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 1-5 of 5 results.

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.

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.

A184827 a(n) = largest k such that A000959(n+1) = A000959(n) + (A000959(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 5, 5, 11, 9, 17, 19, 29, 29, 31, 37, 47, 39, 59, 65, 65, 71, 71, 71, 81, 87, 93, 99, 107, 103, 125, 125, 131, 129, 131, 143, 155, 157, 167, 153, 185, 191, 189, 197, 199, 203, 215, 215, 227, 233, 233, 223, 257, 255, 261, 263
Offset: 1

Views

Author

Rémi Eismann, Jan 23 2011

Keywords

Comments

From the definition, a(n) = A000959(n) - A031883(n) if A000959(n) - A031883(n) > A031883(n), 0 otherwise where A000959 are the lucky numbers and A031883 are the gaps between lucky numbers.

Examples

			For n = 1 we have A000959(1) = 1, A000959(2) = 3; there is no k such that 3 - 1 = 2 = (1 mod k), hence a(1) = 0.
For n = 3 we have A000959(3) = 7, A000959(4) = 9; 5 is the largest k such that 9 - 7 = 2 = (7 mod k), hence a(3) = 5; a(3) = 7 -2 = 5.
For n = 24 we have A000959(24) = 105, A000959(25) = 111; 99 is the largest k such that 111 - 105 = 6 = (105 mod k), hence a(24) = 99; a(24) = 105 - 6 = 99.

Links

Crossrefs

Cf. A000959, A031883, A130889, A184828, A117078, A117563, A001223, A118534.

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 1-5 of 5 results.

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