A001296 -id:A001296 - cp's OEIS frontend

A108650 a(n) = (n+1)^2(n+2)(n+3)(3n+4)/24.

1, 14, 75, 260, 700, 1596, 3234, 6000, 10395, 17050, 26741, 40404, 59150, 84280, 117300, 159936, 214149, 282150, 366415, 469700, 595056, 745844, 925750, 1138800, 1389375, 1682226, 2022489, 2415700, 2867810, 3385200, 3974696, 4643584

Offset: 0

Emeric Deutsch, Jun 13 2005

Kekulé numbers for certain benzenoids.

Bo Gyu Jeong, Table of n, a(n) for n = 0..5000
Somaya Barati, Beáta Bényi, Abbas Jafarzadeh, and Daniel Yaqubi, Mixed restricted Stirling numbers, arXiv:1812.02955 [math.CO], 2018.
S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 230, no. 26).
C. Krishnamachari, The operator (xD)^n, J. Indian Math. Soc., 15 (1923),3-4. [Annotated scanned copy]
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A108645, A108646, A108647, A108648, A108649.

Cf. A001296, A001477.

[(n+1)*StirlingSecond(n+3,n+1): n in [0..40]]; // G. C. Greubel, Oct 19 2023

a:= n-> (n+1)^2*(n+2)*(n+3)*(3*n+4)/24: seq(a(n),n=0..36);
seq((n+1)*stirling2(n+3,n+1), n=0..32); # Zerinvary Lajos, Jan 20 2007

Table[((n+1)^2 (n+2)(n+3)(3n+4))/24,{n,0,40}] (* or *) Table[n StirlingS2[n+2,n],{n,40}] (* Harvey P. Dale, Dec 01 2013 *)

Vec((1 + 8*x + 6*x^2) / (1 - x)^6 + O(x^30)) \\ Colin Barker, Apr 22 2020

[(n+1)*stirling_number2(n+3,n+1) for n in range(41)] # G. C. Greubel, Oct 19 2023

From Zerinvary Lajos, Jan 20 2007: (Start)
a(n) = A001477(n+1)*A001296(n+1) = (n+1)*A001296(n+1).
a(n) = (n+1)*Stirling2(n+3,n+1). (End)
From Colin Barker, Apr 22 2020: (Start)
G.f.: (1 + 8*x + 6*x^2) / (1 - x)^6.
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) for n>5.
(End)
From Amiram Eldar, May 29 2022: (Start)
Sum_{n>=0} 1/a(n) 2*Pi^2 + 54*sqrt(3)*Pi/5 + 486*log(3)/5 - 921/5.
Sum_{n>=0} (-1)^n/a(n) = Pi^2 - 108*sqrt(3)*Pi/5 - 528*log(2)/5 + 909/5. (End)
E.g.f.: (1/24)*(24 +312*x +576*x^2 +304*x^3 +55*x^4 +3*x^5)*exp(x). - G. C. Greubel, Oct 19 2023

A153978 a(n) = n(n-1)(n+1)(3n-2)/12.

Original entry on oeis.org

0, 2, 14, 50, 130, 280, 532, 924, 1500, 2310, 3410, 4862, 6734, 9100, 12040, 15640, 19992, 25194, 31350, 38570, 46970, 56672, 67804, 80500, 94900, 111150, 129402, 149814, 172550, 197780, 225680, 256432, 290224, 327250, 367710, 411810, 459762

Offset: 1

Vladimir Joseph Stephan Orlovsky, Jan 03 2009

Partial sums of A011379.

Antidiagonal sums of the convolution array A213819. - Clark Kimberling, Jul 04 2012

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A003215, A000537, A000578, A005898, A027602, A006007, A153976, A153977, A011379, A052149, A213819.

With[{r=Range[0,50]},Accumulate[r^2+r^3]] (* Harvey P. Dale, Jan 16 2011 *)
Rest[CoefficientList[Series[-2 x^2 * (2 x + 1)/(x - 1)^5, {x, 0, 40}], x]] (* Vincenzo Librandi, Jun 30 2014 *)
LinearRecurrence[{5,-10,10,-5,1}, {0,2,14,50,130}, 25] (* G. C. Greubel, Sep 01 2016 *)

concat(0, Vec(-2*x^2*(2*x+1)/(x-1)^5 + O(x^100))) \\ Colin Barker, Jun 28 2014

a(n) = n*(n-1)*(n+1)*(3*n-2)/12 \\ Charles R Greathouse IV, Sep 01 2016

a(n) = 2 * A001296(n-1) = (n-1)*n*(n+1)*(3*n-2)/12 (n>0). - Bruno Berselli, Apr 21 2010
a(n) = Sum_{i=1..n-1} binomial(i+1,i)*i^2. - Enrique Pérez Herrero, Jun 28 2014
G.f.: 2*x^2*(2*x+1) / (1 - x)^5. - Colin Barker, Jun 28 2014
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4. - Vincenzo Librandi, Jun 30 2014
a(n) = Sum_{k=1..n-1}k*((n-1)*n/2 + k) for n > 1. - J. M. Bergot, Feb 16 2018
From Amiram Eldar, Aug 23 2022: (Start)
Sum_{n>=2} 1/a(n) = 141/5 - 9*sqrt(3)*Pi/5 - 81*log(3)/5.
Sum_{n>=2} (-1)^n/a(n) = 18*sqrt(3)*Pi/5 + 48*log(2)/5 - 129/5. (End)

