A069129 -id:A069129 - cp's OEIS frontend

A275543 A081585 and A069129 interleaved.

1, 1, 9, 17, 33, 49, 73, 97, 129, 161, 201, 241, 289, 337, 393, 449, 513, 577, 649, 721, 801, 881, 969, 1057, 1153, 1249, 1353, 1457, 1569, 1681, 1801, 1921, 2049, 2177, 2313, 2449, 2593, 2737, 2889, 3041, 3201, 3361, 3529, 3697, 3873, 4049, 4233, 4417, 4609

Offset: 0

Daniel Poveda Parrilla, Aug 01 2016

a(A000129(n)) is a square.
(n^2)*a(n) = A275496(n) which is a triangular number.
(A000129(n)^2)*a(A000129(n)) = A275496(A000129(n)) = A001110(n) which is a square triangular number.
a(2n+1)/a(2n) is convergent to 1.

			a(1) = A275496(1) = 1.
a(5) = A275496(5)/25 = 1225/25 = 49.
a(7) = A275496(7)/49 = 4753/49 = 97.
a(12) = A275496(12)/144 = 41616/144 = 289.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..500000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A081585(n) = a(2n), A069129(n) = a(2n + 1).

Cf. A000129, A000217, A000290, A001110, A077221, A093954, A275496.

CoefficientList[Series[(1 - x + 7 x^2 + x^3)/((1 - x)^3 (1 + x)), {x, 0, 48}], x] (* or as defined *)
Riffle[LinearRecurrence[{3, -3, 1}, {1, 9, 33}, #], FoldList[#1 + #2 &, 1, 16 Range@ #]] &@ 25 (* Michael De Vlieger, Aug 01 2016, after Vincenzo Librandi at A081585 and Robert G. Wilson v at A069129 *)

a(n)=(-1)^n + 2*n^2 \\ Charles R Greathouse IV, Aug 03 2016

Vec((1-x+7*x^2+x^3)/((1-x)^3*(1+x)) + O(x^100)) \\ Colin Barker, Aug 21 2016

a(0) = 1; a(n) = A275496(n)/(n^2) for n > 0.
From Colin Barker, Aug 01 2016: (Start)
a(n) = (2*n^2 + (-1)^n).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 3.
G.f.: (1 -x +7*x^2 +x^3) / ((1 - x)^3*(1 + x)).
(End)
From Daniel Poveda Parrilla, Aug 18 2016: (Start)
a(2n) = A077221(2n) + 1.
a(2n + 1) = A077221(2n + 1). (End)
Sum_{n>=0} 1/a(n) = (1 + (tan(c) + coth(c))*c)/2, where c = Pi/(2*sqrt(2)) is A093954. - Amiram Eldar, Aug 21 2022

A035008 Total number of possible knight moves on an (n+2) X (n+2) chessboard, if the knight is placed anywhere.

Original entry on oeis.org

0, 16, 48, 96, 160, 240, 336, 448, 576, 720, 880, 1056, 1248, 1456, 1680, 1920, 2176, 2448, 2736, 3040, 3360, 3696, 4048, 4416, 4800, 5200, 5616, 6048, 6496, 6960, 7440, 7936, 8448, 8976, 9520, 10080, 10656, 11248, 11856, 12480, 13120, 13776

