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.

A005901 Number of points on surface of cuboctahedron (or icosahedron): a(0) = 1; for n > 0, a(n) = 10n^2 + 2. Also coordination sequence for f.c.c. or A_3 or D_3 lattice.

Original entry on oeis.org

1, 12, 42, 92, 162, 252, 362, 492, 642, 812, 1002, 1212, 1442, 1692, 1962, 2252, 2562, 2892, 3242, 3612, 4002, 4412, 4842, 5292, 5762, 6252, 6762, 7292, 7842, 8412, 9002, 9612, 10242, 10892, 11562, 12252, 12962, 13692, 14442, 15212, 16002
Offset: 0

Views

Author

N. J. A. Sloane, R. Vaughan

Keywords

Comments

Sequence found by reading the segment (1, 12) together with the line from 12, in the direction 12, 42, ..., in the square spiral whose vertices are the generalized heptagonal numbers A085787. - Omar E. Pol, Jul 18 2012

References

  • H. S. M. Coxeter, "Polyhedral numbers," in R. S. Cohen et al., editors, For Dirk Struik. Reidel, Dordrecht, 1974, pp. 25-35.
  • Gmelin Handbook of Inorg. and Organomet. Chem., 8th Ed., 1994, TYPIX search code (225) cF4
  • B. Grünbaum, Uniform tilings of 3-space, Geombinatorics, 4 (1994), 49-56. See tiling #1.
  • R. W. Marks and R. B. Fuller, The Dymaxion World of Buckminster Fuller. Anchor, NY, 1973, p. 46.
  • S. Rosen, Wizard of the Dome: R. Buckminster Fuller; Designer for the Future. Little, Brown, Boston, 1969, p. 109.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums give A005902.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.

Programs

  • Magma
    [n eq 0 select 1 else 2*(5*n^2+1): n in [0..55]]; // G. C. Greubel, May 25 2023
    
  • Mathematica
    Join[{1},10*Range[40]^2+2] (* or *) Join[{1},LinearRecurrence[{3,-3,1},{12,42,92},40]] (* Harvey P. Dale, May 28 2014 *)
  • PARI
    a(n)=if(n<0,0,10*n^2+1+(n>0))
    
  • SageMath
    [2*(5*n^2 + 1)-int(n==0) for n in range(56)] # G. C. Greubel, May 25 2023

Formula

G.f.: (1+x)*(1+8*x+x^2)/(1-x)^3. - Simon Plouffe in his 1992 dissertation
G.f. for coordination sequence for A_n lattice is (1-z)^(-n) * Sum_{i=0..n} binomial(n, i)^2*z^i. [Bacher et al.]
a(n+1) = A027599(n+2) + A092277(n+1) - Creighton Dement, Feb 11 2005
a(n) = 2 + A033583(n), n >= 1. - Omar E. Pol, Jul 18 2012
a(n) = 12 + 24*(n-1) + 8*A000217(n-2) + 6*A000290(n-1). The properties of the cuboctahedron, namely, its number of vertices (12), edges (24), and faces as well as face-type (8 triangles and 6 squares), are involved in this formula. - Peter M. Chema, Mar 26 2017
a(n) = A062786(n) + A062786(n+1). - R. J. Mathar, Feb 28 2018
E.g.f.: -1 + 2*(1 + 5*x + 5*x^2)*exp(x). - G. C. Greubel, May 25 2023
Sum{n>=0} 1/a(n) = 3/4 + Pi*sqrt(5)*coth(Pi/sqrt 5)/20 = 1.14624... - R. J. Mathar, Apr 27 2024

A049450 Pentagonal numbers multiplied by 2: a(n) = n*(3*n-1).

