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A003418
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Least common multiple (or LCM) of {1, 2, ..., n} for n >= 1, a(0) = 1.
(Formerly M1590)
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369
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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
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OFFSET
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0,3
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COMMENTS
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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
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+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
For n > 2, (n-1) = Sum_{k=2..n} exp(a(n)*2*i*Pi/k). - Eric Desbiaux, Sep 13 2012
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
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REFERENCES
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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).
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LINKS
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R. E. Crandall and C. Pomerance, Prime numbers: a computational perspective, MR2156291, p. 61.
S. Ramanujan, Highly composite numbers, Proceedings of the London Mathematical Society ser. 2, vol. XIV, no. 1 (1915), pp 347-409. (A variant of a better quality with an additional footnote is available here.)
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FORMULA
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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
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) = exp(Sum_{k=1..n} Sum_{d|k} moebius(d)*log(k/d)). - Peter Luschny, Sep 01 2012
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
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)
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EXAMPLE
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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
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MAPLE
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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:
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MATHEMATICA
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FoldList[ LCM, 1, Range@ 28]
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]]];
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PROG
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(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) 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
(Sage) [lcm(range(1, n)) for n in range(1, 30)] # Zerinvary Lajos, Jun 06 2009
(Haskell)
a003418 = foldl lcm 1 . enumFromTo 2
(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
(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
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))
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CROSSREFS
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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).
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KEYWORD
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nonn,easy,core,nice
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AUTHOR
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Roland Anderson (roland.anderson(AT)swipnet.se)
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STATUS
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approved
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