A001296 -id:A001296 - cp's OEIS frontend

A127778 Triangle T(n,k) = A002411(k) read by rows.

1, 1, 6, 1, 6, 18, 1, 6, 18, 40, 1, 6, 18, 40, 75, 1, 6, 18, 40, 75, 126, 1, 6, 18, 40, 75, 126, 196, 1, 6, 18, 40, 75, 126, 196, 288

Offset: 1

Gary W. Adamson, Jan 28 2007

			First few rows of the triangle are:
1;
1, 6;
1, 6, 18;
1, 6, 18, 40;
1, 6, 18, 40, 75;
...

Cf. A002411, A001296 (row sums), A127773, A002260, A127777.

Equals the matrix product A002260 * A127773 as infinite lower triangular matrices.

A151510 The triangle in A151338 read by rows upwards.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 7, 1, 0, 1, 10, 25, 15, 1, 0, 1, 15, 65, 90, 31, 0, 0, 1, 21, 140, 350, 301, 56, 0, 0, 1, 28, 266, 1050, 1701, 938, 91, 0, 0, 1, 36, 462, 2646, 6951, 7686, 2737, 126, 0, 0, 1, 45, 750, 5880, 22827, 42315, 32725, 7455, 126, 0, 0

Offset: 0

N. J. A. Sloane, May 14 2009

Columns 1-5 are A000012, A000217, A001296, A001297, and A001298. - Nathaniel Johnston, Apr 30 2011

			Triangle begins:
1
1 0
1 1  0
1 3  1   0
1 6  7   1    0
1 10 25  15   1    0
1 15 65  90   31   0   0
1 21 140 350  301  56  0  0
1 28 266 1050 1701 938 91 0 0
...

Extended by Nathaniel Johnston, Apr 30 2011

A259455 n Sum_n Sum_n Sum_n.

Original entry on oeis.org

1, 30, 270, 1400, 5250, 15876, 41160, 95040, 200475, 393250, 726726, 1277640, 2153060, 3498600, 5508000, 8434176, 12601845, 18421830, 26407150, 37191000, 51546726, 70409900, 94902600, 126360000, 166359375, 216751626, 279695430, 357694120, 453635400, 570834000

Offset: 1

N. J. A. Sloane, Jun 30 2015

See the reference for an explanation of the rather cryptic definition.

Alois P. Heinz, Table of n, a(n) for n = 1..1000
C. Krishnamachari, The operator (xD)^n, J. Indian Math. Soc., 15 (1923),3-4. [Annotated scanned copy]

This is the seventh sequence in the sequence A000027, A000217, A002411, A001296, A108650, A001297, ...

a:= n-> n^3*(n+3)*(n+2)*(n+1)^2/48:
seq(a(n), n=1..40);  # Alois P. Heinz, Jul 04 2015

From Alois P. Heinz, Jul 04 2015: (Start)
G.f.: (24*x^3+58*x^2+22*x+1)*x/(x-1)^8.
a(n) = n^3*(n+3)*(n+2)*(n+1)^2/48.
a(n) = n*Stirling2(n+3,n). (End)

More terms from Alois P. Heinz, Jul 04 2015

A351766 a(n) = Sum_{j=1..n} Sum_{i=1..j} (i*j)^4.

Original entry on oeis.org

0, 1, 273, 8211, 98835, 710710, 3659110, 14886186, 50816298, 151416111, 404746111, 990005445, 2248888005, 4798557036, 9703780828, 18730825828, 34711648356, 62053834605, 107439683325, 180766879111, 296393439111, 474761104818, 744484165986, 1145004918190, 1729932641710, 2571200219835

Offset: 0

Roudy El Haddad, Feb 18 2022

a(n) is the sum of all products of two elements from the set {1^4, ..., n^4}.

Roudy El Haddad, Recurrent Sums and Partition Identities, arXiv:2101.09089 [math.NT], 2021.
Roudy El Haddad, A generalization of multiple zeta value. Part 1: Recurrent sums. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 167-199, DOI: 10.7546/nntdm.2022.28.2.167-199.
Index entries for linear recurrences with constant coefficients, signature (11,-55,165,-330,462,-462,330,-165,55,-11,1).

Cf. A000217 (for power 0), A001296 (for power 1), A060493 (for squares), A346642 (for cubes).

Cf. A000583 (fourth powers), A000538 (sum of fourth powers).

