A051190 -id:A051190 - cp's OEIS frontend

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

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

A067911 Product of gcd(k,n) for 1 <= k <= n.

Original entry on oeis.org

1, 2, 3, 8, 5, 72, 7, 128, 81, 800, 11, 41472, 13, 6272, 30375, 32768, 17, 3359232, 19, 20480000, 750141, 247808, 23, 13759414272, 15625, 1384448, 1594323, 5035261952, 29, 30233088000000, 31, 2147483648, 235782657, 37879808

Offset: 1

Sharon Sela (sharonsela(AT)hotmail.com), Mar 10 2002

nonn

T. D. Noe, Table of n, a(n) for n = 1..500
Johann Cigler, Some remarks on the power product expansion of the q-exponential series, arXiv:2006.06242 [math.CO], 2020.
Gottfried Helms, A dream of a (number-) sequence, 2007-2009.
L. Toth, A survey of gcd-sum functions, J. Int. Seq. 13 (2010) # 10.8.1.

Cf. A051190, A051696.
In A018804 the product is replaced by sum.
Product of terms in n-th row of A050873.
Cf. A000010 (comments on product formulas).

with(numtheory): a := n -> mul(d^phi(n/d), d = divisors(n)):
seq(a(i), i = 1..34); # Peter Luschny, Apr 07 2013

a[n_] := Product[d^EulerPhi[n/d], {d, Divisors[n]}];
Array[a, 34] (* Jean-François Alcover, Jun 03 2019 *)

a(n) = prod(k=1, n, gcd(k, n)); \\ Michel Marcus, Aug 23 2016

A067911 = lambda n: mul(gcd(n,i) for i in range(n))
[A067911(n) for n in (1..34)] # Peter Luschny, Apr 07 2013

a(n) = Product_{d|n} d^phi(n/d). - Vladeta Jovovic, Mar 08 2004
a(n) = n*A051190(n). - Peter Luschny, Apr 07 2013
a(n) = Product_{k=1..n} (n/gcd(n,k))^(phi(gcd(n,k))/phi(n/gcd(n,k))) where phi = A000010. - Richard L. Ollerton, Nov 07 2021

Extended and edited by John W. Layman, Mar 14 2002

A224479 a(n) = Product_{k=1..n} Product_{i=1..k-1} gcd(k,i).

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 24, 24, 384, 3456, 276480, 276480, 955514880, 955514880, 428070666240, 866843099136000, 1775294667030528000, 1775294667030528000, 331312591939905257472000, 331312591939905257472000, 339264094146462983651328000000

Offset: 0

Peter Luschny, Apr 07 2013

nonn

The order of the primes in the prime factorization of a(n) is given by
ord_{p}(a(n)) = (1/2)*Sum_{i>=1} floor(n/p^i)*(floor(n/p^i)-1).
Product of all entries of lower-left (excluding main diagonal) triangular submatrix of GCDs. Also the product of all entries of upper-right (excluding main diagonal) triangular submatrix of GCDs, since the matrix is symmetric. - Daniel Forgues, Apr 14 2013
a(n)^2 * n! gives A092287(n), where n! is the product of the main diagonal entries of the matrix. - Daniel Forgues, Apr 14 2013

Charles R Greathouse IV, Table of n, a(n) for n = 0..97
OEIS Wiki, Generalizations of the factorial

Cf. A051190, A092287.

A224479 := proc(n) local h, k, d;
mul(mul(d^phi(k/d), d = divisors(k) minus {k}), k = 1..n) end:
seq(A224479(i), i = 0..20);

a[n_] := Product[ d^EulerPhi[k/d], {k, 1, n}, {d, Divisors[k] // Most}]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jun 27 2013, after Maple *)

a(n)=prod(k=1,n,my(s=1);fordiv(k,d,dCharles R Greathouse IV, Jun 27 2013

def A224479(n):
    R = 1;
    for p in primes(n):
        s = 0; r = n
        while r > 0 :
            r = r//p
            s += r*(r-1)
        R *= p^(s/2)
    return R
[A224479(i) for i in (0..20)]

a(n) = Product_{k=1..n} Product_{d divides k, d < k} d^phi(k/d).
n! * a(n)^2 = A092287(n).
a(n)/a(n-1) = A051190(n) for n >= 1.
a(n) = sqrt(A092287(n) / n!). - Daniel Forgues, Apr 14 2013

A082022 In the following square array a(i,j) = Least Common Multiple of i and j. Sequence contains the product of the terms of the n-th antidiagonal.

Original entry on oeis.org

1, 4, 18, 576, 1200, 518400, 1587600, 180633600, 1646023680, 13168189440000, 461039040000, 229442532802560000, 86553098311680000, 3753113311877529600, 834966920275488000000

Offset: 1

Amarnath Murthy, Apr 06 2003

nonn

If n is even and n+1 is prime, a(n) = n^2 * (n-1)!^2. If n is odd and >3, 2*(n+1)*a(n) is a perfect square, the root of which has the factor 1/2*n*(n-1)*((n-1)/2)!. This was proved by Lawrence Sze. - Ralf Stephan, Nov 16 2004

			1 2 3 4 5...
2 2 6 4 10...
3 6 3 12 15...
4 4 12 4 20...
5 10 15 20 5...
...
The same array in triangular form is
1
2 2
3 2 3
4 6 6 4
5 4 3 4 5
...
Sequence contains the product of the terms of the n-th row.

Cf. A006580, A003990, A082292.