Original entry on oeis.org

0, 2, 10, 24, 44, 70, 102, 140, 184, 234, 290, 352, 420, 494, 574, 660, 752, 850, 954, 1064, 1180, 1302, 1430, 1564, 1704, 1850, 2002, 2160, 2324, 2494, 2670, 2852, 3040, 3234, 3434, 3640, 3852, 4070, 4294, 4524, 4760, 5002, 5250, 5504, 5764
Offset: 0

Views

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

Comments

From Floor van Lamoen, Jul 21 2001: (Start)
Write 1,2,3,4,... in a hexagonal spiral around 0, then a(n) is the sequence found by reading the line from 0 in the direction 0,2,.... The spiral begins:
.
56--55--54--53--52
/ \
57 33--32--31--30 51
/ / \ \
58 34 16--15--14 29 50
/ / / \ \ \
59 35 17 5---4 13 28 49
/ / / / \ \ \ \
60 36 18 6 0 3 12 27 48
/ / / / / . / / / /
61 37 19 7 1---2 11 26 47
\ \ \ \ . / / /
62 38 20 8---9--10 25 46
\ \ \ . / /
63 39 21--22--23--24 45
\ \ . /
64 40--41--42--43--44
\ .
65--66--67--68--69--70
(End)
Starting with offset 1 = binomial transform of [2, 8, 6, 0, 0, 0, ...]. - Gary W. Adamson, Jan 09 2009
Number of possible pawn moves on an (n+1) X (n+1) chessboard (n=>3). - Johannes W. Meijer, Feb 04 2010
a(n) = A069905(6n-1): Number of partitions of 6*n-1 into 3 parts. - Adi Dani, Jun 04 2011
Even octagonal numbers divided by 4. - Omar E. Pol, Aug 19 2011
Partial sums give A011379. - Omar E. Pol, Jan 12 2013
First differences are A016933; second differences equal 6. - Bob Selcoe, Apr 02 2015
For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n-2; {2, 2n-1, 6, 2n-1, 2, 18n-4}]. - Magus K. Chu, Oct 13 2022

Examples

			On a 4 X 4 chessboard pawns at the second row have (3+4+4+3) moves and pawns at the third row have (2+3+3+2) moves so a(3) = 24. - _Johannes W. Meijer_, Feb 04 2010
From _Adi Dani_, Jun 04 2011: (Start)
a(1)=2: the partitions of 6*1-1=5 into 3 parts are [1,1,3] and[1,2,2].
a(2)=10: the partitions of 6*2-1=11 into 3 parts are [1,1,9], [1,2,8], [1,3,7], [1,4,6], [1,5,5], [2,2,7], [2,3,6], [2,4,5], [3,3,5], and [3,4,4].
(End)
.
.                                                         o
.                                                       o o o
.                                      o              o o o o o
.                                    o o o          o o o o o o o
.                       o          o o o o o      o o o o o o o o o
.                     o o o      o o o o o o o    o o o o o o o o o
.            o      o o o o o    o o o o o o o    o o o o o o o o o
.          o o o    o o o o o    o o o o o o o    o o o o o o o o o
.    o     o o o    o o o o o    o o o o o o o    o o o o o o o o o
.    o     o o o    o o o o o    o o o o o o o    o o o o o o o o o
.    2      10         24             44                 70
- _Philippe Deléham_, Mar 30 2013
		

Crossrefs

Cf. A000567.
Bisection of A001859. Cf. A045944, A000326, A033579, A027599, A049451.
Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A035008 (Knight) and A002492 (Bishop).
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488. [Bruno Berselli, Jun 10 2013]
Cf. sequences listed in A254963.

