*n-1) * (5*n-19) * (5*n-9) * (5*n-8) * (6*n-25) * (6*n-13) * (6*n-1) * (2280*n^4 - 18164*n^3 + 49523*n^2 - 53013*n + 18900)*a(n-3) - 3*(2*n-1) * (3*n-7) * (3*n-1) * (5*n-19) * (5*n-14) * (5*n-9) * (5*n-8) * (6*n-25) * (6*n-19) * (6*n-13) * (6*n-1) * (4651*n^4 - 18604*n^3 + 27451*n^2 - 17694*n + 4200)*a(n-1) + 3993*(n-3) * (n-2) * (n-1) * (3*n-4) * (5*n-14) * (5*n-4) * (5*n-3) * (6*n-19) * (6*n-7) * (6*n-1) * (5310*n^5 - 65313*n^4 + 295326*n^3 - 594091*n^2 + 499480*n - 112320)*a(n-4) - 33*(n-1) * (3*n-4) * (5*n-19) * (5*n-14) * (5*n-4) * (5*n-3) * (6*n-25) * (6*n-19) * (6*n-7) * (45306*n^6 - 385101*n^5 + 1267841*n^4 - 2002349*n^3 + 1504595*n^2 - 451668*n + 42120)*a(n-2) - 161051*(n-5) * (n-4) * (n-3) * (n-2) * (n-1) * (3*n-4) * (3*n-1) * (5*n-14) * (5*n-9) * (5*n-4) * (5*n-3) * (6*n-19) * (6*n-13) * (6*n-7) * (6*n-1)*a(n-6) - 43923*(n-4) * (n-3) * (n-2) * (n-1) * (2*n-1) * (3*n-7) * (3*n-1) * (5*n-19) * (5*n-9) * (5*n-8) * (5*n-4) * (6*n-25) * (6*n-13) * (6*n-7) * (6*n-1)*a(n-5) + 5*n*(3*n-7) * (3*n-4) * (5*n-19) * (5*n-14) * (5*n-9) * (5*n-8) * (5*n-4) * (5*n-3) * (5*n-2) * (5*n-1) * (6*n-25) * (6*n-19) * (6*n-13) * (6*n-7)*a(n) = 0.
a(n) ~ 11^n / sqrt(20*Pi*n). (End)
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k * binomial(n, k)*binomial(6*n-11*k-1, n-1). - Peter Bala, Oct 16 2024

Extensions

a(0)=1 prepended by Seiichi Manyama, Dec 14 2018

A201550 Number of arrays of n integers in -6..6 with sum zero.

Original entry on oeis.org

1, 1, 13, 127, 1469, 17151, 204763, 2473325, 30162301, 370487485, 4577127763, 56813989827, 707972099627, 8851373201919, 110976634957761, 1394804756117877, 17567994350713469, 221690794842728445, 2802194053806820153
Offset: 0

Views

Author

R. H. Hardin, Dec 02 2011

Keywords

Comments

Also largest coefficient of (1+x+...+x^12)^n. - Vaclav Kotesovec, Aug 09 2013

Examples

			Some solutions for n=5
.-2...-5...-2...-1....3...-6....0...-3....1....6...-6...-2....5....0...-4...-3
..2...-3...-2....3....0...-1....6...-4....6....1....5....2...-1....3....2....3
..0...-4....4...-6...-4....1...-3....0...-4...-5....0...-6...-3....0....4...-4
..0....6....3....5...-5....6....0....4....3...-4....4....0...-5...-3....3...-1
..0....6...-3...-1....6....0...-3....3...-6....2...-3....6....4....0...-5....5
		

Crossrefs

Column 6 of A201552. Row 13 of A077042.

Programs

  • Maple
    seq(add((-1)^k * binomial(n, k)*binomial(7*n-13*k-1, n-1), k = 0..floor(n/2)), n = 0..20); # Peter Bala, Oct 16 2024
  • Mathematica
    Table[Coefficient[Expand[Sum[x^j,{j,0,12}]^n],x^(6*n)],{n,1,20}] (* Vaclav Kotesovec, Aug 09 2013 *)
  • PARI
    {a(n) = polcoeff((sum(k=0, 12, x^k))^n, 6*n, x)} \\ Seiichi Manyama, Dec 14 2018

Formula

a(n) ~ 13^n / sqrt(28*Pi*n). - Vaclav Kotesovec, Aug 09 2013
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k * binomial(n, k)*binomial(7*n-13*k-1, n-1). - Peter Bala, Oct 16 2024

Extensions

a(0)=1 prepended by Seiichi Manyama, Dec 14 2018

A025013 Central octonomial coefficients: largest coefficient of (1+x+...+x^7)^n.

