A014062 -id:A014062 - cp's OEIS frontend

A060545 a(n) = binomial(n^2, n)/n.

1, 3, 28, 455, 10626, 324632, 12271512, 553270671, 28987537150, 1731030945644, 116068178638776, 8634941152058949, 705873715441872264, 62895036884524942320, 6067037854078498539696, 629921975126394617164575, 70043473196734767582082230

Offset: 1

Henry Bottomley, Apr 02 2001

Harry J. Smith, Table of n, a(n) for n = 1..100
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv preprint arXiv:1111.3057 [math.NT], 2011.

Cf. A005563, A014062, A060543.

A060545:=n->binomial(n^2,n)/n: seq(A060545(n), n=1..20); # Wesley Ivan Hurt, Sep 28 2016

Table[Binomial[n^2, n]/n, {n, 15}] (* Michael De Vlieger, Sep 28 2016 *)

a(n) = binomial(n^2, n)/n; \\ Harry J. Smith, Jul 06 2009

a(n) = A060543(n, n) = A014062(n)/n.
a(n+1) = C(A005563(n), n) for n >= 0. - Fred Daniel Kline, Sep 27 2016
From Peter Bala, Oct 22 2023: (Start)
a(p^r) == 1 (mod p^(3+r)) for all positive integers r and all primes p >= 5 (apply Meštrović, Remark 17, p. 12).
Conjecture: a(2*p^r) == 4*p^r - 1 (mod p^(3+r)) for all positive integers r and all primes p >= 5. (End)

More terms from Fred Daniel Kline, Sep 28 2016

A081370 Numbers k such that binomial(k^2, k) reduced mod k^2 is 0.

Original entry on oeis.org

1, 30, 105, 120, 132, 231, 252, 380, 495, 520, 595, 616, 630, 680, 756, 858, 870, 924, 1040, 1155, 1173, 1365, 1428, 1463, 1547, 1610, 1722, 1768, 1820, 1953, 1976, 1995, 2002, 2016, 2080, 2093, 2170, 2184, 2277, 2310, 2508, 2520, 2530, 2552, 2618, 2622

Offset: 1

Labos Elemer, Mar 20 2003

Amiram Eldar, Table of n, a(n) for n = 1..10000
Max Alekseyev, PARI/GP Scripts for Miscellaneous Math Problems: Binomial coefficients modulo integers (binomod.gp).

Cf. A014062, A081369.

Do[s=Mod[Binomial[n^2, n], n^2]; If[s==0, Print[n]], {n, 1, 10000}]
Select[Range[3000],Mod[Binomial[#^2,#],#^2]==0&] (* Harvey P. Dale, Aug 26 2025 *)

is(k) = binomod(k^2, k, k^2) == 0; \\ Amiram Eldar, Jul 30 2024, using Max Alekseyev's binomod.gp

A091144 a(n) = binomial(n^2, n)/(1+(n-1)*n).

Original entry on oeis.org

1, 1, 2, 12, 140, 2530, 62832, 1997688, 77652024, 3573805950, 190223180840, 11502251937176, 779092434772236, 58448142042957576, 4811642166029230560, 431306008583779517040, 41820546066482630185200

Offset: 0

Paul Barry, Dec 22 2003

Diagonal of array T(n,k) = binomial(kn,n)/(1+(k-1)n).
Number of paths up and left from (0,0) to (n^2-n,n) where x/y <= n-1 for all intermediate points. - Henry Bottomley, Dec 25 2003
Empirical: In the ring of symmetric functions over the fraction field Q(q, t), letting s(1^n) denote the Schur function indexed by (1^n), a(n) is equal to the coefficient of s(n) in nabla^(n)s(1^n) with q=t=1, where nabla denotes the "nabla operator" on symmetric functions, and s(n) denotes the Schur function indexed by the integer partition (n) of n. - John M. Campbell, Apr 06 2018

Vincenzo Librandi, Table of n, a(n) for n = 0..200
D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307-344.

Cf. A002061, A014062, A062993, A070914, A071201, A071202.

