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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A184834 a(n) = A184833(n)/A184832(n) unless A184832(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 0, 2, 1, 1, 3, 3, 6, 1, 1, 5, 1, 10, 7, 5, 1, 14, 1, 1, 16, 11, 11, 18, 1, 1, 20, 1, 10, 15, 1, 7, 17, 1, 28, 19, 19, 30, 1, 32, 13, 13, 34, 23, 23, 36, 1, 38, 11, 19, 27, 27, 42, 17, 17, 29, 1, 46, 31, 31, 3, 50, 1, 1, 52, 35, 35, 54, 1, 1, 56, 1, 28, 39
Offset: 1

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Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "level" of squarefree numbers.
The decomposition of squarefree numbers into weight * level + gap is A005117(n) = A184832(n) * a(n) + A076259(n) if a(n) > 0.
A184833(n) = A005117(n) - A076259(n) if A005117(n) - A076259(n) > A076259(n), 0 otherwise.

Examples

			For n = 1 we have A184832(1) = 0, hence a(1) = 0.
For n = 4 we have A184833(4)/A184832(4)= 4 / 2 = 1; hence a(4) = 2.
For n = 23 we have A184833(23)/A184832(23)= 33 / 3 = 11; hence a(23) = 11.
		

Crossrefs

A184832 a(n) = smallest k such that A005117(n+1) = A005117(n) + (A005117(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 2, 5, 4, 3, 3, 2, 13, 13, 3, 17, 2, 3, 4, 23, 2, 29, 29, 2, 3, 3, 2, 37, 37, 2, 41, 4, 3, 43, 7, 3, 53, 2, 3, 3, 2, 59, 2, 5, 5, 2, 3, 3, 2, 71, 2, 7, 4, 3, 3, 2, 5, 5, 3, 89, 2, 3, 3, 31, 2, 101, 101, 2, 3, 3, 2, 109, 109, 2, 113, 4, 3, 4, 11, 7, 5, 2, 3, 3, 2
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "weight" of squarefree numbers.
The decomposition of squarefree numbers into weight * level + gap is A005117(n) = a(n) * A184834(n) + A076259(n) if a(n) > 0.

Examples

			For n = 1 we have A005117(1) = 1, A005117(2) = 2; there is no k such that 2 - 1 = 1 = (1 mod k), hence a(1) = 0.
For n = 4 we have A005117(4) = 5, A005117(5) = 6; 2 is the smallest k such that 6 - 5 = 1 = (5 mod k), hence a(4) = 2.
For n = 23 we have A005117(23) = 35, A005117(24) = 37; 3 is the smallest k such that 37 - 35 = 2 = (35 mod k), hence a(23) = 3.
		

Crossrefs

Showing 1-2 of 2 results.