A001142 -id:A001142 - cp's OEIS frontend

A371624 a(n) = Product_{k=0..n} (n^2 - k^2)!.

1, 1, 144, 1755758592000, 66052111513207347990207922176000000000

Offset: 0

Vaclav Kotesovec, Mar 30 2024

nonn

The next term has 88 digits.

Cf. A001142, A255322, A296591, A362288, A371603.

Table[Product[((2*n-k)*k)!, {k, 0, n}], {n, 0, 6}]
Table[Product[(n^2 - k^2)!, {k, 0, n}], {n, 0, 6}]

a(n) = (n^2)!^(n+1) / (A255322(n) * A371603(n)).

a(n) ~ c * A^(2*n) * 2^(4*n*(n^2 + 1)/3) * Pi^(n/2) * n^(4*n^3/3 + n^2 + 5*n/6 + 1/4) / exp(16*n^3/9 + n^2/2 + n), where c = 1.291409... = sqrt(2*Pi) / (A255504 * c from A371603) and A is the Glaisher-Kinkelin constant A074962.

A129824 a(n) = Product_{k=0..n} (1 + binomial(n,k)).

Original entry on oeis.org

2, 4, 12, 64, 700, 17424, 1053696, 160579584, 62856336636, 63812936890000, 168895157342195152, 1169048914836855865344, 21209591746609937928524800, 1010490883477487017627972550656, 126641164340871500483202065902080000, 41817338589698457759723104703370865147904

Offset: 0

Henry Gould, Jun 03 2007

A product analog of the binomial expansion.
The sequence is a special case of a(n) = Product_{k=0..n} (1 + binomial(n,k)*x^k).
Let C be a collection of subsets of an n-element set S. Then a(n) is the number of possible shapes K = (k_0, ..., k_n) of C, where k_i is the number of i-element subsets of S in C. - Gabriel Cunningham (oeis(AT)gabrielcunningham.com), Nov 08 2007

			a(4) = (1+1)(1+4)(1+6)(1+4)(1+1) = 2*5*7*5*2 = 700.

H. W. Gould, A product analog of the binomial expansion, unpublished manuscript, Jun 03 2007.

G. C. Greubel, Table of n, a(n) for n = 0..69

Cf. A001142, A055612.

A129824:= func< n | (&*[1 + Binomial(n,k): k in [0..n]]) >;
[A129824(n): n in [0..20]]; // G. C. Greubel, Apr 26 2024

Table[Product[1 + Binomial[n,k], {k,0,n}], {n,0,15}] (* Vaclav Kotesovec, Oct 27 2017 *)

{ a(n) = prod(k=0,n, 1 + binomial(n,k))}
for(n=0,15,print1(a(n),", ")) \\ Paul D. Hanna, Oct 27 2017

def A129824(n): return product(1 + binomial(n,k) for k in range(n+1))
[A129824(n) for n in range(21)] # G. C. Greubel, Apr 26 2024

a(n) = 2*A055612(n). - Reinhard Zumkeller, Jan 31 2015

a(n) ~ exp(n^2/2 + n - 1/12) * A^2 / (n^(n/2 + 1/3) * 2^((n-3)/2) * Pi^((n+1)/2)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Oct 27 2017

Corrected and extended by Vaclav Kotesovec, Oct 27 2017

A229495 Stirling's approximation constant e / sqrt(2*Pi).

Original entry on oeis.org

1, 0, 8, 4, 4, 3, 7, 5, 5, 1, 4, 1, 9, 2, 2, 7, 5, 4, 6, 6, 1, 1, 5, 7, 7, 3, 1, 3, 4, 2, 2, 9, 4, 7, 9, 8, 5, 8, 3, 9, 5, 9, 6, 9, 3, 1, 9, 6, 4, 7, 2, 6, 2, 6, 8, 2, 2, 5, 1, 3, 4, 3, 4, 7, 1, 2, 2, 8, 7, 5, 1, 4, 7, 9, 6, 2, 6, 9, 0, 0, 2, 4, 9, 9, 0, 3, 4, 7, 1, 6, 8, 2, 8, 8, 4, 8, 4, 7, 5, 3, 1, 5, 2, 3, 6, 6, 7, 9, 3, 9, 1, 9, 7, 3, 4, 9, 3, 6, 4, 3, 5, 3, 4, 7, 6, 8, 3, 8, 1, 5, 4, 1, 3, 1, 9, 5, 6, 3, 3, 6, 6, 3, 3, 4, 2, 9, 5, 1, 9, 7

Offset: 1

John W. Nicholson, Sep 24 2013

			1.0844375514192275466115773134229479858...