List([0..20],n->Binomial(n^2,n)/(1+(n-1)*n)); # Muniru A Asiru, Apr 08 2018

[Binomial(n^2, n)/(1+(n-1)*n): n in [0..20]]; // Vincenzo Librandi, Apr 07 2018

A091144 := proc(n)
    binomial(n^2,n)/(1+n*(n-1)) ;
end proc: # R. J. Mathar, Feb 14 2015

Table[Binomial[n^2, n] / (n (n - 1) + 1), {n, 0, 20}] (* Vincenzo Librandi, Apr 07 2018 *)

a(n) = binomial(n^2, n)/(n*(n-1)+1); \\ Altug Alkan, Apr 06 2018

From Henry Bottomley, Dec 25 2003: (Start)
a(n) = A014062(n)/A002061(n);
a(n) = A062993(n-2, n);
a(n) = A070914(n, n-1);
a(n) = A071201(n, n^2-n);
a(n) = A071201(n, n^2-n+1);
a(n) = A071202(n, n^2-n+1). (End)

A107444 a(n) = C(n^3, n).

Original entry on oeis.org

1, 28, 2925, 635376, 234531275, 131513824548, 104200375748469, 110859231254749120, 152494520486567153895, 263409560461970212832400, 558225230412816157198777770, 1424174931670379304734465767920, 4305884331150027666756637066361970, 15224493238177464079881126301239927128

Offset: 1

Zak Seidov, May 26 2005

Vincenzo Librandi, Table of n, a(n) for n = 1..65
Harlan J. Brothers, Pascal's triangle, Sidi polynomials, and powers of e, Missouri J. Math. Sci. (2025) Vol. 37, No. 1, 67-78.

Cf. A014062 (C(n^2, n)), A359665.

[Binomial(n^3, n): n in [1..30]]; // Vincenzo Librandi, Apr 22 2011

Mathematica
```
Table[Binomial[n^3, n], {n, 12}]
```
PARI
```
vector(100,n,binomial(n^3,n))
```

a(n) ~ exp(n) * n^(2*n - 1/2) / sqrt(2*Pi). - Vaclav Kotesovec, Jan 10 2023

A177456 a(n) = binomial(n^2,n+1)/n.

Original entry on oeis.org

2, 42, 1092, 35420, 1391280, 64425438, 3442573064, 208710267480, 14162980464360, 1063958304188780, 87677864005521636, 7865449972066576656, 763126447532235966816, 79629871834780293333510

Offset: 2

Michel Lagneau, May 09 2010

nonn

n divides binomial(n^2,n+1).
Proof 1 :(n+1)*binomial(n^2,n+1) = n*(n-1)*binomial(n^2,n) => n divide binomial(n^2,n+1) because gcd(n,n+1) = 1.
Proof 2 : a(n) = binomial(n^2,n+1)/n = (n-1)*binomial(n^2-2,n-1)=> a(n) is an integer. - Michel Lagneau, May 13 2010

			For n=4, 1092 is in the sequence because binomial(16,5)/4 = 4368/4 = 1092.

G. C. Greubel, Table of n, a(n) for n = 2..335
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Cf. A000984, A014062, A060545, A177234, A177784, A177788.

[Binomial(n^2,n+1)/n: n in [2..30]]; // G. C. Greubel, Apr 29 2024

with(numtheory):n0:=30:T:=array(1..n0-1):for n from 2 to n0 do:T[n-1]:= (binomial(n*n,n+1))/n:od:print(T):

Table[Binomial[n^2,n+1]/n, {n,2,30}] (* G. C. Greubel, Apr 29 2024 *)

[binomial(n^2,n+1)/n for n in range(2,31)] # G. C. Greubel, Apr 29 2024

a(n) = binomial(n^2,n+1)/n.
From G. C. Greubel, Apr 29 2024: (Start)
a(n) = (n-1)*A177234(n).
a(n) = (n-1)*A177788(n)/n.
a(n) = n*(n-1)*A177784(n).
a(n) = A014062(n)/n. (End)

A096130 Triangle read by rows: T(n,k) = binomial(k*n,n), 1 <= k <= n.

Original entry on oeis.org

1, 1, 6, 1, 20, 84, 1, 70, 495, 1820, 1, 252, 3003, 15504, 53130, 1, 924, 18564, 134596, 593775, 1947792, 1, 3432, 116280, 1184040, 6724520, 26978328, 85900584, 1, 12870, 735471, 10518300, 76904685, 377348994, 1420494075, 4426165368, 1, 48620, 4686825, 94143280, 886163135, 5317936260, 23667689815, 85113005120, 260887834350

Offset: 1

Amarnath Murthy, Jul 04 2004

			Triangle begins:
  1;
  1,   6;
  1,  20,   84;
  1,  70,  495,  1820;
  1, 252, 3003, 15504, 53130;
  ...

