A094053 -id:A094053 - cp's OEIS frontend

A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

1, 1, 1, 1, 7, 1, 1, 13, 13, 1, 1, 19, 25, 19, 1, 1, 25, 37, 37, 25, 1, 1, 31, 49, 55, 49, 31, 1, 1, 37, 61, 73, 73, 61, 37, 1, 1, 43, 73, 91, 97, 91, 73, 43, 1, 1, 49, 85, 109, 121, 121, 109, 85, 49, 1, 1, 55, 97, 127, 145, 151, 145, 127, 97, 55, 1, 1, 61, 109, 145, 169, 181, 181, 169, 145, 109, 61, 1

Offset: 0

Kolosov Petro, Aug 31 2017

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(1, n, k), i.e T(n, k) is partial case of L(m, n, k) for m = 1.
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
  ----------------------------------------
  k=    0   1   2   3   4   5   6   7   8
  ----------------------------------------
  n=0:  1;
  n=1:  1,  1;
  n=2:  1,  7,  1;
  n=3:  1, 13, 13,  1;
  n=4:  1, 19, 25, 19,  1;
  n=5:  1, 25, 37, 37, 25,  1;
  n=6:  1, 31, 49, 55, 49, 31,  1;
  n=7:  1, 37, 61, 73, 73, 61, 37,  1;
  n=8:  1, 43, 73, 91, 97, 91, 73, 43,  1;

Georg Fischer, Table of n, a(n) for n = 0..495 [rows 0..10 and 12..30 from Kolosov Petro]
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Columns k=0..6 give A000012, A016921, A017533, A161705, A103214, A128470, A158065.
Column sums k=0..4 give A000027, A000567, A051866, A051872, A255185.
Row sums give A001093.
Various cases of L(m, n, k): This sequence (m=1), A300656(m=2), A300785(m=3). See comments for L(m, n, k).
Differences of cubes n^3 are T(A000124(n), 1).
Cf. A000124, A000578, A003154, A003215, A007318, A008458, A016969.
Cf. A038593, A055012, A068601, A077028, A094053, A166873, A227776.
Cf. A294317, A302971, A304042.

Flat(List([0..11],n->List([0..n],k->6*k*(n-k)+1))); # Muniru A Asiru, Oct 09 2018

/* As triangle */ [[6*k*(n-k) + 1: k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 26 2018

T := (n, k) -> 6*k*(n-k) + 1:
seq(seq(T(n, k), k=0..n), n=0..11); # Muniru A Asiru, Oct 09 2018

T[n_, k_] := 6 k (n - k) + 1; Column[Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Jun 02 2019 *)

t(n, k) = 6*k*(n-k)+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */
trianglerows(9) \\ Felix Fröhlich, Jan 09 2018

def A287326(n,k): return 6*k*(n-k) + 1
flatten([[A287326(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Sep 25 2024

T(n, k) = 6*k*(n-k) + 1.
G.f. of column k: n^k*(1+(6*k-1)*n)/(1-n)^2.
G.f.: (1 - x - x*y + 7*x^2*y)/((1 - x)^2*(1 - x*y)^2). - Stefano Spezia, Oct 09 2018 [Adapted by Stefano Spezia, Sep 25 2024]
From Kolosov Petro, Jun 05 2019: (Start)
T(n, k) = 1/2 * T(A294317(n, k), k) + 1/2.
T(n+1, k) = 2*T(n, k) - T(n-1, k), for n >= k.
T(n, k) = 6*A077028(n, k) - 5.
T(2n, n) = A227776(n).
T(2n+1, n) = A003154(n+1).
T(2n+3, n) = A166873(n+1).
Sum_{k=0..n-1} T(n, k) = Sum_{k=1..n} T(n, k) = A000578(n).
Sum_{k=1..n-1} T(n, k) = A068601(n).
(n+1)^3 - n^3 = T(A000124(n), 1). (End)
Sum_{k=0..n} (-1)^k*T(n, k) = (-1/2)*(1 + (-1)^n)*A016969(floor(n/2) - 1). - G. C. Greubel, Sep 25 2024

A114327 Table T(n,m) = n - m read by upwards antidiagonals.

