A294317 -id:A294317 - 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

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

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

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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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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.