Seiichi Manyama, Rows n = 1..140, flattened

Row-sums give A096131. The leading diagonal is A014062. Cf. A096131.

Cf. A007318.

Flat(List([1..10],n->List([1..n],k->Binomial(k*n,n)))); # Muniru A Asiru, Aug 12 2018

a:=(n,k)->binomial(k*n,n): seq(seq(a(n,k),k=1..n),n=1..10); # Muniru A Asiru, Aug 12 2018

tabl(nrows) = {for (n=1, nrows, for (k=1, n, print1(binomial(k*n, n), ", ");); print(););} \\ Michel Marcus, May 14 2013

T(n, 1) = 1;
T(n, 2) = A000984(n) for n > 1;
T(n, 3) = A005809(n) for n > 2;
T(n, 4) = A005810(n) for n > 3;
T(n, n) = A014062(n).

Corrected and extended by Reinhard Zumkeller, Jan 09 2005

A177788 a(n) = binomial(n^2, n+1)/(n-1).

Original entry on oeis.org

4, 63, 1456, 44275, 1669536, 75163011, 3934369216, 234799050915, 15736644960400, 1170354134607658, 95648578915114512, 8520904136405458044, 821828481957792579648, 85317719822978885714475, 9485883860726883646713600, 1124586875214241546178986915

Offset: 2

Michel Lagneau, May 13 2010

nonn

The entries are integer for n >= 2 because binomial(n^2,n+1)/(n-1) = n*binomial(n^2-2,n-1), which is a product of two integers.

G. C. Greubel, Table of n, a(n) for n = 2..335

Cf. A000108, A014062, A060545, A177234, A177456, A177784.

[Binomial(n^2,n+1)/(n-1): n in [2..30]]; // G. C. Greubel, Apr 28 2024

n0:=30: T:=array(1..n0): T:=array(1..n0-1): for n from 2 to n0 do: T[n-1]:= (binomial(n^2,n+1))/(n-1): od: print(T):

Table[Binomial[n^2,n+1]/(n-1), {n,2,40}] (* G. C. Greubel, Apr 28 2024 *)

a(n) = binomial(n^2, n+1)/(n-1) \\ Charles R Greathouse IV, May 01 2024

[binomial(n^2,n+1)/(n-1) for n in range(2,31)] # G. C. Greubel, Apr 28 2024

a(n) = binomial(n^2,n+1)/(n-1).
a(n) = n * A177234(n).
a(n) = n^2 * A177784(n).

Removed redundant second Maple version - R. J. Mathar, May 14 2010

A276449 Number of 1-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.

Original entry on oeis.org

1, 0, 0, 4, 6, 0, 0, 120, 190, 0, 0, 7140, 11480, 0, 0, 635376, 1028790, 0, 0, 75287520, 122391522, 0, 0, 11143364232, 18161699556, 0, 0, 1978369382080, 3230129794320, 0, 0, 409663695276000, 669741609663270, 0, 0, 96930293990660064, 158625578809472060

Offset: 1

Jason Y.S. Chiu, Chiang, Tung-Ying, Hsiang-An Wang, Hong-Chang Wang, Sep 02 2016

The old name was: Number of ways to choose n points from an n X n grid so that they have 90-degree rotational symmetry.
Consider a square n X n grid with n^2 squares. Each of the n^2 squares comes in two colors.
(E.g., an n X n chessboard with only two black fields, or a binary n X n matrix).
There are N(n) = binomial(n^2,n) = A014062(n) such 2-color grids. We are interested in configurations where n squares are colored in one way, say black, and the remaining ones stay white. Only colored grids modulo rotation around some axis perpendicular to the board through its center are of interest. These rotations represent the cyclic group C_4. Under C_4 operations R(90)^k, k=1..4, there will only be orbits of order 1 (colored grids invariant under R(90)^1, hence any rotation) order 2 (two different grids each not invariant under R(90)^1 but R(90)^2 operation, transforming into each other) and order 4 (four different grids each not invariant under R(90)^k for k=1,2,3, but under R(4)^4, transforming into each other). The orbit structure is denoted by 1^(e(n,1)) 2^(e(n,2)) 4^(e(n,4)) with e(n, 2^j) nonnegative integers for j=0,1,2. One has Sum_{j=0,1,2} 2^j*e(n,2^j) = N(n), and Sum_{j=0,1,2} e(n,2^j) which is the total number of orbits, given in A276454(n).
For example, one of the four 1-orbits of 4 X 4 board. (o) white, (+) black:
   + o o +
   o o o o
   o o o o
   + o o +  ,
an example of a 2-orbit,
   + o + o   o o o +
   o o o o   + o o o
   o o o o   o o o +
   o + o +   + o o o  ,
an example of a 4-orbit,
   + + + +   o o o +   o o o o   + o o o
   o o o o   o o o +   o o o o   + o o o
   o o o o   o o o +   o o o o   + o o o
   o o o o   o o o +   + + + +   + o o o .
The present sequence a(n) gives the number of 1-orbits of such 2-colored boards with n squares of one color under C_4.

