A172634 -id:A172634

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A208446

Duplicate of A172634.

+20
1

1, 1, 7, 31, 175, 991, 5881, 35617, 219871, 1376095, 8710537, 55644337, 358198369, 2320792657, 15120204295, 98984058271, 650725327231, 4293779332927, 28425752310361, 188739799967425, 1256510215733185

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

LINKS

Table of n, a(n) for n = 0..20

KEYWORD

dead

STATUS

approved

A334549

Array read by antidiagonals: T(n,k) is the number of {-1,0,1} n X k matrices with all rows and columns summing to zero.

+10
13

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 7, 7, 1, 1, 1, 1, 19, 31, 19, 1, 1, 1, 1, 51, 175, 175, 51, 1, 1, 1, 1, 141, 991, 2371, 991, 141, 1, 1, 1, 1, 393, 5881, 32611, 32611, 5881, 393, 1, 1, 1, 1, 1107, 35617, 481381, 1084851, 481381, 35617, 1107, 1, 1

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,13

COMMENTS

Equivalently, the number of n X k 0..2 arrays with row sums k and column sums n.

LINKS

Andrew Howroyd, Table of n, a(n) for n = 0..350 (first 26 antidiagonals)

FORMULA

T(n,k) = T(k,n).

EXAMPLE

Array begins:

====================================================================

n\k | 0 1 2 3 4 5 6 7

----|---------------------------------------------------------------

0 | 1 1 1 1 1 1 1 1 ...

1 | 1 1 1 1 1 1 1 1 ...

2 | 1 1 3 7 19 51 141 393 ...

3 | 1 1 7 31 175 991 5881 35617 ...

4 | 1 1 19 175 2371 32611 481381 7343449 ...

5 | 1 1 51 991 32611 1084851 39612501 1509893001 ...

6 | 1 1 141 5881 481381 39612501 3680774301 360255871641 ...

7 | 1 1 393 35617 7343449 1509893001 360255871641 ...

...

The T(3,2) = 7 matrices are:

[0 0] [ 0 0] [ 0 0] [ 1 -1] [-1 1] [ 1 -1] [-1 1]

[0 0] [ 1 -1] [-1 1] [ 0 0] [ 0 0] [-1 1] [ 1 -1]

[0 0] [-1 1] [ 1 -1] [-1 1] [ 1 -1] [ 0 0] [ 0 0]

CROSSREFS

Columns k=0..14 are A000012, A000012, A002426, A172634, A172642, A172639, A172633, A172636, A172638, A172641, A172637, A172644, A172640, A172643, A172635.

Main diagonal is A172645.

Cf. A008300, A333901, A376935, A377063 (up to row permutations).

KEYWORD

nonn,tabl

AUTHOR

Andrew Howroyd, May 09 2020

STATUS

approved

A328747

Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where T(n,k) is Sum_{i=0..n} (-1)^(n-i)*binomial(n,i)*Sum_{j=0..i} binomial(i,j)^k.

+10
8

1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 3, 1, 0, 1, 1, 7, 7, 1, 0, 1, 1, 15, 31, 19, 1, 0, 1, 1, 31, 115, 175, 51, 1, 0, 1, 1, 63, 391, 1255, 991, 141, 1, 0, 1, 1, 127, 1267, 8071, 13671, 5881, 393, 1, 0, 1, 1, 255, 3991, 49399, 161671, 160461, 35617, 1107, 1, 0

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,13

COMMENTS

T(n,k) is the constant term in the expansion of (-1 + Product_{j=1..k-1} (1 + x_j) + Product_{j=1..k-1} (1 + 1/x_j))^n for k > 0.

For fixed k > 0, T(n,k) ~ (2^k - 1)^(n + (k-1)/2) / (2^((k-1)^2/2) * sqrt(k) * (Pi*n)^((k-1)/2)). - Vaclav Kotesovec, Oct 28 2019

LINKS

Seiichi Manyama, Antidiagonals n = 0..100, flattened

EXAMPLE

Square array begins:

1, 1, 1, 1, 1, 1, ...