Original entry on oeis.org

1, 1, 8, 48, 344, 2460, 18152, 134512, 1012664, 7635987, 58199208, 443658688, 3409213016, 26184550496, 202384723528, 1562970918720, 12133130451576, 94094281551304, 732910480638272, 5702603044247504, 44538031693977544
Offset: 0

Views

Author

Keywords

Comments

Generally, largest coefficient of (1+x+...+x^k)^n is asymptotic to (k+1)^n * sqrt(6/(k*(k+2)*Pi*n)). - Vaclav Kotesovec, Aug 09 2013

Crossrefs

Row 8 of A077042.

Programs

  • Maple
    seq(add((-1)^k * binomial(n, k)*binomial(n+floor(7*n/2)-8*k-1, n-1), k = 0..floor(n/2) ), n = 0..20); # Peter Bala, Oct 15 2024
  • Mathematica
    Flatten[{1,Table[Coefficient[Expand[Sum[x^j,{j,0,7}]^n],x^Floor[7*n/2]],{n,1,20}]}] (* Vaclav Kotesovec, Aug 09 2013 *)

Formula

a(n) ~ 8^n * sqrt(2/(21*Pi*n)). - Vaclav Kotesovec, Aug 09 2013
a(n) = Sum_{k = 0..floor(n/2)} (-1)^k * binomial(n, k)*binomial(n+floor(7*n/2)-8*k-1, n-1). - Peter Bala, Oct 15 2024

A225779 Largest coefficient of (1 + x + ... + x^11)^n.

Original entry on oeis.org

1, 1, 12, 108, 1156, 12435, 137292, 1528688, 17232084, 195170310, 2228154512, 25506741084, 293661065788, 3386455204288, 39222848622984, 454745042732160, 5290621952635476, 61590267941514516, 719050614048219912, 8397773337294253140, 98314091309732350656
Offset: 0

Views

Author

Vaclav Kotesovec, Aug 09 2013

Keywords

Comments

Generally, largest coefficient of (1 + x + ... + x^k)^n is asymptotic to (k+1)^n * sqrt(6/(k*(k+2)*Pi*n)).

Crossrefs

Row 12 of A077042.

Programs

  • Maple
    P:= add(x^i,i=0..11):
    seq(coeff(P^n,x,floor(11*n/2)),n=0..50); # Robert Israel, Jan 30 2017
  • Mathematica
    Flatten[{1, Table[Coefficient[Expand[Sum[x^j, {j,0,11}]^n], x^Floor[11*n/2]], {n,1,20}]}]
    f[n_] := Max[CoefficientList[Sum[x^k, {k, 0, 11}]^n, x]]; Array[f, 20, 0] (* Robert G. Wilson v, Jan 29 2017 *)
  • PARI
    a(n) = vecmax(Vec(Pol(vector(12,k,1))^n)); \\ Michel Marcus, Jan 29 2017

Formula

a(n) ~ 12^n * sqrt(6/(143*Pi*n)).

A133458 The size of the largest antichain in the 7-dimensional hypercubic lattice of size n; also the coefficient of x^floor(7*(n-1)/2) in (1 + x + ... + x^(n-1))^7.

Original entry on oeis.org

1, 35, 393, 2128, 8135, 24017, 60691, 134512, 273127, 512365, 908755, 1528688, 2473325, 3852919, 5832765, 8582336, 12354469, 17395119, 24072133, 32726960, 43874139, 57971221, 75715487, 97702640, 124853275, 157924585, 198105727
Offset: 1

Views

Author

Leonid Chindelevitch (leonidus(AT)mit.edu), Dec 22 2007

Keywords

Comments

The middle coefficients for dimension d>=1 are in A000012, A000027, A077043, A005900, A077044, A071816, here, the d-th row in A077042.
For d=8 the sequence starts 1, 70, 1107, 8092, 38165, 135954, 398567, 1012664, 2306025, ... and for d=9 it starts 1, 126, 3139, 30276, 180325, 767394, 2636263, 7635987, 19610233, ... - R. J. Mathar, Sep 04 2011

