A304042 -id:A304042 - cp's OEIS frontend

A316349 Consider coefficients U(m,L,k) defined by the identity Sum_{k=1..L} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = Sum_{k=0..m} (-1)^(m-k) * U(m,L,k) * T^k that holds for all positive integers L,m,T. This sequence gives 3-column table read by rows, where the n-th row lists coefficients U(2,n,k) for k = 0, 1, 2; n >= 1.

31, 60, 30, 512, 540, 150, 2943, 2160, 420, 10624, 6000, 900, 29375, 13500, 1650, 68256, 26460, 2730, 140287, 47040, 4200, 263168, 77760, 6120, 459999, 121500, 8550, 760000, 181500, 11550, 1199231, 261360, 15180, 1821312, 365040, 19500, 2678143, 496860, 24570, 3830624, 661500, 30450

Offset: 1

Kolosov Petro, Jun 29 2018

For L=T, the identity takes form T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k, which holds for all positive integers T and m.

			       column   column  column
   L     k=0      k=1     k=2
  --  -------  -------  ------
   1       31       60      30
   2      512      540     150
   3     2943     2160     420
   4    10624     6000     900
   5    29375    13500    1650
   6    68256    26460    2730
   7   140287    47040    4200
   8   263168    77760    6120
   9   459999   121500    8550
  10   760000   181500   11550
  11  1199231   261360   15180
  12  1821312   365040   19500
  ...

Max Alekseyev, Derivation of the general formula for U(m,n,k), MathOverflow, 2018.
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, More details on derivation of present sequence.
Petro Kolosov, Mathematica program, verifies the identity T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k for m=0,1,...,12.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.
Petro Kolosov, An efficient method of spline approximation for power function, arXiv:2503.07618 [math.GM], 2025.

The case m=1 is A320047.
The case m=3 is A316387.
Column k=0 is A316457.
Column k=1 is A316458.
Column k=2 is A316459.
Cf. A302971, A304042, A287326, A300656, A300785.

(* Define the R[n,k] := A302971(m,j)/A304042(m,j) *)
R[n_, k_] := 0
R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
   Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
   BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
(* Define the U(m,l,t) coefficients *)
U[m_, l_, t_] := (-1)^m Sum[Sum[Binomial[j, t] R[m,j] k^(2 j - t) (-1)^j, {j, t, m}], {k, 1, l}];
(* Define the value of the variable 'm', should be m = 2 for A316349 *)
m = 2;
(* Print first 10 rows of U(m,l,t) coefficients over l: 1 <= l <= 10 *)
Column[Table[U[m, l, t], {l, 1, 10}, {t, 0, m}]]

U(2,n,0) = 6*n^5 + 15*n^4 + 10*n^3; U(2,n,1) = 15*n^4 + 30*n^3 + 15*n^2; U(2,n,2) = 10*n^3 + 15*n^2 + 5*n. - Max Alekseyev, Sep 06 2018
From Colin Barker, Jul 06 2018: (Start)
G.f.: x*(31 + 60*x + 30*x^2 + 326*x^3 + 180*x^4 - 30*x^5 + 336*x^6 - 180*x^7 - 30*x^8 + 26*x^9 - 60*x^10 + 30*x^11 + x^12) / ((1 - x)^6*(1 + x + x^2)^6).
a(n) = 6*a(n-3) - 15*a(n-6) + 20*a(n-9) - 15*a(n-12) + 6*a(n-15) - a(n-18) for n>18. (End)
U(m,L,t) = (-1)^m * Sum_{k=1..L} Sum_{j=t..m} binomial(j,t) * R(m,j) * k^{2j-t} * (-1)^j, where m = 1, L >= 1 and R(m,j) = A302971(m,j)/A304042(m,j); after Max Alekseyev, see links.

Edited by Max Alekseyev, Sep 06 2018

A316387 Consider coefficients U(m,L,k) defined by the identity Sum_{k=1..L} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = Sum_{k=0..m} (-1)^(m-k) * U(m,L,k) * T^k that holds for all positive integers L,m,T. This sequence gives 4-column table read by rows, where the n-th row lists coefficients U(3,n,k) for k = 0, 1, 2, 3; n >= 1.

