A010786 -id:A010786 - cp's OEIS frontend

1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 362880, 362880, 725760, 2177280, 8709120, 43545600, 261273600, 1828915200, 14631321600, 131681894400, 263363788800, 526727577600, 2106910310400, 12641461862400, 101131694899200

Offset: 1

Hieronymus Fischer, Jul 11 2007

			a(12)=dp_10(1)*dp_10(2)*dp_10(3)*...*dp_10(11)*dp_10(12)=1*2*3*4*5*6* 7*8*9*1*(1*1)*(1*2).
a(345)=3*4*5*3^45*4^5*(3-1)!^100*(4-1)!^10*(5-1)!^1*9!^64.
a(1000)=9!^300. a(1111)=9!^321.

Hieronymus Fischer, Table of n, a(n) for n = 1..102

Cf. A131385, A010786, A131387, A007953, A007954, A037131.

with transforms;
f:=proc(n) option remember; if n = 0 then 1 else f(n-1)*digprod0(n); fi; end;[seq(f(n),n=0..40)]; # N. J. A. Sloane, Oct 12 2013

The following formulas are given for general bases p>1:

a(n)=product{1<=k<=n, dp_p(k)} where dp_p(k) = product of the nonzero digits of k in base p.

a(n)=(n mod p)!*product{00}(floor(n/p^j)mod p)^(1+(n mod p^j))*((floor(n/p^j)mod p)-1)!^(p^j).

Recurrence: a(n+k*p^m)=a(n)*k^n*a(k*p^m) for 0<=k

a(n)=n!, for 0<=n

a(k*p^m)=k*(p-1)!^(k*m*p^(m-1))*(k-1)!^(p^m) for 0<=k

a(n)=(p-1)!^((m*p^(m+1)-(m+1)*p^m+1)/(p-1)^2)=(p-1)!^(1+2*p+3*p^2+...+m*p^(m-1)) for n=1+p+p^2+...+p^m.

a(n)=(p-1)!^(k*(m*p^(m+1)-(m+1)*p^m+1)/(p-1)^2)*(k-1)!^(p*(p^m-1)/(p-1))*k^(k*(p^(m+1)-(m+1)*p+m)/(p-1)^2)*k!*k^m, for n=k*(1+p+p^2+...+p^m).

For p=10: a(10^n)=9!^(n*10^(n-1)).

Asymptotic behavior: a(10^n)=10^(0.5559763...*n*10^n). Hence it grows slower than the factorial A000142(10^n) for which we have (10^n)!=10^((n-0.43429448...)*10^n+n/2+0.3990899...+o(1/n)). Example: a(1000) has 1668 digits, whereas 1000! has 2568 digits.

New b-file from Hieronymus Fischer, Sep 10 2007

2 typos in the formula section removed by Hieronymus Fischer, Dec 05 2011

A092143 Cumulative product of all divisors of 1..n.

Original entry on oeis.org

1, 2, 6, 48, 240, 8640, 60480, 3870720, 104509440, 10450944000, 114960384000, 198651543552000, 2582470066176000, 506164132970496000, 113886929918361600000, 116620216236402278400000, 1982543676018838732800000, 11562194718541867489689600000, 219681699652295482304102400000

Offset: 1

Jon Perry, Mar 31 2004

nonn

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). Let b(n) = A006218(n) = Sum_{k=1..n} floor(n/k). The prime factorization of a(n) appears to be given by the following conjectural formula: ordp(a(n),p) = b(floor(n/p)) + b(floor(n/p^2)) + b(floor(n/p^3)) + ... . Compare with the comments in A129365. - Peter Bala, Apr 15 2007

			a(6) = 1*2*3*2*4*5*2*3*6 = 8640.

G. C. Greubel, Table of n, a(n) for n = 1..190
Angelo B. Mingarelli, Abstract factorials, arXiv:0705.4299 [math.NT], 2007-2012.

