A002415 -id:A002415 - cp's OEIS frontend

A241619 T(n,k)=Number of length n+2 0..k arrays with no consecutive three elements summing to more than k.

4, 10, 6, 20, 20, 9, 35, 50, 40, 13, 56, 105, 125, 76, 19, 84, 196, 315, 295, 147, 28, 120, 336, 686, 889, 711, 287, 41, 165, 540, 1344, 2254, 2567, 1730, 556, 60, 220, 825, 2430, 5040, 7586, 7483, 4175, 1077, 88, 286, 1210, 4125, 10242, 19374, 25774, 21631, 10077

Offset: 1

R. H. Hardin, Apr 26 2014

Table starts
...4...10....20.....35......56.......84......120.......165.......220........286
...6...20....50....105.....196......336......540.......825......1210.......1716
...9...40...125....315.....686.....1344.....2430......4125......6655......10296
..13...76...295....889....2254.....5040....10242.....19305.....34243......57772
..19..147...711...2567....7586....19374....44274.....92697....180829.....332761
..28..287..1730...7483...25774....75180...193194....449295....963886....1934647
..41..556..4175..21631...86828...289248...835812...2159025...5093737...11151140
..60.1077.10077..62547..292621..1113348..3617703..10380183..26932543...64309245
..88.2091.24377.181255..988303..4294574.15692003..50011289.142701909..371651553
.129.4057.58928.524877.3335451.16553380.68014233.240772037.755538278.2146210209

			Some solutions for n=5 k=4
..1....0....2....1....0....2....0....1....0....0....0....2....1....0....1....2
..0....0....1....3....0....0....4....2....1....0....1....1....3....0....0....1
..0....3....0....0....0....1....0....1....3....0....0....0....0....2....1....0
..0....0....0....1....2....0....0....0....0....0....1....0....1....1....1....2
..2....0....3....0....0....2....0....0....0....0....2....1....0....0....0....0
..0....1....0....1....2....1....2....0....1....0....0....1....1....0....3....1
..0....0....0....1....0....0....2....4....2....2....0....0....2....0....0....0

R. H. Hardin, Table of n, a(n) for n = 1..10011

Column 1 is A000930(n+4)
Row 1 is A000292(n+1)
Row 2 is A002415(n+2)
Row 3 is A006414
Row 4 is A114244

for m from 1 to 12 do
  r:= [seq(seq([i,j],j=0..m-i),i=0..m)];
  T[m]:= Matrix((m+1)*(m+2)/2,(m+1)*(m+2)/2, proc(i, j) if r[i][1]=r[j][2] and r[i][1]+r[i][2]+r[j][1]<=m then 1 else 0 fi end proc):
  U[m,0]:= Vector((m+1)*(m+2)/2,1);
od:
R:= NULL:
for i from 2 to 12 do
  for j from 1 to i-1 do
    U[i-j,j]:= T[i-j] . U[i-j,j-1];
    R:= R, convert(U[i-j,j],`+`)
od od:
R; # Robert Israel, Sep 04 2019

Empirical for column k, apparently a recurrence of order (k+1)*(k+2)/2:
k=1: a(n) = a(n-1) +a(n-3)
k=2: a(n) = a(n-1) +a(n-2) +2*a(n-3) -a(n-5) -a(n-6)
k=3: a(n) = 2*a(n-1) +4*a(n-3) -3*a(n-4) -a(n-5) -3*a(n-6) +2*a(n-7) +a(n-9) -a(n-10)
k=4: [order 15]
k=5: [order 21]
k=6: [order 28]
k=7: [order 36]
k=8: [order 45]
k=9: [order 55]
k=10: [order 66]
k=11: [order 78]
k=12: [order 91]
Empirical for row n, apparently a polynomial of degree n+2:
n=1: a(n) = (1/6)*n^3 + 1*n^2 + (11/6)*n + 1
n=2: a(n) = (1/12)*n^4 + (2/3)*n^3 + (23/12)*n^2 + (7/3)*n + 1
n=3: a(n) = (1/24)*n^5 + (5/12)*n^4 + (13/8)*n^3 + (37/12)*n^2 + (17/6)*n + 1
n=4: [polynomial of degree 6]
n=5: [polynomial of degree 7]
n=6: [polynomial of degree 8]
n=7: [polynomial of degree 9]
From Robert Israel, Sep 04 2019: (Start)
Column k satisfies a recurrence of order (k+1)*(k+2)/2, since a(n)=e^T T^n e where T is a (k+1)*(k+2)/2 matrix and e the vector of all 1's (see proofs at A241615 and A241618).
Row n is the Ehrhart polynomial of degree n+2 corresponding to the polytope {(x(1),...,x(n+2)): all x(i)>=0, x(i)+x(i+1)+x(i+2)<=1 for i=1..n}, whose vertices have all entries in {0,1}. (End)

A006857 a(n) = binomial(n+5,5) * binomial(n+5,4)/(n+5).

