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 1-3 of 3 results.

A184753 a(n) = A184752(n)/A130650(n) unless A130650(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 4, 1, 13, 2, 1, 10, 1, 5, 16, 1, 15, 16, 22, 5, 37, 2, 4, 2, 1, 24, 11, 10, 2, 28, 23, 11, 41, 20, 2, 3, 73, 13, 76, 12, 1, 20, 13, 85, 34, 1, 21, 2, 46, 62, 5, 3, 2, 2, 2, 1, 2, 78, 39, 80, 81, 122, 3, 63, 51, 32, 88, 1, 1, 1, 69, 70
Offset: 1

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Author

Rémi Eismann, Jan 21 2011

Keywords

Comments

a(n) is the "level" of 3-almost primes.
The decomposition of 3-almost primes into weight * level + gap is A014612(n) = A130650(n) * a(n) + A114403(n) if a(n) > 0.
a(n) = A014612(n) - A114403(n) if A014612(n) - A114403(n) > A114403(n), 0 otherwise.

Examples

			For n = 1 we have A130650(1) = 0, hence a(1) = 0.
For n = 3 we have A184752(3)/A130650(3)= 16 / 4 = 4; hence a(3) = 4.
For n = 21 we have A184752(21)/A130650(21)= 97 / 97 = 28; hence a(21) = 1.

Links

Crossrefs

Cf. A014612, A114403, A130650, A184752, A117078, A117563, A001223, A118534.

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.

Links

Crossrefs

Cf. A014612, A114403, A184753, A184752, A117078, A117563, A001223, A118534.

A184828 a(n) = A184827(n)/A130889(n) unless A130889(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 13, 13, 1, 1, 1, 9, 3, 3, 11, 1, 1, 25, 25, 1, 3, 1, 13, 31, 1, 1, 3, 37, 1, 27, 1, 1, 7, 43, 5, 1, 1, 1, 1, 1, 17, 29, 1, 1, 1, 1, 3, 23, 5, 1, 45, 19, 19, 7, 31, 1, 5, 1, 1, 1, 43, 1, 31, 1, 5, 85, 85, 5, 1, 11, 43, 3
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

a(n) is the "level" of lucky numbers.
The decomposition of lucky numbers into weight * level + gap is A000959(n) = A130889(n) * a(n) + A031883(n) if a(n) > 0.
A184827(n) = A000959(n) - A031883(n) if A000959(n) - A031883(n) > A031883(n), 0 otherwise.

Examples

			For n = 1 we have A130889(1) = 0, hence a(1) = 0.
For n = 3 we have A184752(3)/A130889(3)= 5 / 5 = 1; hence a(3) = 1.
For n = 24 we have A184752(24)/A130889(24)= 99 / 9 = 11; hence a(24) = 11.

Links

Crossrefs

Cf. A000959, A031883, A130889, A184827, A117078, A117563, A001223, A118534.
Showing 1-3 of 3 results.

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