			a(4) = 4, the arrangements are as follows:
   + o o +   o + o o   o o + o   o o o o
   o o o o   o o o +   + o o o   o + + o
   o o o o   + o o o   o o o +   o + + o
   + o o +   o o + o   o + o o   o o o o
a(5) = 6, the arrangements are as follows:
   + o o o +   o + o o o   o o + o o
   o o o o o   o o o o +   o o o o o
   o o + o o   o o + o o   + o + o +
   o o o o o   + o o o o   o o o o o
   + o o o +   o o o + o   o o + o o
   and
   o o o + o   o o o o o   o o o o o
   + o o o o   o + o + o   o O + o o
   O o + o o   o o + o o   o + + + o
   o o o o +   o + O + O   o o + o o
   o + o o o   o o o o o   o o o o o
reformatted - _Wolfdieter Lang_, Oct 02 2016

Hong-Chang Wang, Table of n, a(n) for n = 1..100

Cf. A014062, A276451, A276452, A276454.

seq(op([binomial(2*i*(2*i+1),i),0,0,binomial(4*(i+1)^2,i+1)]),i=0..30); # Robert Israel, Sep 05 2016

Table[If[MemberQ[{2, 3}, #], 0, Function[i, Binomial[(2 i) (2 i + #), i]]@ Floor[n/4]] &@ Mod[n, 4], {n, 37}] (* Michael De Vlieger, Sep 07 2016 *)

import math
def nCr(n,r):
    f = math.factorial
    return f(n) / f(r) / f(n-r)
# main program
for j in range(101):
    if j%4 == 0:
        a = nCr((j*j/4),(j/4))
    elif j%4 == 1:
        a = nCr(((j-1)/2)*((j-1)/2+1),((j-1)/4))
    else:
        a = 0
    print(str(j)+" "+str(a))

a(n) = binomial((2*i)^2,i), for n = 4*i,
a(n) = binomial((2*i)*(2*i+1),i), for n = 4*i+1,
a(n) = 0, for others.

Edited: New name. Old name as a comment. Text substantially changed. Wolfdieter Lang, Oct 02 2016

A276451 Number of 2-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.

Original entry on oeis.org

0, 1, 2, 12, 30, 408, 1012, 17920, 45600, 1059380, 2730756, 78115884, 203235032, 6917206576, 18113945256, 714851008512, 1881039165696, 84449819514060, 223049005408900, 11225502116862880, 29736777118603962, 1658138369930988088, 4403069737450280832

Offset: 1

Chiang, Tung-Ying, Jason Y.S. Chiu, Hong-Chang Wang, Jiangshan Sun, Sep 03 2016

For a definition and examples of this problem see the comment section of A276449.

The present sequence a(n) gives the number of 2-orbits of such 2-color boards with n squares of one color under C_4.

			n = 4: one of the two 2-orbits is (o white, + black)
+ o + o   o o o +
o o o o   + o o o
o o o o   o o o +
o + o +   + o o o,
and one can take the first one as a representative.
For n = 3 there are a(3) = 2 2-orbits, represented by
+ o o       o o o
o + o  and  + + +
o o +       o o o.
The orbit structure for n=3 is 1^0 2^2 4^20; see A276449(3) = 0, a(3) = 2, A276452(3) = 20.
For the 12 2-orbits for n=4, see the representatives given in the link.

Hong-Chang Wang, Table of n, a(n) for n = 1..100
Hong-Chang Wang, Example for n = 4

Cf. A014062, A276449, A276452, A276454.

