A006542 -id:A006542 - cp's OEIS frontend

A169937 a(n) = binomial(m+n-1,n)^2 - binomial(m+n,n+1)*binomial(m+n-2,n-1) with m = 14.

1, 91, 3185, 63700, 866320, 8836464, 71954064, 488259720, 2848181700, 14620666060, 67255063876, 281248448936, 1081724803600, 3863302870000, 12914469594000, 40680579221100, 121443493851225, 345280521733875, 938920716995625, 2451077240157000, 6162708489537600

Offset: 0

N. J. A. Sloane, Aug 28 2010

13th column (and diagonal) of the triangle A001263. - Bruno Berselli, May 07 2012

S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; Prop. 8.4, case n=14.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (25, -300, 2300, -12650, 53130, -177100, 480700, -1081575, 2042975, -3268760, 4457400, -5200300, 5200300, -4457400, 3268760, -2042975, 1081575, -480700, 177100, -53130, 12650, -2300, 300, -25, 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.

Cf. A002378.

[(1/13)*Binomial(n+12,12)^2*(n+13)/(n+1): n in [0..20]]; // Bruno Berselli, Nov 09 2011

f:=m->[seq( binomial(m+n-1,n)^2-binomial(m+n,n+1)*binomial(m+n-2,n-1), n=0..20)]; f(14);

Table[Binomial[13+n,n]^2-Binomial[14+n,n+1]Binomial[12+n,n-1],{n,0,20}] (* Harvey P. Dale, Nov 09 2011 *)

a(n)=binomial(n+12,12)^2*(n+13)/(n+1)/13 \\ Charles R Greathouse IV, Nov 09 2011

a(n) = (1/13)*A010965(n+12)^2*(n+13)/(n+1). - Bruno Berselli, Nov 09 2011
a(n) = Product_{i=1..12} A002378(n+i)/A002378(i). - Bruno Berselli, Sep 01 2016
From Amiram Eldar, Oct 19 2020: (Start)
Sum_{n>=0} 1/a(n) = 45997360927193/23100 - 201753552*Pi^2.
Sum_{n>=0} (-1)^n/a(n) = 16431564019/23100 - 72072*Pi^2. (End)

A004282 a(n) = n(n+1)^2(n+2)^2/12.

Original entry on oeis.org

0, 3, 24, 100, 300, 735, 1568, 3024, 5400, 9075, 14520, 22308, 33124, 47775, 67200, 92480, 124848, 165699, 216600, 279300, 355740, 448063, 558624, 690000, 845000, 1026675, 1238328, 1483524, 1766100, 2090175

Offset: 0

N. J. A. Sloane

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A006542, A107891, A114242.

[n*(n+1)^2*(n+2)^2/12: n in [0..50]]; // Vincenzo Librandi, Feb 09 2012

a:= n-> binomial(2+n, 2)*binomial(2+n, 3): seq(a(n), n=0..31); # Zerinvary Lajos, Apr 26 2007

Table[n*(n+1)^2*(n+2)^2/12,{n,0,40}] (* Vincenzo Librandi, Feb 09 2012 *)

a(n) = binomial(n+2,2)*binomial(n+2,3); \\ Charles R Greathouse IV, Feb 09 2012

a(n) = C(2+n, 2)*C(2+n, 3) = A000217(n+1)*A000292(n). - Zerinvary Lajos, Jan 10 2006
a(n-1) = Sum_{1 <= x_1, x_2 <= n} x_1*(det V(x_1,x_2))^2 = Sum_{1 <= i,j <= n} i*(i-j)^2, where V(x_1,x_2) is the Vandermonde matrix of order 2. - Peter Bala, Sep 21 2007
G.f.: x*(3+6*x+x^2)/(1-x)^6. - Colin Barker, Feb 09 2012
a(n) = Sum_{k=0..n} Sum_{i=0..n} (n-i+1) * C(k+1,k-1). - Wesley Ivan Hurt, Sep 21 2017
a(n) = A004302(n+1) - A000537(n+1). - J. M. Bergot, Mar 28 2018
From Amiram Eldar, May 29 2022: (Start)
Sum_{n>=1} 1/a(n) = 30 - 3*Pi^2.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/2 - 24*log(2) + 12. (End)

A120247 Triangle of Hankel transforms of binomial(n+k, k).