Cf. A006218, A010786, A092144, A117871.
Cf. A129364, A129365, A129439, A345466.
Cf. A000005, A000178, A007955, A175493, A280714.

[(&*[j^Floor(n/j): j in [1..n]]): n in [1..30]]; // G. C. Greubel, Feb 05 2024

seq(sqrt(mul(k^numtheory[tau](k), k=1..n)), n=1..40); # Ridouane Oudra, Oct 31 2024

Reap[For[n = k = 1, k <= 25, k++, Do[n = n*d, {d, Divisors[k]}]; Sow[n]]][[2, 1]] (* Jean-François Alcover, Oct 30 2012 *)
Table[Product[k^Floor[n/k], {k, 1, n}], {n, 1, 25}] (* Vaclav Kotesovec, Jun 24 2021 *)
FoldList[Times, Times @@@ Divisors[Range[25]]] (* Paolo Xausa, Nov 06 2024 *)

my(z=1); for(i=1,25, fordiv(i,j,z*=j); print1(z, ", "))

[product(j^(n//j) for j in range(1,n+1)) for n in range(1,31)] # G. C. Greubel, Feb 05 2024

a(n) = Product_{k=1..n} {floor(n/k)}!. This formula is due to Sebastian Martin Ruiz. - Peter Bala, Apr 15 2007; Formula corrected by R. J. Mathar, May 06 2008
Sum_{n>=1} 1/a(n) = A117871. - Amiram Eldar, Nov 17 2020
log(a(n)) ~ n * log(n)^2 / 2. - Vaclav Kotesovec, Jun 20 2021
a(n) = Product_{k=1..n} k^floor(n/k). - Vaclav Kotesovec, Jun 24 2021
From Ridouane Oudra, Oct 31 2024: (Start)
a(n) = Product_{k=1..n} A007955(k).
a(n) = Product_{k=1..n} k^(tau(k)/2).
a(n) = sqrt(A175493(n)). (End)
a(n) = A000178(n)/A280714(n). - Amiram Eldar, Aug 16 2025

A131385 Product ceiling(n/1)ceiling(n/2)ceiling(n/3)...ceiling(n/n) (the 'ceiling factorial').

Original entry on oeis.org

1, 1, 2, 6, 16, 60, 144, 672, 1536, 6480, 19200, 76032, 165888, 1048320, 2257920, 8294400, 28311552, 126904320, 268738560, 1470873600, 3096576000, 16094453760, 51385466880, 175814737920, 366917713920, 2717245440000, 6782244618240, 22754631352320, 69918208819200

Offset: 0

Hieronymus Fischer, Jul 08 2007

nonn

From R. J. Mathar, Dec 05 2012: (Start)
a(n) = b(n-1) because a(n) = Product_{k=1..n} ceiling(n/k) = Product_{k=1..n-1} ceiling(n/k) = n*Product_{k=2..n-1} ceiling(n/k) = Product_{k=1..1} (1+(n-1)/k)*Product_{k=2..n-1} ceiling(n/k).
The cases of the product are (i) k divides n but does not divide n-1, ceiling(n/k) = n/k = 1 + floor((n-1)/k), (ii) k does not divide n but divides n-1, ceiling(n/k) = 1 + (n-1)/k = 1 + floor((n-1)/k) and (iii) k divides neither n nor n-1, ceiling(n/k) = 1 + floor((n-1)/k).
In all cases, including k=1, a(n) = Product_{k=1..n-1} (1+floor((n-1)/k)) = Product_{k=1..n-1} floor(1+(n-1)/k) = b(n-1).
(End)
a(n) is the number of functions f:D->{1,2,..,n-1} where D is any subset of {1,2,..,n-1} and where f(x) == 0 (mod x) for every x in D. - Dennis P. Walsh, Nov 13 2015