Offset: 0

Ulrich Schimke (ulrschimke(AT)aol.com), Dec 11 1999

16 times the triangular numbers A000217.
Centered 16-gonal numbers A069129, minus 1. Also, sequence found by reading the segment (0, 16) together with the line from 16, in the direction 16, 48, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Apr 26 2008, Nov 20 2008
For n >= 1, number of permutations of n+1 objects selected from 5 objects v, w, x, y, z with repetition allowed, containing n-1 v's. Examples: at n=1, n-1=0 (i.e., zero v's), and a(1)=16 because we have ww, wx, wy, wz, xw, xx, xy, xz, yw, yx, yy, yz, zw, zx, zy, zz; at n=2, n-1=1 (i.e., one v), and there are 3 permutations corresponding to each one in the n=1 case (e.g., the single v can be inserted in any of three places in the 2-object permutation xy, yielding vxy, xvy, and xyv), so a(2) = 3*a(1) = 3*16 = 48; at n=3, n-1=2 (i.e., two v's), and a(3) = C(4,2)*a(1) = 6*16 = 96; etc. - Zerinvary Lajos, Aug 07 2008 (this needs clarification, Joerg Arndt, Feb 23 2014)
Sequence found by reading the line from 0, in the direction 0, 16, ... and the same line from 0, in the direction 0, 48, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - Omar E. Pol, Oct 03 2011
For n > 0, a(n) is the area of the triangle with vertices at ((n-1)^2, n^2), ((n+1)^2, (n+2)^2), and ((n+3)^2, (n+2)^2). - J. M. Bergot, May 22 2014
For n > 0, a(n) is the number of self-intersecting points in star polygon {4*(n+1)/(2*n+1)}. - Bui Quang Tuan, Mar 28 2015
Equivalently: integers k such that k$ / (k/2)! and k$ / (k/2+1)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information). - Bernard Schott, Dec 02 2021

			3 X 3-Board: knight can be placed in 8 positions with 2 moves from each, so a(1) = 16.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Eric Weisstein's World of Mathematics, Star Polygon.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A002492 (Bishop) and A049450 (Pawn).
Cf. A000217, A069129, A027468, A008586 A038231, A002378, A033996, A046092.
Cf. A348692.
Subsequence of A008586 and of A349081.

[8*n*(n+1): n in [0..50]]; // Wesley Ivan Hurt, May 22 2014

seq(binomial(n+1,2)*4^2, n=0..33); # Zerinvary Lajos, Aug 07 2008

CoefficientList[Series[16 x/(1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Feb 24 2014 *)
LinearRecurrence[{3,-3,1},{0,16,48},50] (* or *) 16*Accumulate[ Range[ 0,50]] (* Harvey P. Dale, Aug 05 2018 *)

a(n)=8*n*(n+1) \\ Charles R Greathouse IV, Sep 30 2015

a(n) = 8*n*(n+1).
G.f.: 16*x/(1-x)^3.
a(n) = A069129(n+1) - 1. - Omar E. Pol, Apr 26 2008
a(n) = binomial(n+1,2)*4^2, n >= 0. - Zerinvary Lajos, Aug 07 2008
a(n) = 8*n^2 + 8*n = 16*A000217(n) = 8*A002378(n) = 4*A046092(n) = 2*A033996(n). - Omar E. Pol, Dec 12 2008
a(n) = a(n-1) + 16*n, with a(0)=0. - Vincenzo Librandi, Nov 17 2010
E.g.f.: 8*exp(x)*x*(2 + x). - Stefano Spezia, May 19 2021
From Amiram Eldar, Feb 22 2023: (Start)
Sum_{n>=1} 1/a(n) = 1/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2*log(2) - 1)/8.
Product_{n>=1} (1 - 1/a(n)) = -(8/Pi)*cos(sqrt(3/2)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (8/Pi)*cos(Pi/(2*sqrt(2))). (End)

More terms from Erich Friedman

Minor errors corrected and edited by Johannes W. Meijer, Feb 04 2010

A077221 a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square.

Original entry on oeis.org

0, 1, 8, 17, 32, 49, 72, 97, 128, 161, 200, 241, 288, 337, 392, 449, 512, 577, 648, 721, 800, 881, 968, 1057, 1152, 1249, 1352, 1457, 1568, 1681, 1800, 1921, 2048, 2177, 2312, 2449, 2592, 2737, 2888, 3041, 3200, 3361, 3528, 3697, 3872, 4049, 4232