Original entry on oeis.org

0, 1, -1, 2, 0, -2, 3, 1, -1, -3, 4, 2, 0, -2, -4, 5, 3, 1, -1, -3, -5, 6, 4, 2, 0, -2, -4, -6, 7, 5, 3, 1, -1, -3, -5, -7, 8, 6, 4, 2, 0, -2, -4, -6, -8, 9, 7, 5, 3, 1, -1, -3, -5, -7, -9, 10, 8, 6, 4, 2, 0, -2, -4, -6, -8, -10, 11, 9, 7, 5, 3, 1, -1, -3, -5, -7, -9, -11, 12, 10, 8, 6, 4, 2, 0, -2, -4, -6, -8, -10, -12

Offset: 0

Franklin T. Adams-Watters, Feb 06 2006

From Clark Kimberling, May 31 2011: (Start)
If we arrange A000027 as an array with northwest corner
   1    2    4    7    17 ...
   3    5    8   12    18 ...
   6    9   13   18    24 ...
  10   14   19   25    32 ...
diagonals can be numbered as follows depending on their distance to the main diagonal:
  Diag 0:  1, 5, 13, 25, ...
  Diag 1:  2, 8, 18, 32, ...
  Diag -1: 3, 9, 19, 33, ...,
then a(n) in the flattened array is the number of the diagonal that contains n+1. (End)
Construct the infinite-dimensional matrix representation of angular momentum operators (J_1,J_2,J_3) in Jordan-Schwinger form (cf. Harter, Klee, Schwinger). Triangle terms T(n,k) = T(2j,j-m) satisfy: (1/2) T(2j,j-m) =  = m. Matrix J_3 is diagonal, so this equality determines the only nonzero entries. - Bradley Klee, Jan 29 2016
For the characteristic polynomial of the n X n matrix M_n (Det(x*1_n - M_n)) with elements M_n(i, j) = i-j see the Michael Somos, Nov 14 2002, comment on A002415. - Wolfdieter Lang, Feb 05 2018
The entries of the n-th antidiagonal, T(n,1), T(n-1,2), ... , T(1,n), are the eigenvalues of the Hamming graph H(2,n-1) (or hypercube Q(n-1)). - Miquel A. Fiol, May 21 2024

			From _Wolfdieter Lang_, Feb 05 2018: (Start)
The table T(n, m) begins:
  n\m 0  1  2  3  4  5 ...
  0:  0 -1 -2 -3 -4 -5 ...
  1:  1  0 -1 -2 -3 -4 ...
  2:  2  1  0 -1 -2 -3 ...
  3:  3  2  1  0 -1 -2 ...
  4:  4  3  2  1  0 -1 ...
  5:  5  4  3  2  1  0 ...
  ...
The triangle t(n, k) begins:
  n\k  0  1  2  3  4  5  6  7  8  9  10 ...
  0:   0
  1:   1 -1
  2:   2  0 -2
  3:   3  1 -1 -3
  4:   4  2  0 -2 -4
  5:   5  3  1 -1 -3 -5
  6:   6  4  2  0 -2 -4 -6
  7:   7  5  3  1 -1 -3 -5 -7
  8:   8  6  4  2  0 -2 -4 -6 -8
  9:   9  7  5  3  1 -1 -3 -5 -7 -9
  10: 10  8  6  4  2  0 -2 -4 -6 -8 -10
  ... Reformatted and corrected. (End)

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
W. Harter, Principles of Symmetry, Dynamics, Spectroscopy, Wiley, 1993, Ch. 5, page 345-346.
B. Klee, Quantum Angular Momentum Matrices, Wolfram Demonstrations Project, 2016.
J. Schwinger, On Angular Momentum, Cambridge: Harvard University, Nuclear Development Associates, Inc., 1952.

Apart from signs, same as A049581. Cf. A003056, A025581, A002262, A002260, A004736. J_1,J_2: A094053; J_1^2,J_2^2: A141387, A268759. A002415.

a114327 n k = a114327_tabl !! n !! k
a114327_row n = a114327_tabl !! n
a114327_tabl = zipWith (zipWith (-)) a025581_tabl a002262_tabl
-- Reinhard Zumkeller, Aug 09 2014

seq(seq(i-2*j,j=0..i),i=0..30); # Robert Israel, Jan 29 2016

max = 12; a025581 = NestList[Prepend[#, First[#]+1]&, {0}, max]; a002262 = Table[Range[0, n], {n, 0, max}]; a114327 = a025581 - a002262 // Flatten (* Jean-François Alcover, Jan 04 2016 *)
Flatten[Table[-2 m, {j, 0, 10, 1/2}, {m, -j, j}]] (* Bradley Klee, Jan 29 2016 *)