Original entry on oeis.org

1, 1, -1, 1, -3, -1, 1, -6, -10, 1, 1, -10, -50, 35, 1, 1, -15, -175, 490, 126, -1, 1, -21, -490, 4116, 5292, -462, -1, 1, -28, -1176, 24696, 116424, -60984, -1716, 1, 1, -36, -2520, 116424, 1646568, -3737448, -736164, 6435, 1, 1, -45, -4950, 457380, 16818516, -133613766, -131589315, 9202050, 24310, -1

Offset: 0

Paul Barry, Jun 12 2006

Columns include -A000217, -A006542, A107915.
Row k is the Hankel transform of C(n+k, k).
The matrix inverse starts
          1;
          1,       -1;
         -2,        3,        -1;
        -15,       24,       -10,       1;
        434,     -700,       300,     -35,      1;
      47670,   -76950,     33075,   -3920,    126,  -1;
  -19787592, 31943835, -13733720, 1629936, -52920, 462, -1; - R. J. Mathar, Mar 22 2013

			Triangle begins
  1;
  1,  -1;
  1,  -3,    -1;
  1,  -6,   -10,     1;
  1, -10,   -50,    35,      1;
  1, -15,  -175,   490,    126,     -1;
  1, -21,  -490,  4116,   5292,   -462,    -1;
  1, -28, -1176, 24696, 116424, -60984, -1716,   1;

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

Columns include: A000217, A006542, A107915.

Cf. A120248.

p:= func< m,k | k eq 0 select 1 else (&*[Binomial(m+j, k+1): j in [1..k]]) >;
A120247:= func< n,k | (-1)^Floor((k+1)/2)*p(n,k)/p(k,k) >;
[A120247(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 15 2023

A120247 := proc(n,k)
    (cos(Pi*k/2)-sin(Pi*k/2))*mul(binomial(n+j+1,k+1),j=0..k-1)/mul(binomial(k+j+1,k+1),j=0..k-1) ;
    simplify(%) ;
end proc: # R. J. Mathar, Mar 22 2013

p[m_, k_]:= Product[Binomial[m+j, k+1], {j,k}];
T[n_, k_]:= (-1)^Floor[(k+1)/2]*p[n,k]/p[k,k];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 15 2023 *)

def p(m,k): return product(binomial(m+j+1,k+1) for j in range(k))
def A120247(n,k): return (-1)^((k+1)//2)*p(n,k)/p(k,k)
flatten([[A120247(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Mar 15 2023

T(n, k) = (cos(pi*k/2) - sin(pi*k/2))*( Product{j=0..k-1} C(n+j+1, k+1)/Product{j=0..k-1} C(k+j+1, k+1) ).

A253285 a(n) = RF(n+1,3)*C(n+2,n-1), where RF(a,n) is the rising factorial.

Original entry on oeis.org

0, 24, 240, 1200, 4200, 11760, 28224, 60480, 118800, 217800, 377520, 624624, 993720, 1528800, 2284800, 3329280, 4744224, 6627960, 9097200, 12289200, 16364040, 21507024, 27931200, 35880000, 45630000, 57493800, 71823024, 89011440, 109498200, 133771200, 162370560

Offset: 0

Peter Luschny, Mar 23 2015

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A002378, A006542, A083374, A133754, A253284.

List([0..40], n -> n*((n+1)*(n+2))^2*(n+3)/6); # Bruno Berselli, Mar 06 2018

[n*((n+1)*(n+2))^2*(n+3)/6: n in [0..40]]; // Bruno Berselli, Mar 06 2018

Maple
```
seq(n*((n+1)*(n+2))^2*(n+3)/6,n=0..19);
```

Table[n ((n + 1) (n + 2))^2 (n + 3)/6, {n, 0, 40}] (* Bruno Berselli, Mar 06 2018 *)
LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,24,240,1200,4200,11760,28224},40] (* Harvey P. Dale, Aug 05 2024 *)

[n*((n+1)*(n+2))**2*(n+3)/6 for n in range(40)] # Bruno Berselli, Mar 06 2018

[n*((n+1)*(n+2))^2*(n+3)/6 for n in (0..40)] # Bruno Berselli, Mar 06 2018

G.f.: -24/(x-1)^4 - 144/(x-1)^5 - 240/(x-1)^6 - 120/(x-1)^7. See the comment in A253284 for the general case.
a(n) = n*((n+1)*(n+2))^2*(n+3)/6.
a(n) = (N^3 + 4*N^2 + 4*N)/6 = N*(N + 2)^2/6 with N = n^2 + 3*n.
From Bruno Berselli, Mar 06 2018: (Start)
a(n) = 24*A006542(n+3) for n>0.
a(n) = Sum_{i=0..n} i*(i+1)^3*(i+2). Therefore, the first differences are in A133754. (End)

A296419 Triangle T(i,j) read by rows: Number of plane bipolar orientations with i+1 vertices and j+1 faces.

