cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A092287 a(n) = Product_{j=1..n} Product_{k=1..n} gcd(j,k).

Original entry on oeis.org

1, 1, 2, 6, 96, 480, 414720, 2903040, 5945425920, 4334215495680, 277389791723520000, 3051287708958720000, 437332621360674939863040000, 5685324077688774218219520000, 15974941971638268369709427589120000, 982608696336737613503095822614528000000000
Offset: 0

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Author

N. J. A. Sloane, based on a suggestion from Leroy Quet, Feb 03 2004

Keywords

Comments

Conjecture: Let p be a prime and let ordp(n,p) denote the exponent of the highest power of p that divides n. For example, ordp(48,2)=4, since 48=3*(2^4). Then we conjecture that the prime factorization of a(n) is given by the formula: ordp(a(n),p) = (floor(n/p))^2 + (floor(n/p^2))^2 + (floor(n/p^3))^2 + .... Compare this to the de Polignac-Legendre formula for the prime factorization of n!: ordp(n!,p) = floor(n/p) + floor(n/p^2) + floor(n/p^3) + .... This suggests that a(n) can be considered as generalization of n!. See A129453 for the analog for a(n) of Pascal's triangle. See A129454 for the sequence defined as a triple product of gcd(i,j,k). - Peter Bala, Apr 16 2007
The conjecture is correct. - Charles R Greathouse IV, Apr 02 2013
a(n)/a(n-1) = n, n >= 1, if and only if n is noncomposite, otherwise a(n)/a(n-1) = n * f^2, f > 1. - Daniel Forgues, Apr 07 2013
Conjecture: For a product over a rectangle, f(n,m) = Product_{j=1..n} Product_{k=1..m} gcd(j,k), a factorization similar to the one given above for the square case takes place: ordp(f(n,m),p) = floor(n/p)*floor(m/p) + floor(n/p^2)*floor(m/p^2) + .... By way of directly computing the values of f(n,m), it can be verified that the conjecture holds at least for all 1 <= m <= n <= 200. - Andrey Kaydalov, Mar 11 2019

Links

Crossrefs

Cf. A003989, A018806, A051190, A090494, A129365, A129439, A129453, A129454, A129455, A224479, A224497.

Programs

  • Magma
    [n eq 0 select 1 else (&*[(&*[GCD(j,k): k in [1..n]]): j in [1..n]]): n in [0..30]]; // G. C. Greubel, Feb 07 2024
  • Maple
    f := n->mul(mul(igcd(j,k),k=1..n),j=1..n);
  • Mathematica
    a[0] = 1; a[n_] := a[n] = n*Product[GCD[k, n], {k, 1, n-1}]^2*a[n-1]; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Apr 16 2013, after Daniel Forgues *)
  • PARI
    h(n,p)=if(nCharles R Greathouse IV, Apr 02 2013
  • Sage
    def A092287(n):
    R = 1
    for p in primes(n+1) :
        s = 0; r = n
        while r > 0 :
            r = r//p
            s += r*r
        R *= p^s
    return R
[A092287(i) for i in (0..15)]  # Peter Luschny, Apr 10 2013

Formula

Also a(n) = Product_{k=1..n} Product_{j=1..n} lcm(1..floor(min(n/k, n/j))).
From Daniel Forgues, Apr 08 2013: (Start)
Recurrence: a(0) := 1; for n > 0: a(n) := n * (Product_{j=1..n-1} gcd(n,j))^2 * a(n-1) = n * A051190(n)^2 * a(n-1).
Formula for n >= 0: a(n) = n! * (Product_{j=1..n} Product_{k=1..j-1} gcd(j,k))^2. (End)
a(n) = n! * A224479(n)^2 (the last formula above).
a(n) = n$ * A224497(n)^4, n$ the swinging factorial A056040(n). - Peter Luschny, Apr 10 2013

Extensions

Recurrence formula corrected by Daniel Forgues, Apr 07 2013

A051190 a(n) = Product_{k=1..n-1} gcd(k,n).