T(n,m) = n-m \\ Charles R Greathouse IV, Feb 07 2017

from math import isqrt
def A114327(n): return ((m:=isqrt(k:=n+1<<1))+(k>m*(m+1)))**2+1-k # Chai Wah Wu, Nov 09 2024

G.f. for the table: Sum_{n, m>=0} T(n,m)*x^n*y^n = (x-y)/((1-x)^2*(1-y)^2).
E.g.f. for the table: Sum_{n, m>=0} T(n,m)x^n/n!*y^m/m! = (x-y)*e^{x+y}.
T(n,k) = A025581(n,k) - A002262(n,k).
a(n+1) = A004736(n) - A002260(n) or a(n+1) = ((t*t+3*t+4)/2-n) - (n-t*(t+1)/2), where t=floor((-1+sqrt(8*n-7))/2). - Boris Putievskiy, Dec 24 2012
G.f. as sequence: -(1+x)/(1-x)^2 + Sum_{j>=0} (2*j+1)*x^(j*(j+1)/2) / (1-x). The sum is related to Jacobi theta functions. - Robert Israel, Jan 29 2016
Triangle t(n, k) = n - 2*k, for n >= 0, k = 0..n. (see the Maple program). - Wolfdieter Lang, Feb 05 2018

Formula improved by Reinhard Zumkeller, Aug 09 2014

A300656 Triangle read by rows: T(n,k) = 30k^2(n-k)^2 + 1; n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 31, 1, 1, 121, 121, 1, 1, 271, 481, 271, 1, 1, 481, 1081, 1081, 481, 1, 1, 751, 1921, 2431, 1921, 751, 1, 1, 1081, 3001, 4321, 4321, 3001, 1081, 1, 1, 1471, 4321, 6751, 7681, 6751, 4321, 1471, 1, 1, 1921, 5881, 9721, 12001, 12001, 9721, 5881, 1921, 1

Offset: 0

Kolosov Petro, Mar 10 2018

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(2, n, k).
Fifth power can be expressed as row sum of triangle T(n, k).
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
--------------------------------------------------------------------------
k=    0     1     2      3      4      5      6      7     8     9    10
--------------------------------------------------------------------------
n=0:  1;
n=1:  1,    1;
n=2:  1,   31,    1;
n=3:  1,  121,  121,     1;
n=4:  1,  271,  481,   271,     1;
n=5:  1,  481, 1081,  1081,   481,     1;
n=6:  1,  751, 1921,  2431,  1921,   751,     1;
n=7:  1, 1081, 3001,  4321,  4321,  3001,  1081,     1;
n=8:  1, 1471, 4321,  6751,  7681,  6751,  4321,  1471,    1;
n=9:  1, 1921, 5881,  9721, 12001, 12001,  9721,  5881, 1921,    1;
n=10: 1, 2431, 7681, 13231, 17281, 18751, 17281, 13231, 7681, 2431,   1;

Kolosov Petro, Rows n = 0..2078 of triangle, flattened.
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Various cases of L(m, n, k): A287326(m=1), This sequence (m=2), A300785(m=3). See comments for L(m, n, k).
Row sums give the nonzero terms of A002561.
Cf. A000584, A287326, A007318, A077028, A294317, A068236, A302971, A304042, A002561, A258807, A158558, A094053, A024003, A316349.

T:=Flat(List([0..9],n->List([0..n],k->30*k^2*(n-k)^2+1))); # Muniru A Asiru, Oct 24 2018

[[30*k^2*(n-k)^2+1: k in [0..n]]: n in [0..12]]; // G. C. Greubel, Dec 14 2018

a:=(n,k)->30*k^2*(n-k)^2+1: seq(seq(a(n,k),k=0..n),n=0..9); # Muniru A Asiru, Oct 24 2018

T[n_, k_] := 30 k^2 (n - k)^2 + 1; Column[
Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* Kolosov Petro, Apr 12 2020 *)
f[n_]:=Table[SeriesCoefficient[(1 + 26 y + 336 y^2 + 326 y^3 + 31 y^4 + x^2 (1 + 116 y + 486 y^2 + 116 y^3 + y^4) + x (-2 - 82 y - 882 y^2 - 502 y^3 + 28 y^4))/((-1 + x)^3 (-1 + y)^5), {x, 0, i}, {y, 0, j}], {i, n, n}, {j, 0, n}]; Flatten[Array[f, 11, 0]] (* Stefano Spezia, Oct 30 2018 *)

t(n, k) = 30*k^2*(n-k)^2+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */ trianglerows(9)