Original entry on oeis.org

1, 15, 105, 490, 1764, 5292, 13860, 32670, 70785, 143143, 273273, 496860, 866320, 1456560, 2372112, 3755844, 5799465, 8756055, 12954865, 18818646, 26883780, 37823500, 52474500, 71867250, 97260345, 130179231, 172459665, 226296280, 294296640, 379541184

Offset: 0

Simon Plouffe

Number of permutations of n+5 that avoid the pattern 132 and have exactly 4 descents.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Nov 18 2005
Partial sums of A114242. - Peter Bala, Sep 21 2007
Dimensions of certain Lie algebra (see reference for precise definition).

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 167-169, Table 10.5/II/1).
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 239.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
Brandy Amanda Barnette, Counting Convex Sets on Products of Totally Ordered Sets, Masters Theses & Specialist Projects, Paper 1484, 2015.
G. Kreweras, Traitement simultané du "Problème de Young" et du "Problème de Simon Newcomb", Cahiers du Bureau Universitaire de Recherche Opérationnelle. Institut de Statistique, Université de Paris, 10 (1967), 23-31.
G. Kreweras, Traitement simultané du "Problème de Young" et du "Problème de Simon Newcomb", Cahiers du Bureau Universitaire de Recherche Opérationnelle. Institut de Statistique, Université de Paris, 10 (1967), 23-31. [Annotated scanned copy]
J. M. Landsberg and L. Manivel, The sextonions and E7 1/2, Adv. Math. 201 (2006), 143-179. [Th. 7.3, case a=4]
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 25.
Index entries for linear recurrences with constant coefficients, signature (9,-36,84,-126,126,-84,36,-9,1).

The expression binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1) for the values m = 2 through 14 produces the sequences A000012, A000217, A002415, A006542, A006857, A108679, A134288, A134289, A134290, A134291, A140925, A140935, A169937.
5th column of the table of Narayana numbers A001263.
Cf. A002378, A114242.

A006857:= func< n | Binomial(n+4,3)*Binomial(n+5,5)/4 >;
[A006857(n): n in [0..40]]; // G. C. Greubel, Mar 12 2025

a:=n->(n+1)*(n+2)^2*(n+3)^2*(n+4)^2*(n+5)/2880: seq(a(n),n=0..38); # Emeric Deutsch, Nov 18 2005

Table[Binomial[n+5,5] * Binomial[n+5,4]/(n+5), {n, 0, 50}] (* T. D. Noe, May 29 2012 *)
LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{1,15,105,490,1764,5292,13860,32670,70785},40] (* Harvey P. Dale, Oct 19 2024 *)

a(n) = binomial(n+5,5) * binomial(n+5,4)/(n+5) \\ Charles R Greathouse IV, Jun 11 2015

Vec((1+6*x+6*x^2+x^3)/(1-x)^9 + O(x^99)) \\ Altug Alkan, Sep 01 2016

def A006857(n): return binomial(n+4,3)*binomial(n+5,5)//4
print([A006857(n) for n in range(41)]) # G. C. Greubel, Mar 12 2025

From - Vladeta Jovovic, Jan 29 2003: (Start)
a(n) = (n+4)!*(n+5)!/(2880*n!*(n+1)!).
E.g.f.: 1/2880*(2880 + 40320*x + 109440*x^2 + 105120*x^3 + 45000*x^4 + 9504*x^5 + 1016*x^6 + 52*x^7 + x^8)*exp(x). (End)
From Mike Zabrocki, Aug 26 2004: (Start)
a(n) = C(n+5,8) + 6*C(n+6,8) + 6*C(n+7,8) + C(n+8,8).
a(n) = C(n+4,4)*C(n+5,4)/5.
O.g.f.: (1 + 6*x + 6*x^2 + x^3)/(1-x)^9. (End)
From Wolfdieter Lang, Nov 13 2007: (Start)
a(n) = A001263(n+5,5).
Numerator polynomial of the g.f is the fourth row polynomial of the Narayana triangle. (End)
a(n)= C(n+4,4)^2 - C(n+4,3)*C(n+4,5). - Gary Detlefs, Dec 05 2011
a(n) = Product_{i=1..4} A002378(n+i)/A002378(i). - Bruno Berselli, Sep 01 2016
From Amiram Eldar, Oct 19 2020: (Start)
Sum_{n>=0} 1/a(n) = 25 * (79 - 8*Pi^2).
Sum_{n>=0} (-1)^n/a(n) = 595/3 - 20*Pi^2. (End)

