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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A002182 Highly composite numbers: numbers n where d(n), the number of divisors of n (A000005), increases to a record.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 36, 48, 60, 120, 180, 240, 360, 720, 840, 1260, 1680, 2520, 5040, 7560, 10080, 15120, 20160, 25200, 27720, 45360, 50400, 55440, 83160, 110880, 166320, 221760, 277200, 332640, 498960, 554400, 665280, 720720, 1081080, 1441440, 2162160
Offset: 1

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Author

Keywords

Comments

Where record values of d(n) occur: d(n) > d(k) for all k < n.
A002183 is the RECORDS transform of A000005, i.e., lists the corresponding values d(n) for n in A002182.
Flammenkamp's page also has a copy of the missing Siano paper.
Highly composite numbers are the product of primorials, A002110. See A112779 for the number of primorial terms in the product of a highly composite number. - Jud McCranie, Jun 12 2005
Sigma and tau for highly composite numbers through the 146th entry conform to a power fit as follows: log(sigma)=A*log(tau)^B where (A,B) =~ (1.45,1.38). - Bill McEachen, May 24 2006
a(n) often corresponds to P(n,m) = number of permutations of n things taken m at a time. Specifically, if start=1, pointers 1-6, 9, 10, 13-15, 17-19, 22, 23, 28, 34, 37, 43, 52, ... An example is a(37)=665280, which is P(12,6)=12!/(12-6)!. - Bill McEachen, Feb 09 2009
Concerning the previous comment, if m=1, then P(n,m) can represent any number. So let's assume m > 1. Searching the first 1000 terms, the only indices of terms of the form P(n,m) are 4, 5, 6, 9, 10, 12, 13, 14, 15, 16, 17, 18, 19, 22, 23, 27, 28, 31, 34, 37, 41, 43, 44, 47, 50, 52, and 54. Note that a(44) = 4324320 = P(2079,2). See A163264. - T. D. Noe, Jun 10 2009
A large number of highly composite numbers have 9 as their digit root. - Parthasarathy Nambi, Jun 07 2009
Because 9 divides all highly composite numbers greater than 1680, those numbers have digital root 9. - T. D. Noe, Jul 24 2009
See A181309 for highly composite numbers that are not highly abundant.
a(n) is also defined by the recurrence: a(1) = 1, a(n+1)/sigma(a(n+1)) < a(n) / sigma(a(n)). - Michel Lagneau, Jan 02 2012 [NOTE: This "definition" is wrong (a(20)=7560 does not satisfy this inequality) and incomplete: It does not determine a sequence uniquely, e.g., any subsequence would satisfy the same relation. The intended meaning is probably the definition of the (different) sequence A004394. - M. F. Hasler, Sep 13 2012]
Up to a(1000), the terms beyond a(5) = 12 resp. beyond a(9) = 60 are a multiples of these. Is this true for all subsequent terms? - M. F. Hasler, Sep 13 2012 [Yes: see EXAMPLE in A199337! - M. F. Hasler, Jan 03 2020]
Differs from the superabundant numbers from a(20)=7560 on, which is not in A004394. The latter is not a subsequence of A002182, as might appear from considering the displayed terms: The two sequences have only 449 terms in common, the largest of which is A002182(2567) = A004394(1023). See A166735 for superabundant numbers that are not highly composite, and A004394 for further information. - M. F. Hasler, Sep 13 2012
Subset of A067128 and of A025487. - David A. Corneth, May 16 2016, Jan 03 2020
It seems that a(n) +- 1 is often prime. For n <= 1000 there are 210 individual primes and 17 pairs of twin primes. See link to Lim's paper below. - Dmitry Kamenetsky, Mar 02 2019
There are infinitely many numbers in this sequence and a(n+1) <= 2*a(n), because it is sufficient to multiply a(n) by 2 to get a number having more divisors. (This proves Guess 0 in the Lim paper.) For n = (1, 2, 4, 5, 9, 13, 18, ...) one has equality in this bound, but asymptotically a(n+1)/a(n) goes to 1, cf. formula due to Erdős. See A068507 for the terms such that a(n)+-1 are twin primes. - M. F. Hasler, Jun 23 2019
Conjecture: For n > 7, a(n) is a Zumkeller number (A083207). Verified for n up to and including 48. If this conjecture is true, one may base on it an alternative proof of the fact that for n>7 a(n) is not a perfect square (see Fact 5, Rao/Peng arXiv link at A083207). - Ivan N. Ianakiev, Jun 29 2019
The conjecture above is true (see the proof in the "Links" section). - Ivan N. Ianakiev, Jan 31 2020
The first instance of omega(a(n)) < omega(a(n-1)) (omega = A001221: number of prime divisors) is at a(26) = 45360. Up to n = 10^4, 1759 terms have this property, but omega decreases by 2 only at indices n = 5857, 5914 and 5971. - M. F. Hasler, Jan 02 2020
Inequality (54) in Ramanujan (1915) implies that for any m there is n* such that m | a(n) for all n > n*: see A199337 for the proof. - M. F. Hasler, Jan 03 2020

