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 1-2 of 2 results.

A130533 a(n) = smallest k such that A001358(n+1) = A001358(n) + (A001358(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 2, 6, 13, 9, 2, 19, 2, 19, 2, 3, 4, 37, 8, 43, 47, 47, 53, 2, 6, 59, 61, 8, 71, 6, 79, 2, 5, 83, 89, 2, 3, 12, 101, 107, 4, 3, 3, 2, 11
Offset: 1

Views

Author

Rémi Eismann, Aug 16 2007 - Jan 20 2011

Keywords

Comments

a(n) is the "weight" of semiprimes.
The decomposition of semiprimes into weight * level + gap is A001358(n) = a(n) * A184729(n) + A065516(n) if a(n) > 0.

Examples

			For n = 1 we have A001358(n) = 4, A001358(n+1) = 6; there is no k such that 6 - 4 = 2 = (4 mod k), hence a(1) = 0.
For n = 3 we have A001358(n) = 9, A001358(n+1) = 10; 2 is the smallest k such that 10 - 9 = 1 = (9 mod k), hence a(3) = 2.
For n = 19 we have A001358(n) = 55, A001358(n+1) = 57; 53 is the smallest k such that 57 - 55 = 2 = (55 mod k), hence a(19) = 53.
		

Crossrefs

A184728 a(n) = largest k such that A001358(n+1) = A001358(n) + (A001358(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 8, 6, 13, 9, 20, 19, 24, 19, 32, 33, 32, 37, 32, 43, 47, 47, 53, 56, 54, 59, 61, 64, 71, 72, 79, 84, 85, 83, 89, 92, 93, 84, 101, 107, 112, 117, 117, 120, 121, 117, 125, 132, 127, 140, 141, 141, 144, 137, 152, 157, 157
Offset: 1

Views

Author

Rémi Eismann, Jan 20 2011

Keywords

Comments

a(n) = A001358(n) - A065516(n) if A001358(n) - A065516(n) > A065516(n), 0 otherwise.
A001358(n): semiprimes; A065516(n): first difference of semiprimes.

Examples

			For n = 1 we have A001358(n) = 4, A001358(n+1) = 6; there is no k such that 6 - 4 = 2 = (4 mod k), hence a(1) = 0.
For n = 3 we have A001358(n) = 9, A001358(n+1) = 10; 8 is the largest k such that 10 - 9 = 1 = (9 mod k), hence a(3) = 8; a(3) = A001358(3) - A065516(3) = 8.
For n = 20 we have A001358(n) = 57, A001358(n+1) = 58; 56 is the largest k such that 58 - 57 = 1 = (57 mod k), hence a(20) = 56; a(20) = A001358(20) - A065516(20) = 56.
		

Crossrefs

Showing 1-2 of 2 results.