A025529 -id:A025529 - cp's OEIS frontend

A002805 Denominators of harmonic numbers H(n) = Sum_{i=1..n} 1/i.

1, 2, 6, 12, 60, 20, 140, 280, 2520, 2520, 27720, 27720, 360360, 360360, 360360, 720720, 12252240, 4084080, 77597520, 15519504, 5173168, 5173168, 118982864, 356948592, 8923714800, 8923714800, 80313433200, 80313433200, 2329089562800, 2329089562800, 72201776446800

Offset: 1

N. J. A. Sloane

H(n)/2 is the maximal distance that a stack of n cards can project beyond the edge of a table without toppling.
If n is not in {1, 2, 6} then a(n) has at least one prime factor other than 2 or 5. E.g., a(5) = 60 has a prime factor 3 and a(7) = 140 has a prime factor 7. This implies that every H(n) = A001008(n)/A002805(n), n not from {1, 2, 6}, has an infinite decimal representation. For a proof see the J. Havil reference. - Wolfdieter Lang, Jun 29 2007
a(n) = A213999(n,n-1). - Reinhard Zumkeller, Jul 03 2012
From Wolfdieter Lang, Apr 16 2015: (Start)
a(n)/A001008(n) = 1/H(n) is the solution of the following version of the classical cistern and pipes problem. A cistern is connected to n different pipes of water. For the k-th pipe it takes k time units (say, days) to fill the empty cistern, for k = 1, 2, ..., n. How long does it take for the n pipes together to fill the empty cistern? 1/H(n) gives the answer as a fraction of the time unit.
In general, if the k-th pipe needs d(k) days to fill the empty cistern then all pipes together need 1/Sum_{k=1..n} 1/d(k) = HM(d(1), ..., d(n))/n days, where HM denotes the harmonic mean HM. For the described problem, HM(1, 2, ..., n)/n = A102928(n)/(n*A175441(n)) = 1/H(n).
For a classical cistern and pipes problem see, e.g., the Hunger-Vogel reference (in Greek and German) given in A256101, problem 27, p. 29, where n = 3, and d(1), d(2) and d(3) are 6, 4 and 3 days. On p. 97 of this reference one finds remarks on the history of such problems (called in German 'Brunnenaufgabe'). (End)
From Wolfdieter Lang, Apr 17 2015: (Start)
An example of the above mentioned cistern and pipes problems appears in Chiu Chang Suan Shu (nine books on arithmetic) in book VI, problem 26. The numbers are there 1/2, 1, 5/2, 3 and 5 (days) and the result is 15/75 (day). See the reference (in German) on p. 68.
A historical account on such cistern problems is found in the Johannes Tropfke reference, given in A256101, section 4.2.1.2 Zisternenprobleme (Leistungsprobleme), pp. 578-579.
In Fibonacci's Liber Abaci such problems appear on p. 281 and p. 284 of L. E. Sigler's translation. (End)
All terms > 1 are even while corresponding numerators (A001008) are all odd (proof in Pólya and Szegő). - Bernard Schott, Dec 24 2021

			H(n) = [ 1, 3/2, 11/6, 25/12, 137/60, 49/20, 363/140, 761/280, 7129/2520, ... ] = A001008/A002805.

Chiu Chang Suan Shu, Neun Bücher arithmetischer Technik, translated and commented by Kurt Vogel, Ostwalds Klassiker der exakten Wissenschaften, Band 4, Friedr. Vieweg & Sohn, Braunschweig, 1968, p. 68.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 258-261.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 259.
J. Havil, Gamma, (in German), Springer, 2007, p. 35-6; Gamma: Exploring Euler's Constant, Princeton Univ. Press, 2003.
D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, Vol. 1, p. 615.
G. Pólya and G. Szegő, Problems and Theorems in Analysis, volume II, Springer, reprint of the 1976 edition, 1998, problem 251, p. 154.
L. E. Sigler, Fibonacci's Liber Abaci, Springer, 2003, pp. 281, 284.
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).