			From _Paul D. Hanna_, Nov 26 2012: (Start)
Illustrate initial terms using formula involving the floor function []:
  a(1) = 1;
  a(2) = [2/1] = 2;
  a(3) = [3/1]*[4/2] = 6;
  a(4) = [4/1]*[5/2]*[6/3] = 16;
  a(5) = [5/1]*[5/2]*[7/3]*[8/4] = 60;
  a(6) = [6/1]*[7/2]*[8/3]*[9/4]*[10/5] = 144.
Illustrate another alternative generating method:
  a(1) = 1;
  a(2) = (2/1)^[1/1] = 2;
  a(3) = (2/1)^[2/1] * (3/2)^[2/2] = 6;
  a(4) = (2/1)^[3/1] * (3/2)^[3/2] * (4/3)^[3/3] = 16;
  a(5) = (2/1)^[4/1] * (3/2)^[4/2] * (4/3)^[4/3] * (5/4)^[4/4] = 60.
(End)
For n=3 the a(3)=6 functions f from subsets of {1,2} into {1,2} with f(x) == 0 (mod x) are the following: f=empty set (since null function vacuously holds), f={(1,1)}, f={(1,2)}, f={(2,2)}, f={(1,1),(2,2)}, and f={(1,2),(2,2)}. - _Dennis P. Walsh_, Nov 13 2015

G. C. Greubel, Table of n, a(n) for n = 0..1000
D. P. Walsh, Notes on finite functions from subsets of {1,2,...,n} into {1,2,...,n}

Cf. A010786, A131387, A075885, A131688, A308820, A092143.

a:= n-> mul(ceil(n/k), k=1..n):
seq(a(n), n=0..40); # Dennis P. Walsh, Nov 13 2015

Table[Product[Ceiling[n/k],{k,n}],{n,25}] (* Harvey P. Dale, Sep 18 2011 *)

a(n)=prod(k=1,n-1,floor((n+k-1)/k)) \\ Paul D. Hanna, Feb 01 2013

a(n)=prod(k=1,n-1,((k+1)/k)^floor((n-1)/k))
for(n=1,30,print1(a(n),", ")) \\ Paul D. Hanna, Feb 01 2013

a(n) = Product_{k=1..n} ceiling(n/k).
Formulas from Paul D. Hanna, Nov 26 2012: (Start)
a(n) = Product_{k=1..n-1} floor((n+k-1)/k) for n>1.
a(n) = Product_{k=1..n-1} ((k+1)/k)^floor((n-1)/k) for n>1.
Limits: Let L = limit a(n+1)/a(n) = 3.51748725590236964939979369932386417..., then
(1) L = exp( Sum_{n>=1} log((n+1)/n) / n ) ;
(2) L = 2 * exp( Sum_{n>=1} (-1)^(n+1) * Sum_{k>=2} 1/(n*k^(n+1)) ) ;
(4) L = exp( Sum_{n>=1} (-1)^(n+1) * zeta(n+1)/n ) ;
(5) L = exp( Sum_{n>=1} log(n+1) / (n*(n+1)) ) = exp(c) where c = constant A131688.
Compare L to Alladi-Grinstead constant defined by A085291 and A085361.
(End)
a(n) = A308820(n)/A092143(n-1) for n > 0. - Ridouane Oudra, Sep 28 2024

a(0)=1 prepended by Alois P. Heinz, Oct 30 2023

A138534 Super least prime signatures; LCM of all signatures with n factors.

Original entry on oeis.org

1, 2, 12, 120, 5040, 110880, 43243200, 1470268800, 1173274502400, 269853135552000, 516498901446528000, 32022931889684736000, 3234636350177055183360000, 265240180714518525035520000, 1163343432613878250805790720000, 6014485546613750556665938022400000

Offset: 0

Alford Arnold, Mar 28 2008

nonn

Also the row product of the following table:
     1
     2
     4   3
     8   3   5
    16   9   5   7
    32   9   5   7   11
    64  27  25   7   11   13
   128  27  25   7   11   13   17
   256  81  25  49   11   13   17   19
   512  81 125  49   11   13   17   19   23
  1024 243 125  49  121   13   17   19   23   29
  ...

