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.

Showing 1-3 of 3 results.

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

Views

Author

N. J. A. Sloane

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.

Links

Crossrefs

Cf. A000005 (number of divisors), A002110, A002183, A002473, A004394, A025487, A106037, A108602, A112778, A112779, A112780, A112781, A006218, A126098, A002201, A072938, A094348, A003418, A161184, A037992 (infinitary analog), A108951, A329902, A352418.
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

A217056 Highly composite numbers (A002182) which are the product of 4 consecutive integers (A052762).

Original entry on oeis.org

24, 120, 360, 840, 1680, 5040, 17297280
Offset: 1

Views

Author

Raphie Frank, Sep 25 2012

Keywords

Comments

All elements of this sequence, a subset of A163264, are 24 times an element of binomial(n,4) (A000332) and are, therefore, also 24 times a generalized pentagonal number (A001318) since all elements of binomial(n,4) are generalized pentagonal.
Additionally, sqrt(a(n) + 1) is prime for these 7 terms. It follows that, at least to a(7), the sum of the divisors of sqrt(a(n) + 1) is a pronic number (A002378).
48 = 2*(1*2*3*4)= 7^2 - 1 is the only known highly composite number one less than a square that is not a part of this sequence. 48 is also 24 times a generalized pentagonal number and one less than the square of a prime (see also A072825).
Question: Is this sequence complete?
Next term > A002182(1000) = 3.3826...*10^76. - Joerg Arndt, Oct 07 2012

Examples

			24 = 1*2*3*4 = 5^2 - 1; 24/24 = 1.
120 = 2*3*4*5 = 11^2 - 1; 120/24 = 5.
360 = 3*4*5*6 = 19^2 - 1; 360/24 = 15.
840 = 4*5*6*7 = 29^2 - 1; 840/24 = 35.
1680 = 5*6*7*8 = 41^2 - 1; 1680/24 = 70.
5040 = 7*8*9*10 = 71^2 - 1; 5040/24 = 210.
17297280 = 63*64*65*66 = 4159^2 - 1; 17297280/24 = 720720.
{5, 11, 19, 29, 41, 71, 4159} are all primes one less than a pronic number.
{1, 5, 15, 35, 70, 210, 720720} are all generalized pentagonal numbers.

Crossrefs

Cf. A163264, A002182, A072825.

Formula

A217056 = A002182 intersect A052762. - M. F. Hasler, Oct 18 2013

A325480 a(n) is the largest integer m such that the product of n consecutive integers starting at m is divisible by at most n primes.

Original entry on oeis.org

16, 24, 24, 45, 48, 49, 120, 120, 125, 189, 240, 240, 350, 350, 350, 350, 374, 494, 494, 714, 714, 714, 714, 825, 832, 1078, 1078, 1078, 1078, 1425, 1440, 1440, 1856, 2175, 2175, 2175, 2175, 2175, 2175, 2175, 2870, 2870, 2870, 2871, 2880, 2880, 2880, 3219
Offset: 3

Views

Author

Onno M. Cain, Sep 06 2019

Keywords

Comments

Each term is only conjectured and has been verified up to 10^6.
Note a(2) is undefined if there are infinitely many Mersenne primes.

Examples

			For example, a(3) = 16 because 16 * 17 * 18 = 2^5 * 3^2 * 17 admits only three prime divisors (2, 3, and 17) and appears to be the largest product of three consecutive integers with the property.

Crossrefs

Cf. A002182, A045619, A163264, A164799, A239035, A244656, A006549.

Programs

  • SageMath
    for r in range(3, 100):
  history = []
  M = 0
  for n in range(1, 100000):
    primes = {p for p, _ in factor(n)}
    history.append(primes)
    history = history[-r:]
    total = set()
    for s in history: total |= s
    # Skip if too many primes.
    if len(total) > r: continue
    if n > M: M = n
  print(r, M-r+1)
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Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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