[[30*k^2*(n-k)^2+1 for k in range(n+1)] for n in range(12)] # G. C. Greubel, Dec 14 2018

From Kolosov Petro, Apr 12 2020: (Start)
T(n, k) = 30 * k^2 * (n-k)^2 + 1.
T(n, k) = 30 * A094053(n,k)^2 + 1.
T(n, k) = A158558((n-k) * k).
T(n+2, k) = 3*T(n+1, k) - 3*T(n, k) + T(n-1, k), for n >= k.
Sum_{k=1..n} T(n, k) = A000584(n).
Sum_{k=0..n-1} T(n, k) = A000584(n).
Sum_{k=0..n} T(n, k) = A002561(n).
Sum_{k=1..n-1} T(n, k) = A258807(n).
Sum_{k=1..n-1} T(n, k) = -A024003(n), n > 1.
Sum_{k=1..r} T(n, k) = A316349(2,r,0)*n^0 - A316349(2,r,1)*n^1 + A316349(2,r,2)*n^2. (End)
G.f.: (1 + 26*y + 336*y^2 + 326*y^3 + 31*y^4 + x^2*(1 + 116*y + 486*y^2 + 116*y^3 + y^4) + x*(-2 - 82*y - 882*y^2 - 502*y^3 + 28*y^4))/((-1 + x)^3*(-1 + y)^5). - Stefano Spezia, Oct 30 2018

A300785 Triangle read by rows: T(n,k) = 140k^3(n-k)^3 - 14k(n-k) + 1; n >= 0, 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 127, 1, 1, 1093, 1093, 1, 1, 3739, 8905, 3739, 1, 1, 8905, 30157, 30157, 8905, 1, 1, 17431, 71569, 101935, 71569, 17431, 1, 1, 30157, 139861, 241753, 241753, 139861, 30157, 1, 1, 47923, 241753, 472291, 573217, 472291, 241753, 47923, 1, 1, 71569, 383965, 816229, 1119721, 1119721, 816229, 383965, 71569, 1

Offset: 0

Kolosov Petro, Mar 12 2018

From Kolosov Petro, Apr 12 2020: (Start)
Let A(m, r) = A302971(m, r) / A304042(m, r).
Let L(m, n, k) = Sum_{r=0..m} A(m, r) * k^r * (n - k)^r.
Then T(n, k) = L(3, n, k).
T(n, k) is symmetric: T(n, k) = T(n, n-k). (End)

			Triangle begins:
--------------------------------------------------------------------
k=   0      1       2       3       4       5       6      7     8
--------------------------------------------------------------------
n=0: 1;
n=1: 1,     1;
n=2: 1,   127,      1;
n=3: 1,  1093,   1093,      1;
n=4: 1,  3739,   8905,   3739,      1;
n=5: 1,  8905,  30157,  30157,   8905,      1;
n=6: 1, 17431,  71569, 101935,  71569,  17431,      1;
n=7: 1, 30157, 139861, 241753, 241753, 139861,  30157,     1;
n=8: 1, 47923, 241753, 472291, 573217, 472291, 241753, 47923,    1;

Muniru A Asiru, Rows n=0..100 of triangle, flattened.
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Polynomial identity involving binomial theorem and Faulhaber's formula, 2023.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.

Various cases of L(m, n, k): A287326 (m=1), A300656 (m=2), This sequence (m=3). See comments for L(m, n, k).
Row sums give A258806.
Cf. A000584, A287326, A007318, A077028, A294317, A068236, A300656, A302971, A304042, A001015, A094053, A258808, A024005, A316387.

