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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.

A003418 Least common multiple (or LCM) of {1, 2, ..., n} for n >= 1, a(0) = 1.

Original entry on oeis.org

1, 1, 2, 6, 12, 60, 60, 420, 840, 2520, 2520, 27720, 27720, 360360, 360360, 360360, 720720, 12252240, 12252240, 232792560, 232792560, 232792560, 232792560, 5354228880, 5354228880, 26771144400, 26771144400, 80313433200, 80313433200, 2329089562800, 2329089562800
Offset: 0

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Author

Roland Anderson (roland.anderson(AT)swipnet.se)

Keywords

Comments

The minimal exponent of the symmetric group S_n, i.e., the least positive integer for which x^a(n)=1 for all x in S_n. - Franz Vrabec, Dec 28 2008
Product over all primes of highest power of prime less than or equal to n. a(0) = 1 by convention.
Also smallest number whose set of divisors contains an n-term arithmetic progression. - Reinhard Zumkeller, Dec 09 2002
An assertion equivalent to the Riemann hypothesis is: | log(a(n)) - n | < sqrt(n) * log(n)^2. - Lekraj Beedassy, Aug 27 2006. (This is wrong for n = 1 and n = 2. Should "for n large enough" be added? - Georgi Guninski, Oct 22 2011)
Corollary 3 of Farhi gives a proof that a(n) >= 2^(n-1). - Jonathan Vos Post, Jun 15 2009
Appears to be row products of the triangle T(n,k) = b(A010766) where b = A130087/A130086. - Mats Granvik, Jul 08 2009
Greg Martin (see link) proved that "the product of the Gamma function sampled over the set of all rational numbers in the open interval (0,1) whose denominator in lowest terms is at most n" equals (2*Pi)^(1/2)*a(n)^(-1/2). - Jonathan Vos Post, Jul 28 2009
a(n) = lcm(A188666(n), A188666(n)+1, ..., n). - Reinhard Zumkeller, Apr 25 2011
a(n+1) is the smallest integer such that all polynomials a(n+1)*(1^i + 2^i + ... + m^i) in m, for i=0,1,...,n, are polynomials with integer coefficients. - Vladimir Shevelev, Dec 23 2011
It appears that A020500(n) = a(n)/a(n-1). - Asher Auel, corrected by Bill McEachen, Apr 05 2024
n-th distinct value = A051451(n). - Matthew Vandermast, Nov 27 2009
a(n+1) = least common multiple of n-th row in A213999. - Reinhard Zumkeller, Jul 03 2012
For n > 2, (n-1) = Sum_{k=2..n} exp(a(n)*2*i*Pi/k). - Eric Desbiaux, Sep 13 2012
First column minus second column of A027446. - Eric Desbiaux, Mar 29 2013
For n > 0, a(n) is the smallest number k such that n is the n-th divisor of k. - Michel Lagneau, Apr 24 2014
Slowest growing integer > 0 in Z converging to 0 in Z^ when considered as profinite integer. - Herbert Eberle, May 01 2016
What is the largest number of consecutive terms that are all equal? I found 112 equal terms from a(370261) to a(370372). - Dmitry Kamenetsky, May 05 2019
Answer: there exist arbitrarily long sequences of consecutive terms with the same value; also, the maximal run of consecutive terms with different values is 5 from a(1) to a(5) (see link Roger B. Eggleton). - Bernard Schott, Aug 07 2019
Related to the inequality (54) in Ramanujan's paper about highly composite numbers A002182, also used in A199337: a(A329570(m))^2 is a (not minimal) bound above which all highly composite numbers are divisible by m, according to the right part of that inequality. - M. F. Hasler, Jan 04 2020
For n > 2, a(n) is of the form 2^e_1 * p_2^e_2 * ... * p_m^e_m, where e_m = 1 and e = floor(log_2(p_m)) <= e_1. Therefore, 2^e * p_m^e_m is a primitive Zumkeler number (A180332). Therefore, 2^e_1 * p_m^e_m is a Zumkeller number (A083207). Therefore, for n > 2, a(n) = 2^e_1 * p_m^e_m * r, where r is relatively prime to 2*p_m, is a Zumkeller number (see my proof at A002182 for details). - Ivan N. Ianakiev, May 10 2020
For n > 1, 2|(a(n)+2) ... n|(a(n)+n), so a(n)+2 .. a(n)+n are all composite and (part of) a prime gap of at least n. (Compare n!+2 .. n!+n). - Stephen E. Witham, Oct 09 2021

