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 1-3 of 3 results.

A188135 a(n) = 8*n^2 + 2*n + 1.

Original entry on oeis.org

1, 11, 37, 79, 137, 211, 301, 407, 529, 667, 821, 991, 1177, 1379, 1597, 1831, 2081, 2347, 2629, 2927, 3241, 3571, 3917, 4279, 4657, 5051, 5461, 5887, 6329, 6787, 7261, 7751, 8257, 8779, 9317, 9871, 10441, 11027, 11629, 12247, 12881, 13531, 14197, 14879, 15577, 16291, 17021, 17767
Offset: 0

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Author

Paul Curtz, Mar 30 2011

Keywords

Comments

Bisection of A193867. - Omar E. Pol, Aug 16 2011
Sequence found by reading the line from 1, in the direction 1, 11, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 04 2011

Crossrefs

Programs

Formula

First differences: a(n) - a(n-1) = 16*n - 6 = A113770(n) = 2*A004770(n).
Second differences: a(n) - 2*a(n-1) + a(n-2) = 16.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From R. J. Mathar, Apr 06 2011: (Start)
G.f.: -(1+x)*(7*x+1)/(x-1)^3.
a(n) = A084849(2*n). (End)
E.g.f.: exp(x)*(1 + 10*x + 8*x^2). - Elmo R. Oliveira, Oct 19 2024

Extensions

a(41)-a(47) from Elmo R. Oliveira, Oct 19 2024

A193867 Odd central polygonal numbers.

Original entry on oeis.org

1, 7, 11, 29, 37, 67, 79, 121, 137, 191, 211, 277, 301, 379, 407, 497, 529, 631, 667, 781, 821, 947, 991, 1129, 1177, 1327, 1379, 1541, 1597, 1771, 1831, 2017, 2081, 2279, 2347, 2557, 2629, 2851, 2927, 3161, 3241, 3487, 3571, 3829, 3917, 4187, 4279
Offset: 1

Views

Author

Omar E. Pol, Aug 15 2011

Keywords

Comments

Even triangular numbers plus 1.
Union of A188135 and A185438 without repetitions (A188135 is a bisection of this sequence. Another bisection is A185438 but without its initial term).

Crossrefs

Programs

  • Mathematica
    Select[Accumulate[Range[0,100]],EvenQ]+1 (* or *) LinearRecurrence[{1,2,-2,-1,1},{1,7,11,29,37},50] (* Harvey P. Dale, Nov 29 2014 *)
  • PARI
    Vec(-x*(x^2+1)*(x^2+6*x+1) / ((1+x)^2*(x-1)^3) + O(x^100)) \\ Colin Barker, Jan 27 2016

Formula

a(n) = A000124(A014601(n-1)).
a(n) = 1 + A014494(n-1).
G.f.: -x*(x^2+1)*(x^2+6*x+1) / ( (1+x)^2*(x-1)^3 ). - R. J. Mathar, Aug 25 2011
From Colin Barker, Jan 27 2016: (Start)
a(n) = (4*n^2+2*(-1)^n*n-4*n-(-1)^n+3)/2.
a(n) = 2*n^2-n+1 for n even.
a(n) = 2*n^2-3*n+2 for n odd. (End)
Sum_{n>=1} 1/a(n) = 2*Pi*sinh(sqrt(7)*Pi/4)/(sqrt(7)*(2*cosh(sqrt(7)*Pi/4) - sqrt(2))). - Amiram Eldar, May 11 2025

A241807 Numerators of c(n) = (n^2+n+2)/((n+1)*(n+2)*(n+3)) as defined in A241269.

Original entry on oeis.org

1, 1, 2, 7, 11, 2, 11, 29, 37, 23, 28, 67, 79, 23, 53, 121, 137, 77, 86, 191, 211, 29, 127, 277, 301, 163, 176, 379, 407, 109, 233, 497, 529, 281, 298, 631, 667, 88, 371, 781, 821, 431, 452, 947, 991, 259, 541, 1129, 1177, 613, 638
Offset: 0

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Author

Keywords

Comments

The subsequence 1, 23, 77, 163, 281, 431, 613, 827, ..., with indices congruent to 1 mod 8, is 16n^2+6n+1, that is, A000124(8n+1)/2 or A014206(8n+1)/4. Its second differences are constant: (16n^2+6n+1)'' = 32.
The sequence A014206/A241807 is integral and consists of the 16-periodic sequence (2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2, ...).

Examples

			1/3, 1/6, 2/15, 7/60, 11/105, 2/21, 11/126, 29/360, 37/495, 23/330, ...
		

Crossrefs

Programs

  • Mathematica
    Table[(n^2+n+2)/((n+1)*(n+2)*(n+3)) // Numerator, {n, 0, 50}]

Formula

a(n) = A014206(n)/period 16: repeat 2, 4, 4, 2, 2, 16, 4, 2, 2, 4, 4, 2, 2, 8, 4, 2 (conjectured).
a(4k) = 8*k^2 +2*k +1,
a(4k+2) = 4*k^2 +5*k +2,
a(4k+3) = 8*k^2 +14*k +7,
a(8k+1) = 16*k^2 +6*k +1,
a(16k+5) = 16*k^2 +11*k +2,
a(16k+13) = 32*k^2 + 54*k +23.
Showing 1-3 of 3 results.