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.

Previous Showing 21-24 of 24 results.

A142727 For definition see Comments lines.

Original entry on oeis.org

1, 2, 3, 4, 6, 6, 8, 8, 9, 12, 12, 14, 15, 16, 16, 18, 20, 20, 22, 24, 24, 24, 25, 27, 30, 30, 30, 32, 32, 32, 36, 36, 39, 40, 42, 42, 44, 45, 46, 48, 48, 48, 50, 50, 52, 52, 55, 59, 60, 60, 60, 60, 60, 64, 64, 66, 66, 67, 69, 70, 71, 72, 75, 76, 76, 78, 80, 81, 84
Offset: 1

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Author

David Applegate and N. J. A. Sloane, Oct 03 2008

Keywords

Comments

S is a sequence of numbers with repetitions, sorted in nondecreasing order.
Initially a(1) = 1 and S = N = {1,2,3,4,5,6,...}.
For each n >= 2, let p = prime(n), let a(n) = (p-1)st term of S and set S = S union a(n)*N. Repeat.
A sequence arising from one possible attempt to analyze A141586.

Examples

			The first few stages in the calculation are as follows:
S = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
n = 2, p = 3, so a(2) is the 2nd term of S, which is 2.
Now S = 1 2 2 3 4 4 5 6 6 7 8 8 9 10 10 11 12 12 13 14 14 ...
n = 3, p = 5, so a(3) is the 4th term of S, which is 3.
Now S = 1 2 2 3 3 4 4 5 6 6 6 7 8 8 9 9 10 10 11 12 12 12 13 14 14 ...
n = 4, p = 7, so a(4) is the 6th term of S, which is 4.
Now S = 1 2 2 3 3 4 4 4 5 6 6 6 7 8 8 8 9 9 10 10 11 12 12 12 12 13 14 14 ...
n = 5, p = 11, so a(5) is the 10th term of S, which is 6.
And so on.
		

Formula

A plot of the extended sequence suggests that a(n) ~= c n log(log(n)) + d n for constants c and d. For example, run: $ gnuplot> plot [] [1.27:1.35] a142727.txt using 1:4, 1.12+0.076*log(log(x)).

A209291 Sum of the refactorable numbers less than or equal to n.

Original entry on oeis.org

1, 3, 3, 3, 3, 3, 3, 11, 20, 20, 20, 32, 32, 32, 32, 32, 32, 50, 50, 50, 50, 50, 50, 74, 74, 74, 74, 74, 74, 74, 74, 74, 74, 74, 74, 110, 110, 110, 110, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 150, 206, 206, 206, 206, 266
Offset: 1

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Author

Wesley Ivan Hurt, Jan 16 2013

Keywords

Comments

A number is refactorable if it is divisible by the number of its divisors.
The first 8 terms are odd. The next odd term after 11 is a(225) = 2395.
600 out of the first 1000 terms are odd, including every term from a(625) up to and including a(1000). - Harvey P. Dale, Aug 07 2019

Crossrefs

Programs

  • Maple
    with(numtheory); a:= n -> add(i * (1 + floor(i/tau(i)) - ceil(i/tau(i))), i = 1..n):
  • Mathematica
    Accumulate[Table[If[Divisible[n,DivisorSigma[0,n]],n,0],{n,60}]] (* Harvey P. Dale, Aug 07 2019 *)

Formula

a(n) = Sum_{i=1..n} i*(1 + floor(i/d(i)) - ceiling(i/d(i))) where d(n) is the number of divisors of n (A000005).

A212169 List of highly composite numbers (A002182) with an exponent in its prime factorization that is at least as great as the number of positive exponents; intersection of A002182 and A212165.

Original entry on oeis.org

1, 2, 4, 12, 24, 36, 48, 120, 240, 360, 720, 1680, 5040, 10080, 15120, 20160, 25200, 45360, 50400, 110880, 221760, 332640, 554400, 665280, 2882880, 8648640, 14414400, 17297280, 43243200, 294053760
Offset: 1

Views

Author

Matthew Vandermast, May 22 2012

Keywords

Comments

Sequence can be used to find the largest highly composite number in subsequences of A212165 (of which there are several in the database).
Ramanujan showed that, in the canonical prime factorization of a highly composite number with largest prime factor prime(n), the largest exponent cannot exceed 2*log_2(prime(n+1)). (See formula 54 on page 15 of the Ramanujan paper.) This limit is less than n for all n >= 9 (and prime(n) >= 23).
1. Direct calculation verifies this for 9 <= n <= 11.
2. Nagura proved that, for any integer m >= 25, there is always a prime between m and 1.2*m. Let n = 11, at which point prime(11) = 31 and log_2(prime(n+1)) = log 37/log 2 = 5.209453.... Since log 1.2/log 2 is only 0.263034..., it follows that n must increase by at least 3k before 2*log_2(prime(n+1)) can increase by 2k, for all values of k. Therefore, 2*log_2(prime(n+1)) can never catch up to prime(n) for n > 11.
665280 = 2^6*3^3*5*7*11 is the largest highly composite number whose prime factorization contains an exponent that is strictly greater than the number of positive exponents in that factorization (including the implied 1's).

Examples

			A002182(62) = 294053760 = 2^7*3^3*5*7*11*13*17 has 7 positive exponents in its prime factorization, including 5 implied 1's. The maximal exponent in its prime factorization is also 7. Therefore, 294053760 is a term of this sequence.
		

References

  • S. Ramanujan, Highly composite numbers, Proc. Lond. Math. Soc. 14 (1915), 347-409; reprinted in Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962.

Crossrefs

Programs

  • Mathematica
    okQ[n_] := Module[{f = Transpose[FactorInteger[n]][[2]]}, Max[f] >= Length[f]]; a = 0; t = {}; Do[b = DivisorSigma[0, n]; If[b > a, a = b; If[okQ[n], AppendTo[t, n]]], {n, 10^6}]; t (* T. D. Noe, May 24 2012 *)

A306693 a(n) is the least strongly refactorable multiple of n if any, or a(n) = -1 otherwise.

Original entry on oeis.org

1, 2, 12, 12, 240, 12, 3360, 24, 36, 240, 5280, 12, 6240, 3360, 240, 240, 8160, 36, 9120, 240, 3360, 5280, 11040, 24, 21600, 6240, 4320, 3360, 13920, 240, 14880, 480, 5280, 8160, 3360, 36, 17760, 9120, 6240, 240, 19680, 3360, 20640, 5280, 720, 11040, 22560
Offset: 1

Views

Author

Rémy Sigrist, Mar 05 2019

Keywords

Comments

Strongly refactorable numbers correspond to A141586.
Is a(n) > 0 for any n > 0 ?

Examples

			For n = 3:
- the divisors of 3 are: 1, 3,
- the corresponding numbers of divisors are: 1, 2,
- 2 does not divide 3,
- the divisors of 2*3 are: 1, 2, 3, 6,
- the corresponding numbers of divisors are: 1, 2, 2, 4,
- 4 does not divide 2*3,
- the divisors of 2*2*3 are: 1, 2, 3, 4, 6, 12,
- the corresponding numbers of divisors are: 1, 2, 2, 3, 4, 6,
- they all divide 2*2*3,
- hence a(3) = 2*2*3 = 12.
		

Crossrefs

See A306645 for a similar sequence.

Programs

  • PARI
    a(n) = while (1, my (m=n); fordiv (m, d, m=lcm(m, numdiv(d))); if (n==m, return (n), n=m))

Formula

a(A141586(n)) = A141586(n) for any n > 0.
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