Original entry on oeis.org

125, 406, 420, 140, 9028, 13818, 7140, 1260, 110961, 115836, 41160, 5040, 684176, 545860, 148680, 14000, 2871325, 1858290, 411180, 31500, 9402660, 5124126, 955500, 61740, 25872833, 12182968, 1963920, 109760, 62572096, 25945416, 3684240, 181440, 136972701, 50745870, 6439860, 283500, 276971300, 92745730, 10639860, 423500

Offset: 1

Kolosov Petro, Jul 01 2018

For T=L, the identity takes form T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k, which holds for all positive integers T and m.

			            column      column      column   column
   L          k=0         k=1         k=2      k=3
  --  ------------  ----------  ----------  -------
   1           125         406         420      140
   2          9028       13818        7140     1260
   3        110961      115836       41160     5040
   4        684176      545860      148680    14000
   5       2871325     1858290      411180    31500
   6       9402660     5124126      955500    61740
   7      25872833    12182968     1963920   109760
   8      62572096    25945416     3684240   181440
   9     136972701    50745870     6439860   283500
  10     276971300    92745730    10639860   423500
  11     524988145   160386996    16789080   609840
  12     943023888   264896268    25498200   851760
  13    1618774781   420839146    37493820  1159340
  14    2672907076   646725030    53628540  1543500
  15    4267591425   965662320    74891040  2016000
  16    6616398080  1406064016   102416160  2589440
  17    9995653693  2002403718   137494980  3277260
  18   14757360516  2796022026   181584900  4093740
  19   21343778801  3835983340   236319720  5054000
  20   30303773200  5179983060   303519720  6174000
  ...

Max Alekseyev, Derivation of the general formula for U(m,n,k), MathOverflow, 2018.
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, More details on derivation of present sequence.
Petro Kolosov, Mathematica program, verifies the identity T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k for m=0,1,...,12.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.
Petro Kolosov, An efficient method of spline approximation for power function, arXiv:2503.07618 [math.GM], 2025.

The case m=1 is A320047.
The case m=2 is A316349.
Column k=0 is A317981.
Column k=1 is A317982.
Column k=2 is A317983.
Column k=3 is A317984.
Cf. A287326, A300656, A300785, A302971, A304042.

(* Define the R[n,k] := A302971(m,j)/A304042(m,j) *)
R[n_, k_] := 0
R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
   Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
   BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
(* Define the U(m,l,t) coefficients *)
U[m_, l_, t_] := (-1)^m Sum[Sum[Binomial[j, t] R[m,j] k^(2 j - t) (-1)^j, {j, t, m}], {k, 1, l}];
(* Define the value of the variable 'm' *)
m = 3;
(* Print first 10 rows of U(m,l,t) coefficients for 'm' defined above *)
Column[Table[U[m, l, t], {l, 1, 10}, {t, 0, m}]]

U(3,n,0) = 20*n^7 + 70*n^6 + 70*n^5 - 28*n^3 - 7*n^2; U(3,n,1) = 70*n^6 + 210*n^5 + 175*n^4 - 42*n^2 - 7*n; U(3,n,2) = 84*n^5 + 210*n^4 + 140*n^3 - 14*n; U(3,n,3) = 35*n^4 + 70*n^3 + 35*n^2. - Max Alekseyev, Sep 06 2018
From Colin Barker, Jul 09 2018; corrected by Max Alekseyev, Sep 06 2018: (Start)
G.f.: x*(125 + 406*x + 420*x^2 + 140*x^3 + 8028*x^4 + 10570*x^5 + 3780*x^6 + 140*x^7 + 42237*x^8 + 16660*x^9 - 4200*x^10 - 1120*x^11 + 42272*x^12 - 16660*x^13 - 4200*x^14 + 1120*x^15 + 8007*x^16 - 10570*x^17 + 3780*x^18 - 140*x^19 + 132*x^20 - 406*x^21 + 420*x^22 - 140*x^23 - x^24) / ((1 - x)^8*(1 + x)^8*(1 + x^2)^8).
a(n) = 8*a(n-4) - 28*a(n-8) + 56*a(n-12) - 70*a(n-16) + 56*a(n-20) - 28*a(n-24) + 8*a(n-28) - a(n-32) for n>32. (End)
U(m,L,t) = (-1)^m * Sum_{k=1..L} Sum_{j=t..m} binomial(j,t) * R(m,j) * k^{2j-t} * (-1)^j, where m = 1, L >= 1 and R(m,j) = A302971(m,j)/A304042(m,j); after Max Alekseyev, see links. - Kolosov Petro, Oct 04 2018