Examples

			a(5) = 12 is in the sequence because A000005(12) is larger than any earlier value in A000005. - _M. F. Hasler_, Jan 03 2020
		

References

  • CRC Press Standard Mathematical Tables, 28th Ed, p. 61.
  • J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 180, p. 56, Ellipses, Paris 2008.
  • L. E. Dickson, History of Theory of Numbers, I, p. 323.
  • Ross Honsberger, An introduction to Ramanujan's Highly Composite Numbers, Chap. 14 pp. 193-200 Mathematical Gems III, DME no. 9 MAA 1985
  • Jean-Louis Nicolas, On highly composite numbers, pp. 215-244 in Ramanujan Revisited, Editors G. E. Andrews et al., Academic Press 1988
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 88.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, 128.

Crossrefs

Cf. A261100 (a left inverse).
Cf. A002808. - Peter J. Marko, Aug 16 2018
Cf. A279930 (highly composite and highly Brazilian).
Cf. A068507 (terms such that a(n)+-1 are twin primes).
Cf. A199337 (number of terms not divisible by n).

Programs

  • Mathematica
    a = 0; Do[b = DivisorSigma[0, n]; If[b > a, a = b; Print[n]], {n, 1, 10^7}]
    (* Convert A. Flammenkamp's 779674-term dataset; first, decompress, rename "HCN.txt": *)
    a = Times @@ {Times @@ Prime@ Range@ ToExpression@ First@ #1, If[# == {}, 1, Times @@ MapIndexed[Prime[First@ #2]^#1 &, #]] &@ DeleteCases[-1 + Flatten@ Map[If[StringFreeQ[#, "^"], ToExpression@ #, ConstantArray[#1, #2] & @@ ToExpression@ StringSplit[#, "^"]] &, #2], 0]} & @@ TakeDrop[StringSplit@ #, 1] & /@ Import["HCN.txt", "Data"] (* Michael De Vlieger, May 08 2018 *)
    DeleteDuplicates[Table[{n,DivisorSigma[0,n]},{n,2163000}],GreaterEqual[ #1[[2]],#2[[2]]]&] [[All,1]] (* Harvey P. Dale, May 13 2022 *)
    NestList[Function[last,
      Module[{d = DivisorSigma[0, last]},
       NestWhile[# + 1 &, last, DivisorSigma[0, #] <= d &]]], 1, 40] (* Steven Lu, Mar 30 2023 *)
  • PARI
    print1(r=1); forstep(n=2,1e5,2, if(numdiv(n)>r, r=numdiv(n); print1(", "n))) \\ Charles R Greathouse IV, Jun 10 2011
    
  • PARI
    v002182 = [1]/*vector for memoization*/; A002182(n, i = #v002182) ={ if(n > i, v002182 = Vec(v002182, n); my(k = v002182[i], d, s=1); until(i == n, d = numdiv(k); s<60 && k>=60 && s=60; until(numdiv(k += s) > d,); v002182[i++] = k); k, v002182[n])} \\ Antti Karttunen, Jun 06 2017; edited by M. F. Hasler, Jan 03 2020 and Jun 20 2022
    
  • PARI
    is_A002182(n, a=1, b=1)={while(n>A002182(b*=2), a*=2); until(a>b, my(m=(a+b)\2, t=A002182(m)); if(tn, b=m-1, return(m)))} \\ Also used in other sequences. - M. F. Hasler, Jun 20 2022
    
  • Python
    from sympy import divisor_count
    A002182_list, r = [], 0
    for i in range(1,10**4):
        d = divisor_count(i)
        if d > r:
            r = d
            A002182_list.append(i) # Chai Wah Wu, Mar 23 2015

Formula

Also, for n >= 2, smallest values of p for which A006218(p)-A006318(p-1) = A002183(n). - Philippe LALLOUET (philip.lallouet(AT)wanadoo.fr), Jun 23 2007
a(n+1) < a(n) * (1+log(a(n))^-c) for some positive c (see Erdős). - David A. Corneth, May 16 2016
a(n) = A108951(A329902(n)). - Antti Karttunen, Jan 08 2020
a(n+1) <= 2*a(n). For cases where the equal sign holds, see A072938. - A.H.M. Smeets, Jul 10 2021
Sum_{n>=1} 1/a(n) = A352418. - Amiram Eldar, Mar 24 2022

Extensions

Jun 19 1996: Changed beginning to start at 1.
Jul 10 1996: Matthew Conroy points out that these are different from the super-abundant numbers - see A004394. Last 8 terms sent by J. Lowell; checked by Jud McCranie.
Description corrected by Gerard Schildberger and N. J. A. Sloane, Apr 04 2001
Additional references from Lekraj Beedassy, Jul 24 2001

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

A329570 a(n) is the least prime P such that log(P)/log(p) >= valuation(n,p) for all primes p.

