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

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

Author

Joe Keane (jgk(AT)jgk.org)

Keywords

• nonn
• easy
• nice

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
		

Links

• Ivan Panchenko, Table of n, a(n) for n = 0..1000
• Richard P. Brent, Generalising Tuenter's binomial sums, arXiv:1407.3533 [math.CO], 2014. (page 16)
• Leo Tavares, Illustration: X Hexagons
• Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

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.

Cf. A003215, A016813, A000290, A000217.

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)