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.

A259748 a(n) = (Sum_{0

Original entry on oeis.org

0, 0, 2, 3, 0, 1, 0, 2, 6, 0, 0, 5, 0, 7, 10, 4, 0, 12, 0, 15, 14, 11, 0, 22, 0, 0, 18, 21, 0, 5, 0, 8, 22, 0, 0, 15, 0, 19, 26, 10, 0, 28, 0, 33, 30, 23, 0, 44, 0, 0, 34, 39, 0, 9, 0, 14, 38, 0, 0, 25, 0, 31, 42, 16, 0, 44, 0, 51, 46, 35, 0, 66, 0, 0, 50
Offset: 1

Views

Author

José María Grau Ribas, Jul 04 2015

Keywords

Comments

{a(n)/n: n=1,2,...} = {0, 1/6, 1/4, 5/12, 1/2, 2/3, 3/4, 11/12}.
From Danny Rorabaugh, Oct 22 2015: (Start)
a(n)/n = 0 iff n mod 24 = 1,2,5,7,10,11,13,17,19,23 (A259749);
a(n)/n = 1/6 iff n mod 24 = 6 (A259752);
a(n)/n = 1/4 iff n mod 24 = 8,16 (A259751);
a(n)/n = 5/12 iff n mod 24 = 12 (A073762);
a(n)/n = 1/2 iff n mod 24 = 14,22 (A259750);
a(n)/n = 2/3 iff n mod 24 = 3,9,15,18,21 (A259754);
a(n)/n = 3/4 iff n mod 24 = 4,20 (A259755);
a(n)/n = 11/12 iff n mod 24 = 0 (A008606).
(End)

Links

Crossrefs

Cf. A000914,
A259749 (n such that a(n)=0),
A259750 (n such that n/a(n)=2),
A259751 (n such that n/a(n)=4),
A259752 (n such that n/a(n)=6),
A073762 (n such that n/a(n)=12/5),
A259754 (n such that n/a(n)=3/2),
A259755 (n such that n/a(n)=4/3),
A008606 (n such that n/a(n)=12/11).

Programs

  • Mathematica
    A[n_]:=Sum[a b,{a,1,n},{b,a+1,n}];Table[Mod[A[n],n],{n,1,122}]
  • PARI
    vector(100, n, ((n-1)*n*(n+1)*(3*n+2)/24) % n) \\ Altug Alkan, Oct 22 2015

Formula

a(n) = A000914(n) mod n = (1/24)*(-1 + n)*n*(1 + n)*(2 + 3*n) mod n.
a(24k) = 22k; a(24k+1) = 0; a(24k+2) = 0; a(24k+3) = 16k+2; a(24k+4) = 18k+3; a(24k+5) = 0; a(24k+6) = 4k+1, a(24k+7) = 0; a(24k+8) = 6k+2; a(24k+9) = 16k+6; a(24k+10) = 0; a(24k+11) = 0; a(24k+12) = 10k+5; a(24k+13) = 0; a(24k+14) = 12k+7; a(24k+15) = 16k+10; a(24k+16) = 6k+4; a(24k+17) = 0; a(24k+18) = 16k+12; a(24k+19) = 0; a(24k+20) = 18k+15; a(24k+21) = 16k+14; a(24k+22) = 12k+11; a(24k+23) = 0. - Danny Rorabaugh, Oct 22 2015

A259755 Numbers that are congruent to {4, 20} mod 24.

Original entry on oeis.org

4, 20, 28, 44, 52, 68, 76, 92, 100, 116, 124, 140, 148, 164, 172, 188, 196, 212, 220, 236, 244, 260, 268, 284, 292, 308, 316, 332, 340, 356, 364, 380, 388, 404, 412, 428, 436, 452, 460, 476, 484, 500, 508, 524, 532, 548, 556, 572, 580, 596, 604, 620, 628
Offset: 1

Views

Author

José María Grau Ribas, Jul 18 2015

Keywords

Links

Crossrefs

Cf. A000914, A007310.
Other sequences of numbers k such that A259748(k)/k equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259754.

Programs

  • Magma
    [2*(6*n+(-1)^n-3): n in [1..60]]; // Vincenzo Librandi, Aug 27 2015
  • Mathematica
    A[n_] := A[n] = Sum[a b, {a, 1,n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 3/4 &]
Table[2 (6 n + (-1)^n - 3), {n, 1, 60}] (* Bruno Berselli, Oct 23 2015 *)
LinearRecurrence[{1,1,-1},{4,20,28},60] (* Harvey P. Dale, Jul 19 2016 *)
  • PARI
    vector(100, n, 2*(6*n+(-1)^n-3)) \\ Altug Alkan, Oct 23 2015

Formula

a(n) = 2*(6*n + (-1)^n - 3).
A259748(a(n))/a(n) = 3/4.
a(n) = 4*A007310(n). - Michel Marcus, Sep 22 2015
G.f.: 4*x*(1 + 4*x + x^2) / ((1 + x)*(1 - x)^2). - Bruno Berselli, Oct 23 2015
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/24. - Amiram Eldar, Dec 31 2021
E.g.f.: 2*(2 + (6*x - 3)*exp(x) + exp(-x)). - David Lovler, Sep 05 2022

Extensions

Better name from Danny Rorabaugh, Oct 22 2015

A259749 Numbers that are congruent to {1,2,5,7,10,11,13,17,19,23} mod 24.

Original entry on oeis.org

1, 2, 5, 7, 10, 11, 13, 17, 19, 23, 25, 26, 29, 31, 34, 35, 37, 41, 43, 47, 49, 50, 53, 55, 58, 59, 61, 65, 67, 71, 73, 74, 77, 79, 82, 83, 85, 89, 91, 95, 97, 98, 101, 103, 106, 107, 109, 113, 115, 119, 121, 122, 125, 127, 130, 131, 133, 137, 139, 143, 145
Offset: 1

Views

Author

José María Grau Ribas, Jul 04 2015

Keywords

Comments

Original name: Numbers n such that A259748(n) = 0.

