A177456 -id:A177456 - 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)

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

A177466 a(n) = binomial(n^3, n^2) / (n^2 + n + 1).

Original entry on oeis.org

10, 360525, 23263187479980, 4195317468983232014706855, 3118254010126197540790713959812283024388, 13329519847131745416659896296893907619682838146506167497550

Offset: 2

Michel Lagneau, May 09 2010

nonn

All entries are integers. [Proof: binomial(n^3, n^2) / (n^2 + n + 1) = n^3 (n^3 - 1) (n^3 - 2)*...*(n^3- n^2 +1) / ( (n^2)! *(n^2 + n + 1)). With n^3 - 1 = (n-1)*(n^2 + n + 1), we obtain a(n) = n* binomial(n^3-2, n^2-2) / (n+1). Finally: (n+1) * binomial(n^3, n^2) * 1/ (n^2 + n + 1) = n*binomial(n^3-2, n^2-2). QED]
The step after "finally" seems to demonstrate merely that (n+1)*a(n) is an integer, but not that a(n) is itself an integer. Is the proof incomplete? - R. J. Mathar, Dec 06 2010
So far all that has been shown is that (n+1)*a(n) is an integer. To complete the proof, note that a(n) = n^3*(n-1)*(n^3-2)*...*(n^3-n^2) / (n^2*(n^2-1)!*(n^3-n^2)) = binomial(n^3-2,n^2-1)/n. Hence n*a(n) is also an integer, and so (n+1)*a(n) - n*a(n) = a(n) is an integer. Q.E.D. - N. J. A. Sloane, Dec 09 2010

			For n = 2, a(2) = binomial(8,4)/7 = 70/7 = 10.

G. C. Greubel, Table of n, a(n) for n = 2..23
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.

Cf. A000108, A000984, A060545, A177234, A177456.

[Binomial(n^3,n^2)/(n^2+n+1): n in [2..12]]; // G. C. Greubel, Jul 18 2024

A177466 := proc(n) binomial(n^3,n^2)/(n^2+n+1); end proc:
seq(A177466(n),n=2..10) ; # R. J. Mathar, Dec 06 2010

Table[Binomial[n^3,n^2]/(n^2+n+1),{n,2,7}] (* Harvey P. Dale, Jan 24 2019 *)

[binomial(n^3,n^2)/(n^2+n+1) for n in range(2,13)] # G. C. Greubel, Jul 18 2024

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

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A177466 a(n) = binomial(n^3, n^2) / (n^2 + n + 1).

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