Edited by Max Alekseyev, Sep 06 2018

A320047 Consider coefficients U(m,l,k) defined by the identity Sum_{k=1..l} Sum_{j=0..m} A302971(m,j)/A304042(m,j) * k^j * (T-k)^j = Sum_{k=0..m} (-1)^(m-k) * U(m,l,k) * T^k that holds for all positive integers l,m,T. This sequence gives 2-column table read by rows, where n-th row lists coefficients U(1,n,k) for k = 0, 1 and n >= 1.

Original entry on oeis.org

5, 6, 28, 18, 81, 36, 176, 60, 325, 90, 540, 126, 833, 168, 1216, 216, 1701, 270, 2300, 330, 3025, 396, 3888, 468, 4901, 546, 6076, 630, 7425, 720, 8960, 816, 10693, 918, 12636, 1026, 14801, 1140, 17200, 1260, 19845, 1386, 22748, 1518, 25921, 1656

Offset: 1

Kolosov Petro, Oct 04 2018

For l=T, the identity takes the form T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k, which holds for all positive integers T and m.

			        column  column
   l      k=0     k=1
  ---   ------  ------
   1       5       6
   2      28      18
   3      81      36
   4     176      60
   5     325      90
   6     540     126
   7     833     168
   8    1216     216
   9    1701     270
  10    2300     330
  11    3025     396
  12    3888     468
  ...

Max Alekseyev, Derivation of the general formula for U(m,n,k), MathOverflow, 2018.
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, More details on derivation of present sequence.
Petro Kolosov, Mathematica program, verifies the identity T^(2m+1) = Sum_{k=0..m} (-1)^(m-k)*U(m,T,k)*T^k for m=0,1,...,12.
Petro Kolosov, History and overview of the polynomial P_b^m(x), 2024.
Petro Kolosov, An efficient method of spline approximation for power function, arXiv:2503.07618 [math.GM], 2025.

The case m=2 is A316349.
The case m=3 is A316387.
Column k=0 is A275709.
Column k=1 is A028896.
Cf. A302971, A304042, A287326, A300656, A300785.

(* Define the R[n,k] := A302971(n,k)/A304042(n,k) *)
R[n_, k_] := 0
R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
   Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
   BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
(* Define the U(m,l,t) coefficients *)
U[m_, l_, t_] := (-1)^m Sum[Sum[Binomial[j, t] R[m,j] k^(2 j - t) (-1)^j, {j, t, m}], {k, 1, l}];
(* Define the value of the variable 'm' to be m = 1 for A320047 *)
m = 1;
(* Print first 10 rows of U(m,l,t) coefficients for 'm' defined above *)
Column[Table[U[m, l, t], {l, 1, 10}, {t, 0, m}]]

U(m,l,t) = (-1)^m * Sum_{k=1..l} Sum_{j=t..m} binomial(j,t) * R(m,j) * k^{2j-t} * (-1)^j, where m = 1, l >= 1 and R(m,j) = A302971(m,j)/A304042(m,j); after Max Alekseyev, see links.
Conjectures from Colin Barker, Aug 03 2019: (Start)
G.f.: x*(5 + 6*x + 8*x^2 - 6*x^3 - x^4) / ((1 - x)^4*(1 + x)^4).
a(n) = (4 - 4*(-1)^n - 3*(-5+(-1)^n)*n - 3*(-3+(-1)^n)*n^2 + (1+(-1)^(1+n))*n^3) / 8.
a(n) = 4*a(n-2) - 6*a(n-4) + 4*a(n-6) - a(n-8) for n>8.
(End)

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

Original entry on oeis.org

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

A302971 Triangle read by rows: T(n,k) is the numerator of R(n,k) defined implicitly by the identity Sum_{i=0..l-1} Sum_{j=0..m} R(m,j)(l-i)^ji^j = l^(2*m+1) holding for all l,m >= 0.