{a(n) = n*(n+1)*(n+2)*(2*n+1)*(2*n+3)*(9*n^5+25*n^4-5*n^3-25*n^2+21*n-5)/1800};

a(n) = sum(j=1, n, sum(i=1, j, i^4*j^4));

a(n) = n*(n + 1)*(n + 2)*(2*n + 1)*(2*n + 3)*(9*n^5 + 25*n^4 - 5*n^3 - 25*n^2 + 21*n - 5)/1800.
a(n) = binomial(2*n+4,5) * (9*n^5 + 25*n^4 - 5*n^3 - 25*n^2 + 21*n - 5)/5!.
G.f.: x*(16*x^7 + 1217*x^6 + 12038*x^5 + 30415*x^4 + 23364*x^3 + 5263*x^2 + 262*x + 1)/(1 - x)^11. - Alois P. Heinz, Feb 18 2022

A382225 Triangle read by rows: T(n,k) = Sum_{i=k..n} C(i-1,i-k)*C(i,k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 7, 1, 1, 10, 25, 13, 1, 1, 15, 65, 73, 21, 1, 1, 21, 140, 273, 171, 31, 1, 1, 28, 266, 798, 871, 346, 43, 1, 1, 36, 462, 1974, 3321, 2306, 631, 57, 1, 1, 45, 750, 4326, 10377, 11126, 5335, 1065, 73, 1, 1, 55, 1155, 8646, 28017, 42878, 31795, 11145, 1693, 91, 1

Offset: 0

Vladimir Kruchinin, Mar 19 2025

Triangle T(n,k) of minors of the main diagonal of Pascal's matrix, n -size matrix, k - number of element of diagonal.

			Triangle starts:
  1;
  1,  1;
  1,  3,   1;
  1,  6,   7,   1;
  1, 10,  25,  13,   1;
  1, 15,  65,  73,  21,  1;
  1, 21, 140, 273, 171, 31, 1;
  ...

Paolo Xausa, Table of n, a(n) for n = 0..11475 (rows 0..150 of triangle, flattened).
Wikipedia, Minor (linear algebra)

Columns k=0-3 give: A000012, A000217, A001296(n-1) for n>=1, A107963(n-3) for n>=3.
Row sums give A024718.
T(n+1,n) gives A002061(n+1).
Cf. A007318, A184173.

T:= proc(n, k) option remember; `if`(n<0, 0,
      T(n-1, k)+binomial(n-1, k-1)*binomial(n, k))
    end:
seq(seq(T(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Mar 20 2025

A382225[n_, k_] := A382225[n, k] = If[k == n, 1, A382225[n-1, k] + Binomial[n-1, k-1]*Binomial[n, k]];
Table[A382225[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Mar 22 2025 *)

h[i,j]:=binomial(i+j-3,i-1);
for n:1 thru 7 do
    if n=1 then print([1])
    else (M:genmatrix(h,n,n),
          print(makelist(determinant(minor(M,k,k)),k,1,n))
         );

G.f.: 1/(1-x) * ((1-x*(1-y))/(2*(sqrt((1-x*(1+y))^2-4*x^2*y)))+1/2).

T(n,k) = T(n-1,k)+C(n-1,k-1)*C(n,k).

A273528 Triangle T(n,m) (n >= 1, 0 <= m < n) giving coefficients of (n-1)! P_n, where P_n is the polynomial formula for row n of A213086.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 2, 0, 2, 9, 10, 3, 0, 2, 25, 50, 35, 8, 0, -12, 86, 270, 260, 102, 14, 0, -120, 140, 1344, 2030, 1260, 350, 36, 0, -1248, -1016, 7336, 15862, 13048, 5236, 1024, 78, 0, -9216, -22464, 28528, 124488, 139776, 76104, 22152, 3312, 200, 0, -90720, -322344, 1860, 1036990, 1514205, 1018563, 379890, 80760, 9165, 431

Offset: 1

Jean-François Alcover, May 24 2016

			Row T(5) = {0, 2, 9, 10, 3}, so P_5(k) = (1/4!)(2k + 9k^2 + 10k^3 + 3k^4), which gives 1, 7, 25, 65, 140, 266, ..., that is A001296 (row 5 of A213086), for k >=1.
Triangle begins:
{1},
{0, 1},
{0, 1, 1},
{0, 1, 3, 2},
{0, 2, 9, 10, 3},
{0, 2, 25, 50, 35, 8},
{0, -12, 86, 270, 260, 102, 14},
...

Cf. A213086, A213074.