0, 1, 3, 7, 15, 31, ...

0, 1, 7, 31, 115, 391, ...

0, 1, 19, 175, 1255, 8071, ...

0, 1, 51, 991, 13671, 161671, ...

MATHEMATICA

T[n_, k_] := Sum[(-1)^(n-i) * Binomial[n, i] * Sum[Binomial[i, j]^k, {j, 0, i}], {i, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 06 2021 *)

CROSSREFS

Columns k=0..5 give A019590(n+1), A000012, A002426, A172634, A328725, A328750.

Main diagonal gives A328811.

T(n,n+1) gives A328813.

Cf. A309010, A328748, A328807.

KEYWORD

nonn,tabl

AUTHOR

Seiichi Manyama, Oct 27 2019

STATUS

approved

A328725

Constant term in the expansion of (1 + x + y + z + 1/x + 1/y + 1/z + x*y + y*z + z*x + 1/(x*y) + 1/(y*z) + 1/(z*x) + x*y*z + 1/(x*y*z))^n.

+10
7

1, 1, 15, 115, 1255, 13671, 160461, 1936425, 24071895, 305313415, 3939158905, 51521082405, 681635916325, 9105864515125, 122657982366375, 1664151758259915, 22720725637684215, 311933068664333175, 4303704125389134825, 59640225721889127525, 829774531966386480705

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

LINKS

Seiichi Manyama, Table of n, a(n) for n = 0..854

FORMULA

a(n) = Sum_{i=0..n} (-1)^(n-i)*binomial(n,i)*Sum_{j=0..i} binomial(i,j)^4.

From Vaclav Kotesovec, Oct 28 2019: (Start)

Recurrence: n^3*a(n) = (2*n - 1)^3*a(n-1) + (n-1)*(94*n^2 - 188*n + 93)*a(n-2) + 80*(n-2)*(n-1)*(2*n - 3)*a(n-3) + 75*(n-3)*(n-2)*(n-1)*a(n-4).

a(n) ~ 15^(n + 3/2) / (2^(11/2) * Pi^(3/2) * n^(3/2)). (End)

PROG

(PARI) {a(n) = polcoef(polcoef(polcoef((-1+(1+x)*(1+y)*(1+z)+(1+1/x)*(1+1/y)*(1+1/z))^n, 0), 0), 0)}

(PARI) {a(n) = sum(i=0, n, (-1)^(n-i)*binomial(n, i)*sum(j=0, i, binomial(i, j)^4))}

CROSSREFS

Sum_{i=0..n} (-1)^(n-i)*binomial(n,i)*Sum_{j=0..i} binomial(i,j)^m: A002426 (m=2), A172634 (m=3), this sequence (m=4), A328750 (m=5).

Cf. A002899, A005260, A328713.

KEYWORD

nonn

AUTHOR

Seiichi Manyama, Oct 26 2019

STATUS

approved

A262704

Triangle: Newton expansion of C(n,m)^3, read by rows.

+10
4

1, 0, 1, 0, 6, 1, 0, 6, 24, 1, 0, 0, 114, 60, 1, 0, 0, 180, 690, 120, 1, 0, 0, 90, 2940, 2640, 210, 1, 0, 0, 0, 5670, 21840, 7770, 336, 1, 0, 0, 0, 5040, 87570, 107520, 19236, 504, 1, 0, 0, 0, 1680, 189000, 735210, 407400, 42084, 720, 1, 0, 0, 0, 0, 224700, 2835756, 4280850, 1284360, 83880, 990, 1

(list; table; graph; refs; listen; history; text; internal format)

OFFSET

0,5

COMMENTS

Triangle here T_3(n,m) is such that C(n,m)^3 = Sum_{j=0..n} C(n,j)*T_3(j,m).

Equivalently, lower triangular matrix T_3 such that

|| C(n,m)^3 || = A181583 = P * T_3 = A007318 * T_3.

