A177784 -id:A177784 - cp's OEIS frontend

A014062 a(n) = binomial(n^2, n).

1, 1, 6, 84, 1820, 53130, 1947792, 85900584, 4426165368, 260887834350, 17310309456440, 1276749965026536, 103619293824707388, 9176358300744339432, 880530516383349192480, 91005567811177478095440, 10078751602022313874633200, 1190739044344491048895397910

Offset: 0

N. J. A. Sloane

nonn

Roberts states that Gupta and Khare show that a(n) > A002110(n) for 2 < n < 1794 and that a(n) < A002110(n) for n >= 1794, where A002110(n) is the product of the first n primes. - T. D. Noe, Oct 03 2007
This sequence describes the number of ways to arrange n objects in an n X n array (for example, stars in a flag's field pattern). - Tom Young (mcgreg265(AT)msn.com), Jun 17 2010
It appears that a(n) == n (mod n^3) only if n is 1, an odd prime, the square of an odd prime, or the cube of an odd prime. - Gary Detlefs, Aug 06 2013; corrected by Michel Marcus, May 29 2015

J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 265.

T. D. Noe, Table of n, a(n) for n=0..100
Horst A. Alzer and Jozsef J. B. Šandor, On a binomial coefficient and a product of prime numbers, Appl. An. Disc. Math. 5 (2011) 87-92.
Harlan J. Brothers, Pascal's Prism: Supplementary Material.
Harlan J. Brothers, Pascal's triangle, Sidi polynomials, and powers of e, Missouri J. Math. Sci. (2025) Vol. 37, No. 1, 67-78.
Hansraj Gupta and S. P. Khare, On C(k^2,k) and the product of the first k primes, Publ. Fac. Electrotechn. Belgrade, Ser. Math. Phys. 25-29 (1977) 577-598.

Main diagonal of A060539.

Cf. A177234, A177456, A177784, A177788, A272094, A295773.

[Binomial(n^2,n): n in [0..30]]; // G. C. Greubel, Apr 29 2024

Table[Binomial[n^2,n],{n,0,22}] (* Vladimir Joseph Stephan Orlovsky, Mar 03 2011 *)
Table[SeriesCoefficient[(1+x)^(n^2), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Aug 06 2025 *)

{a(n) = sum(k=0, n, binomial(n, k)*binomial(n^2-n, k))}
for(n=0,20,print1(a(n),", ")) \\ Paul D. Hanna, Nov 18 2015

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

a(n) ~ 1/sqrt(2*Pi) * (e*n)^(n - 1/2). - Charles R Greathouse IV, Jul 07 2007
a(n) = Sum_{k=0..n} binomial(n, k) * binomial(n^2 - n, k). - Paul D. Hanna, Nov 18 2015
a(n) = (n+1)*A177234(n). - R. J. Mathar, Jan 25 2019
From G. C. Greubel, Apr 29 2024: (Start)
a(n) = n*(n+1)*A177784(n).
a(n) = (n+1)*A177456(n)/(n-1).
a(n) = (n+1)*A177788(n)/n. (End)
a(n) = [x^n] (1+x)^(n^2). - Vaclav Kotesovec, Aug 06 2025

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

Original entry on oeis.org

2, 21, 364, 8855, 278256, 10737573, 491796152, 26088783435, 1573664496040, 106395830418878, 7970714909592876, 655454164338881388, 58702034425556612832, 5687847988198592380965, 592867741295430227919600