Offset: 0

Amarnath Murthy, Nov 03 2002

This sequence arises from reading the line from 0, in the direction 0, 1, ... and the same line from 0, in the direction 0, 8, ..., in the square spiral whose vertices are the triangular numbers A000217. Cf. A139591, etc. - Omar E. Pol, May 03 2008
The general formula for alternating sums of powers of odd integers is in terms of the Swiss-Knife polynomials P(n,x) A153641 (P(n,0)-(-1)^k*P(n,2*k))/2. Here n=2, thus a(k) = |(P(2,0)-(-1)^k*P(2,2*k))/2|. - Peter Luschny, Jul 12 2009
Axis perpendicular to A046092 in the square spiral whose vertices are the triangular numbers A000217. See the comment above. - Omar E. Pol, Sep 14 2011
Column 8 of A195040. - Omar E. Pol, Sep 28 2011
Concentric octagonal numbers. A139098 and A069129 interleaved. - Omar E. Pol, Sep 17 2011
Subsequence of A194274. - Bruno Berselli, Sep 22 2011
Partial sums of A047522. - Reinhard Zumkeller, Jan 07 2012
Alternating sum of the first n odd squares in decreasing order, n >= 1. Also number of "ON" cells at n-th stage in simple 2-dimensional cellular automaton. The rules are: on the infinite square grid, start with all cells OFF, so a(0) = 0. Turn a single cell to the ON state, so a(1) = 1. At each subsequent step, the neighbor cells of each cell of the old generation are turned ON, and the cells of the old generation are turned OFF. Here "neighbor" refers to the eight adjacent cells of each ON cell. See example. - Omar E. Pol, Feb 16 2014

			From _Omar E. Pol_, Feb 16 2014: (Start)
Illustration of initial terms as a cellular automaton:
.
.                                   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
.
.     1       8           17              32
.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Bruno Berselli, An origin of A077221 (illustration) (see Pol's comment).
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS.
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Index entries for sequences related to cellular automata.

Cf. A032528, A069129, A077222, A131925, A139098, A194274, A195040, A195041, A195042, A195142.

a077221 n = a077221_list !! n
a077221_list = scanl (+) 0 a047522_list
-- Reinhard Zumkeller, Jan 07 2012

[2*n^2 - (n mod 2): n in [0..50]]; // Vincenzo Librandi, Sep 22 2011

a := n -> 2*n^2 - (n mod 2); # Peter Luschny, Jul 12 2009

a=1;lst={a};Do[b=n^2-a;AppendTo[lst,b];a=b,{n,3,6!,2}];lst (* Vladimir Joseph Stephan Orlovsky, May 18 2009 *)

a(2n) = 8*n^2, a(2n+1) = 8*n(n+1) + 1.
From Ralf Stephan, Mar 31 2003: (Start)
a(n) = 2*n^2 + 4*n + 1 [+1 if n is odd] with a(0)=1.
G.f.: x*(x^2+6*x+1)/(1-x)^3/(1+x). (End)
Row sums of triangle A131925; binomial transform of (1, 7, 2, 4, -8, 16, -32, ...). - Gary W. Adamson, Jul 29 2007
a(n) = a(-n); a(n+1) = A195605(n) - (-1)^n. - Bruno Berselli, Sep 22 2011
a(n) = 2*n^2 + ((-1)^n-1)/2. - Omar E. Pol, Sep 28 2011
Sum_{n>=1} 1/a(n) = Pi^2/48 + tan(Pi/(2*sqrt(2)))*Pi /(4*sqrt(2)). - Amiram Eldar, Jan 16 2023

Extended by Ralf Stephan, Mar 31 2003

A139098 a(n) = 8*n^2.

Original entry on oeis.org

0, 8, 32, 72, 128, 200, 288, 392, 512, 648, 800, 968, 1152, 1352, 1568, 1800, 2048, 2312, 2592, 2888, 3200, 3528, 3872, 4232, 4608, 5000, 5408, 5832, 6272, 6728, 7200, 7688, 8192, 8712, 9248, 9800, 10368, 10952, 11552, 12168, 12800, 13448, 14112, 14792, 15488, 16200

Offset: 0

Omar E. Pol, Apr 25 2008

Opposite numbers to the centered 16-gonal numbers (A069129) in the square spiral whose vertices are the triangular numbers (A000217).
8 times the squares. - Omar E. Pol, Dec 09 2008
a(n-1) is the molecular topological index of the n-wheel graph W_n. - Eric W. Weisstein, Jul 11 2011
An n X n pandiagonal magic square has a(n) orientations. - Kausthub Gudipati, Sep 15 2011
Area of a square with diagonal 4n. - Wesley Ivan Hurt, Jun 19 2014
Sum of all the parts in the partitions of 4n into exactly two parts. - Wesley Ivan Hurt, Jul 23 2014
Equivalently: integers k such that k$ / (k/2-1)! and k$ / (k/2)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information). - Bernard Schott, Dec 02 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..800
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Eric Weisstein's World of Mathematics, Molecular Topological Index.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Subsequence of A008586 and of A349081.
Cf. A000217, A000290, A000178, A001105, A016742, A016766, A033582, A069129, A245058.
Cf. A008590, A348692.