Kenny Lau, Table of n, a(n) for n = 1..2308 [First 200 terms computed by T. D. Noe]
R. M. Dickau, Harmonic numbers and the book stacking problem.
Haight, Frank A., and Robert B. Jones., "A probabilistic treatment of qualitative data with special reference to word association tests." Journal of Mathematical Psychology 11.3 (1974): 237-244. [Annotated scanned copy]
Frank Haight and N. J. A. Sloane, Correspondence, 1975.
Antal Iványi, Leader election in synchronous networks, Acta Univ. Sapientiae, Mathematica, 5, 2 (2013) 54-82.
Fredrik Johansson, How (not) to compute harmonic numbers, Feb 21 2009.
Peter Shiu, The denominators of harmonic numbers, arXiv:1607.02863 [math.NT], 2016.
N. J. A. Sloane, Illustration of initial terms.
Jonathan Sondow and Eric W. Weisstein, MathWorld: Harmonic Number.
Eric Weisstein's World of Mathematics, Book Stacking Problem.
Bing-Ling Wu, Yong-Gao Chen, On the denominators of harmonic numbers, arXiv:1711.00184 [math.NT], 2017.

Cf. A001008 (numerators), A075135, A025529, A203810, A203811, A203812.
Partial sums: A027612/A027611 = 1, 5/2, 13/3, 77/12, 87/10, 223/20,...
The following fractions are all related to each other: Sum 1/n: A001008/A002805, Sum 1/prime(n): A024451/A002110 and A106830/A034386, Sum 1/nonprime(n): A282511/A282512, Sum 1/composite(n): A250133/A296358, Sum 1/n^2: A007406/A007407, Sum 1/n^3: A007408/A007409.

List([1..30],n->DenominatorRat(Sum([1..n],i->1/i))); # Muniru A Asiru, Dec 20 2018

import Data.Ratio ((%), denominator)
a002805 = denominator . sum . map (1 %) . enumFromTo 1
a002805_list = map denominator $ scanl1 (+) $ map (1 %) [1..]
-- Reinhard Zumkeller, Jul 03 2012

[Denominator(HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Apr 16 2015

seq(denom(sum((2*k-1)/k, k=1..n), n=1..30); # Gary Detlefs, Jul 18 2011
f:=n->denom(add(1/k, k=1..n)); # N. J. A. Sloane, Nov 15 2013

Denominator[ Drop[ FoldList[ #1 + 1/#2 &, 0, Range[ 30 ] ], 1 ] ] (* Harvey P. Dale, Feb 09 2000 *)
Table[Denominator[HarmonicNumber[n]], {n, 1, 40}] (* Stefan Steinerberger, Apr 20 2006 *)
Denominator[Accumulate[1/Range[25]]] (* Alonso del Arte, Nov 21 2018 *)

a(n)=denominator(sum(k=2,n,1/k)) \\ Charles R Greathouse IV, Feb 11 2011

from fractions import Fraction
def a(n): return sum(Fraction(1, i) for i in range(1, n+1)).denominator
print([a(n) for n in range(1, 30)]) # Michael S. Branicky, Dec 24 2021

def harmonic(a, b): # See the F. Johansson link.
    if b - a == 1 : return 1, a
    m = (a+b)//2
    p, q = harmonic(a,m)
    r, s = harmonic(m,b)
    return p*s+q*r, q*s
def A002805(n) : H = harmonic(1,n+1); return denominator(H[0]/H[1])
[A002805(n) for n in (1..29)] # Peter Luschny, Sep 01 2012

a(n) = Denominator(Sum_{k=1..n} (2*k-1)/k). - Gary Detlefs, Jul 18 2011
a(n) = n! / gcd(Stirling1(n+1, 2), n!) = n! / gcd(A000254(n),n!). - Max Alekseyev, Mar 01 2018
a(n) = the (reduced) denominator of the continued fraction 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n-1)^2/(2*n-1))))). - Peter Bala, Feb 18 2024

Definition edited by Daniel Forgues, May 19 2010

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

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

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

			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

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).

