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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A018255 Divisors of 30.

Original entry on oeis.org

1, 2, 3, 5, 6, 10, 15, 30
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

For n > 1, These are also numbers m such that k^4 + (k+1)^4 + ... + (k + m - 1)^4 is prime for some k and numbers m such that k^8 + (k+1)^8 + ... + (k + m - 1)^8 is prime for some k. - Derek Orr, Jun 12 2014
These seem to be the numbers m such that tau(n) = n*sigma(n) mod m for all n. See A098108 (mod 2), A126825 (mod 3), and A126832 (mod 5). - Charles R Greathouse IV, Mar 17 2022
The squarefree 5-smooth numbers: intersection of A051037 and A005117. - Amiram Eldar, Sep 26 2023

Examples

			From the second comment: 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 8^3 = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^2 = 729. - _Bruno Berselli_, Dec 28 2014

References

  • Boris A. Kordemsky, The Moscow Puzzles: 359 Mathematical Recreations, C. Scribner's Sons (1972), Chapter XIII, Paragraph 349.

Links

Crossrefs

Cf. A000005, A005117, A051037, A158649, A161715.
Cf. A098108, A126825, A126832.

Programs

Formula

a(n) = A161715(n-1). - Reinhard Zumkeller, Jun 21 2009
Sum_{i=1..8} A000005(a(i))^3 = (Sum_{i=1..8} A000005(a(i)))^2, see Kordemsky in References and Barbeau et al. in Links section. - Bruno Berselli, Dec 28 2014

A070563 a(n) = 0 if 3 divides the Ramanujan number tau(n) (A000594(n)), otherwise 1.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0
Offset: 1

Views

Author

N. J. A. Sloane, May 07 2002

Keywords

Comments

Multiplicative because A000594 is. Conjecture: a(3^k) = 0, if p == 1 mod 3, a(p^2k) = 0 and a(p^(2k+1)) = 1, if p == -1 mod 3, a(p^2k) = 1 and a(p^(2k+1)) = 0. - Christian G. Bower, Jun 10 2005
From Antti Karttunen, Jul 03 2024: (Start)
The above conjecture is not correct. The first counterexample occurs at n = 2401 = 7^4. My improved conjecture is that this is actually a characteristic function of nonmultiples of 3 whose sum of divisors is also a nonmultiple of 3, that is, having a following multiplicative formula: a(3^k) = 0, if p == 1 mod 3, a(p^e) = 1 if e != 2 (mod 3), otherwise 0, and if p == -1 mod 3, a(p^2k) = 1 and a(p^(2k+1)) = 0. This conjecture has now been proved correct by Seiichi Manyama.
Bower's formula is now submitted as A374053.
(End)

Links

Crossrefs

Characteristic function of A374135, nonmultiples of 3 whose sum of divisors is also a nonmultiple of 3.
Cf. A000203, A000594, A011655, A070564, A126825, A353815, A374053.

Programs

Formula

a(n) = A011655(n) * A353815(n), conjectured by Antti Karttunen, proved by Seiichi Manyama, Jul 03 2024
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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