Equals A001044(n) / A051190(n+1).

for(n=1,20,p=1:for(k=1,n,p=p*lcm(k,n+1-k)):print1(p","))

Prod(k=1...n, lcm(k, n+1-k)).

Corrected and extended by Ralf Stephan, Apr 08 2003

A276162 Square array read by antidiagonals: T(n,k) = Product_{i = 1..k} gcd(n, i).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 3, 4, 1, 1, 1, 2, 3, 4, 1, 1, 2, 1, 8, 3, 8, 1, 1, 1, 6, 1, 8, 9, 8, 1, 1, 2, 1, 12, 5, 16, 9, 16, 1, 1, 1, 2, 1, 12, 5, 16, 9, 16, 1, 1, 2, 3, 8, 1, 72, 5, 64, 27, 32, 1, 1, 1, 2, 3, 8, 1, 72, 5, 64, 27, 32, 1, 1, 2, 1, 4

Offset: 1

Peter Kagey, Aug 22 2016

			T(6, 3) = gcd(6, 1) * gcd(6, 2) * gcd(6, 3) = 6.

Peter Kagey, Table of n, a(n) for n = 1..10000
Mohamed Abobakr, Greatest common divisor sequence

Cf. A051190, A067911, A092287.

a276162T n k = product $ map (gcd n) [1..k]
-- Peter Kagey, Aug 23 2016

T(n,k)=prod(i=2,k,gcd(n,i))
for(s=1,15,for(k=1,s-1, print1(T(s-k,k)", "))) \\ Charles R Greathouse IV, Aug 22 2016

A349741 a(n) = Product_{k=1..n-1} phi(gcd(n,k)).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 4, 4, 1, 32, 1, 6, 256, 16, 1, 96, 1, 1024, 2304, 10, 1, 16384, 256, 12, 2304, 13824, 1, 524288, 1, 2048, 102400, 16, 5308416, 14155776, 1, 18, 589824, 134217728, 1, 63700992, 1, 1024000, 86973087744, 22, 1, 8589934592, 46656, 1310720

Offset: 1

Ilya Gutkovskiy, Nov 28 2021

nonn

Cf. A000010, A001088, A029935, A029940, A046022 (positions of 1's), A051190.

Table[Product[EulerPhi[GCD[n, k]], {k, 1, n - 1}], {n, 1, 50}]

a(n) = prod(k=1, n-1, eulerphi(gcd(n, k))); \\ Michel Marcus, Nov 28 2021

a(n) = Product_{d|n, d < n} phi(d)^phi(n/d).

A308944 a(n) = Product_{k=1..n} lcm(n,k) / (k * gcd(n,k)).

Original entry on oeis.org

1, 1, 3, 4, 125, 9, 16807, 1024, 59049, 15625, 2357947691, 5184, 1792160394037, 282475249, 474609375, 17179869184, 2862423051509815793, 3486784401, 5480386857784802185939, 250000000000, 10382917022245341, 5559917313492231481, 39471584120695485887249589623

Offset: 1

Ilya Gutkovskiy, Jul 01 2019

nonn

Cf. A000010, A051190, A056916, A067911, A071248, A119619, A127553.

Table[Product[LCM[n, k]/(k GCD[n, k]), {k, 1, n}], {n, 1, 23}]
Table[Product[d^(EulerPhi[d] - EulerPhi[n/d]), {d, Divisors[n]}], {n, 1, 23}]

a(n) = prod(k=1, n, lcm(n, k)/(k*gcd(n, k))); \\ Michel Marcus, Jul 02 2019

a(n) = Product_{d|n} d^(phi(d)-phi(n/d)).
a(n) = n^n / Product_{d|n} d^(2*phi(n/d)).
a(n) = n^(-n) * Product_{d|n} d^(2*phi(d)).
a(n) = n^n / Product_{k=1..n} gcd(n,k)^2.
a(n) = n^(-n) * Product_{k=1..n} lcm(n,k)^2/k^2.
a(n) = A127553(n)/n!.
a(n) = A056916(n)/A067911(n).
a(p) = p^(p-2), where p is a prime.

A349785 a(n) = 1 + Product_{k=1..n-1} a(gcd(n,k)).

Original entry on oeis.org

2, 3, 5, 13, 17, 181, 65, 1873, 1601, 22033, 1025, 110120401, 4097, 3032641, 46240001, 6563711233, 65537, 61178101070401, 262145, 3771115818193153, 270400000001, 61977830401, 4194305, 27719934186822773256934560001, 87578116097, 8918277230593, 10498871296000001

Offset: 1

Ilya Gutkovskiy, Nov 30 2021

nonn

Cf. A000010, A000058, A051190, A338750, A343390.

a[n_] := a[n] = 1 + Product[a[GCD[n, k]], {k, 1, n - 1}]; Table[a[n], {n, 1, 27}]

a(n) = 1 + Product_{d|n, d < n} a(d)^phi(n/d).

cp's OEIS Frontend

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

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A067911 Product of gcd(k,n) for 1 <= k <= n.

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A224479 a(n) = Product_{k=1..n} Product_{i=1..k-1} gcd(k,i).

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A082022 In the following square array a(i,j) = Least Common Multiple of i and j. Sequence contains the product of the terms of the n-th antidiagonal.

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A276162 Square array read by antidiagonals: T(n,k) = Product_{i = 1..k} gcd(n, i).

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A349741 a(n) = Product_{k=1..n-1} phi(gcd(n,k)).

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A308944 a(n) = Product_{k=1..n} lcm(n,k) / (k * gcd(n,k)).

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