A177234 -id:A177234 - 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

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

Original entry on oeis.org

1, 7, 91, 1771, 46376, 1533939, 61474519, 2898753715, 157366449604, 9672348219898, 664226242466073, 50419551102990876, 4193002458968329488, 379189865879906158731, 37054233830964389244975

Offset: 2

Michel Lagneau, May 13 2010

nonn

All terms are integer because n and n+1 divide the binomial (cf. A060545, A177234).

Empirical: In the ring of symmetric functions over the fraction field Q(q, t), letting s(n) denote the Schur function indexed by n, a(n)*(-1)^(n+1) is equal to the coefficient of s(n) in nabla^(n)s(n) with q=t=1, where nabla denotes the "nabla operator" on symmetric functions. - John M. Campbell, Nov 18 2017

			For n = 3, binomial(9,3)/(3*4) =84/12 = 7.
For example, the coefficient of s(3) in nabla(nabla(nabla(s(3)))) is equal to q^6*t^2+q^5*t^3+q^4*t^4+q^3*t^5+q^2*t^6+q^4*t^3+q^3*t^4, and if we let q and t be equal to 1, this coefficient reduces to 7 = a(3). - _John M. Campbell_, Nov 18 2017

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

Cf. A060545, A177234.

[Binomial(n^2,n)/(2*Binomial(n+1,2)): n in [2..30]]; // G. C. Greubel, Jul 18 2024

A177784 := proc(n)
        binomial(n^2,n)/(n^2+n) ;
end proc:
seq(A177784(n),n=2..20) ; # R. J. Mathar, Nov 07 2011

Table[Binomial[n^2,n]/(2*Binomial[n+1,2]), {n,2,30}] (* G. C. Greubel, Jul 18 2024 *)

[binomial(n^2,n)//(n*(n+1)) for n in range(2,31)] # G. C. Greubel, Jul 18 2024

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

A177424 Exponent of the highest power of 2 dividing binomial(n^2,n).

Original entry on oeis.org

0, 0, 1, 2, 2, 1, 4, 3, 3, 1, 3, 3, 2, 3, 5, 4, 4, 1, 3, 3, 5, 1, 4, 8, 3, 2, 3, 5, 6, 4, 6, 5, 5, 1, 3, 3, 5, 2, 6, 3, 3, 3, 4, 3, 4, 4, 5, 6, 4, 2, 3, 7, 5, 1, 5, 5, 3, 4, 6, 6, 7, 5, 7, 6, 6, 1, 3, 3, 5, 2, 6, 4, 7, 1, 3, 3, 3, 5, 6, 4, 4, 2, 6, 3, 5, 5, 4, 6, 6, 2, 4, 12, 6, 5, 6, 7, 5, 2, 3, 6, 5, 3, 6, 3, 6

Offset: 0

Michel Lagneau, May 07 2010

a(n) is the largest integer such that 2^a(n) divides binomial(n^2,n)=A014062(n).

a(n) is the number of carries when adding n to n^2-n in base 2. - Robert Israel, Oct 23 2019

			For n = 6, binomial(36,6) = 1947792 = 2^4*3*7*11*17*31, the highest power of 2 is 2^4, and the exponent of 2^4 is a(6)=4.

Robert Israel, Table of n, a(n) for n = 0..10000(n=0 to 1000 from Harvey P. Dale)
Wikipedia, Kummer's theorem

Cf. A014062, A000120, A177234

A007814 := proc(n) if type(n,'odd') then 0; else for p in ifactors(n)[2] do if op(1,p) = 2 then return op(2,p); end if; end do: end if; end proc:
A014062 := proc(n) binomial(n^2,n) ; end proc:
A177424 := proc(n) A007814(A014062(n)) ; end proc: seq(A177424(n),n=0..80) ;
# Alternative:
nc:= proc(a,b,c)
  local t;
  if c=0 and (a=0 or b=0) then return 0 fi;
  t:= (a mod 2) + (b mod 2) + c;
  if t < 2 then  procname(floor(a/2),floor(b/2),0)
  else  1 + procname(floor(a/2),floor(b/2),1)
  fi
end proc:
seq(nc(n,n^2-n,0),n=0..100); # Robert Israel, Oct 23 2019

IntegerExponent[Table[Binomial[n^2,n],{n,0,120}],2] (* Harvey P. Dale, Mar 31 2019 *)

valp(n,p=2)=my(s); while(n\=p, s+=n); s
a(n)=valp(n^2)-valp(n^2-n)-valp(n) \\ Charles R Greathouse IV, Jul 08 2022

from math import comb
def A177424(n): return (~(m:=comb(n**2,n))& m-1).bit_length() # Chai Wah Wu, Jul 08 2022

a(n) = A007814(A014062(n)).

Maple program replaced by a structured general version - R. J. Mathar, May 10 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

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

A386879 a(n) = [x^n] 1/(1 - x)^(n*(n-1)/2).

Original entry on oeis.org

1, 0, 1, 10, 126, 2002, 38760, 888030, 23535820, 708930508, 23930713170, 895068996640, 36749279048405, 1643385429346680, 79515468511191440, 4139207762053520646, 230672804560960311000, 13703037308872895467960, 864424422377992704918690, 57711135174726478041405270, 4065392394346039279040037520

Offset: 0

Vaclav Kotesovec, Aug 06 2025

nonn

Cf. A014068, A054688, A116508, A177234, A386880.

Table[SeriesCoefficient[1/(1-x)^(n*(n-1)/2), {x, 0, n}], {n, 0, 25}]
Join[{1}, Table[Binomial[n*(n+1)/2, n] * (n-1) / (n+1), {n, 1, 25}]]

a(n) ~ exp(n) * n^(n - 1/2) / (sqrt(Pi) * 2^(n + 1/2)).

For n > 0, a(n) = binomial(n*(n+1)/2, n) * (n-1)/(n+1).

A386880 a(n) = [x^n] 1/(1 - x)^(n*(n+1)/2).

Original entry on oeis.org

1, 1, 6, 56, 715, 11628, 230230, 5379616, 145008513, 4431613550, 151473214816, 5727160371180, 237377895350076, 10704005376506540, 521748877569771510, 27338999059076777600, 1532576541123942256285, 91527291781199227579626, 5801648509628587739612170, 389032765009190361630625600

Offset: 0

Vaclav Kotesovec, Aug 06 2025

nonn

Cf. A014068, A054688, A116508, A177234, A386879.

Table[SeriesCoefficient[1/(1-x)^(n*(n+1)/2), {x, 0, n}], {n, 0, 25}]
Join[{1}, Table[Binomial[n*(n + 3)/2, n]*(n + 1)/(n + 3), {n, 1, 25}]]

a(n) ~ exp(n+2) * n^(n - 1/2) / (sqrt(Pi) * 2^(n + 1/2)).

For n > 0, a(n) = binomial(n*(n+3)/2, n) * (n+1)/(n+3).

cp's OEIS Frontend

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

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A177784 a(n) = binomial(n^2, n) / ( 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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A177424 Exponent of the highest power of 2 dividing binomial(n^2,n).

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

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