			For n = 3 the signatures are {8, 12, 30} so a(3) = 120.

Alois P. Heinz, Table of n, a(n) for n = 0..140
Angelo B. Mingarelli, Abstract factorials, Notes on Number Theory and Discrete Mathematics, Vol. 19, No. 4 (2013), pp. 43-76; arXiv preprint, arXiv:0705.4299 [math.NT], 2007-2012.

Cf. A006218, A010786, A010786, A010766.
Subsequence of A025487.
LCM of terms in rows of A215366.
Cf. A001222, A006218, A346044.

b:= proc(n, i) option remember; `if`(n=0 or i<2, 2^n,
      ilcm(seq(b(n-i*j, i-1)*ithprime(i)^j, j=0..n/i)))
    end:
a:= n-> b(n$2):
seq(a(n), n=0..17);  # Alois P. Heinz, May 15 2015

b[n_, i_] := b[n, i] = If[n == 0 || i < 2, 2^n, LCM @@ Table[b[n - i j, i - 1] Prime[i]^j, {j, 0, n/i}]];
a[n_] := b[n, n];
a /@ Range[0, 17] (* Jean-François Alcover, Nov 02 2020, after Alois P. Heinz *)
a[n_] := Product[Prime[k]^Floor[n/k], {k, 1, n}]; Array[a, 16, 0] (* Amiram Eldar, Jul 02 2021 *)

a(n) = prod(k=1, n, prime(k)^(n\k)); \\ Michel Marcus, Jul 03 2021

From Amiram Eldar, Jul 02 2021: (Start)
a(n) = Product_{k=1..n} prime(k)^floor(n/k).
A001222(a(n)) = A006218(n). (End)
Sum_{n>=0} 1/a(n) = A346044. - Amiram Eldar, Jul 02 2023

More terms from Reikku Kulon, Oct 02 2008

A345683 a(n) = n! * Sum_{k=1..n} 1/floor(n/k).

Original entry on oeis.org

1, 3, 14, 66, 444, 2880, 25080, 216720, 2247840, 24071040, 304335360, 3752179200, 54965433600, 810550540800, 13176376012800, 219079045785600, 4078723532083200, 75227891042304000, 1550619342784512000, 31871016307113984000, 710529031487987712000, 16180987966182014976000

Offset: 1

Vaclav Kotesovec, Jun 23 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..448
Vaclav Kotesovec, Plot of a(n) / (n*n!) for n = 1..1000000
Mathoverflow, An asymptotic formula for a sum involving powers of floor functions, 2019.

Cf. A006218, A010786, A092143, A117871, A345682, A345684, A345686, A355987.

Table[n! * Sum[1/Floor[n/k], {k, 1, n}], {n, 1, 25}]
Table[n!*(Sum[(Floor[n/j] - Floor[n/(j + 1)])/j, {j, 1, n}]), {n, 1, 25}]

a(n) = n!*sum(k=1, n, 1/(n\k)); \\ Michel Marcus, Jun 24 2021

my(N=30, x='x+O('x^N)); Vec(serlaplace(-sum(k=1, N, (1-x^k)*log(1-x^k))/(1-x))) \\ Seiichi Manyama, Jul 23 2022

from math import factorial, isqrt
def A345683(n): return (m:=factorial(n))*(n-1)+m//n+sum((q:=n//k)*(m//k-m//(k-1))+m//q for k in range(2,isqrt(n)+1)) # Chai Wah Wu, Oct 27 2023

a(n) ~ c * n * n!, where c = Sum_{j>=1} 1/(j^2*(j+1)) = Pi^2/6 - 1 = 0.644934... [proved by Harry Richman, see Mathoverflow link]

E.g.f.: -(1/(1-x)) * Sum_{k>0} (1 - x^k) * log(1 - x^k). - Seiichi Manyama, Jul 23 2022

A345682 a(n) = n! * Sum_{k=1..n} 1/(k*floor(n/k)).