Offset: 2

Michel Lagneau, May 05 2010, May 08 2010

nonn

Theorem: binomial(n^2, n)/(n+1) is an integer for n >= 2.
Proof 1 from William J. Keith, May 08 2010:
binomial(n^2, n) * 1/(n+1)
= (n^2)*(n^2-1)*(n^2-2)!/((n^2-n)!*n*(n-1)*(n-2)!) * 1/(n+1)
= n*(n^2-2)!/((n^2-n)!*(n-2)!) = n * binomial(n^2-2,n-2). QED
Proof 2 from Max Alekseyev, May 08 2010:
Recall that the valuation of m! w.r.t. prime p equals the sum floor(m/p^i) over i=1,2,3,...
Moreover, if m=a+b where a and b are nonnegative integers, then floor(m/p^i) - floor(a/p^i) - floor(b/p^i) >= 0.
Let n>1. To prove that binomial(n^2, n)/(n+1) is an integer, it is enough to show that its valuation w.r.t. any prime p is nonnegative.
It is clear that trouble may come only from primes dividing n+1.
Let valuation(n+1,p)=k > 0, i.e., n+1=p^k*m where prime p does not divide m.
Then n = p^k*m - 1, n^2 = p^(2k)*m^2 - 2*p^k*m + 1 and n^2 - n = p^(2k)*m^2 - 3*p^k*m + 2.
It is easy to check that floor(n^2/p^i) - floor(n/p^i) - floor((n^2-n)/p^i) = 1 for i=1,2,...,k if p>2 and for i=2,3,...,k+1 if p=2, implying that valuation(binomial(n^2, n)/(n+1),p) >= 0. QED

			a(3) = 21 because binomial(9,3)/(3+1) = 84/4 = 21.

H. Gupta and S. P. Khare, On C(k^2,k) and the product of the first k primes, Publ. Fac. Electrotechn. Belgrade, Ser. Math. Phys. 25-29 (1977) 577-598.

G. C. Greubel, Table of n, a(n) for n = 2..335
H. Gupta and S. P. Khare, On C(k^2,k) and the product of the first k primes, Publ. Fac. Electrotechn. Belgrade, Ser. Math. Phys. 25-29 (1977) 577-598. [PDF] [From _R. J. Mathar_, May 09 2010]

Cf. A014062 A123312, A177456, A177784, A177788.

[Binomial(n^2,n)/(n+1): n in [2..30]]; // G. C. Greubel, Apr 27 2024

with(numtheory):n0:=25:T:=array(1..n0-1):for n from 2 to n0 do: T[n-1]:= binomial(n*n,n)/(n+1):od:print(T):

Table[Binomial[n^2,n]/(n+1), {n,2,30}] (* G. C. Greubel, Apr 27 2024 *)

[binomial(n^2,n)/(n+1) for n in range(2,31)] # G. C. Greubel, Apr 27 2024

From G. C. Greubel, Apr 27 2024: (Start)
a(n) = A014062(n)/(n+1).
a(n) = A177456(n)/(n-1).
a(n) = n*A177784(n).
a(n) = A177788(n)/n. (End)

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)

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

A177837 a(n) = binomial(n^3-2, n-2).

Original entry on oeis.org

1, 25, 1891, 302621, 84957251, 37307689133, 23728431347335, 20688443967788245, 23730591032609929084, 34687456062438088435890, 62994291032837018079196115, 139227352512368728514134480110, 368132634640135991872548754745138, 1147827724811251389730308940150980661

Offset: 2

Michel Lagneau, May 14 2010

This is the case p=3 of a(n,p) = binomial(n^p,n) / ( PHI(n,p) * n^(p-1)) where PHI(n,p) = 1 + n + n^2 + ... + n^(p-1) = (n^p - 1) /(n - 1).
These a(n,p) are integer if n, p > = 2. [Proof :
a(n,p) = binomial(n^p,n)* 1 / (n^(p-1)*PHI(n,p))
= n^p *(n^p - 1)*(n^p - 2)...(n^p - n + 1)/((n-2)!*(n-1)*n * n^(p-1)* PHI(n,p)).
Insert PHI(n,p)=(n^p - 1) /(n - 1) and cancel n^p, n-1 and n^p - 1 where n > = 2:
a(n,p) = (n^p - 2)*(n^p - 3)...(n^p - n + 1)/(n - 2)! = binomial (n^p - 2, n - 2). QED]

			a(3) = binomial(3^3-2, 3-2) = binomial(25, 1) = 25.

G. C. Greubel, Table of n, a(n) for n = 2..195

Cf. A177234, A177784 (case p = 2).

