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.

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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

Views

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).

Crossrefs

Row products of A133233.
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

A055874 a(n) = largest m such that 1, 2, ..., m divide n.

Original entry on oeis.org

1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 6, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 4, 1, 2, 1, 2
Offset: 1

Views

Author

Leroy Quet, Jul 16 2000

Keywords

Comments

From Antti Karttunen, Nov 20 2013 & Jan 26 2014: (Start)
Differs from A232098 for the first time at n=840, where a(840)=8, while A232098(840)=7. A232099 gives all the differing positions. See also the comments at A055926 and A232099.
The positions where a(n) is an odd prime is given by A017593 up to A017593(34)=414 (so far all 3's), after which comes the first 7 at a(420). (A017593 gives the positions of 3's.)
(Continued on Jan 26 2014):
Only terms of A181062 occur as values.
A235921 gives such n where a(n^2) (= A235918(n)) differs from A071222(n-1) (= A053669(n)-1). (End)
a(n) is the largest m such that A003418(m) divides n. - David W. Wilson, Nov 20 2014
a(n) is the largest number of consecutive integers dividing n. - David W. Wilson, Nov 20 2014
A051451 gives indices where record values occur. - Gionata Neri, Oct 17 2015
Yuri Matiyasevich calls this the maximum inheritable divisor of n. - N. J. A. Sloane, Dec 14 2023

Examples

			a(12) = 4 because 1, 2, 3, 4 divide 12, but 5 does not.
		

Crossrefs

Programs

  • Haskell
    a055874 n = length $ takeWhile ((== 0) . (mod n)) [1..]
    -- Reinhard Zumkeller, Feb 21 2012, Dec 09 2010
    
  • Maple
    N:= 1000: # to get a(1) to a(N)
    A:= Vector(N,1);
    for m from 2 do
      Lm:= ilcm($1..m);
      if Lm > N then break fi;
      if Lm mod (m+1) = 0 then next fi;
      for k from 1 to floor(N/Lm) do
        A[k*Lm]:=m
      od
    od:
    convert(A,list); # Robert Israel, Nov 28 2014
  • Mathematica
    a[n_] := Module[{m = 1}, While[Divisible[n, m++]]; m - 2]; Array[a, 100] (* Jean-François Alcover, Mar 07 2016 *)
  • PARI
    a(n) = my(m = 1); while ((n % m) == 0, m++); m - 1; \\ Michel Marcus, Jan 17 2014
    
  • Python
    from itertools import count
    def A055874(n):
        for m in count(1):
            if n % m:
                return m-1 # Chai Wah Wu, Jan 02 2022
  • Scheme
    (define (A055874 n) (let loop ((m 1)) (if (not (zero? (modulo n m))) (- m 1) (loop (+ 1 m))))) ;; Antti Karttunen, Nov 18 2013
    

Formula

a(n) = A007978(n) - 1. - Antti Karttunen, Jan 26 2014
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = A064859 (Farhi, 2009). - Amiram Eldar, Jul 25 2022

A007978 Least non-divisor of n.

Original entry on oeis.org

2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 7, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3, 2, 4, 2, 3, 2, 3, 2, 5, 2, 3, 2, 3
Offset: 1

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Author

Keywords

Comments

Least k >= 2 such that sigma(n) divides sigma(n*k), where sigma is A000203. - Benoit Cloitre, Dec 01 2002
Contains all and only the prime powers p^k, k > 0. The first occurrence of p^k is at A003418(p^k-1); so new records occur at indices in A051451. - Franklin T. Adams-Watters, Jun 13 2011

Crossrefs

Cf. also A266620 (least non-divisor of n!).

Programs

  • Haskell
    import Data.List ((\\))
    a007978 = head . ([1..] \\) . a027750_row
    -- Reinhard Zumkeller, May 10 2014
    
  • Maple
    a:= proc(n) local k;
    for k from 2 while n mod k = 0 do od:
    k
    end proc:
    seq(a(n),n=1..100); # Robert Israel, Sep 02 2014
  • Mathematica
    Table[k := 1; While[Mod[n, k] == 0, k++]; k, {n, 2000}]  (* Clark Kimberling, Jun 16 2012 *)
    Join[{2, 3}, Table[Complement[Range[n], Divisors[n]][[1]], {n, 3, 100}]] (* Alonso del Arte, Sep 23 2017 *)
  • PARI
    a(n) = {my(k=2); while(!(n % k), k++); k;} \\ Michel Marcus, Sep 25 2017
    
  • Python
    def a(n):
        k = 2
        while not n%k: k += 1
        return k
    print([a(n) for n in range(1, 101)]) # Michael S. Branicky, Jul 09 2022
    