Edited by Bruno Berselli, Jun 15 2010

Simpler definition as suggested by Wesley Ivan Hurt, Jun 29 2014

A346642 a(n) = Sum_{j=1..n} Sum_{i=1..j} j^3*i^3.

Original entry on oeis.org

0, 1, 73, 1045, 7445, 35570, 130826, 399738, 1063290, 2539515, 5564515, 11362351, 21875503, 40068860, 70321460, 118921460, 194681076, 309689493, 480223005, 727832905, 1080632905, 1574809126, 2256376958, 3183210350, 4427370350, 6077760975, 8243141751

Offset: 0

Roudy El Haddad, Jan 24 2022

a(n) is the sum of all products of two cubes of positive integers up to n, i.e., the sum of all products of two elements from the set of cubes {1^3, ..., n^3}.

			For n=3,
a(3) = (1*1)^3+(2*1)^3+(2*2)^3+(3*1)^3+(3*2)^3+(3*3)^3 = 1045,
a(3) = 1^3*(1^3)+2^3*(1^3+2^3)+3^3*(1^3+2^3+3^3) = 1045.

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 (9,-36,84,-126,126,-84,36,-9,1).

Cf. A000537 (sum of first n cubes), A347107 (for distinct cubes).

Cf. A001296 (for power 1), A060493 (for squares).

CoefficientList[Series[-(8 x^5 + 179 x^4 + 584 x^3 + 424 x^2 + 64 x + 1) x/(x - 1)^9, {x, 0, 26}], x] (* Michael De Vlieger, Feb 04 2022 *)

{a(n) = n*(n+1)*(n+2)*(21n^5+69n^4+45n^3-21n^2-6n+4)/672};

a(n) = sum(i=1, n, sum(j=1, i, i^3*j^3)); \\ Michel Marcus, Jan 27 2022

def A346642(n): return n*(n**2*(n*(n*(n*(n*(21*n + 132) + 294) + 252) + 21) - 56) + 8)//672 # Chai Wah Wu, Feb 17 2022

a(n) = n*(n+1)*(n+2)*(21*n^5+69*n^4+45*n^3-21*n^2-6*n+4)/672 (from the recurrent form of Faulhaber's formula).

G.f.: -(8*x^5+179*x^4+584*x^3+424*x^2+64*x+1)*x/(x-1)^9. - Alois P. Heinz, Jan 27 2022

A143037 Triangle read by rows, A000012 * A127773 * A000012. A000012 is an infinite lower triangular matrix with all 1's, A127773 = (1; 0,3; 0,0,6; 0,0,0,10; ...).

Original entry on oeis.org

1, 3, 4, 6, 9, 10, 10, 16, 19, 20, 15, 25, 31, 34, 35, 21, 36, 46, 52, 55, 56, 28, 49, 64, 74, 80, 83, 84, 36, 64, 85, 100, 110, 116, 119, 120, 45, 81, 109, 130, 145, 155, 161, 164, 165, 55, 100, 136, 164, 185, 200, 210, 216, 219, 220

Offset: 1

Gary W. Adamson & Roger L. Bagula, Jul 18 2008

Right border = tetrahedral numbers, left border = triangular numbers.
Alternatively this is the square array A(n,k)
   1,   3,   6,  10,  15,  21,  28,  36,  45,  55, ...
   4,   9,  16,  25,  36,  49,  64,  81, 100, 121, ...
  10,  19,  31,  46,  64,  85, 109, 136, 166, 199, ...
  20,  34,  52,  74, 100, 130, 164, 202, 244, 290, ...
  35,  55,  80, 110, 145, 185, 230, 280, 335, 395, ...
  56,  83, 116, 155, 200, 251, 308, 371, 440, 515, ...
  ...
read by antidiagonals where A(n,k) = n*(n^2 + 3*k*n + 3*k^2 - 1)/6 is the sum of n triangular numbers starting at A000217(k). - R. J. Mathar, May 06 2015

			First few rows of the triangle:
   1;
   3,  4;
   6,  9, 10;
  10, 16, 19,  20;
  15, 25, 31,  34,  35;
  21, 36, 46,  52,  55,  56;
  28, 49, 64,  74,  80,  83,  84;
  36, 64, 85, 100, 110, 116, 119, 120;
  ...

