A129454 -id:A129454 - cp's OEIS frontend

A129455 An analog of Pascal's triangle based on A129454. T(n, k) = A129454(n+1)/(A129454(n-k+1)*A129454(k+1)).

1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 256, 384, 256, 1, 1, 5, 640, 640, 5, 1, 1, 1146617856, 2866544640, 244611809280, 2866544640, 1146617856, 1, 1, 7, 4013162496, 6688604160, 6688604160, 4013162496, 7, 1, 1, 35184372088832, 123145302310912, 47066867504069920948224, 919274755938865643520, 47066867504069920948224, 123145302310912, 35184372088832, 1

Offset: 0

Peter Bala, Apr 16 2007

It appears that the T(n,k) are always integers. This would follow from the conjectured prime factorization given in A129454. Calculation suggests that the binomial coefficients C(n,k) divide T(n,k) and that T(n,k)/C(n,k) are perfect sixth powers.

			Triangle starts:
  1;
  1,   1;
  1,   2,   1;
  1,   3,   3,   1;
  1, 256, 384, 256,  1;
  1,   5, 640, 640,  5,  1;

G. C. Greubel, Rows n = 0..25 of the triangle, flattened

Cf. A092287, A129453, A129454.

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]]) >;
A129455:= func< n,k | A129454(n+1)/(A129454(n-k+1)*A129454(k+1)) >;
[A129455(n,k): k in [0..n], n in [0..9]]; // G. C. Greubel, Feb 07 2024

A129454[n_]:= Product[GCD[j,k,m], {j,n-1}, {k,n-1}, {m,n-1}];
A129455[n_, k_]:= A129454[n+1]/(A129454[k+1]*A129454[n-k+1]);
Table[A129455[n,k], {n,0,9}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 07 2024 *)

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))
def A129455(n,k): return A129454(n+1)/(A129454(n-k+1)*A129454(k+1))
flatten([[A129455(n,k) for k in range(n+1)] for n in range(10)]) # G. C. Greubel, Feb 07 2024

T(n, k) = Product_{h=1..n} Product_{i=1..n} Product_{j=1..n} gcd(h,i,j)/( (Product_{h=1..n-k} Product_{i=1..n-k} Product_{j=1..n-k} gcd(h,i,j))*(Product_{h=1..k} Product_{i=1..k} Product_{j=1..k} gcd(h,i,j)) ).

T(n, n-k) = T(n, k). - G. C. Greubel, Feb 07 2024

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

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

nonn

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

T. D. Noe, Table of n, a(n) for n = 0..67

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

[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

f := n->mul(mul(igcd(j,k),k=1..n),j=1..n);

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 *)

h(n,p)=if(nCharles R Greathouse IV, Apr 02 2013

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

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

Recurrence formula corrected by Daniel Forgues, Apr 07 2013

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

Peter Bala, Apr 16 2007

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.

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

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A007318, A092287, A129454, A129455.

A092287:= func< n | n eq 0 select 1 else (&*[(&*[GCD(j,k): k in [1..n]]): j in [1..n]]) >;
A129453:= func< n,k | A092287(n)/(A092287(n-k)*A092287(k)) >;
[A129453(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 07 2024

A092287[n_]:= Product[GCD[j,k], {j,n}, {k,n}];
A129453[n_, k_]:= A092287[n]/(A092287[k]*A092287[n-k]);
Table[A129453[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 07 2024 *)

def A092287(n): return product(product( gcd(j,k) for k in range(1,n+1)) for j in range(1,n+1))
def A129453(n,k): return A092287(n)/(A092287(n-k)*A092287(k))
flatten([[A129453(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 07 2024

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

A090494 Product_{j=1..n} Product_{k=1..n} lcm(j,k).

Original entry on oeis.org

1, 1, 8, 7776, 1146617856, 1289945088000000000, 46798828032806092800000000000, 2350577043461005964030008507760640000000000000, 8206262459636402163263383676462776103575725539328000000000000000, 2746781358330240881921653545637784861521126603512175621574459373964492800000000000000000

Offset: 0

N. J. A. Sloane, Feb 03 2004

nonn

Cf. A018806.

Cf. A092143, A092287, A129365, A129454.

f := n->mul(mul(lcm(j,k),k=1..n),j=1..n);

Let p be a prime and let ordp(n,p) denote the exponent of the largest power of p which divides n. For example, ordp(48,2)=4 since 48 = 3*(2^4). Then the prime factorization of a(n) appears to be given by the formula ordp(a(n),p)= sum_{k >= 1} [(2*(p^k)-1)*floor((n/(p^k)))^2] + 2*sum_{k >= 1} [floor(n/(p^k))*mod(n,p^k)], for each prime p. See the comments sections of A092143, A092287, A129365 and A129454 for similar conjectural prime factorizations. - Peter Bala, Apr 23 2007

cp's OEIS Frontend

A129455 An analog of Pascal's triangle based on A129454. T(n, k) = A129454(n+1)/(A129454(n-k+1)*A129454(k+1)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

A090494 Product_{j=1..n} Product_{k=1..n} lcm(j,k).

Views

Author

Keywords

Crossrefs

Programs

Maple

Formula