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.

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A260708 a(2n) = n*(2*n+1), a(2n+7) = a(2n+1) + 12*n + 28, with a(1)=1, a(3)=6, a(5)=16.

Original entry on oeis.org

0, 1, 3, 6, 10, 16, 21, 29, 36, 46, 55, 68, 78, 93, 105, 122, 136, 156, 171, 193, 210, 234, 253, 280, 300, 329, 351, 382, 406, 440, 465, 501, 528, 566, 595, 636, 666, 709, 741, 786, 820, 868, 903, 953, 990, 1042, 1081, 1136, 1176, 1233, 1275, 1334, 1378
Offset: 0

Views

Author

Paul Curtz, Nov 17 2015

Keywords

Comments

Conjecture: this sequence is 0 followed by A264041.
After 3, if a(n) is prime then n == 1 (mod 6).
a(n) is a square for n = 0, 1, 5, 8, 145, 288, 1777, 6533, 9800, 168097, 332928, 2051425, 7539845, ...

Examples

			a(0) = 0*1 = 0,
a(1) = 1,
a(2) = 1*3 = 3,
a(3) = 6,
a(4) = 2*5 = 10,
a(5) = 16,
a(6) = 3*7 = 21,
a(7) = a(1) +12*0 +28 = 29, etc.

Links

Crossrefs

Cf. A000217, A008615, A014105, A260160, A260699, A264041.

Programs

  • Mathematica
    LinearRecurrence[{1, 1, -1, 0, 0, 1, -1, -1, 1}, {0, 1, 3, 6, 10, 16, 21, 29, 36}, 50] (* Bruno Berselli, Nov 18 2015 *)
  • PARI
    concat(0, Vec(-x*(x^6+x^5+3*x^4+2*x^3+2*x^2+2*x+1)/((x-1)^3*(x+1)^2*(x^2-x+1)*(x^2+x+1)) + O(x^100))) \\ Colin Barker, Nov 18 2015
  • Sage
    [n*(n+1)/2+(1-(-1)^n)*floor(n/6+1/3)/2 for n in (0..60)] # Bruno Berselli, Nov 18 2015

Formula

From Colin Barker, Nov 17 2015: (Start)
G.f.: x*(1 + 2*x + 2*x^2 + 2*x^3 + 3*x^4 + x^5 + x^6) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2)*(1 + x + x^2)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-6) - a(n-7) - a(n-8) + a(n-9) for n>8. (End)
a(2*k) = A000217(2*k) by definition; for odd indices:
a(6*k+1) = 18*k^2 + 10*k + 1,
a(6*k+3) = 2*(9*k^2 + 11*k + 3),
a(6*k+5) = 2*(k + 1)*(9*k + 8), that is A178574.
a(n) = A260699(n) + A008615(n).
a(n) = n*(n + 1)/2 + (1 - (-1)^n)*floor(n/6 + 1/3)/2. [Bruno Berselli, Nov 18 2015]

Extensions

Edited by Bruno Berselli, Nov 18 2015

A183110 Period-length of the ultimate periodic behavior of the orbit of a list [1,1,1,...,1] of n 1's under the mapping defined in the comments.

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 4, 4, 1, 5, 5, 5, 5, 1, 6, 6, 6, 6, 6, 1, 7, 7, 7, 7, 7, 7, 1, 8, 8, 8, 8, 8, 8, 8, 1, 9, 9, 9, 9, 9, 9, 9, 9, 1, 10, 10, 10, 10, 10, 10, 10, 10, 10, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 1, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 1, 13, 13
Offset: 1

Views

Author

John W. Layman, Feb 01 2011

Keywords

Comments

We use the list mapping introduced in A092964, whereby one removes the first term of the list, z(1), and adds 1 to each of the next z(1) terms (appending 1's if necessary) to get a new list.
This is also conjectured to be the length of the longest cycle of pebble-moves among the partitions of n (cf. A201144). - Andrew V. Sutherland

Examples

			Under the indicated mapping the list [1,1,1,1,1,1,1] of seven 1's results in the orbit [1,1,1,1,1,1,1], [2,1,1,1,1,1], [2,2,1,1,1], [3,2,1,1], [3,2,2], [3,3,1], [4,2,1], [3,2,1,1], ... which is clearly periodic with period-length 4, so a(7) = 4.

Links

Crossrefs

Cf. A037306, A092964, A178572, A178574.

Programs

  • Mathematica
    f[p_] := Module[{pp, x, lpp, m, i}, pp = p; x = pp[[1]]; pp = Delete[pp,1]; lpp = Length[pp]; m = Min[x, lpp]; For[i = 1, i ≤ m, i++, pp[[i]]++]; For[i = 1, i ≤ x - lpp, i++, AppendTo[pp, 1]]; pp]; orb[p_] := Module[{s, v}, v = p; s = {v}; While[! MemberQ[s, v = f[v]], AppendTo[s, v]]; s]; attractor[p_] := Module[{orbp, pos, len, per}, orbp = orb[p]; pos = Flatten[Position[orbp, f[orbp[[-1]]]]][[1]] - 1; (*pos = steps to enter period*) len = Length[orbp] - pos; per = Take[orbp, -len]; Sort[per]]; a = {}; For[n = 1, n ≤ 80, n++, {rn = Table[1, {k, 1, n}]; orbn = orb[rn]; lenorb = Length[orbn]; lenattr = Length[attractor[rn]]; AppendTo[a, lenattr]}]; Print[a];

Formula

It appears, but has not yet been proved, that a(n)=1 if n=t(k) and a(n)=k if t(k-1) < n < t(k) where t(k) is the k-th triangular number t(k) = k*(k+1)/2.
Showing 1-2 of 2 results.

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