Cf. A001296 (row sums).

A143037 := proc(n,k)
    k*(k^2-3*k*n-3*k+3*n^2+6*n+2) / 6 ;
end proc:
seq(seq(A143037(n,k),k=1..n),n=1..12) ; # R. J. Mathar, Aug 31 2022

T[n_,k_] = k*(k^2-3*k*n-3*k+3*n^2+6*n+2) / 6;Table[T[n,k],{n,10},{k,n}]//Flatten (* James C. McMahon, Aug 13 2025 *)

T(n,k) = k*(k^2-3*k*n-3*k+3*n^2+6*n+2) / 6. - R. J. Mathar, Aug 31 2022

A210440 a(n) = 2n(n+1)*(n+2)/3.

Original entry on oeis.org

0, 4, 16, 40, 80, 140, 224, 336, 480, 660, 880, 1144, 1456, 1820, 2240, 2720, 3264, 3876, 4560, 5320, 6160, 7084, 8096, 9200, 10400, 11700, 13104, 14616, 16240, 17980, 19840, 21824, 23936, 26180, 28560, 31080, 33744, 36556, 39520, 42640, 45920, 49364, 52976

Offset: 0

Michel Marcus, Jan 20 2013

Number of tin boxes necessary to build a tetrahedron with side length n, as shown in the link.

If "0" is prepended, a(n) is the convolution of 2n with itself. - Wesley Ivan Hurt, Mar 14 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Pierre Gallais, Ceci n’est pas une mise en boîte !, Images des Mathématiques, CNRS, 2012.
Pierre Gallais, La vis ... sans fin, Images des Mathématiques, CNRS, 2012.
Jose Manuel Garcia Calcines, Luis Javier Hernandez Paricio, and Maria Teresa Rivas Rodriguez, Semi-simplicial combinatorics of cyclinders and subdivisions, arXiv:2307.13749 [math.CO], 2023. See p. 29.
Pakawut Jiradilok, Some Combinatorial Formulas Related to Diagonal Ramsey Numbers, arXiv:2404.02714 [math.CO], 2024. See p. 19.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000292, A028552, A033488 (partial sums), A046092, A130809.

[2*n*(n+1)*(n+2)/3: n in [0..50]]; // Vincenzo Librandi, Jun 24 2014

A210440:=n->2*n*(n+1)*(n+2)/3; seq(A210440(k), k=0..100); # Wesley Ivan Hurt, Sep 25 2013

Table[2n(n+1)(n+2)/3,{n,0,50}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,4,16,40},50] (* Harvey P. Dale, Feb 13 2013 *)
CoefficientList[Series[4 x/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jun 24 2014 *)

A210440(n):=2*n*(n+1)*(n+2)/3$ makelist(A210440(n),n,0,20); /* Martin Ettl, Jan 22 2013 */

a(n) = 4*A000292(n).
a(n+1)-a(n) = A046092(n+1).
From Bruno Berselli, Jan 20 2013: (Start)
G.f.: 4*x/(1-x)^4.
a(n) = -a(-n-2) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4).
a(n)-a(-n) = A217873(n).
a(n)+a(-n) = A016742(n).
(n-1)*a(n)-n*a(n-1) = A130809(n+1) with n>1. (End)
From Bruno Berselli, Jan 21 2013: (Start)
a(n) = n*A028552(n) - Sum_{i=0..n-1} A028552(i) for n>0.
4*A001296(n) = n*a(n) - Sum_{i=0..n-1} a(i) for n>0. (End)
G.f.: 2*x*W(0), where W(k) = 1 + 1/(1 - x*(k+2)*(k+4)/(x*(k+2)*(k+4) + (k+1)*(k+2)/W(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 24 2013
a(n) = Sum_{i=1..n} i*(2*n-i+3). - Wesley Ivan Hurt, Oct 03 2013
From Amiram Eldar, Apr 30 2023: (Start)
Sum_{n>=1} 1/a(n) = 3/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2) - 15/8. (End)
E.g.f.: 2*exp(x)*x*(6 + 6*x + x^2)/3. - Stefano Spezia, Jul 11 2025

A104633 Triangle T(n,k) = k(k-n-1)(k-n-2)/2 read by rows, 1<=k<=n.

Original entry on oeis.org

1, 3, 2, 6, 6, 3, 10, 12, 9, 4, 15, 20, 18, 12, 5, 21, 30, 30, 24, 15, 6, 28, 42, 45, 40, 30, 18, 7, 36, 56, 63, 60, 50, 36, 21, 8, 45, 72, 84, 84, 75, 60, 42, 24, 9, 55, 90, 108, 112, 105, 90, 70, 48, 27, 10, 66, 110, 135

Offset: 1

Gary W. Adamson, Mar 18 2005

The triangle can be constructed multiplying the triangle A(n,k)=n-k+1 (if 1<=k<=n, else 0) by the triangle B(n,k) =k (if 1<=k<=n, else 0).