Seiichi Manyama, Table of n, a(n) for n = 0..2308 (first 501 terms from T. D. Noe)
R. Anderson and N. J. A. Sloane, Correspondence, 1975.
Dorin Andrica, Sorin Rădulescu, and George Cătălin Ţurcaş, The Exponent of a Group: Properties, Computations and Applications, Disc. Math. and Applications, Springer, Cham (2020), 57-108.
Javier Cilleruelo, Juanjo Rué, Paulius Šarka, and Ana Zumalacárregui, The least common multiple of sets of positive integers, arXiv:1112.3013 [math.NT], 2011.
R. E. Crandall and C. Pomerance, Prime numbers: a computational perspective, MR2156291, p. 61.
Roger B. Eggleton, Least Common Multiple of {1,2,...,n}, Mathematics Magazine, 61(1) (1988), pp. 47-48, Problem 1252.
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(9) (2009), 836-839.
Steven Finch, Cilleruelo's LCM Constants, 2013. [Cached copy, with permission of the author]
V. L. Gavrikov, On property of least common multiple to be a D-magic number, arXiv:1806.09264 [math.NT], 2018.
S. Labbé and E. Pelantová, Palindromic sequences generated from marked morphisms, arXiv:1409.7510 [math.CO], 2014.
J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis, Am. Math. Monthly 109 (6) (2002) 534-543. arXiv:math/0008177 [math.NT], 2000-2001.
Peter Luschny and S. Wehmeier, The lcm(1, 2, ..., n) as a product of sine values sampled over the points in Farey sequences, arXiv:0909.1838 [math.CA], 2009.
Des MacHale and Joseph Manning, Maximal runs of strictly composite integers, The Mathematical Gazette, 99, pp 213-219 (2015).
Greg Martin, A product of Gamma function values at fractions with the same denominator, arXiv:0907.4384 [math.CA], 2009.
M. Nair, On Chebychev-type inequalities for primes Amer. Math. Monthly 89(2) (1982), 126-129.
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.)
E. S. Selmer, On the number of prime divisors of a binomial coefficient, Math. Scand. 39 (1976), no. 2, 271-281 (1977).
Jonathan Sondow, Criteria for irrationality of Euler's constant, Proc. AMS 131 (2003), 3335.
Rosemary Sullivan and Neil Watling, Independent divisibility pairs on the set of integers from 1 to n, INTEGERS 13 (2013) #A65.
M. Tchebichef, Mémoire sur les nombres premiers, J. Math. Pures Appliquées 17 (1852), 366-390.
Helge von Koch, Sur la distribution des nombres premiers, Acta Math. 24 (1) (1901), 159-182.
Eric Weisstein's World of Mathematics, Least Common Multiple, Chebyshev Functions, Mangoldt Function.
Index to divisibility sequences
Index entries for "core" sequences
Index entries for sequences related to lcm's

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).

a003418 = foldl lcm 1 . enumFromTo 2
-- Reinhard Zumkeller, Apr 04 2012, Apr 25 2011

[1] cat [Exponent(SymmetricGroup(n)) : n in [1..28]]; // Arkadiusz Wesolowski, Sep 10 2013

[Lcm([1..n]): n in [0..30]]; // Bruno Berselli, Feb 06 2015

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

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] *)

a(n)=local(t); t=n>=0; forprime(p=2,n,t*=p^(log(n)\log(p))); t

a(n)=if(n<1,n==0,1/content(vector(n,k,1/k)))

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

a(n)=lcm(vector(n,i,i)) \\ Bill Allombert, Apr 18 2012 [via Charles R Greathouse IV]

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

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

# 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

[lcm(range(1,n)) for n in range(1, 30)] # Zerinvary Lajos, Jun 06 2009

(define (A003418 n) (let loop ((n n) (m 1)) (if (zero? n) m (loop (- n 1) (lcm m n))))) ;; Antti Karttunen, Jan 03 2018

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

A110566 a(n) = lcm{1,2,...,n}/denominator of harmonic number H(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 15, 45, 45, 45, 15, 3, 3, 1, 1, 1, 1, 1, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 77, 77, 7, 7, 7, 7, 7, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 9, 9, 9, 27, 27, 27, 9, 9, 9, 3, 3, 3, 3, 3, 33, 33, 33, 33, 11, 11, 11, 11, 11, 11, 11, 1, 1, 1

Offset: 1

Franz Vrabec, Sep 12 2005

nonn

a(n) is always odd.
Unsorted union: 1, 3, 15, 45, 11, 77, 7, 9, 27, 33, 25, 5, 55, 275, 13, 39, 17, 49, 931, 19, 319, 75, ..., . See A112810.
It is conjectured that every odd number occurs in this sequence (see A112822 for the first occurrence of each of them). - Jianing Song, Nov 28 2022

			a(6) = 60/20 = 3 because lcm{1,2,3,4,5,6}=60 and H(6)=49/20.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A001008, A002805, A003418, A025529, A098464, A112810, A112822, A358557.

H:= proc(n) H(n):= 1/n +`if`(n=1, 0, H(n-1)) end:
L:= proc(n) L(n):= ilcm(n, `if`(n=1, 1, L(n-1))) end:
a:= n-> L(n)/denom(H(n)):
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 30 2012

f[n_] := LCM @@ Range[n]/Denominator[HarmonicNumber[n]]; Table[ f[n], {n, 90}] (* Robert G. Wilson v, Sep 15 2005 *)

a(n) = lcm(vector(n, k, k))/denominator(sum(k=1, n, 1/k)); \\ Michel Marcus, Mar 07 2018

from sympy import lcm, harmonic
def A110566(n): return lcm([k for k in range(1,n+1)])//harmonic(n).q # Chai Wah Wu, Mar 06 2021

a(n) = A003418(n)/A002805(n) = A025529(n)/A001008(n).
From Franz Vrabec, Sep 21 2005: (Start)
a(n) = gcd(lcm{1,2,...,n}, H(n)*lcm{1,2,...,n}).
a(n) = gcd(A003418(n), A025529(n)). (End)

More terms from Robert G. Wilson v, Sep 15 2005

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.