Programs

  • GAP
    List([0..50], n-> n*(3*n-1)); # G. C. Greubel, Aug 31 2019
  • Magma
    [n*(3*n-1) : n in [0..50]]; // Wesley Ivan Hurt, Sep 24 2017
    
  • Maple
    seq(n*(3*n-1),n=0..44); # Zerinvary Lajos, Jun 12 2007
  • Mathematica
    Table[n(3n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,2,10},50] (* Harvey P. Dale, Jun 21 2014 *)
    2*PolygonalNumber[5,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 01 2018 *)
  • PARI
    a(n)=n*(3*n-1) \\ Charles R Greathouse IV, Nov 20 2012
    
  • Sage
    [n*(3*n-1) for n in (0..50)] # G. C. Greubel, Aug 31 2019
    

Formula

O.g.f.: A(x) = 2*x*(1+2*x)/(1-x)^3.
a(n) = A049452(n) - A033428(n). - Zerinvary Lajos, Jun 12 2007
a(n) = 2*A000326(n), twice pentagonal numbers. - Omar E. Pol, May 14 2008
a(n) = A022264(n) - A000217(n). - Reinhard Zumkeller, Oct 09 2008
a(n) = a(n-1) + 6*n - 4 (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n) = A014642(n)/4 = A033579(n)/2. - Omar E. Pol, Aug 19 2011
a(n) = A000290(n) + A000384(n) = A000217(n) + A000566(n). - Omar E. Pol, Jan 11 2013
a(n+1) = A014107(n+2) + A000290(n). - Philippe Deléham, Mar 30 2013
E.g.f.: x*(2 + 3*x)*exp(x). - Vincenzo Librandi, Apr 28 2016
a(n) = (2/3)*A000217(3*n-1). - Bruno Berselli, Feb 13 2017
a(n) = A002061(n) + A056220(n). - Bruce J. Nicholson, Sep 21 2017
From Amiram Eldar, Feb 20 2022: (Start)
Sum_{n>=1} 1/a(n) = 3*log(3)/2 - Pi/(2*sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/sqrt(3) - 2*log(2). (End)
From Leo Tavares, Feb 23 2022: (Start)
a(n) = A003215(n) - A016813(n).
a(n) = 2*A000290(n) + 2*A000217(n-1). (End)

A143689 a(n) = (3*n^2 - n + 2)/2.

Original entry on oeis.org

1, 2, 6, 13, 23, 36, 52, 71, 93, 118, 146, 177, 211, 248, 288, 331, 377, 426, 478, 533, 591, 652, 716, 783, 853, 926, 1002, 1081, 1163, 1248, 1336, 1427, 1521, 1618, 1718, 1821, 1927, 2036, 2148, 2263, 2381, 2502, 2626, 2753, 2883, 3016, 3152, 3291
Offset: 0

Views

Author

Gary W. Adamson, Aug 29 2008

Keywords

Comments

Equals left border of triangle A033292.
Equals binomial transform of [1, 1, 3, 0, 0, 0, ...].
A242357(a(n)) = 1. - Reinhard Zumkeller, May 11 2014
These might be called "trisected pentagonal numbers": A figurate pentagonal number is composed of three triangles, of which the central one is the largest, and the removal of the triangular frame (3*n) of the central triangle trisects the figure. This is reflected in the formula a(n) = A000326(n+1) - 3*n. See illustration in links. - John Elias, May 27 2022

Crossrefs

a(n) = A000326(n+1) - 3n. Third column of A107111.

Programs

Formula

a(n) = A000326(n+1) - 3*n. (A000326 are the pentagonal numbers.)
a(n) = (3*n^2 - n + 2)/2 = A027599(n+1)/2. - R. J. Mathar, Sep 03 2008
a(n) = a(n-1) + 3*n - 2 (with a(0)=1). - Vincenzo Librandi, Nov 25 2010
a(n) = 2*a(n-1) - a(n-2) + 3.
O.g.f.: (1-x+3*x^2)/((1-x)^3). - Eric Werley, Jun 27 2011
a(n) = A104249(-n). - Bruno Berselli, Jul 08 2015
a(n) = binomial(n,2) + n^2 + 1 = A152947(n+1) + A000290(n). - Franck Maminirina Ramaharo, Mar 01 2018
E.g.f.: exp(x)*(2 + 2*x + 3*x^2)/2. - Stefano Spezia, Apr 19 2025

Extensions

Index of A000326 in definition, formula and example corrected by R. J. Mathar, Sep 03 2008
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