More terms from Vladeta Jovovic, Jan 29 2003
Better description from Mike Zabrocki, Aug 26 2004
New definition from N. J. A. Sloane, Aug 28 2010
Zabrocki formulas offset corrected by Gary Detlefs, Dec 05 2011

A220212 Convolution of natural numbers (A000027) with tetradecagonal numbers (A051866).

Original entry on oeis.org

0, 1, 16, 70, 200, 455, 896, 1596, 2640, 4125, 6160, 8866, 12376, 16835, 22400, 29240, 37536, 47481, 59280, 73150, 89320, 108031, 129536, 154100, 182000, 213525, 248976, 288666, 332920, 382075, 436480, 496496, 562496, 634865, 714000, 800310, 894216, 996151

Offset: 0

Bruno Berselli, Dec 08 2012

Partial sums of A172073.

Apart from 0, all terms are in A135021: a(n) = A135021(A034856(n+1)) with n>0.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Index to sequences related to pyramidal numbers.

Cf. A135021, A172073.
Cf. convolution of the natural numbers (A000027) with the k-gonal numbers (* means "except 0"):
k= 2 (A000027 ): A000292;
k= 3 (A000217 ): A000332 (after the third term);
k= 4 (A000290 ): A002415 (after the first term);
k= 5 (A000326 ): A001296;
k= 6 (A000384*): A002417;
k= 7 (A000566 ): A002418;
k= 8 (A000567*): A002419;
k= 9 (A001106*): A051740;
k=10 (A001107*): A051797;
k=11 (A051682*): A051798;
k=12 (A051624*): A051799;
k=13 (A051865*): A055268.
Cf. similar sequences with formula n*(n+1)*(n+2)*(k*n-k+2)/12 listed in A264850.

A051866:=func; [&+[(n-k+1)*A051866(k): k in [0..n]]: n in [0..37]];

I:=[0,1,16,70,200]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

A051866[k_] := k (6 k - 5); Table[Sum[(n - k + 1) A051866[k], {k, 0, n}], {n, 0, 37}]
CoefficientList[Series[x (1 + 11 x) / (1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 18 2013 *)

G.f.: x*(1+11*x)/(1-x)^5.
a(n) = n*(n+1)*(n+2)*(3*n-2)/6.
From Amiram Eldar, Feb 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 3*(3*sqrt(3)*Pi + 27*log(3) - 17)/80.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*(6*sqrt(3)*Pi - 64*log(2) + 37)/80. (End)

A006324 a(n) = n(n + 1)(2n^2 + 2n - 1)/6.

Original entry on oeis.org

1, 11, 46, 130, 295, 581, 1036, 1716, 2685, 4015, 5786, 8086, 11011, 14665, 19160, 24616, 31161, 38931, 48070, 58730, 71071, 85261, 101476, 119900, 140725, 164151, 190386, 219646, 252155, 288145, 327856, 371536, 419441, 471835, 528990, 591186

Offset: 1

Albert Rich (Albert_Rich(AT)msn.com), Jun 14 1998

4-dimensional analog of centered polygonal numbers.
Partial sums of A000447. - Zak Seidov, May 19 2006
From Johannes W. Meijer, Jun 27 2009: (Start)
Equals the absolute values of the coefficients that precede the a(n-1) factors of the recurrence relations RR(n) of A162011.
This sequence enabled the analysis of A162012 and A162013. (End)
Equals the number of integer quadruples (x,y,z,w) such that min(x,y) < min(z,w), max(x,y) < max(z,w), and 0 <= x,y,z,w <= n. - Andrew Woods, Apr 21 2014
For n>3 a(n)=twice the area of an irregular quadrilateral with vertices at the points (C(n,4),C(n+1,4)), (C(n+1,4),C(n+2,4)), (C(n+2,4),C(n+3,4)), and (C(n+3,4),C(n+4,4)). - J. M. Bergot, Jun 14 2014

Delbert L. Johnson, Table of n, a(n) for n = 1..20000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000447, A000539, A000217, A002415, A053134.

Cf. A162011, A162012, a(n-2), and A162013, a(n-3). - Johannes W. Meijer, Jun 27 2009

[ n*(n + 1)*(2*n^2 + 2*n - 1)/6 : n in [1..30] ]; // Wesley Ivan Hurt, Jun 14 2014

A006324:=n->n*(n + 1)*(2*n^2 + 2*n - 1)/6; seq(A006324(n), n=1..30); # Wesley Ivan Hurt, Jun 14 2014

Table[Sum[k^5,{k,n}]/Sum[k,{k,n}], {n,40}] (* Alexander Adamchuk, Apr 12 2006 *)

a(n) = 8*C(n + 2, 4) + C(n + 1, 2).
a(n) = (Sum_{k=1..n} k^5) / (Sum_{k=1..n} k) = A000539(n) / A000217(n). - Alexander Adamchuk, Apr 12 2006
From Johannes W. Meijer, Jun 27 2009: (Start)
Recurrence relation 0 = Sum_{k=0..5} (-1)^k*binomial(5,k)*a(n-k).
G.f.: (1+6*z+z^2)/(1-z)^5. (End)
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Wesley Ivan Hurt, May 02 2021
Sum_{n>=1} 1/a(n) = 6 + 2*sqrt(3)*Pi*tan(sqrt(3)*Pi/2). - Amiram Eldar, Aug 23 2022
a(n) = A053134(n-1) - 4*A002415(n). - Yasser Arath Chavez Reyes, Feb 12 2024

Simpler definition from Alexander Adamchuk, Apr 12 2006

More terms from Zak Seidov

A008911 a(n) = n^2*(n^2 - 1)/6.

Original entry on oeis.org

0, 0, 2, 12, 40, 100, 210, 392, 672, 1080, 1650, 2420, 3432, 4732, 6370, 8400, 10880, 13872, 17442, 21660, 26600, 32340, 38962, 46552, 55200, 65000, 76050, 88452, 102312, 117740, 134850, 153760, 174592, 197472, 222530, 249900, 279720, 312132

Offset: 0

N. J. A. Sloane, R. K. Guy

Number of equilateral triangles in rhombic portion of side n+1 in hexagonal lattice.
The hexagonal lattice is the familiar 2-dimensional lattice in which each point has 6 neighbors. This is sometimes called the triangular lattice.
Sum of squared distances on n X n board between n queens each on its own row and column. - Zak Seidov, Sep 04 2002
For queens "each on its column and row" the sum of squared distances does not depend on configuration - while sum of distances does.
Number of cycles of length 3 in the bishop's graph associated with an (n+1) X (n+1) chessboard. - Anton Voropaev (anton.n.voropaev(AT)gmail.com), Feb 01 2009
a(n) is number of ways to place 3 queens on an (n+1) X (n+1) chessboard so that they diagonally attack each other exactly 3 times. The maximal possible attack number, p=binomial(k,2)=3 for k=3 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form the corresponding complete graph. - Antal Pinter, Dec 27 2015
From a(1), convolution of the oblong numbers (A002378) with the odd numbers (A005408). - Bruno Berselli, Oct 24 2016
Consider the partitions of 2n into two parts (p,q) where p <= q. Then a(n) is the total volume of the family of rectangular prisms with dimensions p, p and |q-p|. - Wesley Ivan Hurt, Apr 15 2018

			a(2)=2 because on 2 X 2 board queens "each on its column and row" may take only two angular cells, then squared distance is 1^2+1^2=2. a(3)=12 because on 3 X 3 board queens "each on its column and row" make only two essentially distinct configurations: {1,2,3}, {1,3,2} and in both cases the sum of three squared distances is 12.
G.f.: 2*x^2 + 12*x^3 + 40*x^4 + 100*x^5 + 210*x^6 + 392*x^7 + 672*x^8 + ...

James Propp, Enumeration of matchings: problems and progress, pp. 255-291 in L. J. Billera et al., eds, New Perspectives in Algebraic Combinatorics, Cambridge, 1999 (see Problem 6).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Gabriele Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2
James Propp, Enumeration of matchings: problems and progress, in L. J. Billera et al. (eds.), New Perspectives in Algebraic Combinatorics, 1999.
James Propp, Updated article, 2009.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A002415, A006011, A047928, A083374.
Cf. A002378, A005408.
Convolution of the oblong numbers with the even numbers: A033488.