Original entry on oeis.org

2, 2, 3, 5, 5, 3, 7, 11, 11, 5, 11, 5, 13, 7, 5, 17, 17, 11, 19, 5, 7, 11, 23, 11, 29, 13, 29, 7, 29, 5, 31, 37, 11, 17, 7, 11, 37, 19, 13, 11, 41, 7, 43, 11, 11, 23, 47, 17, 53, 29, 17, 13, 53, 29, 11, 11, 19, 29, 59, 5, 61, 31, 11, 67, 13, 11, 67, 17, 23, 7, 71, 11, 73, 37, 29, 19, 11
Offset: 1

Views

Author

M. F. Hasler, Jan 03 2020

Keywords

Comments

Related to the inequality (54) in Ramanujan's paper about highly composite numbers A002182, also used in A199337: This is the largest prime factor of the bound A329571(n)^2 above which all highly composite numbers are divisible by n.

Crossrefs

Programs

  • Mathematica
    a[n_] := NextPrime[Max[Power @@@ FactorInteger[n]] - 1]; a[1] = 2; Array[a, 100] (* Amiram Eldar, Jan 17 2025 *)
  • PARI
    apply( {A329570(n,f=Col(factor(max(n,2))), P=nextprime(vecmax([log(f[1])*f[2] | f<-f])))=[while( logint(P,f[1]) < f[2], P=nextprime(P+1)) | f<-f]; P}, [1..99])

Formula

a(n) = A007918(A034699(n)). - Amiram Eldar, Jan 17 2025

A329571 a(n) = Product_{prime p} p^floor(log_p P) with P = A329570(n) the least prime with log_p P >= valuation(n,p) for all primes p.

Original entry on oeis.org

2, 2, 6, 60, 60, 6, 420, 27720, 27720, 60, 27720, 60, 360360, 420, 60, 12252240, 12252240, 27720, 232792560, 60, 420, 27720, 5354228880, 27720, 2329089562800, 360360, 2329089562800, 420, 2329089562800, 60, 72201776446800, 5342931457063200, 27720, 12252240, 420, 27720, 5342931457063200, 232792560, 360360, 27720, 219060189739591200, 420, 9419588158802421600, 27720
Offset: 1

Views

Author

M. F. Hasler, Jan 03 2020

Keywords

Comments

Related to the inequality (54) in Ramanujan's paper about highly composite numbers (HCN) A002182, also used in A199337: This is the square root of the (not minimal) bound a(n)^2 above which all HCN are divisible by n, according to the right part of that inequality.
Like the highly composite numbers A002182, all terms in this sequence are a product of primorials.

Crossrefs

Cf. A329570, A002182 (highly composite numbers), A199337 (number of HCN not divisible by n), A003418 (lcm(1..n)), A056604 (lcm(1..prime(n))), A025487.

Programs

  • Mathematica
    a[n_] := Module[{P = NextPrime[Max[Power @@@ FactorInteger[n]] - 1], p}, p = Select[Range[P], PrimeQ]; Times @@ (p^Floor[Log[p, P]])]; a[1] = 2; Array[a, 50] (* Amiram Eldar, Jan 17 2025 *)
  • PARI
    apply( {A329571(n)=vecprod([p^logint(n,p)|p<-primes([2,n=A329570(n)])])}, [1..44])

Formula

a(n) = lcm([1..P]) = A003418(P) = A056604(i) with P = A329570(n), i = A000720(P).

A330737 a(n) is the first index k in A002182 (highly composite numbers) from which onward all terms A002182(i), i >= k, are multiples of the n-th prime, a(0) = 1 by convention.