Ovidiu Furdui, Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis, New York: Springer, 2013. See Problem 1.5, pages 2 and 27-28.

G. C. Greubel, Table of n, a(n) for n = 1..10000
Rafael Jakimczuk, Two Topics in Number Theory: Products Related with the e Number and Sum of Subscripts in Prime Numbers, ResearchGate, May 2025. See p. 2, the constant exp(C) in eq. (2.1).
Wikipedia, Stirling's approximation.

Cf. A001113 (e), A019727 (sqrt(2*Pi)), A001142, A110544.

SetDefaultRealField(RealField(100)); R:= RealField(); Exp(1)/Sqrt(2*Pi(R)); // G. C. Greubel, Oct 06 2018

evalf(exp(1)/sqrt(2*Pi),120); # Muniru A Asiru, Oct 07 2018

RealDigits[E/Sqrt[2Pi],10,120][[1]] (* Harvey P. Dale, Jan 21 2017 *)

exp(1)/sqrt(2*Pi) \\ Ralf Stephan, Sep 26 2013

Equals exp(1)/sqrt(2*Pi).
Equals lim_{n->oo} (A001142(n)^(1/n)*sqrt(n)/(exp(n/2))) (Furdui, 2013). - Amiram Eldar, Mar 26 2022
Equals Product_{n>=1} (1 + 1/n)^(n+1/2)/e. - Amiram Eldar, Jul 08 2023
Equals exp(A110544). - Amiram Eldar, May 30 2025

More terms from Ralf Stephan, Sep 26 2013

Corrected and extended by Harvey P. Dale, Jan 21 2017

A249422 Transpose of square array A249421.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 2, 5, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 4, 0, 0, 0, 0, 0, 4, 4, 0, 0, 0, 0, 0, 0, 0, 3, 2, 17, 0, 0, 0, 0, 0, 0, 0, 2, 0, 10, 0, 0, 0, 0, 0, 0, 0, 6, 1, 14, 12, 0, 0, 0, 0, 0, 0, 0, 0, 5, 0, 10, 4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 8, 6, 18, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 3, 6, 13, 8, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 4, 8, 11

Offset: 1

Antti Karttunen, Oct 28 2014

See comments at A249421.

Cf. A249421, A060175, A001142.

(define (A249422 n) (A249421bi (A004736 n) (A002260 n))) ;;
Code for A249421bi given in A249421.

A168510 Products across consecutive rows of the denominators of the Leibniz harmonic triangle (A003506).

Original entry on oeis.org

1, 4, 54, 2304, 300000, 116640000, 133413966000, 444110104166400, 4267295479315169280, 117595223746560000000000, 9245836018244425723200000000, 2065215715357207851951980544000000

Offset: 1

Harlan J. Brothers, Nov 27 2009

As in A001142, lim_{n->inf} (a(n)a(n+2))/a(n+1)^2 = e, demonstrating an underlying relation between A003506 and Pascal's triangle A007318. Unlike A001142, in this case the function is asymptotic from above.

			For n=3, row 3 of A003506 = {3, 6, 3} and a(3)=54.
a(5) = 5^5 * 4^3 * 3^1 * 2^-1 * 1^-3 = 5^5 * 3 * 2^5 = 300000. - _Peter Munn_, Mar 07 2018

A. Bogomolny, Cut The Knot: Leibniz and Pascal Triangles
H. J. Brothers, Pascal's prism, The Mathematical Gazette, 96 (July 2012), 213-220.

Cf. A003506, A001142, A007318. For n >= 1, a(n) = n!*A001142(n).

Table[n! Product[k^(2 k - n - 1), {k, 1, n}], {n, 1, 12}]
Table[Product[Product[(1 - 1/k)^-k, {k, 2, j}], {j, 1, n}], {n, 1, 12}]
(* or *)
a[1] = 1; a[n_] := a[n - 1] Product[(1 - 1/k)^-k, {k, 2, n}]

a(n) = n!*Product_{k=1..n} k^(2k-n-1).
a(n) = Product_{j=1..n} Product_{k=2..j} ((1-1/k)^-k).
a(1) = 1; a(n) = a(n-1)*Product_{k=2..n} ((1-1/k)^-k).
a(n) ~ A^2 * exp(n^2/2 - 1/12) * n^(n/2 + 1/6) / (2*Pi)^(n/2), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Oct 22 2017
a(n) = Product_{k=0..n-1} (n-k)^(n-2k). - Peter Munn, Mar 07 2018

A208651 Number of paths through the subset array whose trace is a permutation of (1,2,...,n); see Comments.