Swapping the two triangles of this matrix product would generate A104634.

			First few rows of the triangle:
  1;
  3,  2;
  6,  6,  3;
 10, 12,  9,  4;
 15, 20, 18, 12,  5;
 21, 30, 30, 24, 15,  6;
 28, 42, 45, 40, 30, 18,  7;
 36, 56, 63, 60, 50, 36, 21, 8;
 ...
e.g. Col. 3 = 3 * (1, 3, 6, 10, 15...) = 3, 9, 18, 30, 45...

G. C. Greubel, Rows n=1..100 of triangle, flattened
Isabel Cação, Helmuth R. Malonek, Maria Irene Falcão, and Graça Tomaz, Intrinsic Properties of a Non-Symmetric Number Triangle, J. Int. Seq., Vol. 26 (2023), Article 23.4.8.
Joaquín Figueroa, Ivan Gonzalez, and Daniel Salinas-Arizmendi, A Novel Transfer Matrix Framework for Multiple Dirac Delta Potentials, arXiv:2503.23134 [quant-ph], 2025. See pp. 4, 9.

Cf. A062707, A158824, A104634, A001296, A000332 (row sums).

[[k*(k-n-1)*(k-n-2)/2: k in [1..n]]: n in [1..20]]; // G. C. Greubel, Aug 12 2018

A104633 := proc(n,k) k*(k-n-1)*(k-n-2)/2 ; end proc:
seq(seq(A104633(n,k),k=1..n),n=1..16) ; # R. J. Mathar, Mar 03 2011

Table[k*(k-n-1)*(k-n-2)/2, {n, 1, 20}, {k, 1, n}] // Flatten (* G. C. Greubel, Aug 12 2018 *)

for(n=1,20, for(k=1,n, print1(k*(k-n-1)*(k-n-2)/2, ", "))) \\ G. C. Greubel, Aug 12 2018

G.f.: x*y/((1 - x)^3*(1 - x*y)^2). - Stefano Spezia, May 22 2023

A164948 Fibonacci matrix read by antidiagonals. (Inverse of A136158.)

Original entry on oeis.org

1, 1, -1, 3, -4, 1, 9, -15, 7, -1, 27, -54, 36, -10, 1, 81, -189, 162, -66, 13, -1, 243, -648, 675, -360, 105, -16, 1, 729, -2187, 2673, -1755, 675, -153, 19, -1, 2187, -7290, 10206, -7938, 3780, -1134, 210, -22, 1, 6561, -24057, 37908, -34020, 19278, -7182, 1764, -276, 25, -1, 19683, -78732, 137781, -139968, 91854, -40824, 12474, -2592, 351, -28, 1

Offset: 0

Mark Dols, Sep 01 2009

Triangle, read by rows, given by [1,2,0,0,0,0,0,0,0,...] DELTA [-1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Sep 02 2009

			As triangle:
    1;
    1,   -1;
    3,   -4,    1;
    9,  -15,    7,   -1;
   27,  -54,   36,  -10,    1;
   81, -189,  162,  -66,   13,   -1;
  243, -648,  675, -360,  105,  -16,    1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000007, A000326, A001296, A001519, A002411, A003688, A006234.
Cf. A016777, A038763, A080420, A080421, A081294, A084938, A098399.
Cf. A133494, A136158, A164942.

A164948:= func< n,k | n eq 0 select 1 else (-1)^k*3^(n-k-1)*(n+2*k)*Binomial(n,k)/n >;
[A164948(n, k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 26 2023

A164948[n_,k_]:= If[n==0,1,(-1)^k*3^(n-k-1)*(n+2*k)*Binomial[n,k]/n];
Table[A164948[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 26 2023 *)

def A164948(n,k): return 1 if (n==0) else (-1)^k*3^(n-k-1)*((n+2*k)/n)*binomial(n, k)
flatten([[A164948(n, k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Dec 26 2023

Sum_{k=0..n} T(n, k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A001519(n).
From Philippe Deléham, Oct 09 2011: (Start)
T(n,k) = 3*T(n-1,k) - T(n-1,k-1) with T(0,0)=1, T(1,0)=1, T(1,1)=-1.
Row n: Expansion of (1-x)*(3-x)^(n-1), n>0. (End)
G.f.: (1-2*x)/(1-3*x+x*y). - R. J. Mathar, Aug 12 2015
From G. C. Greubel, Dec 26 2023: (Start)
T(n, k) = (-1)^k * A136158(n, k).
T(n, k) = (-1)^k*3^(n-k-1)*((n+2*k)/n)*binomial(n, k), for n > 0, with T(0, 0) = 1.
T(n, 0) = A133494(n).
T(n, 1) = -A006234(n+2), n >= 1.
T(n, 2) = A080420(n-2), n >= 2.
T(n, 3) = -A080421(n-3), n >= 3.
T(2*n, n) = 4*(-1)^n*A098399(n-1) - (1/3)*[n=0].
T(n, n-4) = 27*(-1)^n*A001296(n-3), n >= 4.
T(n, n-3) = 9*(-1)^(n-1)*A002411(n-2), n >= 3.
T(n, n-2) = 3*(-1)^n*A000326(n-1) = (-1)^n*A062741(n-1), n >= 2.
T(n, n-1) = (-1)^(n-1)*A016777(n-1), n >= 1.
T(n, n) = (-1)^n.
Sum_{k=0..n} (-1)^k*T(n, k) = A081294(n).
Sum_{k=0..n} (-1)^k*T(n-k, k) = A003688(n). (End)

