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 1-5 of 5 results.

A129635 Decimal expansion of 1 + A117871.

Original entry on oeis.org

2, 6, 9, 1, 7, 9, 9, 2, 0, 9, 8, 2, 1, 7, 1, 2, 3, 5, 1, 3, 3, 9, 2, 6, 1, 8, 0, 6, 7, 8, 7, 6, 3, 1, 8, 6, 9, 8, 2, 3, 6, 9, 3, 7, 6, 2, 9, 2, 5, 8, 1, 9, 1, 3, 4, 5, 5, 6, 9
Offset: 1

Views

Author

Jonathan Vos Post, May 31 2007

Keywords

Examples

			2.6917992098217123513392618067876318698236937629258191345569...
		

Crossrefs

Cf. A092143. See A117871 for a better version.

Extensions

More terms from R. J. Mathar, Sep 02 2007
Edited by N. J. A. Sloane, Sep 16 2007 and May 06 2008

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

Views

Author

Jon Perry, Mar 31 2004

Keywords

Comments

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

Examples

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

Crossrefs

Programs

  • Magma
    [(&*[j^Floor(n/j): j in [1..n]]): n in [1..30]]; // G. C. Greubel, Feb 05 2024
    
  • Maple
    seq(sqrt(mul(k^numtheory[tau](k), k=1..n)), n=1..40); # Ridouane Oudra, Oct 31 2024
  • Mathematica
    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 *)
  • PARI
    my(z=1); for(i=1,25, fordiv(i,j,z*=j); print1(z, ", "))
    
  • SageMath
    [product(j^(n//j) for j in range(1,n+1)) for n in range(1,31)] # G. C. Greubel, Feb 05 2024

Formula

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

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

Views

Author

Vaclav Kotesovec, Jun 23 2021

Keywords

Crossrefs

Programs

  • Mathematica
    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}]
  • PARI
    a(n) = n!*sum(k=1, n, 1/(n\k)); \\ Michel Marcus, Jun 24 2021
    
  • PARI
    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
    
  • Python
    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

Formula

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

Views

Author

Vaclav Kotesovec, Jun 23 2021

Keywords

Crossrefs

Programs

  • Mathematica
    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}]
  • PARI
    a(n) = n!*sum(k=1, n, 1/(k*(n\k))); \\ Michel Marcus, Jun 24 2021
    
  • PARI
    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

Formula

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

A345684 a(n) = n! * Sum_{k=1..n} k/floor(n/k).

Original entry on oeis.org

1, 5, 32, 198, 1584, 12480, 122520, 1214640, 14011200, 166924800, 2274894720, 31135104000, 485667705600, 7710089587200, 133974352512000, 2386854434764800, 46621903994265600, 918384939343872000, 19760215067873280000, 430137075045629952000, 10042411264251125760000
Offset: 1

Views

Author

Vaclav Kotesovec, Jun 23 2021

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n!*Sum[k/Floor[n/k], {k, 1, n}], {n, 1, 25}]
    Table[n!*Sum[(Floor[n/j] - Floor[n/(1 + j)])*((1 + Floor[n/j] + Floor[n/(1 + j)])/2/j), {j, 1, n}], {n, 1, 25}]
  • PARI
    a(n) = n!*sum(k=1, n, k/(n\k)); \\ Michel Marcus, Jun 23 2021
    
  • PARI
    my(N=30, x='x+O('x^N)); Vec(serlaplace(-sum(k=1, N, k*(1-x^k)*log(1-x^k))/(1-x))) \\ Seiichi Manyama, Jul 23 2022

Formula

a(n) ~ c * n^2 * n!, where c = Sum_{j>=1} (2*j + 1) / (2*j^3*(j+1)^2) = Pi^2/12 + zeta(3)/2 - 1 = 0.423495...
E.g.f.: -(1/(1-x)) * Sum_{k>0} k * (1 - x^k) * log(1 - x^k). - Seiichi Manyama, Jul 23 2022
Showing 1-5 of 5 results.