T:=Flat(List([0..9], n->List([0..n], k->140*k^3*(n-k)^3 - 14*k*(n-k)+1))); # G. C. Greubel, Dec 14 2018

/* As triangle */ [[140*k^3*(n-k)^3-14*k*(n-k)+1: k in [0..n]]: n in [0..10]]; // Bruno Berselli, Mar 21 2018

T:=(n,k)->140*k^3*(n-k)^3-14*k*(n-k)+1: seq(seq(T(n,k),k=0..n),n=0..9); # Muniru A Asiru, Dec 14 2018

T[n_, k_] := 140*k^3*(n - k)^3 - 14*k*(n - k) + 1; Column[
Table[T[n, k], {n, 0, 10}, {k, 0, n}], Center] (* From Kolosov Petro, Apr 12 2020 *)

t(n, k) = 140*k^3*(n-k)^3-14*k*(n-k)+1
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(t(x, y), ", ")); print(""))
/* Print initial 9 rows of triangle as follows */ trianglerows(9)

[[140*k^3*(n-k)^3 - 14*k*(n-k)+1 for k in range(n+1)] for n in range(12)] # G. C. Greubel, Dec 14 2018

From Kolosov Petro, Apr 12 2020: (Start)
T(n, k) = 140*k^3*(n-k)^3 - 14*k*(n-k) + 1.
T(n, k) = 140*A094053(n, k)^3 + 0*A094053(n, k)^2 - 14*A094053(n, k)^1 + 1.
T(n+3, k) = 4*T(n+2, k) - 6*T(n+1, k) + 4*T(n, k) - T(n-1, k), for n >= k.
Sum_{k=1..n} T(n, k) = A001015(n).
Sum_{k=0..n} T(n, k) = A258806(n).
Sum_{k=0..n-1} T(n, k) = A001015(n).
Sum_{k=1..n-1} T(n, k) = A258808(n).
Sum_{k=1..n-1} T(n, k) = -A024005(n).
Sum_{k=1..r} T(n, k) = -A316387(3, r, 0)*n^0 + A316387(3, r, 1)*n^1 - A316387(3, r, 2)*n^2 + A316387(3, r, 3)*n^3. (End)
G.f.: (1 + 127*x^6*y^3 - 3*x*(1 + y) + 585*x^5*y^2*(1 + y) + 129*x^4*y*(1 + 17*y + y^2) + 3*x^2*(1 + 45*y + y^2) - x^3*(1 - 579*y - 579*y^2 + y^3))/((1 - x)^4*(1 - x*y)^4). - Stefano Spezia, Sep 14 2024

A141387 Triangle read by rows: T(n,m) = n + 2m(n - m) (0 <= m <= n).

Original entry on oeis.org

0, 1, 1, 2, 4, 2, 3, 7, 7, 3, 4, 10, 12, 10, 4, 5, 13, 17, 17, 13, 5, 6, 16, 22, 24, 22, 16, 6, 7, 19, 27, 31, 31, 27, 19, 7, 8, 22, 32, 38, 40, 38, 32, 22, 8, 9, 25, 37, 45, 49, 49, 45, 37, 25, 9, 10, 28, 42, 52, 58, 60, 58, 52, 42, 28, 10

Offset: 0

Roger L. Bagula, Aug 03 2008

Construct the infinite-dimensional matrix representation of angular momentum operators (J_1,J_2,J_3) in Jordan-Schwinger form (cf. Harter, Klee, Schwinger). The triangle terms T(n,k)=T(2j,j+m) satisfy:(1/4)T(2j,j+m) =  = . Matrices for J_1^2 and J_2^2 are sparse. These diagonal equalities and the off-diagonal equalities of A268759 determine the only nonzero entries. Comments on A268759 provide a conjecture for the clear interpretation of these numbers in the context of binomial coefficients and other geometrical sequences. - Bradley Klee, Feb 20 2016
This sequence appears in the probability of the coin tossing "Gambler's Ruin". Call the probability of winning a coin toss = p, and the probability of losing the toss is 1-p = q, and call z = q/p. A gambler starts with $1, and tosses for $1 stakes till he has $0 (ruin) or has $n (wins). The average time T_win_lose(n) of a game (win OR lose) is a well-known function of z and n. The probability of the gambler winning P_win(n) is also known, and is equal to (1-z)/(1-z^n). T_win(n) defined as the average time it takes the gambler to win such a game is not so well known (I have not found it in the literature). I calculated T_win(n) and found it to be T_win(n) = P_win(n) * Sum_{m=0..n} T(n,m) * z^m. - Steve Newman, Oct 24 2016
As a square array A(n,m), gives the odd number's index of the product of n-th and m-th odd number. See formula. - Rainer Rosenthal, Sep 07 2022