Examples

			LCM of {1,2,3,4,5,6} = 60. The primes up to 6 are 2, 3 and 5. floor(log(6)/log(2)) = 2 so the exponent of 2 is 2.
floor(log(6)/log(3)) = 1 so the exponent of 3 is 1.
floor(log(6)/log(5)) = 1 so the exponent of 5 is 1. Therefore, a(6) = 2^2 * 3^1 * 5^1 = 60. - _David A. Corneth_, Jun 02 2017

References

  • J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 365.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

Crossrefs

Row products of A133233.
Cf. A000142, A000793, A002110, A002182, A002201, A002944, A014963, A020500, A025527, A038610, A051173, A064446, A064859, A069513, A072938, A093880, A094348, A096179, A099996, A102910, A106037, A119682, A179661, A193181, A225558, A225630, A225632, A225640, A225642.
Cf. A025528 (number of prime factors of a(n) with multiplicity).
Cf. A275120 (lengths of runs of consecutive equal terms), A276781 (ordinal transform from term a(1)=1 onward).

Programs

  • Haskell
    a003418 = foldl lcm 1 . enumFromTo 2
-- Reinhard Zumkeller, Apr 04 2012, Apr 25 2011
  • Magma
    [1] cat [Exponent(SymmetricGroup(n)) : n in [1..28]]; // Arkadiusz Wesolowski, Sep 10 2013
  • Magma
    [Lcm([1..n]): n in [0..30]]; // Bruno Berselli, Feb 06 2015
  • Maple
    A003418 := n-> lcm(seq(i,i=1..n));
HalfFarey := proc(n) local a,b,c,d,k,s; a := 0; b := 1; c := 1; d := n; s := NULL; do k := iquo(n + b, d); a, b, c, d := c, d, k*c - a, k*d - b; if 2*a > b then break fi; s := s,(a/b); od: [s] end: LCM := proc(n) local i; (1/2)*mul(2*sin(Pi*i),i=HalfFarey(n))^2 end: # Peter Luschny
# next Maple program:
a:= proc(n) option remember; `if`(n=0, 1, ilcm(n, a(n-1))) end:
seq(a(n), n=0..33);  # Alois P. Heinz, Jun 10 2021
  • Mathematica
    Table[LCM @@ Range[n], {n, 1, 40}] (* Stefan Steinerberger, Apr 01 2006 *)
FoldList[ LCM, 1, Range@ 28]
A003418[0] := 1; A003418[1] := 1; A003418[n_] := A003418[n] = LCM[n,A003418[n-1]]; (* Enrique Pérez Herrero, Jan 08 2011 *)
Table[Product[Prime[i]^Floor[Log[Prime[i], n]], {i, PrimePi[n]}], {n, 0, 28}] (* Wei Zhou, Jun 25 2011 *)
Table[Product[Cyclotomic[n, 1], {n, 2, m}], {m, 0, 28}] (* Fred Daniel Kline, May 22 2014 *)
a1[n_] := 1/12 (Pi^2+3(-1)^n (PolyGamma[1,1+n/2] - PolyGamma[1,(1+n)/2])) // Simplify
a[n_] := Denominator[Sqrt[a1[n]]];
Table[If[IntegerQ[a[n]], a[n], a[n]*(a[n])[[2]]], {n, 0, 28}] (* Gerry Martens, Apr 07 2018 [Corrected by Vaclav Kotesovec, Jul 16 2021] *)
  • PARI
    a(n)=local(t); t=n>=0; forprime(p=2,n,t*=p^(log(n)\log(p))); t
  • PARI
    a(n)=if(n<1,n==0,1/content(vector(n,k,1/k)))
  • PARI
    a(n)=my(v=primes(primepi(n)),k=sqrtint(n),L=log(n+.5));prod(i=1,#v,if(v[i]>k,v[i],v[i]^(L\log(v[i])))) \\ Charles R Greathouse IV, Dec 21 2011
  • PARI
    a(n)=lcm(vector(n,i,i)) \\ Bill Allombert, Apr 18 2012 [via Charles R Greathouse IV]
  • PARI
    n=1; lim=100; i=1; j=1; until(n==lim, a=lcm(j,i+1); i++; j=a; n++; print(n" "a);); \\ Mike Winkler, Sep 07 2013
  • Python
    from functools import reduce
from operator import mul
from sympy import sieve
def integerlog(n,b): # find largest integer k>=0 such that b^k <= n
    kmin, kmax = 0,1
    while b**kmax <= n:
        kmax *= 2
    while True:
        kmid = (kmax+kmin)//2
        if b**kmid > n:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return kmin
def A003418(n):
    return reduce(mul,(p**integerlog(n,p) for p in sieve.primerange(1,n+1)),1) # Chai Wah Wu, Mar 13 2021
  • Python
    # generates initial segment of sequence
from math import gcd
from itertools import accumulate
def lcm(a, b): return a * b // gcd(a, b)
def aupton(nn): return [1] + list(accumulate(range(1, nn+1), lcm))
print(aupton(30)) # Michael S. Branicky, Jun 10 2021
  • Sage
    [lcm(range(1,n)) for n in range(1, 30)] # Zerinvary Lajos, Jun 06 2009
  • Scheme
    (define (A003418 n) (let loop ((n n) (m 1)) (if (zero? n) m (loop (- n 1) (lcm m n))))) ;; Antti Karttunen, Jan 03 2018