The first formulas (stripped of factorials) :
1,
k,
k + k^2,
k + 3 k^2 + 2 k^3,
2 k + 9 k^2 + 10 k^3 + 3 k^4,
2 k + 25 k^2 + 50 k^3 + 35 k^4 + 8 k^5,
-12 k + 86 k^2 + 270 k^3 + 260 k^4 + 102 k^5 + 14 k^6,
-120 k + 140 k^2 + 1344 k^3 + 2030 k^4 + 1260 k^5 + 350 k^6 + 36 k^7,
...

A301972 a(n) = n(n^2 - 2n + 4)binomial(2n,n)/((n + 1)*(n + 2)).

Original entry on oeis.org

0, 1, 4, 21, 112, 570, 2772, 13013, 59488, 266526, 1175720, 5123426, 22108704, 94645460, 402503220, 1702300725, 7165821120, 30043474230, 125523450360, 522857438070, 2172127120800, 9002522512620, 37233403401480, 153704429299746, 633442159732032, 2606543487445100, 10710790748646352, 43957192722175908

Offset: 0

Ilya Gutkovskiy, Mar 29 2018

nonn

For n > 2, a(n) is the n-th term of the main diagonal of iterated partial sums array of n-gonal numbers (in other words, a(n) is the n-th (n+2)-dimensional n-gonal number, see also example).

			For n = 5 we have:
----------------------------
0   1    2    3     4    [5]
----------------------------
0,  1,   5,  12,   22,   35,  ... A000326 (pentagonal numbers)
0,  1,   6,  18,   40,   75,  ... A002411 (pentagonal pyramidal numbers)
0,  1,   7,  25,   65,  140,  ... A001296 (4-dimensional pyramidal numbers)
0,  1,   8,  33,   98,  238,  ... A051836 (partial sums of A001296)
0,  1,   9,  42,  140,  378,  ... A051923 (partial sums of A051836)
0,  1,  10,  52,  192, [570], ... A050494 (partial sums of A051923)
----------------------------
therefore a(5) = 570.

Index to sequences related to polygonal numbers

Cf. A000984, A002457, A006484, A057145, A060354, A080851, A080852, A100119, A180266, A275490, A292551.

Table[n (n^2 - 2 n + 4) Binomial[2 n, n]/((n + 1) (n + 2)), {n, 0, 27}]
nmax = 27; CoefficientList[Series[(-4 + 31 x - 66 x^2 + 28 x^3 + (4 - 7 x) (1 - 4 x)^(3/2))/(2 x^2 (1 - 4 x)^(3/2)), {x, 0, nmax}], x]
nmax = 27; CoefficientList[Series[Exp[2 x] (4 - x + 2 x^2) BesselI[1, 2 x]/x - 2 Exp[2 x] (2 - x) BesselI[0, 2 x], {x, 0, nmax}], x] Range[0, nmax]!
Table[SeriesCoefficient[x (1 - 3 x + n x)/(1 - x)^(n + 3), {x, 0, n}], {n, 0, 27}]

O.g.f.: (-4 + 31*x - 66*x^2 + 28*x^3 + (4 - 7*x)*(1 - 4*x)^(3/2))/(2*x^2*(1 - 4*x)^(3/2)).
E.g.f.: exp(2*x)*(4 - x + 2*x^2)*BesselI(1,2*x)/x - 2*exp(2*x)*(2 - x)*BesselI(0,2*x).
a(n) = [x^n] x*(1 - 3*x + n*x)/(1 - x)^(n+3).
a(n) ~ 4^n*sqrt(n)/sqrt(Pi).
D-finite with recurrence: -(n+2)*(961*n-3215)*a(n) +4*(2081*n^2-4414*n-4668)*a(n-1) -28*(320*n-389)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Jan 27 2020

A301973 a(n) = (n^2 - 3n + 6)binomial(n+2,3)/4.

Original entry on oeis.org

0, 1, 4, 15, 50, 140, 336, 714, 1380, 2475, 4180, 6721, 10374, 15470, 22400, 31620, 43656, 59109, 78660, 103075, 133210, 170016, 214544, 267950, 331500, 406575, 494676, 597429, 716590, 854050, 1011840, 1192136, 1397264, 1629705, 1892100, 2187255, 2518146, 2887924, 3299920, 3757650, 4264820

Offset: 0

Ilya Gutkovskiy, Mar 29 2018

For n > 2, a(n) is the n-th term of the partial sums of n-gonal pyramidal numbers (in other words, a(n) is the n-th 4-dimensional n-gonal number).

Index to sequences related to pyramidal numbers
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000292, A000332, A001296, A002415, A006484, A060354, A080852, A256859, A291288 (partial sums), A292551.