Original entry on oeis.org

1, 1, 6, 1, 0, 30, 1, -14, 0, 140, 1, -120, 0, 0, 630, 1, -1386, 660, 0, 0, 2772, 1, -21840, 18018, 0, 0, 0, 12012, 1, -450054, 491400, -60060, 0, 0, 0, 51480, 1, -11880960, 15506040, -3712800, 0, 0, 0, 0, 218790, 1, -394788954, 581981400, -196409840, 8817900, 0, 0, 0, 0, 923780, 1, -16172552880, 26003271294, -10863652800, 1031151660, 0, 0, 0, 0, 0, 3879876

Offset: 0

Kolosov Petro, Apr 16 2018

			Triangle begins:
------------------------------------------------------------------------
k=   0          1         2         3    4     5      6      7       8
------------------------------------------------------------------------
n=0: 1;
n=1: 1,         6;
n=2: 1,         0,       30;
n=3: 1,       -14,        0,      140;
n=4: 1,      -120,        0,        0, 630;
n=5: 1,     -1386,      660,        0,   0, 2772;
n=6: 1,    -21840,    18018,        0,   0,    0, 12012;
n=7: 1,   -450054,   491400,   -60060,   0,    0,     0, 51480;
n=8: 1, -11880960, 15506040, -3712800,   0,    0,     0,     0, 218790;

P.-Y. Huang, S.-C. Liu, and Y.-N. Yeh, Congruences of Finite Summations of the Coefficients in certain Generating Functions, The Electronic Journal of Combinatorics, 21 (2014), #P2.45.
C. Jordan, Calculus of Finite Differences, Röttig and Romwalter, Budapest, 1939. [Annotated scans of pages 448-450 only]
Petro Kolosov, On the link between binomial theorem and discrete convolution, arXiv:1603.02468 [math.NT], 2016-2025.
Petro Kolosov, Definition and table of values.
Petro Kolosov, An unusual identity for odd-powers, arXiv:2101.00227 [math.GM], 2021.
Petro Kolosov, 106.37 An unusual identity for odd-powers, The Mathematical Gazette, 2022.
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.
Petro Kolosov, A novel proof of power rule in calculus, GitHub, 2024. See p. 5.
Petro Kolosov, Odd-power identity via multiplication of certain matrices, GitHub, 2024. See p. 4.
Petro Kolosov, An efficient method of spline approximation for power function, arXiv:2503.07618 [math.GM], 2025.
Petro Kolosov, Discussion on coefficients of odd polynomial identity, GitHub, 2025.
MathOverflow, Discussion of these coefficients, 2018.

Items of second row are the coefficients in the definition of A287326.
Items of third row are the coefficients in the definition of A300656.
Items of fourth row are the coefficients in the definition of A300785.
T(n,n) gives A002457(n).
Denominators of R(n,k) are shown in A304042.
Row sums return A000079(2n+1) - 1.
Cf. A007318, A027641, A027642, A055012, A077028, A000146, A002882, A003245, A127187, A127188, A074909, A164555.

R := proc(n, k) if k < 0 or k > n then return 0 fi; (2*k+1)*binomial(2*k, k);
if n = k then % else -%*add((-1)^j*R(n, j)*binomial(j, 2*k+1)*
bernoulli(2*j-2*k)/(j-k), j=2*k+1..n) fi end: T := (n, k) -> numer(R(n, k)):
seq(print(seq(T(n, k), k=0..n)), n=0..12);
# Numerical check that S(m, n) = n^(2*m+1):
S := (m, n) -> add(add(R(m, j)*(n-k)^j*k^j, j=0..m), k=0..n-1):
seq(seq(S(m, n) - n^(2*m+1), n=0..12), m=0..12); # Peter Luschny, Apr 30 2018

R[n_, k_] := 0
R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
   Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
   BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n;
T[n_, k_] := Numerator[R[n, k]];
(* Print Fifteen Initial rows of Triangle A302971 *)
Column[ Table[ T[n, k], {n, 0, 15}, {k, 0, n}], Center]