Links

Crossrefs

Cf. A000914.
Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259750, A259751, A259752, A259754, A259755.

Programs

  • Mathematica
    A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]  == 0 & ]
Rest@ CoefficientList[Series[x (1 + x^2) (1 + 2 x^2 - x^3 + 2 x^4 - 2 x^5 + 3 x^6 + x^7)/((1 - x)^2*(1 - x + x^2 - x^3 + x^4) (1 + x + x^2 + x^3 + x^4)), {x, 0, 61}], x] (* Michael De Vlieger, Aug 25 2016 *)
Select[Range[150],MemberQ[{1,2,5,7,10,11,13,17,19,23},Mod[#,24]]&] (* or *) LinearRecurrence[{2,-2,2,-2,2,-2,2,-2,2,-1},{1,2,5,7,10,11,13,17,19,23},70] (* Harvey P. Dale, Jan 15 2022 *)
  • PARI
    Vec(x*(1+x^2)*(1+2*x^2-x^3+2*x^4-2*x^5+3*x^6+x^7)/((1-x)^2*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)) + O(x^100)) \\ Colin Barker, Aug 25 2016

Formula

A259748(a(n)) = Sum_{x*y: x,y in Z/a(n)Z, x<>y} = 0.
G.f.: x*(1+x^2)*(1+2*x^2-x^3+2*x^4-2*x^5+3*x^6+x^7) / ((1-x)^2*(1-x+x^2-x^3+x^4)*(1+x+x^2+x^3+x^4)). - Colin Barker, Aug 25 2016

Extensions

Better name from Danny Rorabaugh, Oct 22 2015

A259751 Numbers that are congruent to {8, 16} mod 24.

Original entry on oeis.org

8, 16, 32, 40, 56, 64, 80, 88, 104, 112, 128, 136, 152, 160, 176, 184, 200, 208, 224, 232, 248, 256, 272, 280, 296, 304, 320, 328, 344, 352, 368, 376, 392, 400, 416, 424, 440, 448, 464, 472, 488, 496, 512, 520, 536, 544, 560, 568, 584, 592, 608, 616, 632
Offset: 1

Views

Author

José María Grau Ribas, Jul 06 2015

Keywords

Comments

Original name: Numbers n such that n/A259748(n) = 4.

Links

Crossrefs

Cf. A000914, A001651.
Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259752, A259754, A259755.

Programs

  • Mathematica
    A[n_] := A[n] = Sum[a b, {a, 1,  n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/4 & ]
  • PARI
    Vec(8*x*(1+x+x^2)/((1-x)^2*(1+x)) + O(x^100)) \\ Colin Barker, Aug 26 2016

Formula

A259748(a(n))/a(n) = 1/4.
a(n) = 8*A001651. - Danny Rorabaugh, Oct 22 2015
From Colin Barker, Aug 26 2016: (Start)
a(n) = 12*n-2*(-1)^n-6.
a(n) = a(n-1)+a(n-2)-a(n-3) for n>3.
G.f.: 8*x*(1+x+x^2) / ((1-x)^2*(1+x)).
(End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/72. - Amiram Eldar, Dec 31 2021
E.g.f.: 2*(4 + (6*x - 3)*exp(x) - exp(-x)). - David Lovler, Sep 05 2022

Extensions

Better name from Danny Rorabaugh, Oct 22 2015

A259754 Numbers that are congruent to {3,9,15,18,21} mod 24.

Original entry on oeis.org

3, 9, 15, 18, 21, 27, 33, 39, 42, 45, 51, 57, 63, 66, 69, 75, 81, 87, 90, 93, 99, 105, 111, 114, 117, 123, 129, 135, 138, 141, 147, 153, 159, 162, 165, 171, 177, 183, 186, 189, 195, 201, 207, 210, 213, 219, 225, 231, 234, 237, 243, 249, 255, 258, 261, 267
Offset: 1

Views

Author

José María Grau Ribas, Jul 12 2015

Keywords

Comments

Original name: Numbers n such that n/A259748(n) = 3/2.

Links

Crossrefs

Cf. A000914.
Other sequences of numbers n such that A259748(n)/n equals a constant: A008606, A073762, A259749, A259750, A259751, A259752, A259755.

Programs

  • Mathematica
    A[n_] := A[n] = Sum[a b, {a, 1, n}, {b, a + 1, n}]; Select[Range[200], Mod[A[#], #]/# == 2/3 &]
Rest@ CoefficientList[Series[3 x (1 + x) (1 + x + x^2 + x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)), {x, 0, 56}], x] (* Michael De Vlieger, Aug 25 2016 *)
LinearRecurrence[{1,0,0,0,1,-1},{3,9,15,18,21,27},60] (* Harvey P. Dale, Aug 30 2016 *)
  • PARI
    Vec(3*x*(1+x)*(1+x+x^2+x^4)/((1-x)^2*(1+x+x^2+x^3+x^4)) + O(x^100)) \\ Colin Barker, Aug 25 2016

Formula

A259748(a(n))/a(n) = 2/3.
a(n) = 3*A047584(n). - Michel Marcus, Jul 18 2015
From Colin Barker, Aug 25 2016: (Start)
a(n) = a(n-1)+a(n-5)-a(n-6) for n>6.
G.f.: 3*x*(1+x)*(1+x+x^2+x^4) / ((1-x)^2*(1+x+x^2+x^3+x^4)).
(End)

Extensions

Better name from Danny Rorabaugh, Oct 22 2015
Showing 1-5 of 5 results.

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