[8*n^2: n in [0..50]]; // Vincenzo Librandi, Apr 26 2011

A139098:=n->8*n^2; seq(A139098(n), n=0..50); # Wesley Ivan Hurt, Jun 19 2014

8 Range[0, 50]^2 (* Wesley Ivan Hurt, Jun 19 2014 *)
LinearRecurrence[{3,-3,1},{0,8,32},50] (* Harvey P. Dale, Oct 05 2023 *)

a(n)=8*n^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*A000290(n) = 4*A001105(n) = 2*A016742(n). - Omar E. Pol, Dec 13 2008
G.f.: -8*x*(1+x)/(x-1)^3. - R. J. Mathar, Nov 27 2015
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/48 (A245058).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/96.
Product_{n>=1} (1 + 1/a(n)) = sqrt(8)*sinh(Pi/sqrt(8))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(8)*sin(Pi/sqrt(8))/Pi. (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Wesley Ivan Hurt, Dec 03 2021
From Elmo R. Oliveira, Dec 01 2024: (Start)
E.g.f.: 8*x*(1 + x)*exp(x).
a(n) = n*A008590(n) = A001105(2*n). (End)

A051870 18-gonal (or octadecagonal) numbers: a(n) = n(8n-7).

Original entry on oeis.org

0, 1, 18, 51, 100, 165, 246, 343, 456, 585, 730, 891, 1068, 1261, 1470, 1695, 1936, 2193, 2466, 2755, 3060, 3381, 3718, 4071, 4440, 4825, 5226, 5643, 6076, 6525, 6990, 7471, 7968, 8481, 9010, 9555, 10116, 10693, 11286, 11895, 12520

Offset: 0

N. J. A. Sloane, Dec 15 1999

Also, sequence found by reading the segment (0, 1) together with the line from 1, in the direction 1, 18, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Apr 26 2008
This sequence does not contain any triangular numbers other than 0 and 1. See A188892. - T. D. Noe, Apr 13 2011
Also sequence found by reading the line from 0, in the direction 0, 18, ... and the parallel line from 1, in the direction 1, 51, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - Omar E. Pol, Jul 18 2012
Partial sums of 16n + 1 (with offset 0), compare A005570. - Jeremy Gardiner, Aug 04 2012
All x values for Diophantine equation 32*x + 49 = y^2 are given by this sequence and A139278. - Bruno Berselli, Nov 11 2014
This is also a star enneagonal number: a(n) = A001106(n) + 9*A000217(n-1). - Luciano Ancora, Mar 30 2015

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 189.
Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing, 2012, page 6.

Jeremy Gardiner, Table of n, a(n) for n = 0..999
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Wikipedia, Polygonal number.
Index to sequences related to polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A014634, A014635, A033585, A033586, A033587, A035008, A069129, A085250, A129271-A129278, A139278.

A051870 := proc(n) n*(8*n-7) ; end proc: seq(A051870(n),n=0..30) ; # R. J. Mathar, Feb 05 2011

Table[n (8 n - 7), {n, 0, 40}] (* Bruno Berselli, Nov 11 2014 *)

a(n)=n*(8*n-7) \\ Charles R Greathouse IV, Jul 19 2011

G.f.: x*(1+15*x)/(1-x)^3. - Bruno Berselli, Feb 04 2011
a(n) = 16*n + a(n-1) - 15, with n > 0, a(0) = 0. - Vincenzo Librandi, Aug 06 2010
a(16*a(n)+121*n+1) = a(16*a(n)+121*n) + a(16*n+1). - Vladimir Shevelev, Jan 24 2014
E.g.f.: (8*x^2 + x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = ((1+sqrt(2))*Pi + 2*sqrt(2)*arccoth(sqrt(2)) + 8*log(2))/14. - Amiram Eldar, Oct 20 2020
Product_{n>=2} (1 - 1/a(n)) = 8/9. - Amiram Eldar, Jan 22 2021

