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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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A117874 Primes for which the level is equal to 5 in A117563.

Original entry on oeis.org

17, 61, 131, 151, 271, 523, 541, 571, 751, 797, 971, 991, 997, 1291, 1321, 1361, 1741, 1901, 1913, 2011, 2179, 2297, 2341, 2441, 2447, 2551, 2791, 2851, 3301, 3511, 3761, 3803, 4051, 4391, 4397, 4423, 4441, 4561, 4651, 4703, 4759, 5101, 5471, 5483, 5521
Offset: 1

Views

Author

Rémi Eismann, May 02 2006

Keywords

Examples

			19=17+17 mod(3)=17+17 mod(15), level=5
157=151+151 mod(29)=151+151 mod(145) level=5
2203=2179+2179 mod(431)=2179+2179 mod(2155), level=5

Links

Crossrefs

Cf. A117078.

Programs

  • Mathematica
    f[n_] := Block[{d, j = 2, p = Prime@n}, d = Prime[n + 1] - p; While[j < p && Mod[p, j] != d, j++ ]; If[j == p, 0, j]]; g[n_] := Block[{d, k = p = Prime@n}, d = Prime[n + 1] - p; While[k > 0 && Mod[p, k] != d, k-- ]; If[k == 0, 0, k]]; h[n_] := Block[{a = f@n, b = g@n}, If[a == 0, 0, b/a]]; Prime@Select[ Range@763, h@# == 5 &] (* Robert G. Wilson v *)

Extensions

More terms from Robert G. Wilson v, May 06 2006
Edited by N. J. A. Sloane, May 14 2006

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.

A133151 a(n) = smallest k such that A000326(n+1) = A000326(n) + (A000326(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 19, 32, 24, 67, 89, 38, 71, 173, 69, 61, 71, 109, 373, 211, 79, 529, 587, 72, 89, 779, 283, 461, 499, 359, 1159, 311, 111, 1423, 1517, 269, 857, 1817, 641, 127, 134, 251, 2377, 1249, 138, 2749, 2879, 251, 787, 173, 381, 1787, 1861, 1291
Offset: 1

Views

Author

Rémi Eismann, Sep 22 2007 - Jan 21 2011

Keywords

Comments

a(n) is the "weight" of pentagonal numbers (A000326).
The decomposition of pentagonal numbers into weight * level + gap is A000326(n) = a(n) * A184751(n) + A016777(n) if a(n) > 0.

Examples

			For n = 1 we have A000326(n) = 1, A000326(n+1) = 5; there is no k such that 5 - 1 = 4 = (1 mod k), hence a(1) = 0.
For n = 5 we have A000326(n) = 35, A000326(n+1) = 51; 19 is the smallest k such that 51 - 35 = 16 = (35 mod k), hence a(5) = 19.
For n = 18 we have A000326(n) = 477, A000326(n+1) = 532; 211 is the smallest k such that 532 - 477 = 55 = (477 mod k), hence a(18) = 211.

Links

Crossrefs

Cf. A000326, A016777, A184751, A184750, A117078, A117563, A001223, A118534.

A179621 a(n) = A179620(n)/A130882(n) unless A130882(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 1, 1, 4, 2, 2, 1, 7, 2, 4, 1, 10, 5, 1, 12, 5, 13, 2, 7, 1, 16, 11, 17, 2, 1, 19, 2, 10, 1, 22, 11, 1, 24, 7, 25, 10, 1, 27, 11, 28, 14, 2, 1, 31, 21, 32, 16, 1, 34, 17, 14, 1, 37, 25, 38, 19, 1, 40, 20, 1, 42, 17, 43, 2, 1, 45, 13, 46, 31, 47, 2, 1, 49, 14, 25, 1, 52, 26, 2, 1
Offset: 1

Views

Author

Rémi Eismann, Jan 09 2011

Keywords

Comments

a(n) is the "level" of composite numbers.
The decomposition of composite numbers into weight * level + gap is A002808(n) = A130882(n) * a(n) + A073783(n) if a(n) > 0.
A179620(n) = A002808(n) - A073783(n) if A002808(n) - A073783(n) > A073783(n), 0 otherwise.

Examples

			For n = 1 we have A130882(1) = 0, hence a(1) = 0.
For n = 3 we have A179620(3)/A130882(3)= 7 / 7 = 1; hence a(3) = 1.
For n = 24 we have A179620(24)/A130882(24)= 34 / 17 = 2; hence a(24) = 2.

Links

Crossrefs

Cf. A002808, A073783, A130882, A179620, A117078, A117563, A001223, A118534.

