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.

A166273 Bisection 2n+1 of A165605.

Original entry on oeis.org

3, 3, 3, 6, 6, 9, 3, 6, 9, 6, 12, 9, 9, 9, 12, 9, 6, 6, 18, 0, 9, 15, 9, 15, 12, 15, 9, 12, 15, 12, 15, 12, 15, 21, 18, 21, 9, 9, 0, 12, 24, 12, 12, 15, 6, 18, 15, 18, 21, 24, 18, 12, 18, 27, 18, 24, 18, 12, 12, 12, 33, 21, 12, 33, 12, 27, 18, 21, 24, 6, 33, 15, 15, 24, 24, 30
Offset: 0

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Author

Antti Karttunen, Oct 13 2009

Keywords

Comments

Conjecture: a(n) = -3 * A166269(n).

Crossrefs

a(n) = A165605(A005408(n)).

A165601 Midpoint height of Jacobi-bridge, computed for 4n+3. a(n) = Sum_{i=0..(2n+1)} J(i,4n+3), where J(i,m) is the Jacobi symbol.

Original entry on oeis.org

1, 1, 3, 2, 3, 3, 1, 3, 6, 4, 3, 5, 6, 4, 9, 2, 3, 7, 2, 5, 9, 6, 6, 8, 0, 5, 9, 8, 6, 10, 6, 5, 15, 2, 9, 10, 0, 7, 12, 10, 3, 11, 6, 2, 15, 8, 6, 13, 12, 9, 12, 0, 9, 14, 12, 7, 15, 12, 6, 15, 1, 6, 21, 12, 12, 13, 6, 11, 0, 6, 9, 14, 12, 8, 24, 10, 9, 19, 0, 10, 12, 12, 9, 18, 18, 1, 15
Offset: 0

Views

Author

Antti Karttunen, Oct 06 2009

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[JacobiSymbol[i, 4n + 3], {i, 0, 2n + 1}], {n, 0, 100}] (* Indranil Ghosh, May 13 2017 *)
  • PARI
    a(n) = sum(i=0, 2*n + 1, kronecker(i, 4*n + 3)); \\ Indranil Ghosh, May 13 2017
    
  • Python
    from sympy import jacobi_symbol as J
    def a(n): return sum([J(i, 4*n + 3) for i in range(2*n + 2)]) # Indranil Ghosh, May 13 2017

A165460 The height at the 1/3 point of Jacobi-bridge, computed for 12n+7. a(n) = Sum_{i=0..(4n+2)} J(i,12n+7), where J(i,m) is the Jacobi symbol.

Original entry on oeis.org

2, 2, 6, 2, 8, 2, 10, 4, 10, 4, 10, 6, 14, 2, 4, 4, 18, 6, 14, 4, 12, 8, 22, 6, 16, 6, 20, 6, 2, 8, 18, 6, 28, 4, 20, 4, 30, 12, 14, 0, 14, 6, 28, 10, 28, 6, 32, 10, 16, 8, 26, 10, 26, 6, 24, 8, 36, 10, 28, 8, 26, 10, 30, 8, 0, 10, 32, 14, 18, 12, 0, 14, 44, 6, 32, 6, 38, 0, 32, 8, 22
Offset: 0

Views

Author

Antti Karttunen, Oct 06 2009

Keywords

Comments

Conjecture: a(2n) = 2*A165605(2n) and a(2n+1) = (2/3)*A165605(2n+1). - Antti Karttunen, Oct 05 2009. (If true, then implies also the truth of conjecture in A165462.)

Crossrefs

Programs

  • Mathematica
    Table[Sum[JacobiSymbol[i, 12n + 7], {i, 0, 4n + 2}], {n, 0, 100}] (* Indranil Ghosh, May 13 2017 *)
  • PARI
    a(n) = sum(i=0, 4*n + 2, kronecker(i, 12*n + 7)); \\ Indranil Ghosh, May 13 2017
    
  • Python
    from sympy import jacobi_symbol as J
    def a(n): return sum([J(i, 12*n + 7) for i in range(4*n + 3)]) # Indranil Ghosh, May 13 2017

A165462 a(n) = (A165463(n)-3)/4.

Original entry on oeis.org

118, 193, 211, 232, 379, 493, 568, 574, 673, 868, 925, 943, 1243, 1261, 1300, 1318, 1372, 1408, 1471, 1618, 1693, 1702, 1816, 1993, 2068, 2290, 2323, 2368, 2389, 2395, 2437, 2443, 2512, 2731, 2743, 2797, 2818, 2968, 3106, 3118, 3193, 3235
Offset: 0

Views

Author

Antti Karttunen, Oct 06 2009

Keywords

Comments

Conjecture: These are all those terms of A165602 which = 1 modulo 3. If this is true, then A165461 gives also the positions of zeros in A165605. - Antti Karttunen, Oct 05 2009

Crossrefs

Cf. A165461-A165463. See also the conjecture in A165460.

A165461 Positions of zeros in A165460.

Original entry on oeis.org

39, 64, 70, 77, 126, 164, 189, 191, 224, 289, 308, 314, 414, 420, 433, 439, 457, 469, 490, 539, 564, 567, 605, 664, 689, 763, 774, 789, 796, 798, 812, 814, 837, 910, 914, 932, 939, 989, 1035, 1039, 1064, 1078, 1106, 1112, 1155, 1164, 1189, 1253, 1280
Offset: 0

Views

Author

Antti Karttunen, Oct 06 2009

Keywords

Comments

Also the positions of zeros in A165605 if the conjecture given in A165462 is true.

Crossrefs

A166268 Bisection 2n of A166050.

Original entry on oeis.org

1, 3, 4, 5, 5, 5, 7, 2, 9, 7, 6, 11, 8, 10, 1, 9, 14, 10, 15, 7, 7, 14, 14, 16, 8, 13, 13, 12, 18, 14, 13, 15, 0, 16, 9, 0, 22, 16, 19, 16, 11, 17, 16, 23, 19, 9, 22, 18, 16, 2, 18, 27, 12, 23, 11, 15, 24, 4, 27, 9, 23, 23, 20, 0, 19, 15, 22, 24, 22, 17, 18, 28, 26, 27, 17, 12
Offset: 0

Views

Author

Antti Karttunen, Oct 13 2009

Keywords

Comments

Conjecture: a(n) = A166270(n)/2, and also A165605(A005843(n)).

Crossrefs

a(n) = A166050(A005843(n)).
Showing 1-6 of 6 results.