A101321 Table T(n,m) = 1 + nm(m+1)/2 read by antidiagonals: centered polygonal numbers.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 7, 4, 1, 1, 11, 13, 10, 5, 1, 1, 16, 21, 19, 13, 6, 1, 1, 22, 31, 31, 25, 16, 7, 1, 1, 29, 43, 46, 41, 31, 19, 8, 1, 1, 37, 57, 64, 61, 51, 37, 22, 9, 1, 1, 46, 73, 85, 85, 76, 61, 43, 25, 10, 1, 1, 56, 91, 109, 113, 106, 91, 71, 49, 28, 11, 1, 1, 67

Offset: 0

Eugene McDonnell (eemcd(AT)mac.com), Dec 24 2004

Row n gives the centered figurate numbers of the n-gon.

Antidiagonal sums are in A101338.

			The upper left corner of the infinite array T is
|0| 1   1   1   1   1   1   1   1   1   1 ... A000012
|1| 1   2   4   7  11  16  22  29  37  46 ... A000124
|2| 1   3   7  13  21  31  43  57  73  91 ... A002061
|3| 1   4  10  19  31  46  64  85 109 136 ... A005448
|4| 1   5  13  25  41  61  85 113 145 181 ... A001844
|5| 1   6  16  31  51  76 106 141 181 226 ... A005891
|6| 1   7  19  37  61  91 127 169 217 271 ... A003215
|7| 1   8  22  43  71 106 148 197 253 316 ... A069099
|8| 1   9  25  49  81 121 169 225 289 361 ... A016754
|9| 1  10  28  55  91 136 190 253 325 406 ... A060544

Cf. A000217, A001263, A101338.

A101321 := proc(n,k)
        n*k*(k+1)/2+1 ;
end proc: # R. J. Mathar, Jul 28 2016

T[n_, m_] := 1 + n m (m + 1)/2;
Table[T[n - m, m], {n, 0, 12}, {m, n, 0, -1}] // Flatten (* Jean-François Alcover, Mar 23 2020 *)

T(n,m) = 1 + n*m*(m+1)/2 \\ Charles R Greathouse IV, Jul 28 2016

T(n,2) = A016777(n). T(n,3) = A016921(n). T(n,4) = A017281(n).
T(10,m) = A062786(m+1).
T(11,m) = A069125(m+1).
T(12,m) = A003154(m+1).
T(13,m) = A069126(m+1).
T(14,m) = A069127(m+1).
T(15,m) = A069128(m+1).
T(16,m) = A069129(m+1).
T(17,m) = A069130(m+1).
T(18,m) = A069131(m+1).
T(19,m) = A069132(m+1).
T(20,m) = A069133(m+1).
T(n+1,m) = T(n,m) + m*(m+1)/2. - Gary W. Adamson and Michel Marcus, Oct 13 2015

Edited by R. J. Mathar, Oct 21 2009

A069190 Centered 24-gonal numbers.

Original entry on oeis.org

1, 25, 73, 145, 241, 361, 505, 673, 865, 1081, 1321, 1585, 1873, 2185, 2521, 2881, 3265, 3673, 4105, 4561, 5041, 5545, 6073, 6625, 7201, 7801, 8425, 9073, 9745, 10441, 11161, 11905, 12673, 13465, 14281, 15121, 15985, 16873, 17785, 18721, 19681, 20665, 21673

Offset: 1

Terrel Trotter, Jr., Apr 10 2002

Sequence found by reading the line from 1, in the direction 1, 25, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. Semi-axis opposite to A135453 in the same spiral. - Omar E. Pol, Sep 16 2011

			a(5) = 241 because 12*5^2 - 12*5 + 1 = 300 - 60 + 1 = 241.

Ivan Panchenko, Table of n, a(n) for n = 1..1000
John Elias, Illustration: Odd Ordered Star Perimeters.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
R. Yin, J. Mu, and T. Komatsu, The p-Frobenius Number for the Triple of the Generalized Star Numbers, Preprints 2024, 2024072280. See p. 2.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. centered k-gonal numbers with k=3..25: A005448, A001844, A005891, A003215, A069099, A016754, A060544, A062786, A069125, A003154, A069126, A069127, A069128, A069129, A069130, A069131, A069132, A069133, A069178, A069173, A069174, A069190, A262221.