More terms from Philippe Deléham, Oct 09 2011

A261721 Fourth-dimensional figurate numbers.

Original entry on oeis.org

1, 1, 5, 1, 6, 15, 1, 7, 20, 35, 1, 8, 25, 50, 70, 1, 9, 30, 65, 105, 126, 1, 10, 35, 80, 140, 196, 210, 1, 11, 40, 95, 175, 266, 336, 330, 1, 12, 45, 110, 210, 336, 462, 540, 495, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715, 1, 14, 55, 140, 280, 476, 714, 960, 1155, 1210, 1001, 1

Offset: 1

Gary W. Adamson, Aug 30 2015

Generating polygons for the sequences are: Triangle, Square, Pentagon, Hexagon, Heptagon, Octagon, ... .
n-th row sequence is the binomial transform of the fourth row of Pascal's triangle (1,n) followed by zeros; and the fourth partial sum of (1, n, n, n, ...).
n-th row sequence is the binomial transform of:
((n-1) * (0, 1, 3, 3, 1, 0, 0, 0) + (1, 4, 6, 4, 1, 0, 0, 0)).
Given the n-th row of the array (1, b, c, d, ...), the next row of the array is (1, b, c, d, ...) + (0, 1, 5, 15, 35, ...)

			The array as shown in A257200:
  1,  5, 15,  35,  70, 126, 210,  330, ... A000332
  1,  6, 20,  50, 105, 196, 336,  540, ... A002415
  1,  7, 25,  65, 140, 266, 462,  750, ... A001296
  1,  8, 30,  80, 175, 336, 588,  960, ... A002417
  1,  9, 35,  95, 210, 406, 714, 1170, ... A002418
  1, 10, 40, 110, 245, 476, 840, 1380, ... A002419
  ...
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of the fourth row of Pascal's triangle (1,3) followed by zeros: (1, 6, 12, 10, 3, 0, 0, 0, ...); and the fourth partial sum of (1, 3, 3, 3, 0, 0, 0).
(1, 7, 25, 65, 140, ...) is the third row of the array and is the binomial transform of: (2 * (0, 1, 3, 3, 1, 0, 0, 0, ...) + (1, 4, 6, 4, 1, 0, 0, 0, ...)); that is, the binomial transform of (1, 6, 12, 10, 3, 0, 0, 0, ...).
Row 2 of the array is (1, 5, 15, 35, 70, ...) + (0, 1, 5, 15, 35, ...), = (1, 6, 20, 50, 105, ...).

Albert H. Beiler, "Recreations in the Theory of Numbers"; Dover, 1966, p. 195 (Table 80).

Alois P. Heinz, Antidiagonals n = 1..141, flattened
Index to sequences related to pyramidal numbers

Cf. A257200, A261720 (pyramidal numbers), A000332, A002415, A001296, A002417, A002418, A002419.
Similar to A080852 but without row n=0.
Main diagonal gives A256859.

A:= (n, k)-> binomial(k+3, 3) + n*binomial(k+3, 4):
seq(seq(A(d-k, k), k=0..d-1), d=1..13);  # Alois P. Heinz, Aug 31 2015

row[1] = LinearRecurrence[{5, -10, 10, -5, 1}, {1, 5, 15, 35, 70}, m = 10];
row1 = Join[{0}, row[1] // Most]; row[n_] := row[n] = row[n-1] + row1;
Table[row[n-k+1][[k]], {n, 1, m}, {k, 1, n}] // Flatten (* Jean-François Alcover, May 26 2016 *)

A(n, k) = binomial(k+3, 3) + n*binomial(k+3, 4)
table(n, k) = for(x=1, n, for(y=0, k-1, print1(A(x, y), ", ")); print(""))
/* Print initial 6 rows and 8 columns as follows: */
table(6, 8) \\ Felix Fröhlich, Jul 28 2016

G.f. for row n: (1 + (n-1)*x)/(1 - x)^5.
A(n,k) = C(k+3,3) + n * C(k+3,4) = A080852(n,k).
E.g.f. as array: exp(y)*(exp(x)*(24 + 24*(3 + x)*y + 36*(1 + x)*y^2 + 4*(1 + 3*x)*y^3 + x*y^4) - 4*(6 + 18*y + 9*y^2 + y^3))/24. - Stefano Spezia, Aug 15 2024