Original entry on oeis.org

1, 1, 1, 2, 1, 12, 1, 16, 9, 80, 1, 3456, 1, 448, 2025, 2048, 1, 186624, 1, 1024000, 35721, 11264, 1, 573308928, 625, 53248, 59049, 179830784, 1, 1007769600000, 1, 67108864, 7144929, 1114112, 37515625, 160489808068608, 1, 4980736, 89813529
Offset: 1

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Author

Antti Karttunen, Oct 21 1999

Keywords

Comments

a(n) > 1 if and only if n is composite. - Charles R Greathouse IV, Jan 04 2013

Links

Crossrefs

Cf. A067911, A006579, A224479, A092287.

Programs

  • Haskell
    a051190 n = product $ map (gcd n) [1..n-1]
-- Reinhard Zumkeller, Nov 22 2011
  • Maple
    A051190 := proc(n) local i; mul(igcd(n, i ), i = 1..(n-1)) end;
  • Mathematica
    a[n_] := If[PrimeQ[n], 1, Times @@ (GCD[n, #]& /@ Range[n-1])]; Table[a[n], {n, 1, 39}]  (* Jean-François Alcover, Jul 18 2012 *)
Table[Times @@ GCD[n, Range[n-1]], {n, 50}] (* T. D. Noe, Apr 12 2013 *)
Table[Product[GCD[k,n],{k,n-1}],{n,50}] (* Harvey P. Dale, Jan 29 2025 *)
  • PARI
    a(n)=my(f=factor(n)); prod(i=1, #f[,1], prod(j=1, f[i,2], f[i,1]^(n\f[i,1]^j)))/n \\ Charles R Greathouse IV, Jan 04 2013
  • PARI
    a(n) = prod(k=1,n-1,gcd(k,n)); /* Joerg Arndt, Apr 14 2013 */
  • Sage
    A051190 = lambda n: mul(gcd(n,i) for i in (1..n-1))
[A051190(n) for n in (1..39)] # Peter Luschny, Apr 07 2013
  • Sage
    # A second, faster version, based on the prime factorization of a(n):
def A051190(n):
    R = 1
    if not is_prime(n) :
        for p in primes(n//2+1):
            s = 0; r = n; t = n-1
            while r > 0 :
                r = r//p; t = t//p
                s += (r-t)*(r+t-1)
            R *= p^(s/2)
    return R
[A051190(i) for i in (1..1000)]  # Peter Luschny, Apr 08 2013

Formula

a(n) = Product_{ d divides n, d < n } d^phi(n/d). - Peter Luschny, Apr 07 2013
a(n) = A067911(n) / n. - Peter Luschny, Apr 07 2013
Product_{j=1..n} Product_{k=1..j-1} gcd(j,k), n >= 1. - Daniel Forgues, Apr 11 2013
a(n) = sqrt( (1/n) * (A092287(n) / A092287(n-1)) ). - Daniel Forgues, Apr 13 2013

A224497 a(n) = sqrt(floor(n/2)! * Product_{k=1..n} Product_{i=1..k-1} gcd(k,i)).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 12, 12, 96, 288, 5760, 5760, 829440, 829440, 46448640, 2090188800, 267544166400, 267544166400, 346737239654400, 346737239654400, 1109559166894080000, 209706682542981120000, 73816752255129354240000, 73816752255129354240000
Offset: 0

Views

Author

Peter Luschny, Apr 08 2013

Keywords

Comments

The order of the primes in the prime factorization of a(n) is given by
ord_{p}(a(n)) = (1/4)*Sum_{i>=1} floor(n/p^i)*(floor(n/p^i)-1) + (1/2)*Sum_{i>=1} floor(floor(n/2)/p^i).
For n > 1: a(n) = a(n-1) if and only if n is prime.

Crossrefs

Cf. A224479.

Programs

  • Maple
    A224497 := n -> sqrt(iquo(n,2)!*mul(mul(igcd(k,i), i=1..k-1), k=1..n)):
seq(A224497(i), i = 0..23);
  • Sage
    def A224497(n):
    R = 1;
    for p in primes(n):
        s = 0; t = 0
        r = n; u = n//2
        while r > 0 :
            r = r//p; u = u//p
            t += u; s += r*(r-1)
        R *= p^((t+s/2)/2)
    return R
[A224497(i) for i in (0..23)]

Formula

a(n) = sqrt(floor(n/2)! * A224479(n)).
A092287(n) = A056040(n) * a(n)^4.
Showing 1-3 of 3 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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