			As a triangle:
  { 0},
  { 1,  1},
  { 2,  4,  2},
  { 3,  7,  7,  3},
  { 4, 10, 12, 10,  4},
  { 5, 13, 17, 17, 13,  5},
  { 6, 16, 22, 24, 22, 16,  6},
  { 7, 19, 27, 31, 31, 27, 19,  7},
  { 8, 22, 32, 38, 40, 38, 32, 22,  8},
  { 9, 25, 37, 45, 49, 49, 45, 37, 25,  9},
  {10, 28, 42, 52, 58, 60, 58, 52, 42, 28, 10}
From _Peter Munn_, Sep 28 2022: (Start)
Square array A(n,m) starts:
  0,  1,  2,  3,  4,  5,  6,  7, ...
  1,  4,  7, 10, 13, 16, 19, 22, ...
  2,  7, 12, 17, 22, 27, 32, 37, ...
  3, 10, 17, 24, 31, 38, 45, 52, ...
  4, 13, 22, 31, 40, 49, 58, 67, ...
  5, 16, 27, 38, 49, 60, 71, 82, ...
  6, 19, 32, 45, 58, 71, 84, 97, ...
...
(End)

R. N. Cahn, Semi-Simple Lie Algebras and Their Representations, Dover, NY, 2006, ISBN 0-486-44999-8, p. 139.

Michael De Vlieger, Table of n, a(n) for n = 0..10010 (rows 0 <= n <= 70, flattened)
Tomislav Došlić, On the Laplacian Szeged Spectrum of Paths, Iranian J. Math. Chem. (2020) Vol. 11, No. 1, 57-63.
W. Harter, Principles of Symmetry, Dynamics, Spectroscopy, Wiley, 1993, Ch. 5, page 345-346.
B. Klee, Quantum Angular Momentum Matrices, Wolfram Demonstrations Project, 2016.
J. Schwinger, On Angular Momentum, Cambridge: Harvard University, Nuclear Development Associates, Inc., 1952.

[0, 0] together with the row sums give A007290.

Cf. A003991, A268759, A094053, A098353, A114327.

T[n_, m_] = n + 2* m *(-m + n);
a = Table[Table[T[n, m], {m, 0, n}], {n, 0, 10}];
Flatten[a]
(* second program: *)
Flatten[ Table[2 j + 2 j^2 - 2 m^2, {j, 0, 10, 1/2}, {m, -j, j}]] (* Bradley Klee, Feb 20 2016 *)

{T(n, m) = if( m<0 || nMichael Somos, May 28 2017

T(n,m) = n + 2*m*(n-m).

Square array A(n,m) = 2*n*m + n + m, read by antidiagonals, satisfying 2*A(n,m) + 1 = (2*n+1)*(2*m+1) = A005408(n)*A005408(m) = A098353(n+1,m+1). - Rainer Rosenthal, Oct 01 2022

Edited by N. J. A. Sloane, Feb 21 2016

A268759 Triangle T(n,k) read by rows: T(n,k) = (1/4)(1 + k)(2 + k)(k - n)(1 + k - n).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 3, 3, 0, 0, 6, 9, 6, 0, 0, 10, 18, 18, 10, 0, 0, 15, 30, 36, 30, 15, 0, 0, 21, 45, 60, 60, 45, 21, 0, 0, 28, 63, 90, 100, 90, 63, 28, 0, 0, 36, 84, 126, 150, 150, 126, 84, 36, 0, 0, 45, 108, 168, 210, 225, 210, 168, 108, 45, 0, 0, 55, 135, 216, 280, 315

Offset: 0

Bradley Klee, Feb 20 2016

Off-diagonal elements of angular momentum matrices J_1^2 and J_2^2.
Construct the infinite-dimensional matrix representation of angular momentum operators (J_1,J_2,J_3) in the block-diagonal, Jordan-Schwinger form (cf. Harter, Klee, Schwinger). The triangle terms T(n,k) satisfy:(1/2)T(n,k)^(1/2) =  =  = -  = - . In the Dirac notation, we write elements m_{ij} of matrix M as =m_{ij}. Matrices for J_1^2 and J_2^2 are sparse. These equalities and the central-diagonal equalities of A141387 determine the only nonzero entries.
Notice that a(n) = T(n,k) is always a multiple of the triangular numbers, up to an offset. Conjecture: the triangle tabulating matrix elements  is determined entirely by the coefficients: binomial(n,p) (cf. A094053). Various sequences along the diagonals of matrix J_1^p lead to other numbers with geometric interpretations (Cf. A000567, A100165).