A293616 Array of generalized Eulerian number triangles read by ascending antidiagonals, with m >= 0, n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 0, 0, 1, 6, 0, 1, 0, 1, 10, 0, 7, 1, 0, 1, 15, 0, 25, 4, 0, 0, 1, 21, 0, 65, 10, 0, 1, 0, 1, 28, 0, 140, 20, 0, 15, 4, 0, 1, 36, 0, 266, 35, 0, 90, 30, 1, 0, 1, 45, 0, 462, 56, 0, 350, 120, 5, 0, 0, 1, 55, 0, 750, 84, 0, 1050, 350, 15, 0, 1, 0

Offset: 0

Peter Luschny, Oct 14 2017

			Array starts:
m\j| 0   1  2     3    4  5       6       7    8  9      10      11      12
---|----------------------------------------------------------------------------
m=0| 1,  0, 0,    0,   0, 0,      0,      0,   0, 0,      0,      0,      0, ...
m=1| 1,  1, 0,    1,   1, 0,      1,      4,   1, 0,      1,     11,     11, ...
m=2| 1,  3, 0,    7,   4, 0,     15,     30,   5, 0,     31,    146,     91, ...
m=3| 1,  6, 0,   25,  10, 0,     90,    120,  15, 0,    301,    896,    406, ...
m=4| 1, 10, 0,   65,  20, 0,    350,    350,  35, 0,   1701,   3696,   1316, ...
m=5| 1, 15, 0,  140,  35, 0,   1050,    840,  70, 0,   6951,  11886,   3486, ...
m=6| 1, 21, 0,  266,  56, 0,   2646,   1764, 126, 0,  22827,  32172,   8022, ...
m=7| 1, 28, 0,  462,  84, 0,   5880,   3360, 210, 0,  63987,  76692,  16632, ...
m=8| 1, 36, 0,  750, 120, 0,  11880,   5940, 330, 0, 159027, 165792,  31812, ...
m=9| 1, 45, 0, 1155, 165, 0,  22275,   9900, 495, 0, 359502, 331617,  57057, ...
   A000217, A001296,A000292,A001297,A027789,A000332,A001298,A293610,A293611, ...
.
m\j| ...    13  14      15       16       17      18      19 20
---|----------------------------------------------------------------
m=0| ...,    0, 0,       0,       0,       0,      0,      0, 0, ...  [A000007]
m=1| ...,    1, 0,       1,      26,      66,     26,      1, 0, ...  [A173018]
m=2| ...,    6, 0,      63,     588,     868,    238,      7, 0, ...  [A062253]
m=3| ...,   21, 0,     966,    5376,    5586,   1176,     28, 0, ...  [A062254]
m=4| ...,   56, 0,    7770,   30660,   24570,   4200,     84, 0, ...  [A062255]
m=5| ...,  126, 0,   42525,  129780,   84630,  12180,    210, 0, ...
m=6| ...,  252, 0,  179487,  446292,  245322,  30492,    462, 0, ...
m=7| ...,  462, 0,  627396, 1315776,  625086,  68376,    924, 0, ...
m=8| ...,  792, 0, 1899612, 3444012, 1440582, 140712,   1716, 0, ...
m=9| ..., 1287, 0, 5135130, 8198190, 3063060, 270270,   3003, 0, ...
          A000389, A112494, A293612, A293613,A293614,A000579.
.
The parameter m runs over the triangles and j indexes the triangles by reading them by rows. Let T(m, n) denote the row [T(m, n, k) for 0 <= k <= n] and T(m) denote the triangle [T(m, n) for n >= 0]. Then for instance T(2) is the triangle A062253, T(4, 2) is row 2 of A062255 (which is [65, 20, 0]) and T(4, 2, 1) = 20.

Eric Weisstein's World of Mathematics, Nielsen Generalized Polylogarithm.

A000217(n) = T(n, 1, 0), A001296(n) = T(n, 2, 0), A000292(n) = T(n, 2, 1),
A001297(n) = T(n, 3, 0), A027789(n) = T(n, 3, 1), A000332(n) = T(n, 3, 2),
A001298(n) = T(n, 4, 0), A293610(n) = T(n, 4, 1), A293611(n) = T(n, 4, 2),
A000389(n) = T(n, 4, 3), A112494(n) = T(n, 5, 0), A293612(n) = T(n, 5, 1),
A293613(n) = T(n, 5, 2), A293614(n) = T(n, 5, 3), A000579(n) = T(n, 5, 4),
A144969(n) = T(n, 6, 0), A000580(n) = T(n, 6, 5), A000295(n) = T(1, n, 1),
A000460(n) = T(1, n, 2), A000498(n) = T(1, n, 3), A000505(n) = T(1, n, 4),
A000514(n) = T(1, n, 5), A001243(n) = T(1, n, 6), A001244(n) = T(1, n, 7),
A126646(n) = T(2, n, 0), A007820(n) = T(n, n, 0).
Cf. A173018, A062253, A062254, A062255, A293609.

