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

Views

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

A129454 a(n) = Product{i=1..n-1} Product{j=1..n-1} Product{k=1..n-1} gcd(i,j,k).

Original entry on oeis.org

1, 1, 1, 2, 6, 1536, 7680, 8806025134080, 61642175938560, 2168841254587541957294161920, 7562281854741110985626291951024209920, 1362299589723309231779453337910253309054734620740812800000000
Offset: 0

Views

Author

Peter Bala, Apr 16 2007

Keywords

Comments

Conjecture: Let p be a prime and let ordp(n,p) denote the largest power of p which 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 ordp(a(n),p)=(floor(n/p))^3 + (floor(n/p^2))^3 + (floor(n/p^3))^3 + . . .. Compare with the comments in A092287.

Links

Crossrefs

Cf. A092287, A129455.

Programs

  • Magma
    A129454:= func< n | n le 1 select 1 else (&*[(&*[(&*[GCD(GCD(j,k),m): k in [1..n-1]]): j in [1..n-1]]): m in [1..n-1]]) >;
[A129454(n): n in [0..20]]; // G. C. Greubel, Feb 07 2024
  • Mathematica
    A129454[n_]:= Product[GCD[j,k,m], {j,n-1}, {k,n-1}, {m,n-1}];
Table[A129454[n], {n,0,20}] (* G. C. Greubel, Feb 07 2024 *)
  • SageMath
    def A129454(n): return product(product(product(gcd(gcd(j,k),m) for k in range(1,n)) for j in range(1,n)) for m in range(1,n))
[A129454(n) for n in range(21)] # G. C. Greubel, Feb 07 2024

Formula

a(n) = Product{i=1..n-1} Product{j=1..n-1} Product{k=1..n-1} gcd(i,j,k), for n > 2, otherwise a(n) = 1.

A129453 An analog of Pascal's triangle based on A092287. T(n,k) = A092287(n)/(A092287(n-k)*A092287(k)), 0 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 16, 24, 16, 1, 1, 5, 40, 40, 5, 1, 1, 864, 2160, 11520, 2160, 864, 1, 1, 7, 3024, 5040, 5040, 3024, 7, 1, 1, 2048, 7168, 2064384, 645120, 2064384, 7168, 2048, 1, 1, 729, 746496, 1741824, 94058496, 94058496, 1741824, 746496, 729, 1
Offset: 0

Views

Author

Peter Bala, Apr 16 2007

Keywords

Comments

It appears that the T(n,k) are always integers. This would follow from the conjectured prime factorization given in A092287. Calculation suggests that the binomial coefficients C(n,k) divide T(n,k) and indeed that T(n,k)/C(n,k) are perfect squares.

Examples

			Triangle starts:
  1;
  1,  1;
  1,  2,  1;
  1,  3,  3,  1;
  1, 16, 24, 16,  1;
  1,  5, 40, 40,  5,  1;

Links

Crossrefs

Cf. A007318, A092287, A129454, A129455.

Programs

Formula

T(n, k) = (Product_{i=1..n} Product_{j=1..n} gcd(i,j)) / ( (Product_{i=1..n-k} Product_{j=1..n-k} gcd(i,j)) * ( Product_{i=1..k} Product_{j=1..k} gcd(i,j)) ), note that empty products equal to 1.
T(n, n-k) = T(n, k). - G. C. Greubel, Feb 07 2024
Showing 1-3 of 3 results.

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

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