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.

Previous Showing 11-14 of 14 results.

A195045 Concentric 13-gonal numbers.

Original entry on oeis.org

0, 1, 13, 27, 52, 79, 117, 157, 208, 261, 325, 391, 468, 547, 637, 729, 832, 937, 1053, 1171, 1300, 1431, 1573, 1717, 1872, 2029, 2197, 2367, 2548, 2731, 2925, 3121, 3328, 3537, 3757, 3979, 4212, 4447, 4693, 4941, 5200, 5461, 5733, 6007, 6292, 6579, 6877, 7177, 7488, 7801, 8125
Offset: 0

Views

Author

Omar E. Pol, Sep 27 2011

Keywords

Comments

Also concentric tridecagonal numbers or concentric triskaidecagonal numbers.
Partial sums of A175886. - Reinhard Zumkeller, Jan 07 2012

Crossrefs

Programs

Formula

a(n) = 13*n^2/4+9*((-1)^n-1)/8.
From R. J. Mathar, Sep 28 2011: (Start)
G.f.: -x*(1+11*x+x^2) / ( (1+x)*(x-1)^3 ).
a(n)+a(n+1) = A069126(n+1). (End)
a(n) = 2*a(n-1)-2*a(n-3)+a(n-4) for n>3. - Wesley Ivan Hurt, Nov 22 2015
Sum_{n>=1} 1/a(n) = Pi^2/78 + tan(3*Pi/(2*sqrt(13)))*Pi/(3*sqrt(13)). - Amiram Eldar, Jan 16 2023

A195046 Concentric 15-gonal numbers.

Original entry on oeis.org

0, 1, 15, 31, 60, 91, 135, 181, 240, 301, 375, 451, 540, 631, 735, 841, 960, 1081, 1215, 1351, 1500, 1651, 1815, 1981, 2160, 2341, 2535, 2731, 2940, 3151, 3375, 3601, 3840, 4081, 4335, 4591, 4860, 5131, 5415, 5701, 6000, 6301, 6615, 6931, 7260, 7591
Offset: 0

Views

Author

Omar E. Pol, Sep 27 2011

Keywords

Crossrefs

Programs

  • Mathematica
    Table[15n^2/4+11((-1)^n-1)/8,{n,0,50}] (* or *) LinearRecurrence[ {2,0,-2,1},{0,1,15,31},50] (* Harvey P. Dale, Feb 23 2012 *)
  • PARI
    a(n)=15*n^2/4+11*((-1)^n-1)/8 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = 15*n^2/4+11*((-1)^n-1)/8.
From Harvey P. Dale, Feb 23 2012: (Start)
a(0)=0, a(1)=1, a(2)=15, a(3)=31, a(n)=2*a(n-1)-2*a(n-3)+a(n-4).
G.f.: -((x*(1+x*(13+x)))/((-1+x)^3*(1+x))). (End)
Sum_{n>=1} 1/a(n) = Pi^2/90 + tan(sqrt(11/15)*Pi/2)*Pi/sqrt(165). - Amiram Eldar, Jan 16 2023

Extensions

a(1)=1 added by Harvey P. Dale, Feb 23 2012

A195314 Centered 28-gonal numbers.

Original entry on oeis.org

1, 29, 85, 169, 281, 421, 589, 785, 1009, 1261, 1541, 1849, 2185, 2549, 2941, 3361, 3809, 4285, 4789, 5321, 5881, 6469, 7085, 7729, 8401, 9101, 9829, 10585, 11369, 12181, 13021, 13889, 14785, 15709, 16661, 17641, 18649, 19685, 20749, 21841, 22961, 24109, 25285, 26489
Offset: 1

Views

Author

Omar E. Pol, Sep 16 2011

Keywords

Comments

Sequence found by reading the line from 1, in the direction 1, 29, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Semi-axis opposite to A144555 in the same spiral.

Crossrefs

Programs

  • Magma
    [(14*n^2-14*n+1): n in [1..50]]; // Vincenzo Librandi, Sep 19 2011
    
  • Mathematica
    Table[14n^2-14n+1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,29,85},50]
  • PARI
    a(n)=14*n^2-14*n+1 \\ Charles R Greathouse IV, Oct 07 2015

Formula

a(n) = 14*n^2 - 14*n + 1.
G.f.: -x*(1 + 26*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 18 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Oct 01 2011
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(5/7)*Pi/2)/(2*sqrt(35)). - Amiram Eldar, Feb 11 2022
From Elmo R. Oliveira, Nov 14 2024: (Start)
E.g.f.: exp(x)*(14*x^2 + 1) - 1.
a(n) = 2*A069127(n) - 1. (End)

A270693 Alternating sum of centered 25-gonal numbers.

Original entry on oeis.org

1, -25, 51, -100, 151, -225, 301, -400, 501, -625, 751, -900, 1051, -1225, 1401, -1600, 1801, -2025, 2251, -2500, 2751, -3025, 3301, -3600, 3901, -4225, 4551, -4900, 5251, -5625, 6001, -6400, 6801, -7225, 7651, -8100, 8551, -9025, 9501, -10000, 10501
Offset: 0

Views

Author

Ilya Gutkovskiy, Mar 21 2016

Keywords

Comments

The absolute value alternating sum of centered k-gonal numbers gives concentric k-gonal numbers.
More generally, the ordinary generating function for the alternating sum of centered k-gonal numbers is (1 - (k - 2)*x + x^2)/((1 - x)*(1 + x)^3).

Crossrefs

Programs

  • Magma
    [((-1)^n*(50*n^2 + 100*n + 29) - 21)/8 : n in [0..40]]; // Wesley Ivan Hurt, Mar 21 2016
  • Maple
    A270693:=n->((-1)^n*(50*n^2 + 100*n + 29) - 21)/8: seq(A270693(n), n=0..100); # Wesley Ivan Hurt, Sep 18 2017
  • Mathematica
    LinearRecurrence[{-2, 0, 2, 1}, {1, -25, 51, -100}, 41]
    Table[((-1)^n (50 n^2 + 100 n + 29) - 21)/8, {n, 0, 40}]
  • PARI
    x='x+O('x^100); Vec((1-23*x+x^2)/((1-x)*(1+x)^3)) \\ Altug Alkan, Mar 21 2016
    

Formula

G.f.: (1 - 23*x + x^2)/((1 - x)*(1 + x)^3).
E.g.f.: (1/8)*(-21*exp(x) + (29 - 150*x + 50*x^2)*exp(-x)).
a(n) = -2*a(n-1) + 2*a(n-3) + a(n-4).
a(n) = ((-1)^n*(50*n^2 + 100*n + 29) - 21)/8.
Previous Showing 11-14 of 14 results.