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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A184831 a(n) = A184830(n)/A184829(n) unless A184829(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 1, 1, 1, 3, 1, 1, 3, 2, 5, 5, 3, 7, 1, 5, 9, 15, 3, 3, 13, 3, 15, 9, 1, 19, 2, 1, 9, 23, 1, 11, 1, 11, 9, 3, 33, 11, 35, 21, 7, 13, 41, 63, 25, 5, 45, 3, 49, 5, 1, 3, 55, 33, 1, 59, 9, 63, 27, 65, 11, 1, 3, 75, 45, 1, 79, 1, 5, 41, 85, 1, 1, 89, 5, 39, 93, 1, 57, 9
Offset: 1

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Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "level" of prime powers.
The decomposition of prime powers into weight * level + gap is A000961(n) = A184829(n) * a(n) + A057820(n) if a(n) > 0.
A184830(n) = A000961(n) - A057820(n) if A000961(n) - A057820(n) > A057820(n), 0 otherwise.

Examples

			For n = 1 we have A184829(1) = 0, hence a(1) = 0.
For n = 3 we have A184830(3)/A184829(3)= 2 / 2 = 1; hence a(3) = 1.
For n = 24 we have A184830(24)/A184829(24)= 45 / 5 = 9; hence a(24) = 9.
		

Crossrefs

A184830 a(n) = largest k such that A000961(n+1) = A000961(n) + (A000961(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 2, 3, 3, 6, 7, 7, 9, 10, 15, 15, 15, 21, 23, 25, 27, 30, 27, 33, 39, 39, 45, 45, 47, 57, 58, 61, 63, 69, 67, 77, 79, 77, 81, 93, 99, 99, 105, 105, 105, 117, 123, 126, 125, 125, 135, 129, 147, 145, 151, 159, 165, 165, 167, 177, 171, 189, 189, 195
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

From the definition, a(n) = A000961(n) - A057820(n) if A000961(n) - A057820(n) > A057820(n), 0 otherwise where A000961 are the prime powers and A057820 are the gaps between prime powers.

Examples

			For n = 1 we have A000961(1) = 1, A000961(2) = 2; for all k >= 2, 2 = 1 + (1 mod k), hence the largest k does not exist and a(1) = 0.
For n = 3 we have A000961(3) = 3, A000961(4) = 4; 2 is the largest k such that 4 = 3 + (3 mod k), hence a(3) = 2.
For n = 24 we have A000961(24) = 49, A000961(25) = 53; 45 is the largest k such that 53 = 49 + (49 mod k), hence a(24) = 45.
		

Crossrefs

Programs

Showing 1-2 of 2 results.