			0;
0,  0;
1,  0,  0;
3,  3,  0,  0;
6,  9,  6,  0,  0;
10, 18, 18, 10, 0,  0;
15, 30, 36, 30, 15, 0, 0;
...

W. Harter, Principles of Symmetry, Dynamics, Spectroscopy, Wiley, 1993, Ch. 5, page 345-346.
B. Klee, Quantum Angular Momentum Matrices, Wolfram Demonstrations Project, 2016.
J. Schwinger, On Angular Momentum , Cambridge: Harvard University, Nuclear Development Associates, Inc., 1952.

Cf. A114327, A094053, A141387.

Flatten[Table[(1/4) (1 + k) (2 + k) (k - n) (1 + k - n), {n, 0, 10, 1}, {k, 0, n, 1}]]

T(n,k) = (1/4)*(1 + k)*(2 + k)*(k - n)*(1 + k - n).

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

A128139 Triangle read by rows: matrix product A004736 * A128132.

Original entry on oeis.org

1, 1, 2, 1, 3, 3, 1, 4, 5, 4, 1, 5, 7, 7, 5, 1, 6, 9, 10, 9, 6, 1, 7, 11, 13, 13, 11, 7, 1, 8, 13, 16, 17, 16, 13, 8, 1, 9, 15, 19, 21, 21, 19, 15, 9, 1, 10, 17, 22, 25, 26, 25, 22, 17, 10

Offset: 0

Gary W. Adamson, Feb 16 2007

A077028 with the final term in each row omitted.
Interchanging the factors in the matrix product leads to A128140 = A128132 * A004736.
From Gary W. Adamson, Jul 01 2012: (Start)
Alternatively, antidiagonals of an array A(n,k) of sequences with arithmetic progressions as follows:
  1, 2, 3,  4,  5,  6, ...
  1, 3, 5,  7,  9, 11, ...
  1, 4, 7, 10, 13, 16, ...
  1, 5, 9, 13, 17, 21, ...
  ... (End)
From Gary W. Adamson, Jul 02 2012: (Start)
A summation generalization for Sum_{k>=1} 1/(A(n,k)*A(n,k+1)) (formulas copied from A002378, A000466, A085001, A003185):
  1 = 1/(1)*(2) + 1/(2)*(3) + 1/(3)*(4)  + ...
  1 = 2/(1)*(3) + 2/(3)*(5) + 2/(5)*(7)  + ...
  1 = 3/(1)*(4) + 3/(4)*(7) + 3/(7)*(10) + ...
  1 = 4/(1)*(5) + 4/(5)*(9) + 4/(9)*(13) + ...
  ...
As a summation of terms equating to a definite integral:
  Integral_{0..1} dx/(1+x) = ... 1 - 1/2 + 1/3 - 1/4 + ... = log(2).
  Integral_{0..1} dx/(1+x^2) = 1 - 1/3 + 1/5 - 1/7  + ... = Pi/4 (see A157142)
  Integral_{0..1} dx/(1+x^3) = 1 - 1/4 + 1/7 - 1/10 + ... (see A016777)
  Integral_{0..1} dx/(1+x^4) = 1 - 1/5 + 1/9 - 1/13 + ... (see A016813). (End)

			First few rows of the triangle:
  1;
  1,  2;
  1,  3,  3;
  1,  4,  5,  4;
  1,  5,  7,  7,  5;
  1,  6,  9, 10,  9,  6;
  1,  7, 11, 13, 13, 11,  7;
  1,  8, 13, 16, 17, 16, 13,  8;
  1,  9, 15, 19, 21, 21, 19, 15,  9;
  1, 10, 17, 22, 25, 26, 25, 22, 17, 10;
  ...

Cf. A004736, A128132, A128140, A004006 (row sums).

A004736 * A128132 as infinite lower triangular matrices.

T(n,k) = k*(1+n-k)+1 = 1 + A094053(n+1,1+n-k). - R. J. Mathar, Jul 09 2012

A127738 Triangle read by rows: the matrix product A004736 * A127701 of two triangular matrices.