Original entry on oeis.org

1, 2, 12, 216, 11520, 1800000, 816480000, 1067311728000, 3996990937497600, 42672954793151692800, 1293547461212160000000000, 110950032218933108678400000000, 26847804299643702075375747072000000

Offset: 1

Clark Kimberling, Mar 01 2012

nonn

See A208650. The trace of a path is a permutation of (1,2,...,n) if and only if the range of the path is {1,2,...,n}.

			Taking n=3:
row 1:  {1},{2},{3} ---------> 1,2,3
row 2:  {1,2},{1,3},{2,3} ---> 1,1,2,2,3,3
row 3:  {1,2,3} -------------> 1,2,3
3 ways to choose a number from row 1,
4 ways to choose a different number from row 2,
1 way to choose remaining number from row 3.
Total:  a(3) = 1*4*3 = 12 paths.

Nick Early, Generalized Permutohedra, Scattering Amplitudes, and a Cubic Three-Fold, arXiv:1709.03686 [math.CO], 2017.

Cf. A208650.

p[n_]:=Product[Binomial[n-1,k],{k,1,n-1}]
Table[p[n],{n,1,20}]    (* A001142(n-1) *)
Table[p[n]*n,{n,1,20}]  (* A208650 *)
Table[p[n]*n!,{n,1,20}] (* A208651 *)

A276710 Composite numbers m such that Product_{k=0..m} binomial(m,k) is divisible by m^(m-1).

Original entry on oeis.org

36, 40, 63, 80, 84, 90, 105, 108, 132, 144, 150, 154, 160, 165, 168, 175, 176, 180, 182, 195, 198, 200, 208, 210, 220, 260, 264, 270, 273, 275, 280, 286, 288, 297, 300, 306, 308, 312, 315, 320, 324, 330, 340, 351, 357, 360, 364, 374, 378, 380, 385, 390

Offset: 1

Stanislav Sykora, Sep 15 2016

nonn

The numbers Product_{k=0..m} binomial(m,k) form the sequence A001142(m). When m is a prime, the m-1 factors for 0 < k < m are all divisible by m and therefore A001142(m) is divisible by m^(m-1). When m is a composite, this is generally not so, except for the numbers listed here (a variety of pseudoprimes).
Conjecture, tested so far up to m = 3828: "When m belongs to this list, Product_{k=0..m} binomial(m,k) is divisible also by m^m". Since this is impossible for prime m (see, e.g., A109874), the conjecture is equivalent to the statement "m is prime if and only if Product_{k=0..m} binomial(m,k) is divisible by m^(m-1) but not by m^m".
From Hagen von Eitzen, Jul 31 2022: (Start)
This conjecture has been proved, see corollary 3 in PDF link below.
The set of all numbers in this sequence has natural density 1 - log(2), see theorem 2 in PDF link below. (End)

			Since 36 is composite and 36^35 divides Product_{k=1..36} binomial(36,k), 36 is a member. In addition, it turns out that 36^36 also divides the product.

Stanislav Sykora and Chai Wah Wu, Table of n, a(n) for n = 1..10000, terms for n = 1..797 from Stanislav Sykora
Hagen von Eitzen, C source code for fast computation
Hagen von Eitzen, Some Results About Sequence A276710

Cf. A000169 (n^(n-1)), A001142, A109873, A109874.

Select[DeleteCases[Range[2, 400], p_ /; PrimeQ@ p], Divisible[Product[Binomial[#, k], {k, 0, #}], #^(# - 1)] &] (* Michael De Vlieger, Sep 16 2016 *)

generator() = { /* Operates on two pre-defined integer vectors a, b of the same size. a(n) receives the terms of this sequence, while b(n) receives 0 if n^n|Product(binomial(n,k)), or 1 otherwise, and serves exclusively to test the conjecture. */
  my (m=1,n=0,p);for(k=1,#a,a[k]=0;b[k]=0);
  while(1,m++;p=prod(k=1,m-1,binomial(m,k));
    if((p%m^(m-1)==0)&&(!isprime(m)),n++;a[n]=m;
    if(p%m^m==0,b[n]=0,b[n]=1);if(n==#a,break)));
}
  a=vector(1000);b=a;generator();
  a /* Displays the result.
  Note: execution was interrupted due to excessive execution time */

from itertools import islice
from math import comb
from sympy import nextprime
def A276710_gen(): # generator of terms
    p, q = 3, 5
    while True:
        for m in range(p+1,q):
            r = m**(m-1)
            c = 1
            for k in range(m+1):
                c = c*comb(m,k) % r
            if c == 0:
                yield m
        p, q = q, nextprime(q)
A276710_list = list(islice(A276710_gen(),40)) # Chai Wah Wu, Jun 08 2022

A308943 a(n) = Product_{d|n} binomial(n,d).