List([0..40], n-> n^2*(n^2-1)/6); # G. C. Greubel, Sep 13 2019

[n^2*(n^2-1)/6: n in [0..40]]; // Vincenzo Librandi, Sep 14 2011

A008911 := n->n^2*(n^2-1)/6; seq(A008911(n), n=0..40);

a[m_]:= m^2(m^2-1)/6;
Binomial[Range[0,40]^2, 2]/3 (* G. C. Greubel, Sep 13 2019 *)

PARI
```
{a(n) = n^2*(n^2-1)/6};
```

[n^2*(n^2-1)/6 for n in (0..40)] # G. C. Greubel, Sep 13 2019

G.f.: 2*x^2*(1+x)/(1-x)^5.
a(n) = 2*A002415(n) = A047928(n-1)/6 = A083374(n-1)/3 = A006011(n)*2/3. - Zerinvary Lajos, May 09 2007
a(n) = n*binomial(n+1,3). - Martin Renner, Apr 03 2011
a(n+1) = (n+1)*A000292(n). - Tom Copeland, Sep 13 2011
From G. C. Greubel, Sep 13 2019: (Start)
a(n) = binomial(n^2,2)/3.
E.g.f.: x^2*(6 + 6*x + x^2)*exp(x)/6. (End)
From Amiram Eldar, Nov 02 2021: (Start)
Sum_{n>=2} 1/a(n) = 21/2 - Pi^2.
Sum_{n>=2} (-1)^n/a(n) = (Pi^2 - 9)/2. (End)
a(n) = Sum_{j=0..n-1} binomial(n,2) + binomial(n,3). - Detlef Meya, Jan 20 2024

A103905 Square array T(n,k) read by antidiagonals: number of tilings of an hexagon.

Original entry on oeis.org

1, 1, 2, 1, 6, 3, 1, 20, 20, 4, 1, 70, 175, 50, 5, 1, 252, 1764, 980, 105, 6, 1, 924, 19404, 24696, 4116, 196, 7, 1, 3432, 226512, 731808, 232848, 14112, 336, 8, 1, 12870, 2760615, 24293412, 16818516, 1646568, 41580, 540, 9, 1, 48620, 34763300

Offset: 1

Ralf Stephan, Feb 22 2005

As a square array, T(n,k) = number of all k-watermelons without a wall of length n. - Steven Finch, Mar 30 2008

			Array begins:
  1,   2,     3,      4,        5,         6, ...
  1,   6,    20,     50,      105,       196, ...
  1,  20,   175,    980,     4116,     14112, ...
  1,  70,  1764,  24696,   232848,   1646568, ...
  1, 252, 19404, 731808, 16818516, 267227532, ...
  ...

Peter J. Forrester and Alex Gamburd, Counting formulas associated with some random matrix averages, arXiv:math/0503002 [math.CO], 2005.
Anthony J. Guttmann, Aleksandr L. Owczarek and Xavier G. Viennot, Vicious walkers and Young tableaux. I. Without walls, J. Phys. A 31 (1998) 8123-8135.
Harald Helfgott and Ira M. Gessel, Enumeration of tilings of diamonds and hexagons with defects, arXiv:math/9810143 [math.CO], 1998.
Christian Krattenthaler, Advanced Determinant Calculus: A Complement, Linear Algebra Appl. 411 (2005), 68-166; arXiv:math/0503507 [math.CO], 2005.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 28.
Percy A. MacMahon, Combinatory Analysis, vol. 2, Cambridge University Press, 1916; reprinted by Chelsea, New York, 1960.

Rows include A002415, A047819, A047835, A047831.
Columns include A000984 and A000891.
Main diagonal is A008793.
Cf. A120258, A133112.

t[n_, k_] := Product[j!*(j + 2*n)!/(j + n)!^2, {j, 0, k - 1}]; Join[{1}, Flatten[ Table[ t[n - k , k], {n, 1, 10}, {k, 1, n}]]] (* Jean-François Alcover, May 16 2012, from 2nd formula *)