Original entry on oeis.org

1, 3, 2, 5, 5, 3, 7, 8, 7, 4, 9, 11, 11, 9, 5, 11, 14, 15, 14, 11, 6, 13, 17, 19, 19, 17, 13, 7, 15, 20, 23, 24, 23, 20, 15, 8, 17, 23, 27, 29, 29, 27, 23, 17, 9, 19, 26, 31, 34, 35, 34, 31, 26, 19, 10

Offset: 1

Gary W. Adamson, Jan 27 2007

Left column = A028387: (1, 5, 11, 19, 29, 41, 55, ...).

			First few rows of the triangle:
   1;
   3,  2;
   5,  5,  3;
   7,  8,  7,  4;
   9, 11, 11,  9,  5;
  11, 14, 15, 14, 11,  6;
  13, 17, 19, 19, 17, 13,  7;
  ...

Cf. A004736, A127701, A008778 (row sums), A028387.

T(n,k) = Sum_{j=k..n} A004736(n,j)*A127701(j,k). - R. J. Mathar, Aug 31 2022

T(n,k) = k+(k+1)*(n-k) = n+k*(n-k) = n +A094053(n,k) = A059036(n,k). - R. J. Mathar, Aug 31 2022

A357503 a(n) is the hafnian of the 2n X 2n symmetric matrix whose element (i,j) equals abs(i-j).

Original entry on oeis.org

1, 1, 8, 174, 7360, 512720, 53245824, 7713320944, 1486382446592, 367691598791424, 113570289012090880

Offset: 0

Stefano Spezia, Oct 01 2022

			a(2) = M_{1,2}*M_{3,4} + M_{1,3}*M_{2,4} + M_{1,4}*M_{2,3} = 8 is the hafnian of
    0, 1, 2, 3;
    1, 0, 1, 2;
    2, 1, 0, 1;
    3, 2, 1, 0.

Wikipedia, Hafnian
Wikipedia, Symmetric matrix

Cf. A049581, A085750 (determinant of M(n)), A085807 (permanent of M(n)), A094053 (super- and subdiagonal sums of M(n) in reversed order), A144216 (row- and column sums of M(n)), A338456.

M[i_, j_, n_]:=Part[Part[Table[Abs[r-c], {r, n}, {c, n}], i], j]; a[n_]:=Sum[Product[M[Part[PermutationList[s, 2n], 2i-1], Part[PermutationList[s, 2n], 2i], 2n], {i, n}], {s, SymmetricGroup[2n]//GroupElements}]/(n!*2^n); Array[a, 6, 0]

tm(n) = matrix(n, n, i, j, abs(i-j));
a(n) = my(m = tm(2*n), s=0); forperm([1..2*n], p, s += prod(j=1, n, m[p[2*j-1], p[2*j]]); ); s/(n!*2^n); \\ Michel Marcus, May 02 2023

a(6) from Michel Marcus, May 02 2023

a(7)-a(10) from Pontus von Brömssen, Oct 15 2023

cp's OEIS Frontend

A287326 Triangle read by rows: T(n, k) = 6*k*(n-k) + 1; n >= 0, 0 <= k <= n.

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A114327 Table T(n,m) = n - m read by upwards antidiagonals.

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A300656 Triangle read by rows: T(n,k) = 30*k^2*(n-k)^2 + 1; n >= 0, 0 <= k <= n.

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A300785 Triangle read by rows: T(n,k) = 140*k^3*(n-k)^3 - 14*k*(n-k) + 1; n >= 0, 0 <= k <= n.

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A141387 Triangle read by rows: T(n,m) = n + 2*m*(n - m) (0 <= m <= n).

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A268759 Triangle T(n,k) read by rows: T(n,k) = (1/4)*(1 + k)*(2 + k)*(k - n)*(1 + k - n).

A287326 Triangle read by rows: T(n, k) = 6k(n-k) + 1; n >= 0, 0 <= k <= n.

A300656 Triangle read by rows: T(n,k) = 30k^2(n-k)^2 + 1; n >= 0, 0 <= k <= n.

A300785 Triangle read by rows: T(n,k) = 140k^3(n-k)^3 - 14k(n-k) + 1; n >= 0, 0 <= k <= n.

A141387 Triangle read by rows: T(n,m) = n + 2m(n - m) (0 <= m <= n).

A268759 Triangle T(n,k) read by rows: T(n,k) = (1/4)(1 + k)(2 + k)(k - n)(1 + k - n).