T_3(n,m) = 0 for n < m and for 3*m < n. In fact:

C(x,m)^q and C(x,m), with m nonnegative and q positive integer, are polynomial in x of degree m*q and m respectively, and C(x,m) is a divisor of C(x,m)^q.

Therefore the Newton series will give C(x,m)^q = T_q(m,m)*C(x,m) + T_q(m+1,m)*C(x,m+1) + ... + T_q(q*m,m)*C(x,q*m), where T_q(n,m) is the n-th forward finite difference of C(x,m)^q at x = 0.

Example:

C(x,2)^3 = x^3*(x-1)^3 / 8 = 1*C(x,2) + 24*C(x,3) + 114*C(x,4) + 180*C(x,5) + 90*C(x,6);

C(5,2)^3 = C(5,3)^3 = 1000 = 1*C(5,2) + 24*C(5,3) + 114*C(5,4) + 180*C(5,5) = 1*C(5,3) + 60*C(5,4) + 690*C(5,5).

So we get the expansion of the 3rd power of the binomial coefficient in terms of the binomial coefficients on the same row.

T_1 is the unitary matrix,

T_2 is the transpose of A109983,

T_3 is this sequence,

T_4, T_5 are A262705, A262706.

LINKS

Gheorghe Coserea, Rows n = 0..200, flattened

P. Blasiak, K. A. Penson and A. I. Solomon, The general boson normal ordering problem, arXiv:quant-ph/0402027, 2004.

FORMULA

T_3(n,m) = Sum_{j=0..n} (-1)^(n-j)*C(n,j)*C(j,m)^3.

Also, let S(r,s)(n,m) denote the Generalized Stirling2 numbers as defined in the link above,then T_3(n,m) = n! / (m!)^3 * S(m,m)(3,n).

EXAMPLE

Triangle starts:

n\m [0] [1] [2] [3] [4] [5] [6] [7] [8]

[0] 1;

[1] 0, 1;

[2] 0, 6, 1;

[3] 0, 6, 24, 1;

[4] 0, 0, 114, 60, 1;

[5] 0, 0, 180, 690, 120, 1;

[6] 0, 0, 90, 2940, 2640, 210, 1;

[7] 0, 0, 0, 5670, 21840, 7770, 336, 1;

[8] 0, 0, 0, 5040, 87570, 107520, 19236, 504, 1;

[9] ...

MATHEMATICA

T3[n_, m_] := Sum[(-1)^(n - j) * Binomial[n, j] * Binomial[j, m]^3, {j, 0, n}]; Table[T3[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Oct 01 2015 *)

PROG

(MuPAD)

// as a function

T_3:=(n, m)->_plus((-1)^(n-j)*binomial(n, j)*binomial(j, m)^3 $ j=0..n):

// as a matrix h x h

_P:=h->matrix([[binomial(n, m) $m=0..h]$n=0..h]):

_P_3:=h->matrix([[binomial(n, m)^3 $m=0..h]$n=0..h]):

_T_3:=h->_P(h)^-1*_P_3(h):

(PARI) T_3(nmax) = {for(n=0, nmax, for(m=0, n, print1(sum(j=0, n, (-1)^(n-j)*binomial(n, j)*binomial(j, m)^3), ", ")); print())} \\ Colin Barker, Oct 01 2015

(Magma) [&+[(-1)^(n-j)*Binomial(n, j)*Binomial(j, m)^3: j in [0..n]]: m in [0..n], n in [0..10]]; // Bruno Berselli, Oct 01 2015

(PARI) t3(n, m) = sum(j=0, n, (-1)^((n-j)%2)* binomial(n, j)*binomial(j, m)^3);

concat(vector(11, n, vector(n, k, t3(n-1, k-1)))) \\ Gheorghe Coserea, Jul 14 2016

CROSSREFS

Row sums are A172634, the inverse binomial transform of the Franel numbers (A000172).

Column sums are the A126086, per the comment given thereto by Brendan McKay.

Second diagonal (T_3(n+1,n)) is A007531 (n+2).

Column T_3(n,2) is A122193(3,n).