  • Python
    def A007978(n): return next(filter(lambda d:n%d,range(2,n))) if n>2 else n+1 # Chai Wah Wu, Feb 22 2023

Formula

a(n) = A053669(n) + A061853(n) = A055874(n) + 1. - Henry Bottomley, May 10 2001
G.f.: sum(k >= 2, -k*(x^A003418(k) - x^A003418(k-1))/((x^A003418(k) - 1)*(x^A003418(k-1) - 1))). - Robert Israel, Sep 02 2014
From Alonso del Arte, Sep 23 2017: (Start)
a(n) < n for all n > 2.
a(2n + 1) = 2, a(2n) >= 3.
a(2^k) = 3 for k > 0.
a(n!) = prime(pi(n) + 1) for n >= 0, except for a(3!) = 4. (End)
Asymptotic mean: lim_{n->oo} Sum_{k=1..n} a(k) = 1 + A064859 (Farhi, 2009). - Amiram Eldar, Jun 29 2021

A064857 Numerators of partial sums of reciprocals of lcm(1..n) = A003418(n).

Original entry on oeis.org

1, 3, 5, 7, 53, 107, 25, 1501, 563, 901, 12389, 16519, 322121, 644243, 53687, 1288489, 3650719, 4380863, 18917363, 3557111, 104045497, 416181989, 2393046437, 455818369, 23930464373, 47860928747, 10255913303, 11044829711
Offset: 1

Views

Author

Labos Elemer, Oct 08 2001

Keywords

Examples

			n = 7: LCM values: 1, 2, 6, 12, 60, 60, 420; partial sum = 1 + 1/2 + 1/6 + 1/12 + 1/60 + 1/60 + 1/420 = (420 + 210 + 70 + 35 + 7 + 7 + 1)/420 = 750/420 = 25/14, so a(7) = 25.
		

Crossrefs

Programs

  • Maple
    R:= 1: m:= 1: for n from 2 to 100 do m:= ilcm(m,n); R:= R,1/m od:
    P:= ListTools:-PartialSums([R]):
    map(numer,P); # Robert Israel, Jan 30 2025
  • Mathematica
    q[x_] := Apply[LCM, Table[j, {j, 1, x}]] Table[Numerator[Apply[Plus, Table[1/q[w], {w, 1, m}]]], {m, 1, 30}]
    Accumulate[Table[1/LCM@@Range[n],{n,30}]]//Numerator (* Harvey P. Dale, Aug 08 2021 *)

Formula

a(n) = numerator(Sum_{j=1..n} 1/lcm(1..n)).

A064858 Denominators of partial sums of reciprocals of lcm(1..n) = A003418(n).

Original entry on oeis.org

1, 2, 3, 4, 30, 60, 14, 840, 315, 504, 6930, 9240, 180180, 360360, 30030, 720720, 2042040, 2450448, 10581480, 1989680, 58198140, 232792560, 1338557220, 254963280, 13385572200, 26771144400, 5736673800, 6177956400, 291136195350
Offset: 1

Views

Author

Labos Elemer, Oct 08 2001

Keywords

Examples

			n = 7: LCM values: 1, 2, 6, 12, 60, 60, 420; partial sum = 1 + 1/2 + 1/6 + 1/12 + 1/60 + 1/60 + 1/420 = (420 + 210 + 70 + 35 + 7 + 7 + 1)/420 = 750/420 = 25/14, so a(7) = 14.
		

Crossrefs

Programs

  • Maple
    R:= 1: m:= 1: for n from 2 to 100 do m:= ilcm(m,n); R:= R,1/m od:
    P:= ListTools:-PartialSums([R]):
    map(denom, P); # Robert Israel, Jan 30 2025
  • Mathematica
    q[x_] := Apply[LCM, Table[j, {j, 1, x}]] Table[Denominator[Apply[Plus, Table[1/q[w], {w, 1, m}]]], {m, 1, 30}]

Formula

a(n) = denominator(Sum_{j=1..n} 1/lcm(1..j)).

A064890 Decimal expansion of the sum of reciprocals of A051451, which includes 1 and values of lcm(1,...,x), where x is a prime power (A000961).

Original entry on oeis.org

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

Views

Author

Labos Elemer, Oct 11 2001

Keywords

Examples

			c = 1.7706752443255802279197960076426608022331837672728335205224589644122203381...
c = 1 + (1/2) + (1/6) + (1/12) + (1/60) + (1/420) + ... = 743/420 + ... = 1.7690 + ... = 1.7706752... Compare with A064859, an analogous constant obtained from A003418, where the constant is larger than c: 1.7877805 > 1.7706752. Repeated occurrences of LCM values in A003418 is responsible for the 1.78778... - 1.77067... = 0.0171... excess.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := LCM @@ Range@ n; RealDigits[ Plus @@ (1/Union@ Array[f, 251]), 10, 111][[1]] (* Robert G. Wilson v, Jul 11 2011 *)

A064888 Numerators of partial sums of 1/A051451.