A293616 := proc(m, n, k) option remember:
if m = 0 then m^n elif k < 0 or k > n then 0 elif n = 0 then 1 else
(k+m)*A293616(m,n-1,k) + (n-k)*A293616(m,n-1,k-1) + A293616(m-1,n,k) fi end:
for m in [$0..4] do for n in [$0..6] do print(seq(A293616(m, n, k), k=0..n)) od od;
# Sample uses:
A001298 := n -> A293616(n, 4, 0): A293614 := n -> A293616(n, 5, 3):
# Flatten:
a := proc(n) local w; w := proc(k) local t, s; t := 1; s := 1;
while t <= k do s := s + 1; t := t + s od; [s - 1, s - t + k] end:
seq(A293616(n - k, w(k)[1], w(k)[2]), k=0..n) end: seq(a(n), n = 0..11);

GenEulerianRow[0, n_] := Table[If[n==0 && j==0,1,0], {j,0,n}];
GenEulerianRow[m_, n_] := If[n==0,{1},Join[CoefficientList[x^(-m) (1 - x)^(n+m)
    PolyLog[-n-m, m, x] /. Log[1-x] -> 0, x], {0}]];
(* Sample use: *)
A173018Row[n_] := GenEulerianRow[1, n]; Table[A173018Row[n], {n, 0, 6}]

T(m, n, k) = (k + m)*T(m, n-1, k) + (n - k)*T(m, n-1, k-1) + T(m-1, n, k) with boundary conditions T(0, n, k) = 0^n; T(m, n, k) = 0 if k < 0 or k > n; and T(m, 0, k) = 0^k.

Let h(m, n) = x^(-m)*(1 - x)^(n + m)*PolyLog(-n - m, m, x) and p(m, n) the polynomial given by the expansion of h(m, n) after replacing log(1 - x) by 0. Then T(m, n, k) is the k-th coefficient of p(m, n) for 0 <= k < n.

A293617 Array of triangles read by ascending antidiagonals, T(m, n, k) = Pochhammer(m, k) * Stirling2(n + m, k + m) with m >= 0, n >= 0 and 0 <= k <= n.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 2, 1, 0, 1, 10, 3, 7, 3, 0, 1, 15, 4, 25, 12, 2, 0, 1, 21, 5, 65, 30, 6, 1, 0, 1, 28, 6, 140, 60, 12, 15, 7, 0, 1, 36, 7, 266, 105, 20, 90, 50, 12, 0, 1, 45, 8, 462, 168, 30, 350, 195, 60, 6, 0, 1, 55, 9, 750, 252, 42, 1050, 560, 180, 24, 1, 0

Offset: 0

Peter Luschny, Oct 20 2017

			Array starts:
m\j| 0   1  2     3       4       5       6       7       8       9      10
---|-----------------------------------------------------------------------
m=0| 1,  0, 0,    0,      0,      0,      0,      0,      0,      0,      0
m=1| 1,  1, 1,    1,      3,      2,      1,      7,     12,      6,      1
m=2| 1,  3, 2,    7,     12,      6,     15,     50,     60,     24,     31
m=3| 1,  6, 3,   25,     30,     12,     90,    195,    180,     60,    301
m=4| 1, 10, 4,   65,     60,     20,    350,    560,    420,    120,   1701
m=5| 1, 15, 5,  140,    105,     30,   1050,   1330,    840,    210,   6951
m=6| 1, 21, 6,  266,    168,     42,   2646,   2772,   1512,    336,  22827
m=7| 1, 28, 7,  462,    252,     56,   5880,   5250,   2520,    504,  63987
m=8| 1, 36, 8,  750,    360,     72,  11880,   9240,   3960,    720, 159027
m=9| 1, 45, 9, 1155,    495,     90,  22275,  15345,   5940,    990, 359502
   A000217, A001296,A027480,A002378,A001297,A293475,A033486,A007531,A001298
.
m\j| ...      11      12      13      14
---|-----------------------------------------
m=0| ...,      0,      0,      0,      0, ... [A000007]
m=1| ...,     15,     50,     60,     24, ... [A028246]
m=2| ...,    180,    390,    360,    120, ... [A053440]
m=3| ...,   1050,   1680,   1260,    360, ... [A294032]
m=4| ...,   4200,   5320,   3360,    840, ...
m=5| ...,  13230,  13860,   7560,   1680, ...
m=6| ...,  35280,  31500,  15120,   3024, ...
m=7| ...,  83160,  64680,  27720,   5040, ...
m=8| ..., 178200, 122760,  47520,   7920, ...
m=9| ..., 353925, 218790,  77220,  11880, ...
         A293476,A293608,A293615,A052762, ...
.
The parameter m runs over the triangles and j indexes the triangles by reading them by rows. Let T(m, n) denote the row [T(m, n, k) for 0 <= k <= n] and T(m) denote the triangle [T(m, n) for n >= 0]. Then for instance T(2) is the triangle A053440, T(3, 2) is row 2 of A294032 (which is [25, 30, 12]) and T(3, 2, 1) = 30.
.
Remark: To adapt the sequences A028246 and A053440 to our enumeration use the exponential generating functions exp(x)/(1 - y*(exp(x) - 1)) and exp(x)*(2*exp(x) - y*exp(2*x) + 2*y*exp(x) - 1 - y)/(1 - y*(exp(x) - 1))^2 instead of those indicated in their respective entries.