Table[(Function[j, Binomial[2 j (j + Boole@ OddQ@ n), j]]@ Floor[n/2] - If[MemberQ[{2, 3}, #], 0, Function[i, Binomial[(2 i) (2 i + #), i]]@ Floor[n/4]] &@ Mod[n, 4])/2, {n, 23}] (* Michael De Vlieger, Sep 07 2016 *)

import math
def nCr(n,r):
    f = math.factorial
    return f(n) / f(r) / f(n-r)
# main program
for j in range(101):
    i = j/2
    if j%2==0:
        b = nCr(2*i*i,i)
    else:
        b = nCr(2*i*(i+1),i)
    if j%4==0:
        c = nCr((j*j/4),(j/4))
    elif j%4==1:
        c = nCr(((j-1)/2)*((j-1)/2+1),((j-1)/4))
    else:
        c = 0
    print(str(j)+" "+str((b-c)/2))

a(n) = (binomial(2*i*i,i) - A276449(n))/2, for n = 2*i.

a(n) = (binomial(2*i*(i+1),i) - A276449(n))/2, for n = 2*i+1.

Edited: Wolfdieter Lang, Oct 02 2016

A276452 Number of 4-orbits of the cyclic group C_4 for a bi-colored square n X n grid with n squares of one color.

Original entry on oeis.org

0, 1, 20, 448, 13266, 486744, 21474640, 1106532352, 65221935740, 4327576834420, 319187489891256, 25904823417117120, 2294089575084464472, 220132629092378694832, 22751391952785312551232, 2519687900505221042995200, 297684761086121821704009432, 37370623083548749203599933004

Offset: 1

Hong-Chang Wang, Jason Y.S. Chiu, Chiang, Tung-Ying, Sep 03 2016

For a definition and examples of this problem see the comment section of A276449. The present sequence a(n) gives the number of 4-orbits under C_4 of such 2-colored n X n grids with n squares of one color.

			a(2) = 1: the 4-orbit is
+ +   o +   o o   + o
o o   o +   + +   + o  ,
and one can take the first one as representative.
For n = 3 there are a(3) = 20 4-orbits, represented by
+ + +   + + o   + + o   + + o   + + o
o o o   + o o   o + o   o o +   o o o
o o o   o o o   o o o   o o o   + o o
--------------------------------------
+ + o   + + o   + o +   + o +   + o +
o o o   o o o   + o o   o + o   o o o
o + o   o o +   o o o   o o o   + o o
--------------------------------------
+ o +   + o o   + o o   + o o   + o o
o o o   + + o   + o +   + o o   + o o
o + o   o o o   o o o   o + o   o o +
--------------------------------------
+ o o   + o o   + o o   o + o   o + o
o + +   o + o   o o +   + + o   + o +
o o o   o + o   o + o   o o o   o o o .
--------------------------------------
The complete orbit structure for n=3 is 1^0 2^2 4^20, see A276449(3) = 0, A276451(3) = 2, a(3) = 20

Hong-Chang Wang, Table of n, a(n) for n = 1..69

Cf. A014062, A276449, A276451, A276454.

f[n_] := If[MemberQ[{2, 3}, #], 0, Function[i, Binomial[(2 i) (2 i + #), i]]@ Floor[n/4]] &@ Mod[n, 4]; g[n_] := (Function[j, Binomial[2 j (j + Boole@ OddQ@ n), j]]@ Floor[n/2] - f@ n)/2; Table[(Binomial[n^2, n] - 2 g@ n - f@ n)/4, {n, 18}] (* Michael De Vlieger, Sep 07 2016 *)

import math
def nCr(n,r):
    f = math.factorial
    return f(n) / f(r) / f(n-r)
# main program
for j in range(101):
    a = nCr(j*j,j)
    i = j/2
    if j%2==0:
        b = nCr(2*i*i,i)
    else:
        b = nCr(2*i*(i+1),i)
    print(str(j)+" "+str((a-b)/4))

a(n) = (A014062(n) - A276451(n)*2 - A276449(n))/4 for n = 1, 2, 3, ...

cp's OEIS Frontend

A060545 a(n) = binomial(n^2, n)/n.

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A081370 Numbers k such that binomial(k^2, k) reduced mod k^2 is 0.

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A091144 a(n) = binomial(n^2, n)/(1+(n-1)*n).

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A107444 a(n) = C(n^3, n).

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A177456 a(n) = binomial(n^2,n+1)/n.

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A096130 Triangle read by rows: T(n,k) = binomial(k*n,n), 1 <= k <= n.

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A177788 a(n) = binomial(n^2, n+1)/(n-1).

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