Original entry on oeis.org

1, 2, 4, 9, 15, 28, 38, 55, 71, 92, 110, 125, 146, 167, 183, 206, 225, 258, 281, 313, 339, 363, 399, 425, 453, 488, 515, 550, 585, 618, 657, 705, 739, 794, 830, 866, 902, 950, 999, 1036, 1074, 1113, 1151, 1198, 1234, 1270, 1306, 1347, 1393, 1436, 1479, 1528, 1571, 1615, 1671, 1719, 1774, 1824, 1875, 1925, 1975, 2026, 2087, 2170, 2235
Offset: 0

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Author

Antti Karttunen, Dec 29 2019

Keywords

Comments

Equivalently, a(n) is the first index k in A002182 from which onward all terms A002182(i), i >= k, are multiples of A002110(n), the n-th primorial number.
Question: Is this sequence well-defined for any n > 1? For all n? See also A199337.
Note that this differs from A072846 at n = 5, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, ...
Yes, the sequence is well defined for all n, see A199337 for proof that all A002182(k) >= A329571(n)^2 are divisible by n. - M. F. Hasler, Jan 07 2020

Examples

			a(0) = 1 as A002110(0) = 1, and A002182(1) = 1, and as all integers are divisible by 1, including all terms of A002182.
A002182(9) = 60, and because from then onward all highly composite numbers are multiples of 30 (= A002110(3) = prime(1)*prime(2)*prime(3)), we have a(3) = 9.
		

Crossrefs

Programs

  • PARI
    \\ v002182 contains the terms of A002182 up to some suitably big value:
    A330737(n) = if(!n,1,my(x=prime(n)); forstep(k=#v002182,1,-1,if(v002182[n]%x,return(1+k))));

A328520 GCD of terms in A002182 that have n prime factors counted with multiplicity.

Original entry on oeis.org

1, 2, 2, 12, 12, 12, 60, 120, 2520, 2520, 2520, 55440, 55440, 720720, 720720, 12252240, 36756720, 698377680, 3491888400, 80313433200, 160626866400, 160626866400, 9316358251200, 288807105787200, 2021649740510400, 74801040398884800, 74801040398884800, 3066842656354276800, 131874234223233902400
Offset: 0

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Author

David A. Corneth, Jan 04 2020

Keywords

Comments

If for every term t > 1 in A002182 there exists a term in A002182 of the form t/p for some prime p|t (Cf. A328523) then (1): a(n+1) is a multiple of a(n) for n >= 1 and (2): this sequence can be used to find terms to A199337 and A330737.
Proof of (1): Let Generation(n) be the terms in A002182 with n prime factors counted with multiplicity. Then a(n) = GCD(Generation(n)). As each term in Generation(n) is of the form a(n) * t for some t and each term in Generation(n + 1) is p * g for some g in Generation(n), a(n + 1) is a multiple of a(n).
Proof of (2): As a(n + 1) is a multiple of a(n) we have that if m | a(n), we also have m | a(n + k), k >= 0 hence the largest number not divisible by m can have at most n - 1 prime factors counted with multiplicity.
Given Generation(n) we can find all candidates for Generation(n + 1) and from there on find terms to A002182, A199337 and A330737.

Examples

			The terms in A002182 with n = 4 prime divisors counted with multiplicity are 24, 36 and 60. Their GCD is 12 hence a(4) = 12.
Furthermore, If for every term t > 1 in A002182 there exists a term in A002182 of the form t/p for some prime p|t then we have that each term with more than 4 prime divisors counted with multiplicity is a multiple of at least one of 24, 36 or 60 hence is divisible by 12.
		

Crossrefs

Programs

  • Mathematica
    (* First, load the function f at A025487, then: *)
    Block[{s = Union@ Flatten@ f@ 20, t}, t = DivisorSigma[0, s]; s = Map[s[[FirstPosition[t, #][[1]] ]] &, Union@ FoldList[Max, t]]; t = PrimeOmega[s]; Drop[Array[GCD @@ s[[Position[t, #][[All, 1]] ]] &, Max@ t + 1, 0], -3] ] (* Michael De Vlieger, Jan 12 2020 *)

A328523 a(n) = number of terms in A002182 that are of the form A002182(n)/p for some prime p|n.

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 2, 3, 3, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 1, 1, 2, 3, 2, 2, 3, 2, 3, 3, 2, 2, 1, 3, 2, 1, 3, 1, 2, 2, 2, 2, 3, 2, 2, 4, 3, 1, 4, 2, 2, 2, 2, 1, 2, 2, 2, 2, 4, 3, 4, 3, 2, 3, 1, 1, 1, 1, 2, 2, 3
Offset: 1

Views

Author

David A. Corneth, Jan 04 2020

Keywords

Comments

Conjecture: a(n) > 0 for n > 1. (This is false.)
It seems that a(815) = 0. - M. F. Hasler, Jan 08 2020

Crossrefs

Cf. A112778(n) = A001222(A002182(n)) (bigomega of HCN's).

Programs

  • PARI
    apply( {a(n)=#[0|p<-factor(n=A2182[n])[,1], setsearch(A2182,n/p)]}, [1..99]) \\ Uses global var. A2182 which must be assigned to a vector of at least n elements of that sequence. - M. F. Hasler, Jan 08 2020
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