T(n, k) = [V(2n+k-1)V(k-1)V(n-1)^2]/[V(2n-1)V(n+k-1)^2], with V(n) the superfactorial numbers (A000178).
T(n, k) = Prod[j=0..k-1, j!(j+2n)!/(j+n)!^2 ].
T(n, k) = Prod[h=1..n, Prod[i=1..k, Prod[j=1..n, (h+i+j-1)/(h+i+j-2) ]]].
T(n, k) = Prod[i=1..k, Prod[j=n+1..2n+1, i+j]/Prod[j=0..n, i+j]]; - Paul Barry, Jun 13 2006
Conjectural formula as a sum of squares of Vandermonde determinants: T(n,k) = 1/((1!*2! ... *(n-1)!)^2*n!)* sum {1 <= x_1, ..., x_n <= k} (det V(x_1, ..., x_n))^2, where V(x_1, ..., x_n) is the Vandermonde matrix of order n. Compare with A133112. - Peter Bala, Sep 18 2007
For k >= 1, T(n,k)=det(binomial(2*n,n+i-j))1<=i,j<=k [Krattenhaller, Theorem 4].
Let H(n) = product {k = 1..n-1} k!. Then for a,b,c nonnegative integers (H(a)*H(b)*H(c)*H(a+b+c))/(H(a+b)*H(b+c)*H(c+a)) is an integer [MacMahon, Section 4.29 with x -> 1]. Setting a = b = n and c = k gives the entries for this table. - Peter Bala, Dec 22 2011

A110813 A triangle of pyramidal numbers.

Original entry on oeis.org

1, 3, 1, 5, 4, 1, 7, 9, 5, 1, 9, 16, 14, 6, 1, 11, 25, 30, 20, 7, 1, 13, 36, 55, 50, 27, 8, 1, 15, 49, 91, 105, 77, 35, 9, 1, 17, 64, 140, 196, 182, 112, 44, 10, 1, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1, 23, 121, 385, 825

Offset: 0

Paul Barry, Aug 05 2005

Triangle A029653 less first column. In general, the product (1/(1-x),x/(1-x))*(1+m*x,x) yields the Riordan array ((1+(m-1)x)/(1-x)^2,x/(1-x)) with general term T(n,k)=(m*n-(m-1)*k+1)*C(n+1,k+1)/(n+1). This is the reversal of the (1,m)-Pascal triangle, less its first column. - Paul Barry, Mar 01 2006
The column sequences give, for k=0..10: A005408 (odd numbers), A000290 (squares), A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.
Linked to Chebyshev polynomials by the fact that this triangle with interpolated zeros in the rows and columns is a scaled version of A053120.
Row sums are A033484. Diagonal sums are A001911(n+1) or F(n+4)-2. Factors as (1/(1-x),x/(1-x))*(1+2x,x). Inverse is A110814 or (-1)^(n-k)*A104709.
This triangle is a subtriangle of the [2,1] Pascal triangle A029653 (omit there the first column).
Subtriangle of triangles in A029653, A131084, A208510. - Philippe Deléham, Mar 02 2012
This is the iterated partial sums triangle of A005408 (odd numbers). Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given). - Wolfdieter Lang, Mar 23 2015

			The number triangle T(n, k) begins
n\k  0   1   2   3    4    5    6   7   8  9 10 11
0:   1
1:   3   1
2:   5   4   1
3:   7   9   5   1
4:   9  16  14   6    1
5:  11  25  30  20    7    1
6:  13  36  55  50   27    8    1
7:  15  49  91 105   77   35    9   1
8:  17  64 140 196  182  112   44  10   1
9:  19  81 204 336  378  294  156  54  11  1
10: 21 100 285 540  714  672  450 210  65 12  1
11: 23 121 385 825 1254 1386 1122 660 275 77 13  1
... reformatted by _Wolfdieter Lang_, Mar 23 2015
As a number square S(n, k) = T(n+k, k), rows begin
  1,   1,   1,   1,   1,   1, ...
  3,   4,   5,   6,   7,   8, ...
  5,   9,  14,  20,  27,  35, ...
  7,  16,  30,  50,  77, 112, ...
  9,  25,  55, 105, 182, 294, ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened

Cf. A000290, A000330, A002415, A005408, A005585, A029655, A040977, A050486, A053347, A054333, A054334, A057788.

Table[2*Binomial[n + 1, k + 1] - Binomial[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 19 2017 *)

for(n=0,10, for(k=0,n, print1(2*binomial(n+1, k+1) - binomial(n,k), ", "))) \\ G. C. Greubel, Oct 19 2017

