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 1-6 of 6 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

Views

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.

A130882 a(n) = smallest k such that A002808(n+1) = A002808(n) + (A002808(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 4, 7, 2, 4, 5, 13, 2, 7, 4, 19, 2, 4, 23, 2, 5, 2, 13, 4, 31, 2, 3, 2, 17, 37, 2, 19, 4, 43, 2, 4, 47, 2, 7, 2, 5, 53, 2, 5, 2, 4, 29, 61, 2, 3, 2, 4, 67, 2, 4, 5, 73, 2, 3, 2, 4, 79, 2, 4, 83, 2, 5, 2, 43, 89, 2, 7, 2, 3, 2, 47, 97
Offset: 1

Views

Author

Rémi Eismann, Aug 21 2007 - Jan 09 2011

Keywords

Comments

a(n) is the "weight" of composite numbers.
The decomposition of composite numbers into weight * level + gap is A002808(n) = a(n) * A179621(n) + A073783(n) if a(n) > 0.

Examples

			For n = 1 we have A002808(n) = 4, A002808(n+1) = 6; there is no k such that 6 - 4 = 2 = (4 mod k), hence a(1) = 0.
For n = 3 we have A002808(n) = 8, A002808(n+1) = 9; 7 is the smallest k such that 9 - 8 = 1 = (8 mod k), hence a(3) = 7.
For n = 24 we have A002808(n) = 36, A002808(n+1) = 38; 17 is the smallest k such that 38 - 36 = 2 = (36 mod k), hence a(24) = 17.

Links

Crossrefs

Cf. A020639, A117078, A117563, A001223, A118534, A090369, A090368, A130533, A130650, A130703.

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.

Links

Crossrefs

Cf. A014612, A114403, A130650, A184753, A117078, A117563, A001223, A118534.

A133150 a(n) = smallest k such that A000290(n+1) = A000290(n) + (A000290(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 14, 23, 17, 47, 31, 79, 49, 119, 71, 167, 97, 223, 127, 41, 46, 359, 199, 439, 241, 527, 82, 89, 337, 727, 391, 839, 449, 137, 73, 1087, 577, 1223, 647, 1367, 103, 217, 94, 1679, 881, 1847, 967, 119, 151, 2207, 1151, 2399, 1249, 113, 193, 401, 1457
Offset: 1

Views

Author

Rémi Eismann, Sep 22 2007 - Jan 10 2011

Keywords

Comments

a(n) is the "weight" of squares (A000290).
The decomposition of squares into weight * level + gap is A000217(n) = a(n) * A184221(n) + A005408(n) if a(n) > 0.

Examples

			For n = 1 we have A000290(n) = 1, A000290(n+1) = 4; there is no k such that 4 - 1 = 3 = (1 mod k), hence a(1) = 0.
For n = 5 we have A000290(n) = 25, A000290(n+1) = 36; 14 is the smallest k such that 36 - 25 = 11 = (25 mod k), hence a(5) = 14.
For n = 18 we have A000290(n) = 324, A000290(n+1) = 361; 41 is the smallest k such that 361 - 324 = 37 = (324 mod k), hence a(18) = 41.

Links

Crossrefs

Cf. A020639, A117078, A117563, A001223, A118534, A090369, A090368, A130533, A130650, A130703, A130889, A130882.

A133346 a(n) = smallest k such that prime(n+2) = prime(n) + (prime(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 0, 7, 11, 0, 15, 21, 21, 31, 7, 11, 35, 9, 17, 17, 61, 9, 21, 23, 23, 77, 7, 19, 97, 101, 91, 19, 13, 41, 25, 127, 47, 139, 21, 17, 31, 11, 167, 13, 37, 11, 61, 25, 39, 7, 13, 73, 9, 227, 25, 239, 35, 15, 9, 29, 271, 269, 37, 25, 7, 61, 59, 27, 21, 13, 11, 113, 113
Offset: 1

Views

Author

Rémi Eismann, Oct 20 2007

Keywords

Examples

			For n = 1 we have prime(n) = 2, prime(n+2) = 5; there is no k such that 5 - 2 = 3 = (2 mod k), hence a(1) = 0.
For n = 6 we have prime(n) = 13, prime(n+2) = 19; 7 is the smallest k such that 19 - 13 = 6 = (13 mod k), hence a(6) = 7.
For n = 30 we have prime(n) = 113, prime(n+2) = 131; 19 is the smallest k such that 131 - 113 = 18 = (113 mod k), hence a(30) = 19.

Links

Crossrefs

Cf. A000040, A117078, A117563, A001223, A118534, A020639, A090369, A090368, A130533, A130650, A130703, A130889, A130882.

A133347 a(n) = smallest k such that prime(n+3) = prime(n) + (prime(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 17, 19, 27, 29, 27, 33, 39, 47, 49, 55, 59, 19, 61, 65, 15, 29, 31, 31, 29, 29, 89, 23, 113, 41, 121, 15, 27, 47, 21, 17, 31, 15, 33, 61, 25, 57, 57, 193, 71, 43, 31, 43, 221, 73, 233, 27, 83, 257, 37, 29, 51, 51, 21, 11, 97, 289, 41, 313, 107, 67
Offset: 1

Views

Author

Rémi Eismann, Oct 20 2007

Keywords

Examples

			For n = 1 we have prime(n) = 2, prime(n+3) = 7; there is no k such that 7 - 2 = 5 = (2 mod k), hence a(1) = 0.
For n = 10 we have prime(n) = 29, prime(n+3) = 41; 17 is the smallest k such that 41 - 29 = 12 = (29 mod k), hence a(10) = 17.
For n = 53 we have prime(n) = 241, prime(n+3) = 263; 73 is the smallest k such that 263 - 241 = 22 = (241 mod k), hence a(53) = 73.

Links

Crossrefs

Cf. A000040, A117078, A117563, A001223, A118534, A020639, A090369, A090368, A130533, A130650, A130703, A130889, A130882.
Showing 1-6 of 6 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.