A184219 a(n) = A184218(n)/A130703(n) unless A130703(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 1, 1, 2, 3, 1, 1, 5, 4, 1, 5, 4, 5, 7, 1, 5, 9, 5, 1, 9, 10, 1, 9, 7, 8, 11, 1, 14, 11, 1, 7, 13, 10, 1, 13, 10, 11, 15, 1, 11, 20, 7, 5, 18, 14, 5, 17, 22, 14, 19, 11, 14, 19, 1, 1, 27, 16, 13, 27, 16, 26, 23
Offset: 1

Views

Author

Rémi Eismann, Jan 10 2011

Keywords

Comments

a(n) is the "level" of triangular numbers.
The decomposition of triangular numbers into weight * level + gap is A000217(n) = A130703(n) * a(n) + (n + 1) if a(n) > 0.
A184218(n) = A000217(n) - (n + 1) if A000217(n) - (n + 1) > (n + 1), 0 otherwise.

Examples

			For n = 3 we have A130703(3) = 0, hence a(3) = 0.
For n = 5 we have A184218(5)/A130703(5) = 9 / 9 = 1, hence a(5) = 1.
For n = 24 we have A184218(24)/A130703(24) = 275 / 55 = 5, hence a(24) = 5.

Links

Crossrefs

Cf. A000217, A000027, A130703, A184218, A118534, A117078, A117563, A001223.

A184221 a(n) = A184220(n)/A133150(n) unless A133150(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 7, 7, 1, 2, 1, 2, 1, 7, 7, 2, 1, 2, 1, 2, 7, 14, 1, 2, 1, 2, 1, 14, 7, 17, 1, 2, 1, 2, 17, 14, 1, 2, 1, 2, 23, 14, 7, 2, 1, 2, 17, 2, 7, 14, 1, 17, 1, 23, 1, 14, 7, 2, 1, 31, 1, 2, 7, 34, 1, 2, 1
Offset: 1

Views

Author

Rémi Eismann, Jan 10 2011

Keywords

Comments

a(n) is the "level" of squares (A000290).
The decomposition of squares into weight * level + gap is A000217(n) = A133150(n) * a(n) + A005408(n) if a(n) > 0.
A184220(n) = A000290(n) - A005408(n) if A000217(n) - A005408(n) > A005408(n), 0 otherwise.

Examples

			For n = 3 we have A133150(3) = 0, hence a(3) = 0.
For n = 5 we have A184220(5)/A133150(5) = 14 / 14 = 1, hence a(5) = 1.
For n = 25 we have A184220(25)/A133150(25) = 574 / 82 = 5, hence a(25) = 7.

Links

Crossrefs

Cf. A000290, A005408, A133150, A184220, A118534, A117078, A117563, A001223.

A184728 a(n) = largest k such that A001358(n+1) = A001358(n) + (A001358(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 8, 6, 13, 9, 20, 19, 24, 19, 32, 33, 32, 37, 32, 43, 47, 47, 53, 56, 54, 59, 61, 64, 71, 72, 79, 84, 85, 83, 89, 92, 93, 84, 101, 107, 112, 117, 117, 120, 121, 117, 125, 132, 127, 140, 141, 141, 144, 137, 152, 157, 157
Offset: 1

Views

Author

Rémi Eismann, Jan 20 2011

Keywords

Comments

a(n) = A001358(n) - A065516(n) if A001358(n) - A065516(n) > A065516(n), 0 otherwise.
A001358(n): semiprimes; A065516(n): first difference of semiprimes.

Examples

			For n = 1 we have A001358(n) = 4, A001358(n+1) = 6; there is no k such that 6 - 4 = 2 = (4 mod k), hence a(1) = 0.
For n = 3 we have A001358(n) = 9, A001358(n+1) = 10; 8 is the largest k such that 10 - 9 = 1 = (9 mod k), hence a(3) = 8; a(3) = A001358(3) - A065516(3) = 8.
For n = 20 we have A001358(n) = 57, A001358(n+1) = 58; 56 is the largest k such that 58 - 57 = 1 = (57 mod k), hence a(20) = 56; a(20) = A001358(20) - A065516(20) = 56.

Links

Crossrefs

Cf. A001358, A065516, A130533, A184729, A117078, A117563, A001223, A118534.

A184729 a(n) = A184728(n)/A130533(n) unless A130533(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 4, 1, 1, 1, 10, 1, 12, 1, 16, 11, 8, 1, 4, 1, 1, 1, 1, 28, 9, 1, 1, 8, 1, 12, 1, 42, 17, 1, 1, 46, 31, 7, 1, 1, 28, 39, 39, 60, 11, 13, 25, 66, 1, 70, 47, 47, 72, 1, 38, 1, 1, 26, 1, 7, 88, 1, 1, 61, 20, 17, 100, 67, 67, 102, 29, 41, 106
Offset: 1

Views

Author

Rémi Eismann, Jan 20 2011

Keywords

Comments

a(n) is the "level" of semiprimes.
The decomposition of semiprimes into weight * level + gap is A001358(n) = A130533(n) * a(n) + A065516(n) if a(n) > 0.
A184728(n) = A001358(n) - A065516(n) if A001358(n) - A065516(n) > A065516(n), 0 otherwise.