Original entry on oeis.org

1, 1, 4, 1, 10, 50, 1, 20, 175, 980, 1, 35, 490, 4116, 24696, 1, 56, 1176, 14112, 116424, 731808, 1, 84, 2520, 41580, 457380, 3737448, 24293412, 1, 120, 4950, 108900, 1557270, 16195608, 131589315, 877262100, 1, 165, 9075, 259545, 4723719, 61408347, 614083470, 4971151900, 33803832920

Offset: 1

R. J. Mathar, Feb 25 2018

			The triangle starts in row 1 as
  1;
  1,   4;
  1,  10,   50;
  1,  20,  175,    980;
  1,  35,  490,   4116,   24696;
  1,  56, 1176,  14112,  116424,   731808;
  1,  84, 2520,  41580,  457380,  3737448,  24293412;
  1, 120, 4950, 108900, 1557270, 16195608, 131589315, 877262100;

E. Fusy, D. Poulalhon, and G. Schaeffer, Bijective counting of plane bipolar orientations, El. Notes Discr. Math. 29 (2007) 283-287.

Cf. rows/columns: A006542, A047819, A107915, A140901, A140903, A140907.

A296419 := proc(i,j)
    2*(i+j-2)!*(i+j-1)!*(i+j)!/(i-1)!/i!/(i+1)!/(j-1)!/j!/(j+1)! ;
end proc:
seq(seq(A296419(i,j),j=1..i),i=1..10) ;

T(i,j) = T(j,i) = 2*(i+j-2)!*(i+j-1)!*(i+j)!/((i-1)!*i!*(i+1)!*(j-1)!*j!*(j+1)!).

A071910 a(n) = t(n)t(n+1)t(n+2), where t() are the triangular numbers.

Original entry on oeis.org

0, 18, 180, 900, 3150, 8820, 21168, 45360, 89100, 163350, 283140, 468468, 745290, 1146600, 1713600, 2496960, 3558168, 4970970, 6822900, 9216900, 12273030, 16130268, 20948400, 26910000, 34222500, 43120350, 53867268, 66758580, 82123650, 100328400, 121777920

Offset: 0

N. J. A. Sloane, Jun 13 2002

nonn

a(n) is also the number of three-dimensional cage assemblies such that the assembly is not a cube. See also A052149 for the two-dimensional version and to A059827 for the non-exclusive version. - Alejandro Rodriguez, Oct 20 2020

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A006542, (first differences of a(n) /18) A006414, (second differences of a(n) /18) A006322, (third differences of a(n) /18) A004068, (fourth differences of a(n) /18) A005891, (fifth differences of a(n) /18) A008706.

Join[{0},Times@@@Partition[Accumulate[Range[40]],3,1]] (* or *) LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,18,180,900,3150,8820,21168},40] (* Harvey P. Dale, Aug 08 2025 *)

t(n) = n*(n+1)/2;
a(n) = t(n)*t(n+1)*t(n+2); \\ Michel Marcus, Oct 21 2015

a(n) = 18*A006542(n+3). - Vladeta Jovovic, Jun 14 2002
G.f.: 18*x*(1+3*x+x^2)/(1-x)^7. - Vladeta Jovovic, Jun 14 2002
a(n) = ((n+1)*(n+2))^3/8 - Sum_{i=1..n+1} i^3. - Jon Perry, Feb 13 2004
a(n) = C(2+n, n)*C(3+n, 1+n)*C(4+n, 2+n). - Zerinvary Lajos, Jul 29 2005
a(n) = A059827(n+1) - A000537(n+1). - Michel Marcus, Oct 21 2015

A256551 Triangle read by rows, T(n,k) matrix inverse of A256550, for n>=0 and 0<=k<=n.