Original entry on oeis.org

1, 2, 3, 24, 5, 1800, 7, 15680, 756, 113400, 11, 79693891200, 13, 4372368, 20495475, 44972928000, 17, 2028339316523520, 19, 52737518268864000, 3247700400, 3585005424, 23, 38135556819759802035135799296, 1328250, 87885070000, 370142004375, 10293527616645873600000, 29

Offset: 1

Ilya Gutkovskiy, Jul 01 2019

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..419

Cf. A001142, A008578 (fixed points), A056045 (similar, with Sum), A098710, A135396.

Cf. A000010 (comments on product formulas).

Table[Product[Binomial[n, d], {d, Divisors[n]}], {n, 1, 29}]

a(n) = my(p=1); fordiv(n, d, p *= binomial(n, d)); p; \\ Michel Marcus, Jul 02 2019

from math import prod, comb
from sympy import divisors
def A308943(n): return prod(comb(n,d) for d in divisors(n,generator=True)) # Chai Wah Wu, Jul 22 2024

a(n) = Product_{k=1..n} binomial(n,gcd(n,k))^(1/phi(n/gcd(n,k))) = Product_{k=1..n} binomial(n,n/gcd(n,k))^(1/phi(n/gcd(n,k))) where phi = A000010. - Richard L. Ollerton, Nov 08 2021

A371646 a(n) = Product_{k=0..n} binomial(n^3, k^3).

Original entry on oeis.org

1, 1, 8, 59942025, 239830737497318918172122578944, 788243862228623056807478850630904903414781894638966172447366478063616699218750

Offset: 0

Vaclav Kotesovec, Mar 31 2024

nonn

Cf. A001142, A371603.

Cf. A007685, A255358, A371468.

Table[Product[Binomial[n^3, k^3], {k, 0, n}], {n, 0, 6}]

a(n) ~ c * exp((9/4 - sqrt(3)*Pi/8)*n^4 + (3*zeta(3)/(4*Pi^2) - Pi/(4*sqrt(3)) + 3)*n) / ((2*Pi)^(n/2) * A^(3*n^2) * 3^(9*n^4/8 - n^2/4 + 3*n/4) * n^(n^2/4 + 3*n/2 - 8/15)), where c = 0.498332919... and A is the Glaisher-Kinkelin constant A074962.

A061778 a(n) = Product_{j=0..floor(n/2)} binomial(n,j).

Original entry on oeis.org

1, 2, 3, 24, 50, 1800, 5145, 878080, 3429216, 2857680000, 15219319500, 63117561830400, 457937132487120, 9577928124440387712, 94609025993497640625, 10077943267571584204800000

Offset: 1

Labos Elemer, Jun 22 2001

			n=5: a(5) = 1*5*10 = 50;
n=6: a(6) = 1*6*15*20 = 1800. [corrected by _Jon E. Schoenfield_, Jul 01 2018]

Harry J. Smith, Table of n, a(n) for n = 1..97

Cf. A001142, A001405.

List([1..20],n->Product([0..Int(n/2)],j->Binomial(n,j))); # Muniru A Asiru, Jul 01 2018

Table[Apply[Times, Table[Binomial[n, j], {j, 0, Floor[n/2]}]], {w, 1, 20}]
Table[Product[Binomial[n,j],{j,0,Floor[n/2]}],{n,20}] (* Harvey P. Dale, Dec 06 2018 *)

{ for (n=1, 97, write("b061778.txt", n, " ", prod(j=0, n\2, binomial(n, j))) ) } \\ Harry J. Smith, Jul 27 2009

For odd n, a(n) = sqrt(A001142(n)); for even n, (a(n)^2)/A001405(n) = A001142(n).

cp's OEIS Frontend

A371624 a(n) = Product_{k=0..n} (n^2 - k^2)!.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Formula

A129824 a(n) = Product_{k=0..n} (1 + binomial(n,k)).

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A229495 Stirling's approximation constant e / sqrt(2*Pi).

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

Extensions

A249422 Transpose of square array A249421.

Views

Author

Keywords

Comments

Crossrefs

Programs

Scheme

A168510 Products across consecutive rows of the denominators of the Leibniz harmonic triangle (A003506).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A208651 Number of paths through the subset array whose trace is a permutation of (1,2,...,n); see Comments.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A276710 Composite numbers m such that Product_{k=0..m} binomial(m,k) is divisible by m^(m-1).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica