A025537 -id:A025537 - cp's OEIS frontend

A002944 a(n) = LCM(1,2,...,n) / n.

1, 1, 2, 3, 12, 10, 60, 105, 280, 252, 2520, 2310, 27720, 25740, 24024, 45045, 720720, 680680, 12252240, 11639628, 11085360, 10581480, 232792560, 223092870, 1070845776, 1029659400, 2974571600, 2868336900, 80313433200, 77636318760, 2329089562800

Offset: 1

N. J. A. Sloane

Equals LCM of all numbers of (n-1)-st row of Pascal's triangle [Montgomery-Breusch]. - J. Lowell, Apr 16 2014. Corrected by N. J. A. Sloane, Sep 04 2019

Williams proves that a(n+1) = A034386(n) for n=2, 11 and 23 only. This is trivially the case for n=0 and 1, too. - Michel Marcus, Apr 16 2020

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..500
Bakir Farhi, An identity involving the least common multiple of binomial coefficients and its application, arXiv:0906.2295 [math.NT], 2009.
Bakir Farhi, An identity involving the least common multiple of binomial coefficients and its application, Amer. Math. Monthly, 116 (2009), 836-839.
Peter L. Montgomery (proposer) and Robert Breusch (solver), LCM of Binomial Coefficients, Problem E2686, American Mathematical Monthly, Vol. 84 (1977), p. 820.
Peter L. Montgomery (proposer) and Robert Breusch (solver), LCM of Binomial Coefficients, Solution to Problem E2686, American Mathematical Monthly, Vol. 86 (1979), p. 131.
Ian S. Williams, On a problem of Kurt Mahler concerning binomial coefficents (sic), Bulletin of the Australian Mathematical Society, Volume 14, Issue 2, April 1976, pp. 299-302.
Index entries for sequences related to lcm's

Cf. A025527 and A025537.
Cf. A100561, A003418, A001142, A001405. - Franklin T. Adams-Watters, Jul 05 2009
Cf. A056606 (squarefree kernel).

a002944 n = a003418 n `div` n  -- Reinhard Zumkeller, Mar 16 2015

A003418 := n-> lcm(seq(i,i=1..n)); f:=n->A003418(n)/n;
BB:=n->sum(1/sqrt(k), k=1..n): a:=n->floor(denom(BB(n))/n): seq(a(n), n=1..29); # Zerinvary Lajos, Mar 29 2007

Table[Apply[LCM,Range[n]]/n,{n,1,30}]  (* Geoffrey Critzer, Feb 10 2013 *)

a(n) = lcm(vector(n, i, i))/n; \\ Michel Marcus, Apr 16 2014

a(n) = A003418(n) / n.
a(n) = LCM of C(n-1, 0), C(n-1, 1), ..., C(n-1, n-1). [Montgomery-Breusch] [Corrected by N. J. A. Sloane, Jun 11 2008]
Equally, a(n+1) = LCM_{k=0..n} binomial(n,k). - Franklin T. Adams-Watters, Jul 05 2009

More terms from Jud McCranie, Jan 17 2000

Edited by N. J. A. Sloane, Jun 11 2008 and Sep 04 2019

A025527 a(n) = n!/lcm{1,2,...,n} = (n-1)!/lcm{C(n-1,0), C(n-1,1), ..., C(n-1,n-1)}.

Original entry on oeis.org

1, 1, 1, 2, 2, 12, 12, 48, 144, 1440, 1440, 17280, 17280, 241920, 3628800, 29030400, 29030400, 522547200, 522547200, 10450944000, 219469824000, 4828336128000, 4828336128000, 115880067072000, 579400335360000, 15064408719360000

Offset: 1

Clark Kimberling, Dec 11 1999

nonn

a(n) = a(n-1) iff n is prime. Thus a(1)=a(2)=a(3)=1 is the only triple in this sequence. - Franz Vrabec, Sep 10 2005
a(k) = a(k+1) for k in A006093. - Lekraj Beedassy, Aug 03 2006
Partial products of A048671. - Peter Luschny, Sep 09 2009

			a(5) = 2 as 5!/lcm(1..5) = 120/60 = 2.

Alois P. Heinz, Table of n, a(n) for n = 1..500
Liam Solus, Simplices for Numeral Systems, arXiv:1706.00480 [math.CO], 2017. Mentions this sequence.
Index entries for sequences related to lcm's

Cf. A000142, A002541, A006093, A003418, A048671, A025529, A205957.

See also A002944, A025537.

List([1..30],n->Factorial(n)/Lcm([1..n])); # Muniru A Asiru, Apr 02 2018

seq(n!/lcm($1..n), n=1..30);
A025527 := proc(n) option remember; `if`(n < 3, 1, ilcm(op(numtheory[divisors](n) minus{1,n}))*A025527(n-1)) end:
seq(A025527(i),i=1..26); # Peter Luschny, Mar 23 2011

Table[n!/Apply[LCM,Range[n]],{n,1,26}] (* Geoffrey Critzer, Jun 17 2013 *)

a(n)=n!/lcm([2..n]) \\ Charles R Greathouse IV, Mar 06 2014

def A025527(n) :
    if n < 2 : return 1
    else :
        D = divisors(n); D.pop()
    return lcm(D)*A025527(n-1)
[A025527(i) for i in (1..26)] # Peter Luschny, Feb 03 2012

a(n) = A000142(n)/A003418(n) = A000254(n)/A025529(n). - Franz Vrabec, Sep 13 2005
log a(n) = n log n - 2n + O(n/log^4 n). (The error term can be improved. On the Riemann Hypothesis it is O(n^k) for any k > 1/2.) - Charles R Greathouse IV, Oct 16 2012
a(n) = A205957(n), 1 <= n <= 11. - Daniel Forgues, Apr 22 2014
Conjecture: a(A006093(n)) = phi(A000142(A006093(n))) / phi(A003418(A006093(n))), where phi is the Euler totient function. - Fred Daniel Kline, Jun 03 2017

cp's OEIS Frontend

A002944 a(n) = LCM(1,2,...,n) / n.

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