Original entry on oeis.org

1, 2, 7, 26, 148, 804, 6228, 47424, 441936, 4288320, 50437440, 560373120, 7723935360, 106618256640, 1614841401600, 25127582054400, 446784010444800, 7727747269939200, 152873884406476800, 2966599550251008000, 62987912790921216000, 1378192085174919168000

Offset: 1

Vaclav Kotesovec, Jun 23 2021

nonn

Seiichi Manyama, Table of n, a(n) for n = 1..449
Vaclav Kotesovec, Plot of a(n)/n! for n = 1..1000000

Cf. A006218, A010786, A024917, A092143, A117871, A345683, A345684.

Table[n! * Sum[1/(k*Floor[n/k]), {k, 1, n}], {n, 1, 25}]
Table[n! * Sum[(HarmonicNumber[Floor[n/j]] - HarmonicNumber[Floor[n/(1 + j)]])/j, {j, 1, n}], {n, 1, 25}]

a(n) = n!*sum(k=1, n, 1/(k*(n\k))); \\ Michel Marcus, Jun 24 2021

my(N=30, x='x+O('x^N)); Vec(serlaplace(-sum(k=1, N, (1-x^k)*log(1-x^k)/k)/(1-x))) \\ Seiichi Manyama, Jul 23 2022

a(n) ~ c * n!, where c = Sum_{j>=1} log(1 + 1/j)/j = A131688 = 1.25774...

E.g.f.: -(1/(1-x)) * Sum_{k>0} (1 - x^k) * log(1 - x^k)/k. - Seiichi Manyama, Jul 23 2022

A131387 Product of the nonzero digital products of n for all the bases 1 to n (a 'total digital-product factorial').

Original entry on oeis.org

1, 1, 1, 1, 2, 4, 12, 48, 48, 192, 2880, 2880, 25920, 552960, 3265920, 1935360, 116121600, 278691840, 9405849600, 26754416640, 94058496000, 3210529996800, 869100503040000, 423789959577600, 927040536576000, 135612787064832000

Offset: 1

Hieronymus Fischer, Jul 08 2007

a(n) = {p = 1; for (b=2, n, digs = digits(n, b); p *= prod(k=1, #digs, if (digs[k], digs[k], 1));); return (p);} \\ Michel Marcus, Jul 15 2013

a(n)=product{1<=p<=n, dp_p(n)} where dp_p(n) = product of the nonzero digits of n in base p.

Cf. A131385, A010786, A007954.

Showing 1-10 of 25 results. Next

cp's OEIS Frontend

A208448 Greatest common divisors of consecutive floor-factorial numbers (A010786).

Views

Author

Keywords

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Formula

A208449 Numerator of A010786(n+1) / A010786(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Haskell

Mathematica

PARI

Formula

A208450 Denominator of A010786(n+1) / A010786(n).

Views

Author

Keywords

Links

Crossrefs

Programs

Haskell

Mathematica

Formula

A131451 Product of the nonzero digital products of all the numbers 1 to n (a 'total digital-product factorial' in base 10).

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Maple

Formula

Extensions

A092143 Cumulative product of all divisors of 1..n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

SageMath

Formula

A131385 Product ceiling(n/1)*ceiling(n/2)*ceiling(n/3)*...*ceiling(n/n) (the 'ceiling factorial').

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

Formula

Extensions

A138534 Super least prime signatures; LCM of all signatures with n factors.

Views

Author

Keywords

Comments

Examples

A131385 Product ceiling(n/1)ceiling(n/2)ceiling(n/3)...ceiling(n/n) (the 'ceiling factorial').