A027446 Triangle read by rows: square of the lower triangular mean matrix.

Original entry on oeis.org

1, 3, 1, 11, 5, 2, 25, 13, 7, 3, 137, 77, 47, 27, 12, 147, 87, 57, 37, 22, 10, 1089, 669, 459, 319, 214, 130, 60, 2283, 1443, 1023, 743, 533, 365, 225, 105, 7129, 4609, 3349, 2509, 1879, 1375, 955, 595, 280, 7381, 4861, 3601, 2761, 2131, 1627, 1207, 847, 532, 252

Offset: 1

Olivier Gérard

Numerators of nonzero elements of A^2, written as rows using the least common denominator, where A[i,j] = 1/i if j <= i, 0 if j > i. [Edited by M. F. Hasler, Nov 05 2019]

			Triangle starts
     1
     3,    1
    11,    5,    2
    25,   13,    7,    3
   137,   77,   47,   27,   12
   147,   87,   57,   37,   22,   10
  1089,  669,  459,  319,  214,  130,  60
  2283, 1443, 1023,  743,  533,  365, 225, 105
  7129, 4609, 3349, 2509, 1879, 1375, 955, 595, 280
  ... - _Joerg Arndt_, Mar 29 2013

L. Bendersky, Sur la fonction gamma généralisée, Acta Math. 61 (1933), p. 263-322. See p. 295.

The row sums give A081528(n), n>=1.
The column sequences give A025529, A027457, A027458 for j=1..3.
The diagonal sequences give A002944, A027449, A027450.
Cf. A027447, A027448.

rows = 10;
M = MatrixPower[Table[If[j <= i, 1/i, 0], {i, 1, rows}, {j, 1, rows}], 2];
T = Table[M[[n]]*LCM @@ Denominator[M[[n]]], {n, 1, rows}];
Table[T[[n, k]], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Mar 05 2013, updated May 06 2022 *)

A027446_upto(n)={my(M=matrix(n, n, i, j, (j<=i)/i)^2); vector(n,r,M[r,1..r]*denominator(M[r,1..r]))} \\ M. F. Hasler, Nov 05 2019

The rational matrix A^2, where the matrix A has elements a[i,j] = 1/A002024(i,j), is equal to A119947(i,j)/A119948(i,j).

a(i,j) = lcm(seq(A119948(i,m),m=1..i))*A119947(i,j)/A119948(i,j), 1 <= j =< i and zero otherwise.

Edited by M. F. Hasler, Nov 05 2019

A096617 Numerator of n*HarmonicNumber(n).

Original entry on oeis.org

1, 3, 11, 25, 137, 147, 363, 761, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 42822903, 275295799, 279175675, 56574159, 19093197, 444316699, 1347822955, 34052522467, 34395742267, 312536252003

Offset: 1

Eric W. Weisstein, Jul 01 2004

a(1) = 1, a(n) = Numerator( H(n) / H(n-1) ), where H(n) = HarmonicNumber(n) = A001008(n)/A002805(n). - Alexander Adamchuk, Oct 29 2004
Sampling a population of n distinct elements with replacement, n HarmonicNumber(n) is the expectation of the sample size for the acquisition of all n distinct elements. - Franz Vrabec, Oct 30 2004
p^2 divides a(p-1) for prime p>3. - Alexander Adamchuk, Jul 16 2006
It appears that a(n) = b(n) defined by b(n+1) = b(n)*(n+1)/g(n) + f(n), f(n) = n*f(n-1)/g(n) and g(n) = gcd(b(n)*(n+1), n*f(n-1)), b(1) = f(1) = g(1) = 1, i.e., the recurrent formula of A000254(n) where both terms are divided by their GCD at each step (and remain divided by this factor in the sequel). Is this easy to prove? - M. F. Hasler, Jul 04 2019

			1, 3, 11/2, 25/3, 137/12, 147/10, 363/20, 761/35, 7129/280, ...

W. Feller, An Introduction to Probability Theory and Its Applications, Vol. I, 2nd Ed. 1957, p. 211, formula (3.3)

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Complete Set

Cf. A027611, A001008, A002805.
Differs from A025529 at 7th term.
Cf. A193758.

