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.

A130650 a(n) = smallest k such that A014612(n+1) = A014612(n) + (A014612(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 4, 13, 2, 13, 18, 4, 43, 8, 3, 41, 4, 4, 3, 13, 2, 37, 16, 43, 97, 4, 9, 10, 53, 4, 5, 10, 3, 6, 61, 43, 2, 11, 2, 12, 163, 8, 13, 2, 5, 173, 8, 89, 4, 3, 37, 61, 101, 101, 107, 229, 113
Offset: 1

Views

Author

Rémi Eismann, Aug 16 2007 - Jan 21 2011

Keywords

Comments

a(n) is the "weight" of 3-almost primes.
The decomposition of 3-almost primes into weight * level + gap is A014612(n) = a(n) * A184753(n) + A114403(n) if a(n) > 0.

Examples

			For n = 1 we have A014612(1) = 8, A014612(2) = 12; there is no k such that 12 - 8 = 4 = (8 mod k), hence a(1) = 0.
For n = 3 we have A014612(3) = 18, A014612(4) = 20; 4 is the smallest k such that 20 - 18 = 2 = (18 mod k), hence a(3) = 4.
For n = 21 we have A014612(21) = 98, A014612(22) = 99; 97 is the smallest k such that 99 - 98 = 1 = (97 mod k), hence a(21) = 97.
		

Crossrefs

A184752 a(n) = largest k such that A014612(n+1) = A014612(n) + (A014612(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 16, 13, 26, 26, 18, 40, 43, 40, 48, 41, 60, 64, 66, 65, 74, 74, 64, 86, 97, 96, 99, 100, 106, 112, 115, 110, 123, 120, 122, 129, 146, 143, 152, 144, 163, 160, 169, 170, 170, 173, 168, 178, 184, 186, 185, 183, 202, 202, 214
Offset: 1

Views

Author

Rémi Eismann, Jan 21 2011

Keywords

Comments

From the definition, a(n) = A014612(n) - A114403(n) if A014612(n) - A114403(n) > A114403(n), 0 otherwise where A014612 are the 3-almost primes and A114403 are the gaps between 3-almost primes.

Examples

			For n = 1 we have A014612(1) = 8, A014612(2) = 12; there is no k such that 12 - 8 = 4 = (8 mod k), hence a(1) = 0.
For n = 3 we have A014612(3) = 18, A014612(4) = 20; 16 is the largest k such that 20 - 18 = 2 = (18 mod k), hence a(3) = 16.
For n = 21 we have A014612(21) = 98, A014612(22) = 99; 97 is the largest k such that 99 - 98 = 1 = (97 mod k), hence a(21) = 97.
		

Crossrefs

Showing 1-2 of 2 results.