Original entry on oeis.org

1, 0, 1, 0, -1, 1, 0, 1, -3, 1, 0, 1, 6, -6, 1, 0, -15, 10, 20, -10, 1, 0, 48, -225, 50, 50, -15, 1, 0, 581, 1008, -1575, 175, 105, -21, 1, 0, -11069, 16268, 9408, -7350, 490, 196, -28, 1, 0, 20784, -398484, 195216, 56448, -26460, 1176, 336, -36, 1

Offset: 0

Peter Luschny, Apr 01 2015

			Triangle starts:
[1]
[0,   1]
[0,  -1,    1]
[0,   1,   -3,     1]
[0,   1,    6,    -6,   1]
[0, -15,   10,    20, -10,   1]
[0,  48, -225,    50,  50, -15,   1]
[0, 581, 1008, -1575, 175, 105, -21, 1]

Cf. A000217, A002415, A006542, A256550.

T(n+1,n) = -A000217(n).
T(n+2,n) = A002415(n+1).
For T(n,1) compare A006542.

A107981 Triangle read by rows: T(n,k) = (k+1)(k+2)(n+2)(n+3)(6n^2 - 8n*k + 18n + 3k^2 - 11k + 12)/144 for 0<=k<=n.

Original entry on oeis.org

1, 6, 10, 20, 40, 50, 50, 110, 155, 175, 105, 245, 371, 455, 490, 196, 476, 756, 980, 1120, 1176, 336, 840, 1380, 1860, 2220, 2436, 2520, 540, 1380, 2325, 3225, 3975, 4515, 4830, 4950, 825, 2145, 3685, 5225, 6600, 7700, 8470, 8910, 9075, 1210, 3190

Offset: 0

Emeric Deutsch, Jun 12 2005

Kekulé numbers for certain benzenoids. Column 0 yields A002415. Main diagonal yields A006542.

			Triangle begins:
1;
6,10;
20,40,50;
50,110,155,175;

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 237, K{F(n,3,l)}).

Cf. A002415, A006542.

T:=proc(n,k) if k<=n then 1/144*(k+1)*(k+2)*(n+2)*(n+3)*(6*n^2-8*n*k+18*n+3*k^2-11*k+12) else 0 fi end: for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

A107983 Triangle read by rows: T(n,k) = (k+1)(n+2)(n+3)(n-k+2)(n-k+1)/12 for 0<=k<=n.

Original entry on oeis.org

1, 6, 4, 20, 20, 10, 50, 60, 45, 20, 105, 140, 126, 84, 35, 196, 280, 280, 224, 140, 56, 336, 504, 540, 480, 360, 216, 84, 540, 840, 945, 900, 750, 540, 315, 120, 825, 1320, 1540, 1540, 1375, 1100, 770, 440, 165, 1210, 1980, 2376, 2464, 2310, 1980, 1540

Offset: 0

Emeric Deutsch, Jun 12 2005

Kekulé numbers for certain benzenoids. Column 0 yields A002415. Main diagonal yields A000292. Row sums yield A006542.

T(n,k) = number of Dyck (n+4)-paths with 4 peaks (UDs) and last descent of length k+1. For example, T(1,1)=4 counts UUDUDUDUDD, UDUUDUDUDD, UDUDUUDUDD, UDUDUDUUDD. - David Callan, Jun 26 2006

			Triangle begins:
1;
6,4;
20,20,10;
50,60,45,20;

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 237, K{F(n,3,-l)}).

Cf. A002415, A000292, A006542.

T:=proc(n,k) if k<=n then (k+1)*(n+2)*(n+3)*(n-k+2)*(n-k+1)/12 else 0 fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

Flatten[Table[((k+1)(n+2)(n+3)(n-k+2)(n-k+1))/12,{n,0,10},{k,0,n}]] (* Harvey P. Dale, Aug 08 2013 *)

cp's OEIS Frontend

A169937 a(n) = binomial(m+n-1,n)^2 - binomial(m+n,n+1)*binomial(m+n-2,n-1) with m = 14.

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A004282 a(n) = n*(n+1)^2*(n+2)^2/12.

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A120247 Triangle of Hankel transforms of binomial(n+k, k).

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A253285 a(n) = RF(n+1,3)*C(n+2,n-1), where RF(a,n) is the rising factorial.

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A296419 Triangle T(i,j) read by rows: Number of plane bipolar orientations with i+1 vertices and j+1 faces.

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A071910 a(n) = t(n)*t(n+1)*t(n+2), where t() are the triangular numbers.

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A256551 Triangle read by rows, T(n,k) matrix inverse of A256550, for n>=0 and 0<=k<=n.

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A004282 a(n) = n(n+1)^2(n+2)^2/12.

A071910 a(n) = t(n)t(n+1)t(n+2), where t() are the triangular numbers.