[Numerator(n*HarmonicNumber(n)): n in [1..40]]; // Vincenzo Librandi, Feb 19 2014

ZL:=n->sum(sum(1/i, i=1..n), j=1..n): a:=n->floor(numer(ZL(n))): seq(a(n), n=1..27); # Zerinvary Lajos, Jun 14 2007

Numerator[Table[(Sum[(1/k), {k, 1, n}]/Sum[(1/k), {k, 1, n-1}]), {n, 1, 20}]] (* Alexander Adamchuk, Oct 29 2004 *)
Table[n*HarmonicNumber[n] // Numerator, {n, 1, 27}]  (* Jean-François Alcover, Feb 17 2014 *)

{h(n) = sum(k=1,n,1/k)};
for(n=1,50, print1(numerator(n*h(n)), ", ")) \\ G. C. Greubel, Sep 01 2018

A=List(f=1); for(k=1,999, t=[A[k]*(k+1),f*=k]; t/=gcd(t); listput(A,t[1]+f=t[2])) \\ Illustrate conjectured equality. - M. F. Hasler, Jul 04 2019

a(n) = abs(Stirling1(n+1, 2))/(n-1)!. - Vladeta Jovovic, Jul 06 2004

a(n) = numerator of Integral_{t=0..oo} 1-(1-exp(-t/n))^n dt. - Jean-François Alcover, Feb 17 2014

A174341 a(n) = Numerator of Bernoulli(n, 1) + 1/(n+1).

Original entry on oeis.org

2, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, -37, 1, 37, 1, -211, 1, 2311, 1, -407389, 1, 37153, 1, -1181819909, 1, 76977929, 1, -818946931, 1, 277930363757, 1, -84802531453217, 1, 90219075042851, 1, -711223555487930419, 1, 12696640293313423, 1, -6367871182840222481, 1, 35351107998094669831, 1, -83499808737903072705023, 1, 12690449182849194963361, 1

Offset: 0

Paul Curtz, Mar 16 2010

a(n) is numerator of (A164555(n)/A027642(n) + 1/(n+1)).
1/(n+1) and Bernoulli(n,1) are autosequences in the sense that they remain the same (up to sign) under inverse binomial transform. This feature is kept for their sum, a(n)/A174342(n) = 2, 1, 1/2, 1/4, 1/6, 1/6, 1/6, 1/8, 7/90, 1/10, ...
Similar autosequences are also A000045, A001045, A113405, A000975 preceded by two zeros, and A140096.
Conjecture: the numerator of (A164555(n)/(n+1) + A027642(n)/(n+1)^2) is a(n) and the denominator of this fraction is equal to 1 if and only if n+1 is prime or 1. Cf. A309132. - Thomas Ordowski, Jul 09 2019
The "if" part of the conjecture is true: see the theorems in A309132 and A326690. The values of the numerator when n+1 is prime are A327033. - Jonathan Sondow, Aug 15 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..300
OEIS Wiki, Autosequence

Cf. A164555, A027642, A174342 (denominators), A025529, A003506, A309132, A326690, A327033.

[2,1] cat [Numerator(Bernoulli(n)+1/(n+1)): n in [2..40]]; // Vincenzo Librandi, Jul 18 2019

A174341 := proc(n) bernoulli(n,1)+1/(n+1); numer(%) end proc: # R. J. Mathar, Nov 19 2010

a[n_] := Numerator[BernoulliB[n, 1] + 1/(n + 1)];
Table[a[n], {n, 0, 47}] (* Peter Luschny, Jul 13 2019 *)

B(n)=if(n!=1, bernfrac(n), -bernfrac(n));
a(n)=numerator(B(n) + 1/(n + 1));
for(n=0, 50, print1(a(n),", ")) \\ Indranil Ghosh, Jun 19 2017

a(n)=numerator(bernpol(n, 1) + 1/(n + 1)); \\ Michel Marcus, Jun 26 2025

from sympy import bernoulli, Integer
def a(n): return (bernoulli(n) + 1/Integer(n + 1)).numerator # Indranil Ghosh, Jun 19 2017

Reformulation of the name by Peter Luschny, Jul 13 2019

cp's OEIS Frontend

A002805 Denominators of harmonic numbers H(n) = Sum_{i=1..n} 1/i.

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

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A110566 a(n) = lcm{1,2,...,n}/denominator of harmonic number H(n).

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

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A027446 Triangle read by rows: square of the lower triangular mean matrix.

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