Table[(n^2 - 3 n + 6) Binomial[n + 2, 3]/4, {n, 0, 40}]
nmax = 40; CoefficientList[Series[x (1 - 2 x + 6 x^2)/(1 - x)^6, {x, 0, nmax}], x]
nmax = 40; CoefficientList[Series[Exp[x] x (24 + 24 x + 24 x^2 + 10 x^3 + x^4)/24, {x, 0, nmax}], x] Range[0, nmax]!
Table[SeriesCoefficient[x (1 - 3 x + n x)/(1 - x)^5, {x, 0, n}], {n, 0, 40}]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 4, 15, 50, 140}, 41]

O.g.f.: x*(1 - 2*x + 6*x^2)/(1 - x)^6.
E.g.f.: exp(x)*x*(24 + 24*x + 24*x^2 + 10*x^3 + x^4)/24.
a(n) = [x^n] x*(1 - 3*x + n*x)/(1 - x)^5.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).

A351770 a(n) = Sum_{j=1..n} Sum_{i=1..j} (i*j)^5.

Original entry on oeis.org

0, 1, 1057, 68125, 1399325, 15227450, 110102426, 597639882, 2621915850, 9756511275, 31839011275, 93340522951, 250280856007, 622316813300, 1450471654100, 3196426654100, 6706824221076, 13476181309557, 26055415288725, 48670370285425, 88136930285425, 155187254126926

Offset: 0

Roudy El Haddad, Feb 18 2022

a(n) is the sum of all products of two elements from the set {1^5, ..., n^5}.

Roudy El Haddad, Recurrent Sums and Partition Identities, arXiv:2101.09089 [math.NT], 2021.
Roudy El Haddad, A generalization of multiple zeta value. Part 1: Recurrent sums. Notes on Number Theory and Discrete Mathematics, 28(2), 2022, 167-199, DOI: 10.7546/nntdm.2022.28.2.167-199.
Index entries for linear recurrences with constant coefficients, signature (13,-78,286,-715,1287,-1716,1716,-1287,715,-286,78,-13,1).

Cf. A001296 (for power 1), A060493 (for squares), A346642 (for cubes), A351766 (for fourth powers).

Cf. A000584 (fifth powers), A000539 (sum of fifth powers).

seq(n*(n+1)*(n+2)*(44*n^9+276*n^8+492*n^7-48*n^6-609*n^5+207*n^4+487*n^3-291*n^2-90*n+60)/3168,
n=0..30);# Robert Israel, Feb 18 2022

{a(n) = n*(n+1)*(n+2)*(44*n^9+276*n^8+492*n^7-48*n^6-609*n^5+207*n^4+487*n^3-291*n^2-90*n+60)/3168};

a(n) = sum(j=1, n, sum(i=1, j, i^5*j^5));

a(n) = n*(n+1)*(n+2)*(44*n^9 + 276*n^8 + 492*n^7 - 48*n^6 - 609*n^5 + 207*n^4 + 487*n^3 - 291*n^2 - 90*n + 60)/3168.

G.f.: x*(1 + 1044*x + 54462*x^2 + 595860*x^3 + 2048388*x^4 + 2563644*x^5 + 1193226*x^6 + 188508*x^7 + 7635*x^8 + 32*x^9)/(1-x)^13. - Robert Israel, Feb 18 2022

cp's OEIS Frontend

A127778 Triangle T(n,k) = A002411(k) read by rows.

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A151510 The triangle in A151338 read by rows upwards.

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A259455 n Sum_n Sum_n Sum_n.

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A351766 a(n) = Sum_{j=1..n} Sum_{i=1..j} (i*j)^4.

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A382225 Triangle read by rows: T(n,k) = Sum_{i=k..n} C(i-1,i-k)*C(i,k).

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Maxima

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A273528 Triangle T(n,m) (n >= 1, 0 <= m < n) giving coefficients of (n-1)! P_n, where P_n is the polynomial formula for row n of A213086.

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A301972 a(n) = n*(n^2 - 2*n + 4)*binomial(2*n,n)/((n + 1)*(n + 2)).

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Mathematica

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A301973 a(n) = (n^2 - 3*n + 6)*binomial(n+2,3)/4.

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A351770 a(n) = Sum_{j=1..n} Sum_{i=1..j} (i*j)^5.

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A301972 a(n) = n(n^2 - 2n + 4)binomial(2n,n)/((n + 1)*(n + 2)).

A301973 a(n) = (n^2 - 3n + 6)binomial(n+2,3)/4.