Number triangle T(n, k) = C(n, k)*(2n-k+1)/(k+1) = 2*C(n+1, k+1) - C(n, k); Riordan array ((1+x)/(1-x)^2, x/(1-x)); As a number square read by antidiagonals, T(n, k)=C(n+k, k)(2n+k+1)/(k+1).
Equals A007318 * an infinite bidiagonal matrix with 1's in the main diagonal and 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal, all 2's in the subdiagonal and the rest zeros. - Gary W. Adamson, Dec 12 2007
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=T(1,1)=1, T(1,0)=3, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 30 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 9*x + 5*x^2/2! + x^3/3!) = 7 + 16*x + 30*x^2/2! + 50*x^3/3! + 77*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
T(n, k) = ps(1, 2; k, n-k) with ps(a, d; k, n) = sum(ps(a, d; k-1, j), j=0..n) and input ps(a, d; 0, j) = a + d*j. See the iterated partial sums comment from Mar 23 2015 above. - Wolfdieter Lang, Mar 23 2015
From Franck Maminirina Ramaharo, May 21 2018: (Start)
T(n,k) = coefficients in the expansion of ((x + 2)*(x + 1)^n - 2)/x.
T(n,k) = A135278(n,k) + A135278(n-1,k).
T(n,k) = A097207(n,n-k).
G.f.: (y + 1)/((y - 1)*(x*y + y - 1)).
E.g.f.: ((x + 2)*exp(x*y + y) - 2*exp(y))/x.
(End)

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

A132366 Partial sum of centered tetrahedral numbers A005894.

Original entry on oeis.org

1, 6, 21, 56, 125, 246, 441, 736, 1161, 1750, 2541, 3576, 4901, 6566, 8625, 11136, 14161, 17766, 22021, 27000, 32781, 39446, 47081, 55776, 65625, 76726, 89181, 103096, 118581, 135750, 154721, 175616, 198561, 223686, 251125, 281016, 313501, 348726, 386841

Offset: 0

Jonathan Vos Post, Nov 09 2007

From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) is the number of achiral colorings of the 5 tetrahedral facets (or vertices) of a regular 4-dimensional simplex using n or fewer colors. An achiral arrangement is identical to its reflection. The 4-dimensional simplex is also called a 5-cell or pentachoron. Its Schläfli symbol is {3,3,3}.
There are 60 elements in the automorphism group of the 4-dimensional simplex that are not in its rotation group. Each is an odd permutation of the vertices and can be associated with a partition of 5 based on the conjugacy class of the permutation. The first formula for a(n-1) is obtained by averaging their cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
  Partition  Count  Odd Cycle Indices
  41         30     x_1x_4^1
  32         20     x_2^1x_3^1
  2111       10     x_1^3x_2^1 (End)

Nathaniel Johnston, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000292, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
Cf. A337895 (oriented), A000389(n+4) (unoriented), A000389 (chiral), A331353 (5-cell edges, faces), A337955 (8-cell vertices, 16-cell facets), A337958 (16-cell vertices, 8-cell facets), A338951 (24-cell), A338967 (120-cell, 600-cell).
a(n-1) = A325001(4,n).

Do[Print[n, " ", (n^4 + 4 n^3 + 11 n^2 + 14 n + 6)/6 ], {n, 0, 10000}]
Accumulate[Table[(2n+1)(n^2+n+3)/3,{n,0,40}]] (* or *) LinearRecurrence[ {5,-10,10,-5,1},{1,6,21,56,125},40] (* Harvey P. Dale, Feb 26 2020 *)

a(n) = (n^4 + 4*n^3 + 11*n^2 + 14*n + 6)/6 = (n^2+2*n+6)*(n+1)^2/6.
G.f.: -(x+1)*(x^2+1) / (x-1)^5. - Colin Barker, May 04 2013
From Robert A. Russell, Oct 09 2020: (Start)
a(n-1) = n^2 * (5 + n^2) / 6.
a(n-1) = binomial(n+4,5) - binomial(n,5) = A000389(n+4) - A000389(n).
a(n-1) = 1*C(n,1) + 4*C(n,2) + 6*C(n,3) + 4*C(n,4), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n-1) = 2*A000389(n+4) - A337895(n) = A337895(n) - 2*A000389(n) .
G.f. for a(n-1): x * (x+1) * (x^2+1) / (1-x)^5. (End)
From Amiram Eldar, Feb 14 2023: (Start)
Sum_{n>=0} 1/a(n) = Pi^2/5 + 3/25 - 3*Pi*coth(sqrt(5)*Pi)/(5*sqrt(5)).
Sum_{n>=0} (-1)^n/a(n) = Pi^2/10 - 3/25 + 3*Pi*cosech(sqrt(5)*Pi)/(5*sqrt(5)). (End)
a(n) = A006007(n) + A006007(n+1) = A002415(n) + A002415(n+2). - R. J. Mathar, Jun 05 2025

Corrected offset, Mathematica program by Tomas J. Bulka (tbulka(AT)rodincoil.com), Sep 02 2009

A057788 Expansion of (1+x)/(1-x)^12.