Eric Weisstein's World of Mathematics, Nørlund Polynomial.

A000217(n) = T(n, 1, 0), A001296(n) = T(n, 2, 0), A027480(n) = T(n, 2, 1),
A002378(n) = T(n, 2, 2), A001297(n) = T(n, 3, 0), A293475(n) = T(n, 3, 1),
A033486(n) = T(n, 3, 2), A007531(n) = T(n, 3, 3), A001298(n) = T(n, 4, 0),
A293476(n) = T(n, 4, 1), A293608(n) = T(n, 4, 2), A293615(n) = T(n, 4, 3),
A052762(n) = T(n, 4, 4), A052787(n) = T(n, 5, 5), A000225(n) = T(1, n, 1),
A028243(n) = T(1, n, 2), A028244(n) = T(1, n, 3), A028245(n) = T(1, n, 4),
A032180(n) = T(1, n, 5), A228909(n) = T(1, n, 6), A228910(n) = T(1, n, 7),
A000225(n) = T(2, n, 0), A007820(n) = T(n, n, 0).
A028246(n,k) = T(1, n, k), A053440(n,k) = T(2, n, k), A294032(n,k) = T(3, n, k),
A293926(n,k) = T(n, n, k), A124320(n,k) = T(n, k, k), A156991(n,k) = T(k, n, n).
Cf. A293616.

A293617 := proc(m, n, k) option remember:
if m = 0 then 0^n elif k < 0 or k > n then 0 elif n = 0 then 1 else
(k+m)*A293617(m,n-1,k) + k*A293617(m,n-1,k-1) + A293617(m-1,n,k) fi end:
for m in [$0..4] do for n in [$0..6] do print(seq(A293617(m, n, k), k=0..n)) od od;
# Sample uses:
A027480 := n -> A293617(n, 2, 1): A293608 := n -> A293617(n, 4, 2):
# Flatten:
a := proc(n) local w; w := proc(k) local t, s; t := 1; s := 1;
while t <= k do s := s + 1; t := t + s od; [s - 1, s - t + k] end:
seq(A293617(n - k, w(k)[1], w(k)[2]), k=0..n) end: seq(a(n), n = 0..11);

T[m_, n_, k_] := Pochhammer[m, k] StirlingS2[n + m, k + m];
For[m = 0, m < 7, m++, Print[Table[T[m, n, k], {n,0,6}, {k,0,n}]]]
A293617Row[m_, n_] := Table[T[m, n, k], {k,0,n}];
(* Sample use: *)
A293926Row[n_] := A293617Row[n, n];

T(m,n,k) = (k + m)*T(m, n-1, k) + k*T(m, n-1, k-1) + T(m-1, n, k) with boundary conditions T(0, n, k) = 0^n; T(m, n, k) = 0 if k<0 or k>n; and T(m, 0, k) = 0^k.

T(m,n,k) = Pochhammer(m, k)*binomial(n + m, k + m)*NorlundPolynomial(n - k, -k - m).

cp's OEIS Frontend

A108650 a(n) = (n+1)^2*(n+2)*(n+3)*(3*n+4)/24.

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A153978 a(n) = n*(n-1)*(n+1)*(3*n-2)/12.

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A346642 a(n) = Sum_{j=1..n} Sum_{i=1..j} j^3*i^3.

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A143037 Triangle read by rows, A000012 * A127773 * A000012. A000012 is an infinite lower triangular matrix with all 1's, A127773 = (1; 0,3; 0,0,6; 0,0,0,10; ...).

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A210440 a(n) = 2*n*(n+1)*(n+2)/3.

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A104633 Triangle T(n,k) = k*(k-n-1)*(k-n-2)/2 read by rows, 1<=k<=n.

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A108650 a(n) = (n+1)^2(n+2)(n+3)(3n+4)/24.

A153978 a(n) = n(n-1)(n+1)(3n-2)/12.

A210440 a(n) = 2n(n+1)*(n+2)/3.

A104633 Triangle T(n,k) = k(k-n-1)(k-n-2)/2 read by rows, 1<=k<=n.