[Binomial(n^3-2, n-2): n in [2..30]]; // G. C. Greubel, Jul 18 2024

with(numtheory): n0:=30: T:=array(1..n0): T:=array(1..n0-1):
for n from 2 to n0 do: p:=3: T[n-1]:= (n-1)*(binomial(n^p,n))/((n^(p-1))*(n^p-1)): od: print(T):

Table[Binomial[n^3-2,n-2],{n,2,20}] (* Harvey P. Dale, Aug 08 2013 *)

[binomial(n^3-2,n-2) for n in range(2,31)] # G. C. Greubel, Jul 18 2024

a(n) = binomial(n^3, n) / (n^2 * (n^2 + n + 1) ).

Swapped general and specific definitions - R. J. Mathar, Oct 12 2010

A351501 a(n) = binomial(n^2 + n - 1, n) / (n^2 + n - 1).

Original entry on oeis.org

1, 2, 15, 204, 4095, 109668, 3689595, 149846840, 7141879503, 391139588190, 24218296445200, 1673538279265020, 127715832778905150, 10670643284149377480, 968929726650218004435, 95024894699780159868144, 10011211830149283223044015

Offset: 1

F. Chapoton, May 03 2022

nonn

Empirical: In the ring of symmetric functions over the fraction field Q(q, t), let s(n) denote the Schur function indexed by n. Then (up to sign) a(n) is the coefficient of s(1^n) in nabla^(n) s(n) with q=t=1, where nabla denotes the "nabla operator" on symmetric functions.

Closely related to A177784. See also A091144.

Diagonal of A162382. Multiple of A182316.

Table[With[{c=n^2+n-1},Binomial[c,n]/c],{n,20}] (* Harvey P. Dale, Jan 01 2024 *)

from math import comb
def A351501(n): return comb(m := n**2+n-1,n)//m # Chai Wah Wu, May 07 2022

[binomial(n*n+n-1,n)/(n*n+n-1) for n in range(1,29)]

a(n) ~ c*exp(n-1/(6*n))*n^(n-5/2), where c = sqrt(e/(2*Pi)). - Stefano Spezia, May 04 2022

a(n) = n * A182316(n - 1). - F. Chapoton, Sep 22 2023

A386558 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals: A(n,k) = binomial((k+1)*n+k-1,n)/(n+1).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 2, 0, 1, 3, 7, 5, 0, 1, 4, 15, 30, 14, 0, 1, 5, 26, 91, 143, 42, 0, 1, 6, 40, 204, 612, 728, 132, 0, 1, 7, 57, 385, 1771, 4389, 3876, 429, 0, 1, 8, 77, 650, 4095, 16380, 32890, 21318, 1430, 0, 1, 9, 100, 1015, 8184, 46376, 158224, 254475, 120175, 4862, 0

Offset: 0

Seiichi Manyama, Jul 26 2025

			Square array begins:
  1,   1,    1,     1,      1,      1,       1, ...
  0,   1,    2,     3,      4,      5,       6, ...
  0,   2,    7,    15,     26,     40,      57, ...
  0,   5,   30,    91,    204,    385,     650, ...
  0,  14,  143,   612,   1771,   4095,    8184, ...
  0,  42,  728,  4389,  16380,  46376,  109668, ...
  0, 132, 3876, 32890, 158224, 548340, 1533939, ...

Wikipedia, Fuss-Catalan number

Columns k=0..10 give A000007, A000108, A006013, A006632, A118971, A130564(n+1), A130565(n+1), A234466, A234513, A234573, A235340.
Main diagonal gives A177784(n+1).
Cf. A162382.

a(n, k) = binomial((k+1)*n+k-1, n)/(n+1);

For k > 0, A(n,k) = r * binomial(n*p+r,n)/(n*p+r), the Fuss-Catalan number with p=k+1 and r=k.

G.f. of column k: (1/x) Series_Reverion( x*(1-x)^k ).

cp's OEIS Frontend

A014062 a(n) = binomial(n^2, n).

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

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

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

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A177837 a(n) = binomial(n^3-2, n-2).

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A351501 a(n) = binomial(n^2 + n - 1, n) / (n^2 + n - 1).

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