Original entry on oeis.org

1, 13, 90, 442, 1729, 5733, 16744, 44200, 107406, 243542, 520676, 1058148, 2057510, 3848222, 6953544, 12183560, 20764055, 34512075, 56071470, 89224590, 139299615, 213696795, 322561200, 479634480, 703323660, 1018031196, 1455797448, 2058314440, 2879378332

Offset: 0

N. J. A. Sloane, Nov 04 2000

1/2^10 of twelfth unsigned column of triangle A053120 (T-Chebyshev, rising powers, zeros omitted).
If a 2-set Y and an (n-3)-set Z are disjoint subsets of an n-set X then a(n-12) is the number of 12-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007
11-dimensional square numbers, tenth partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = sum_{i=0..n} C(n+10,i+10)*b(i), where b(i)=[1,2,0,0,0,...]. - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
2*a(n) is number of ways to place 10 queens on an (n+10) X (n+10) chessboard so that they diagonally attack each other exactly 45 times. The maximal possible attack number, p=binomial(k,2) =45 for k=10 queens, is achievable only when all queens are on the same diagonal. In graph-theory representation they thus form the corresponding complete graph. - Antal Pinter, Dec 27 2015

T. D. Noe, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 15.
Index entries for linear recurrences with constant coefficients, signature (12,-66, 220,-495,792,-924,792,-495,220,-66,12,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A053120, A054334, A054333, A053347, A002415, A005585, A040977, A050486.
Partial sums of A054334.
Sixth column of A111125 (s=5, without leading zeros). - Wolfdieter Lang, Oct 18 2012

List([0..30], n -> (2*n+11)*Binomial(n+10, 10)/11); # G. C. Greubel, Dec 02 2018

[Binomial(n+10,10)*(2*n+11)/11: n in [0..40]]; // Vincenzo Librandi, Feb 14 2016

A057788 := proc(n)
        1/39916800*(2*n+11) *(n+10) *(n+9) *(n+8) *(n+7) *(n+6) *(n+5) *(n+4) *(n+3) *(n+2) *(n+ 1) ; end proc: # R. J. Mathar, Mar 22 2011

Table[(2*n+11)*Binomial[n+10, 10]/11, {n,0,40}] (* G. C. Greubel, Dec 02 2018 *)
CoefficientList[Series[(1 + x) / (1 - x)^12, {x, 0, 40}], x] (* Vincenzo Librandi, Feb 14 2016 *)
LinearRecurrence[{12,-66,220,-495,792,-924,792,-495,220,-66,12,-1},{1,13,90,442,1729,5733,16744,44200,107406,243542,520676,1058148},30] (* Harvey P. Dale, Sep 07 2022 *)

Vec((1+x)/(1-x)^12+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

[(2*n+11)*binomial(n+10, 10)/11 for n in range(40)] # G. C. Greubel, Dec 02 2018

a(n) = 2*C(n+11, 11) - C(n+10, 10). - Paul Barry, Mar 04 2003
a(n) = C(n+10,10) + 2*C(n+10,11). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
a(n) = C(n+10,10)*(2n+11)/11. - Antal Pinter, Dec 27 2015
a(n) = 12*a(n-1)-66*a(n-2)+220*a(n-3)-495*a(n-4)+792*a(n-5)-924*a(n-6)+792*a(n-7)-495*a(n-8)+220*a(n-9)-66*a(n-10)+12*a(n-11)-a(n-12) for n >11. - Vincenzo Librandi, Feb 14 2016
a(n) = (2*n+11)*binomial(n+10, 10)/11. - G. C. Greubel, Dec 02 2018
From Amiram Eldar, Jan 26 2022: (Start)
Sum_{n>=0} 1/a(n) = 419751541/13230 - 2883584*log(2)/63.
Sum_{n>=0} (-1)^n/a(n) = 720896*Pi/63 - 237793798/6615. (End)

cp's OEIS Frontend

A241619 T(n,k)=Number of length n+2 0..k arrays with no consecutive three elements summing to more than k.

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A006857 a(n) = binomial(n+5,5) * binomial(n+5,4)/(n+5).

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A220212 Convolution of natural numbers (A000027) with tetradecagonal numbers (A051866).

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A006324 a(n) = n*(n + 1)*(2*n^2 + 2*n - 1)/6.

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A008911 a(n) = n^2*(n^2 - 1)/6.

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A103905 Square array T(n,k) read by antidiagonals: number of tilings of an hexagon.

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A006324 a(n) = n(n + 1)(2n^2 + 2n - 1)/6.