Formula

The prime number theorem implies that lcm(1,2,...,n) = exp(n(1+o(1))) as n -> infinity. In other words, log(lcm(1,2,...,n))/n -> 1 as n -> infinity. - Jonathan Sondow, Jan 17 2005
a(n) = Product (p^(floor(log n/log p))), where p runs through primes not exceeding n (i.e., primes 2 through A007917(n)). - Lekraj Beedassy, Jul 27 2004
Greg Martin showed that a(n) = lcm(1,2,3,...,n) = Product_{i = Farey(n), 0 < i < 1} 2*Pi/Gamma(i)^2. This can be rewritten (for n > 1) as a(n) = (1/2)*(Product_{i = Farey(n), 0 < i <= 1/2} 2*sin(i*Pi))^2. - Peter Luschny, Aug 08 2009
Recursive formula useful for computations: a(0)=1; a(1)=1; a(n)=lcm(n,a(n-1)). - Enrique Pérez Herrero, Jan 08 2011
From Enrique Pérez Herrero, Jun 01 2011: (Start)
a(n)/a(n-1) = A014963(n).
if n is a prime power p^k then a(n)=a(p^k)=p*a(n-1), otherwise a(n)=a(n-1).
a(n) = Product_{k=2..n} (1 + (A007947(k)-1)*floor(1/A001221(k))), for n > 1. (End)
a(n) = A079542(n+1, 2) for n > 1.
a(n) = exp(Sum_{k=1..n} Sum_{d|k} moebius(d)*log(k/d)). - Peter Luschny, Sep 01 2012
a(n) = A025529(n) - A027457(n). - Eric Desbiaux, Mar 14 2013
a(n) = exp(Psi(n)) = 2 * Product_{k=2..A002088(n)} (1 - exp(2*Pi*i * A038566(k+1) / A038567(k))), where i is the imaginary unit, and Psi the second Chebyshev's function. - Eric Desbiaux, Aug 13 2014
a(n) = A064446(n)*A038610(n). - Anthony Browne, Jun 16 2016
a(n) = A000142(n) / A025527(n) = A000793(n) * A225558(n). - Antti Karttunen, Jun 02 2017
log(a(n)) = Sum_{k>=1} (A309229(n, k)/k - 1/k). - Mats Granvik, Aug 10 2019
From Petros Hadjicostas, Jul 24 2020: (Start)
Nair (1982) proved that 2^n <= a(n) <= 4^n for n >= 9. See also Farhi (2009). Nair also proved that
a(n) = lcm(m*binomial(n,m): 1 <= m <= n) and
a(n) = gcd(a(m)*binomial(n,m): n/2 <= m <= n). (End)
Sum_{n>=1} 1/a(n) = A064859. - Bernard Schott, Aug 24 2020

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