Crossrefs

Programs

  • Magma
    [-25*(-1)^n/512 +2261*n^2/23040 +25/512 +5887*n^6/11520 -77*(-1)^n*n^4/1536 +847*n^4/4608 -91*(-1)^n*n^2/1536 : n in [1..40]]; // Vincenzo Librandi, Sep 07 2011
  • Maple
    f:=(L,d)->(sum(x^k,k=0..L-1))^d; A:=[seq(coeff(f(j,7),x,floor(7*(j-1)/2)),j=1..25)];
    A133458 := proc(n) -25/512*(-1)^n +2261/23040*n^2 -91/1536*(-1)^n*n^2 -77/1536*(-1)^n*n^4 +847/4608*n^4 +5887/11520*n^6 +25/512 ; end proc: # R. J. Mathar, Sep 05 2011

Formula

From R. J. Mathar, Feb 19 2010: (Start)
a(n)= 2*a(n-1) +4*a(n-2) -10*a(n-3) -5*a(n-4) +20*a(n-5) -20*a(n-7) +5*a(n-8) +10*a(n-9) -4*a(n-10) -2*a(n-11) +a(n-12).
G.f.: x*(1+33*x +319*x^2 +1212*x^3 +2662*x^4 +3320*x^5 +2662*x^6 +1212*x^7 +319*x^8 +33*x^9 +x^10)/ ((1+x)^5 * (1-x)^7).
a(n) = -25*(-1)^n/512 +2261*n^2/23040 +25/512 +5887*n^6/11520 -77*(-1)^n*n^4/1536 +847*n^4/4608 -91*(-1)^n*n^2/1536. (End)

Extensions

More terms from R. J. Mathar, Feb 19 2010

A329020 Square array T(n,k), n>=0, k>=0, read by antidiagonals, where T(n,k) is the constant term in the expansion of ( Sum_{j=1..k} x_j^(2*j-1) + x_j^(-(2*j-1)) )^(2*n).

Original entry on oeis.org

1, 1, 0, 1, 2, 0, 1, 4, 6, 0, 1, 6, 44, 20, 0, 1, 8, 146, 580, 70, 0, 1, 10, 344, 4332, 8092, 252, 0, 1, 12, 670, 18152, 135954, 116304, 924, 0, 1, 14, 1156, 55252, 1012664, 4395456, 1703636, 3432, 0, 1, 16, 1834, 137292, 4816030, 58199208, 144840476, 25288120, 12870, 0
Offset: 0

Views

Author

Seiichi Manyama, Nov 02 2019

Keywords

Examples

			(x^3 + x + 1/x + 1/x^3)^2 = x^6 + 2*x^4 + 3*x^2 + 4 + 3/x^2 + 2/x^4 + 1/x^6. So T(1,2) = 4.
Square array begins:
   1,   1,      1,       1,        1,         1, ...
   0,   2,      4,       6,        8,        10, ...
   0,   6,     44,     146,      344,       670, ...
   0,  20,    580,    4332,    18152,     55252, ...
   0,  70,   8092,  135954,  1012664,   4816030, ...
   0, 252, 116304, 4395456, 58199208, 432457640, ...
		

Crossrefs

Columns k=0-3 give A000007, A000984, A005721, A063419.
Rows n=0-2 give A000012, A005843, 2*A143166.
Main diagonal gives A329021.
Cf. A077042.

Programs

  • Mathematica
    T[n_, 0] = Boole[n == 0]; T[n_, k_] := Sum[(-1)^j * Binomial[2*n, j] * Binomial[(2*k + 1)*n - 2*k*j - 1, (2*k - 1)*n - 2*k*j], {j, 0, Floor[(2*k - 1)*n/(2*k)]}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, May 06 2021 *)

Formula

T(n,k) = Sum_{j=0..floor((2*k-1)*n/(2*k))} (-1)^j * binomial(2*n,j) * binomial((2*k+1)*n-2*k*j-1,(2*k-1)*n-2*k*j) for k > 0.
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