Examples

			For n = 1 we have A130533(1) = 0, hence a(1) = 0.
For n = 3 we have A184728(3)/A130533(3)= 8 / 2 = 4; hence a(3) = 4.
For n = 20 we have A184728(20)/A130533(20)= 56 / 2 = 28; hence a(20) = 28.

Links

Crossrefs

Cf. A001358, A065516, A130533, A184728, A117078, A117563, A001223, A118534.

A184750 a(n) = largest k such that A000326(n+1) = A000326(n) + (A000326(n) mod k), or 0 if no such k exists.

Original entry on oeis.org

0, 0, 0, 0, 19, 32, 48, 67, 89, 114, 142, 173, 207, 244, 284, 327, 373, 422, 474, 529, 587, 648, 712, 779, 849, 922, 998, 1077, 1159, 1244, 1332, 1423, 1517, 1614, 1714, 1817, 1923, 2032, 2144, 2259, 2377, 2498, 2622, 2749
Offset: 1

Views

Author

Rémi Eismann, Jan 21 2011

Keywords

Comments

From the definition, a(n) = A000326(n) - A016777(n) if A000326(n) - A016777(n) > A016777(n), 0 otherwise, where A000326 are the pentagonal numbers and A016777 are the gaps between pentagonal numbers: 3n + 1.

Examples

			For n = 3 we have A000326(3) = 12, A000326(4) = 22; there is no k such that 22 - 12 = 10 = (12 mod k), hence a(3) = 0.
For n = 5 we have A000326(5) = 35, A000326(6) = 51; 19 is the largest k such that 51 - 35 = 16 = (35 mod k), hence a(5) = 19; a(5) = (75-35-2)/2 = 19.
For n = 25 we have A000326(25) = 925, A000326(26) = 1001; 849 is the largest k such that 1001 - 925 = 76 = (925 mod k), hence a(25) = 849; a(25) = (1875-175-2)/2 = 849.

Links

Crossrefs

Cf. A000326, A016777, A133151, A184751, A117078, A117563, A001223, A118534.

Programs

  • Maple
    A184750:=n->(3*n^2 - 7*n - 2)*signum(floor(n/5))/2; seq(A184750(n), n=1..50); # Wesley Ivan Hurt, Apr 05 2014
  • Mathematica
    Table[(3 n^2 - 7 n - 2) Sign[Floor[n/5]]/2, {n, 50}] (* Wesley Ivan Hurt, Apr 05 2014 *)
  • PARI
    concat([0,0,0,0], Vec(-x^5*(9*x^2-25*x+19)/(x-1)^3 + O(x^100))) \\ Colin Barker, Apr 05 2014

Formula

a(n) = (3n^2-7n-2)/2 for n >= 5 and a(n) = 0 for n <= 4.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>7. G.f.: x^5*(9*x^2-25*x+19) / (1-x)^3. - Colin Barker, Apr 05 2014
a(n) = A000326(n) - A016777(n), n>=5, (see a comment above). - Wolfdieter Lang, Apr 19 2014

Extensions

Edited - Wolfdieter Lang, Apr 19 2014

A184751 a(n) = A184750(n)/A133151(n) unless A133151(n) = 0 in which case a(n) = 0.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 1, 1, 3, 2, 1, 3, 4, 4, 3, 1, 2, 6, 1, 1, 9, 8, 1, 3, 2, 2, 3, 1, 4, 12, 1, 1, 6, 2, 1, 3, 16, 16, 9, 1, 2, 19, 1, 1, 12, 4, 19, 9, 2, 2, 3, 1, 8, 24, 1, 1, 18, 23, 1, 3, 19, 4, 3, 1, 23, 19, 1, 1, 32, 16, 1, 3, 2, 2, 27, 1, 4, 12, 1, 19, 23
Offset: 1

Views

Author

Rémi Eismann, Jan 21 2011

Keywords

Comments

a(n) is the "level" of pentagonal numbers (A000326).
The decomposition of pentagonal numbers into weight * level + gap is A000326(n) = A133151(n) * a(n) + A016777(n) if a(n) > 0.
A184750(n) = A000326(n) - A016777(n) if A000326(n) - A016777(n) > A016777(n), 0 otherwise.

Examples

			For n = 3 we have A133151(3) = 0, hence a(3) = 0.
For n = 5 we have A184750(5)/A133151(5) = 19 / 19 = 1, hence a(5) = 1.
For n = 25 we have A184750(25)/A133151(25) = 849 / 283 = 5, hence a(25) = 3.

Links

Crossrefs

Cf. A000326, A016777, A133151, A184750, A117078, A117563, A001223, A118534.
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