Cf. A109983 (transpose of), A262705, A262706.

Cf. A078739, A078741, A274786.

KEYWORD

nonn,tabl,easy

AUTHOR

Giuliano Cabrele, Sep 27 2015

STATUS

approved

A377065

Number of 3 X n 0..2 matrices with row sums n and column sums 3 up to permutations of rows.

+10
4

1, 1, 2, 6, 30, 166, 981, 5937, 36646, 229350, 1451757, 9274057, 59699729, 386798777, 2520034050, 16497343046, 108454221206, 715629888822, 4737625385061, 31456633327905, 209418369288865, 1397521222483385, 9346484009527370, 62632803958053870, 420481623373564025

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

Also, the number of 3 X n {-1,0,1} matrices with all rows and columns summing to zero up to permutations of rows.

LINKS

Andrew Howroyd, Table of n, a(n) for n = 0..500

FORMULA

a(n) = (A172634(n) - 1)/6 + 1.

a(n) = (5 + Sum_{i=0..n} Sum_{j=0..i} (-1)^(n-i)*binomial(n, i)*binomial(i, j)^3)/6.

EXAMPLE

The a(2) = 2 matrices are:

[1 1] [2 0]

[1 1] [0 2]

[1 1] [1 1]

The a(3) = 6 matrices are:

[1 1 1] [2 1 0] [2 0 1] [1 2 0] [2 1 0] [2 0 1]

[1 1 1] [0 1 2] [0 2 0] [1 0 2] [1 0 2] [1 2 0]

[1 1 1] [1 1 1] [1 1 1] [1 1 1] [0 2 1] [0 1 2]

PROG

(PARI) a(n)={(5+sum(i=0, n, sum(j=0, i, (-1)^(n-i)*binomial(n, i)*binomial(i, j)^3)))/6}

CROSSREFS

Row n=3 of A377063.

Cf. A172634, A377066, A377067.

KEYWORD

nonn

AUTHOR

Andrew Howroyd, Oct 15 2024

STATUS

approved

A377067

Number of n X 3 0..2 matrices with row sums 3 and column sums n up to permutations of rows.

+10
4

1, 1, 4, 6, 12, 18, 30, 42, 63, 85, 118, 154, 204, 258, 330, 408, 507, 615, 748, 892, 1066, 1254, 1476, 1716, 1995, 2295, 2640, 3010, 3430, 3880, 4386, 4926, 5529, 6171, 6882, 7638, 8470, 9352, 10318, 11340, 12453, 13629, 14904, 16248, 17700, 19228, 20872, 22600, 24453, 26397, 28476

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,3

COMMENTS

Also, the number of n X 3 {-1,0,1} matrices with all rows and columns summing to zero up to permutations of rows.

LINKS

Andrew Howroyd, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,1,-3,-1,1,3,-1,-2,1).

FORMULA

G.f.: (2/(1 - x^3) - 1)/((1 - x)*(1 - x^2)^3).

G.f.: (1 - x + x^2)/((1 - x)^5*(1 + x)^2*(1 + x + x^2)).

EXAMPLE

The a(2) = 4 matrices are:

[1 1 1] [2 1 0] [2 0 1] [1 2 0]

[1 1 1] [0 1 2] [0 2 0] [1 0 2]

The a(3) = 6 matrices are:

[1 1 1] [2 1 0] [2 0 1] [1 2 0] [2 1 0] [2 0 1]

[1 1 1] [0 1 2] [0 2 0] [1 0 2] [1 0 2] [1 2 0]

[1 1 1] [1 1 1] [1 1 1] [1 1 1] [0 2 1] [0 1 2]

PROG

(PARI) Vec((1 - x + x^2)/((1 - x)^5*(1 + x)^2*(1 + x + x^2)) + O(x^51))

CROSSREFS

Column k=3 of A377063.

Cf. A172634, A377065, A377068.

KEYWORD

nonn,easy

AUTHOR

Andrew Howroyd, Oct 15 2024

STATUS

approved

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