Original entry on oeis.org

1, 2, 3, 23, 323, 647, 971, 7121, 6943, 185147, 363173, 179407463, 412637165, 20631858251, 140035237, 3716297449, 27822060851939, 2853544702763, 58823785801243, 568431869527163, 279209361758899949, 4944838261875010691
Offset: 1

Views

Author

Labos Elemer, Oct 10 2001

Keywords

Comments

A051451 unlike A003418 contains distinct values of lcm(1..m).

Examples

			n=6, s(n) = 1/2 + 1/6 + 1/12 + 1/60 + 1/420 + 1/840 = 646/840 = 323/420, the numerator is a(6)=323.
		

Crossrefs

Programs

  • Mathematica
    s=Select[Range[a, b], Equal[Length[FactorInteger[ # ]], 1]&] a(m) = Apply[LCM, Table[Part[s, w], {w, 1, m}]]

Formula

a(n) = Sum_{k=1..n} 1/A051451(k), where A051451 is the lcm(1...m), m is a prime or true power of prime. The sequence is convergent.

A181121 As n increases, the reciprocal of a(n) = asymptotic fraction of positive integers whose longest string of consecutive divisors is A181062(n).

Original entry on oeis.org

2, 3, 12, 15, 70, 840, 1260, 2772, 30030, 720720, 765765, 12932920, 243374040, 6692786100, 40156716600, 83181770100, 2406725881560, 144403552893600, 148414762696200, 5476504743489780, 224275908542914800
Offset: 1

Views

Author

Matthew Vandermast, Oct 07 2010

Keywords

Comments

The asymptotic average, as n increases, of n's longest string of consecutive divisors is the constant 1.787780456..., given in A064859.

Examples

			A number's longest string of consecutive divisors is a(5)=6 iff the integer is a multiple of 60 but not of 7. As n increases, the asymptotic fraction of positive integers satisfying those conditions equals 1/60 * 6/7 = 1/70. Therefore a(5) = 70.
		

Formula

a(n) = A051451(n) * A025473(n+1)/(A025473(n+1)-1).
If A181062(n) = 2^(e-1), then a(n) = A003418(2^e) = A051451(n+1).

A346155 Partial sums of A007978.

Original entry on oeis.org

2, 5, 7, 10, 12, 16, 18, 21, 23, 26, 28, 33, 35, 38, 40, 43, 45, 49, 51, 54, 56, 59, 61, 66, 68, 71, 73, 76, 78, 82, 84, 87, 89, 92, 94, 99, 101, 104, 106, 109, 111, 115, 117, 120, 122, 125, 127, 132, 134, 137, 139, 142, 144, 148, 150, 153, 155, 158, 160, 167
Offset: 1

Views

Author

Shobhit Tewari, Jul 09 2021

Keywords

Examples

			If b(n) is least non-divisor of n then, a(6) = b(1)+b(2)+b(3)+b(4)+b(5)+b(6) => 2+3+2+3+2+4 = 16.
		

Crossrefs

Programs

  • Mathematica
    lnd[n_] := Module[{k = 2}, While[Divisible[n, k], k++]; k]; Accumulate @ Array[lnd, 100] (* Amiram Eldar, Jul 11 2021 *)
  • PARI
    f(n) = {my(k=2); while(!(n % k), k++); k; } \\ A007978
    a(n) = sum(k=1, n, f(k)); \\ Michel Marcus, Jul 11 2021

Formula

a(n) ~ (1 + A064859) * n. - Amiram Eldar, Jul 11 2021

A064889 Denominators of partial sums of reciprocals of A051451 (A051451 includes lcm(1,...,x), x=power of prime from A000961 and also contains 1).

Original entry on oeis.org

1, 2, 3, 4, 30, 420, 840, 1260, 9240, 9009, 240240, 471240, 232792560, 535422888, 26771144400, 181704600, 4822131600, 36100888223400, 3702655202400, 76327592243760, 737576396429600, 362291852261631600, 6416241209619040800
Offset: 1

Views

Author

Labos Elemer, Oct 11 2001

Keywords

Examples

			n=6, sum = (1/1) + (1/2) + (1/6) + (1/12) + (1/60) + (1/420) = 743/420, the denominator = a(6) = 743. Remark: If 1 is omitted from A051451, the denominators of partial sums of reciprocals do not change; however, numerators are changing, not compatible with these denominators (see A064888).
		

Crossrefs

Formula

a(n) = denominator(Sum_{k=1..n} 1/A051451(k)).
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