T(n, k) = if ((n>k) || (n<0), 0, if (k==n, (2*n+1)*binomial(2*n, n), if (2*n+1>k, 0, if (n==0, 1, (2*n+1)*binomial(2*n, n)*sum(j=2*n+1, k+1, T(j, k)*binomial(j, 2*n+1)*(-1)^(j-1)/(j-n)*bernfrac(2*j-2*n))))));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(numerator(T(k,n)), ", ")); print); \\ Michel Marcus, Apr 27 2018

Recurrence given by Max Alekseyev (see the MathOverflow link):
R(n, k) = 0 if k < 0 or k > n.
R(n, k) = (2k+1)*binomial(2k, k) if k = n.
R(n, k) = (2k+1)*binomial(2k, k)*Sum_{j=2k+1..n} R(n, j)*binomial(j, 2k+1)*(-1)^(j-1)/(j-k)*Bernoulli(2j-2k), otherwise.
T(n, k) = numerator(R(n, k)).

A306548 Triangle T(n,k) read by rows, where the k-th column is the shifted self-convolution of the power function n^k, n >= 0, 0 <= k <= n.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 1, 0, 0, 3, 4, 1, 0, 0, 4, 10, 8, 1, 0, 0, 5, 20, 34, 16, 1, 0, 0, 6, 35, 104, 118, 32, 1, 0, 0, 7, 56, 259, 560, 418, 64, 1, 0, 0, 8, 84, 560, 2003, 3104, 1510, 128, 1, 0, 0, 9, 120, 1092, 5888, 16003, 17600, 5554, 256, 1, 0, 0, 10, 165, 1968, 14988, 64064, 130835, 101504, 20758, 512, 1, 0, 0

Offset: 0

Kolosov Petro, Feb 23 2019

For n > 0 an odd-power identity n^(2m+1)+1, m >= 0 can be found using the current sequence. The sum of the n-th diagonal of T(n,k) over 0 <= k <= m multiplied by A(m,k) gives n^(2m+1)-1, where A(m,k) = A302971(m,k)/A304042(m,k). For example, consider the case n=4, m=2: the n-th diagonal of T(n, 0 <= k <= m) is {5, 10, 34}, and the m-th row of triangle A(m, 0 <= k <= m) is {1, 0, 30}, thus (3+1)^5 + 1 = 5*1 + 10*0 + 34*30 = 1025.

			==================================================================
k=    0     1     2     3      4      5     6    7    8    9    10
==================================================================
n=0:  2;
n=1:  2,    0;
n=2:  3,    0,    0;
n=3:  4,    1,    0,    0;
n=4:  5,    4,    1,    0,     0;
n=5:  6,   10,    8,    1,     0,     0;
n=6:  7,   20,   34,   16,     1,     0,    0;
n=7:  8,   35,  104,  118,    32,     1,    0,   0;
n=8:  9,   56,  259,  560,   418,    64,    1,   0,   0;
n=9:  10,  84,  560, 2003,  3104,  1510,  128,   1,   0,   0;
n=10: 11, 120, 1092, 5888, 16003, 17600, 5554, 256,   1,   0;   0;
...

D. V. Widder et al., The Convolution Transform, Bull. Amer. Math. Soc. 60 (1954), 444-456.
Wikipedia, Convolution.
Wikipedia, Convolution power.

Nonzero terms of columns k=0..5 give: A000027, A000292, A033455, A145216, A145217, A145218.
Partial sums of columns k=1..2 give: A000332, A259181.
Cf. A302971, A304042, A198633, A000079.

f[m_, s_] := Piecewise[{{s^m, s >= 0}, {0, True}}];
F[n_, m_] := Sum[f[m, n - k]*f[m, k], {k, -Infinity, +Infinity}];
T[n_, k_] := F[n - k, k];
Column[Table[T[n, k], {n, 0, 12}, {k, 0, n}], Left]

f(m, s) = s^m, if s >= 0;
f(m, s) = 0, otherwise.
F(n,m) = Sum_{k} f(m, n-k) * f(m, k), -oo < k < +oo;
T(n,k) = F(n-k, k).

Edited by Kolosov Petro, Mar 13 2019

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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.

A302971 Triangle read by rows: T(n,k) is the numerator of R(n,k) defined implicitly by the identity Sum_{i=0..l-1} Sum_{j=0..m} R(m,j)(l-i)^ji^j = l^(2*m+1) holding for all l,m >= 0.