FoldList[#1 + #2 &, 1, 24 Range@ 45] (* Robert G. Wilson v *)
Table[12n^2-12n+1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,25,73},50] (* Harvey P. Dale, Jul 17 2011 *)

a(n)=12*n^2-12*n+1 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 12*n^2 - 12*n + 1.
a(n) = 24*n + a(n-1) - 24 with a(1)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=1, a(2)=25, a(3)=73. - Harvey P. Dale, Jul 17 2011
G.f.: x*(1+22*x+x^2)/(1-x)^3. - Harvey P. Dale, Jul 17 2011
Binomial transform of [1, 24, 24, 0, 0, 0, ...] and Narayana transform (cf. A001263) of [1, 24, 0, 0, 0, ...]. - Gary W. Adamson, Jul 26 2011
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} 1/a(n) = Pi*tan(Pi/sqrt(6))/(4*sqrt(6)).
Sum_{n>=1} a(n)/n! = 13*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 13/e - 1. (End)
E.g.f.: exp(x)*(1 + 12*x^2) - 1. - Stefano Spezia, May 31 2022

More terms from Harvey P. Dale, Jul 17 2011

A077591 Maximum number of regions into which the plane can be divided using n (concave) quadrilaterals.

Original entry on oeis.org

1, 2, 18, 50, 98, 162, 242, 338, 450, 578, 722, 882, 1058, 1250, 1458, 1682, 1922, 2178, 2450, 2738, 3042, 3362, 3698, 4050, 4418, 4802, 5202, 5618, 6050, 6498, 6962, 7442, 7938, 8450, 8978, 9522, 10082, 10658, 11250, 11858, 12482, 13122, 13778

Offset: 0

Joshua Zucker and the Castilleja School MathCounts club, Nov 07 2002

Sequence found by reading the segment (1, 2) together with the line from 2, in the direction 2, 18, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 05 2011
For a(n) > 1, a(n) are the numbers such that phi(sum of the odd divisors of a(n)) = phi(sum of even divisors of a(n)). - Michel Lagneau, Sep 14 2011
Apart from first term, subsequence of A195605. - Bruno Berselli, Sep 21 2011
Engel expansion of 1F2(1; 1/2, 1/2; 1/8). - Benedict W. J. Irwin, Jun 21 2018
Let f(n) = 4*n^2 - 5, then (x, y, z) = (a(n+1), -f(n), -f(n + 1)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 512. - XU Pingya, Apr 25 2022

			a(2) = 18 if you draw two concave quadrilaterals such that all four sides of one cross all four sides of the other.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Keyang Li, when n=1, number of regions is 2; when n=2, maximum number of regions is 18
Keyang Li, when n=3, maximum number of regions is 50
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Cf. A006752, A077588, A239186.
Cf. A071974, A071975.
Cf. A016754, A069129.

Concatenation([1], List([1..2000], n->8*n^2 - 8*n + 2)); # Muniru A Asiru, Jan 29 2018

A077591:=n->`if`(n=0, 1, 8*n^2 - 8*n + 2); seq(A077591(n), n=0..50); # Wesley Ivan Hurt, Mar 12 2014

Table[2*(4*n^2 - 4*n + 1), {n,0,50}] (* G. C. Greubel, Jul 15 2017 *)

isok(n) = (sod = sumdiv(n, d, (d%2)*d)) && (sed = sumdiv(n, d, (1 - d%2)*d)) && (eulerphi(sod) == eulerphi(sed)); \\ from Michel Lagneau comment; Michel Marcus, Mar 15 2014

a(n) = 8*n^2 - 8*n + 2 = 2*(2*n-1)^2, n > 0, a(0)=1.
Proof from Keyang Li, Jun 18 2022: (Start)
Represent the configuration of n concave quadrilaterals by a planar graph with a node for each vertex of the quadrilaterals and for each intersection point. Let there be v_n nodes and e_n edges. By Euler's formula for planar graphs, a(n) = e_n - v_n + 2. When we go from n to n+1 quadrilaterals, each side of the new quadrilateral can meet each side of the existing quadrilaterals at most 4 times, so Dv_n := v_{n+1} - v_n <= 4*4n = 16n.
Each of these intersection points increases the number of edges in the graph by 2, so De_n := e_{n+1} - e_n = 4 + 2*Dv_n, Da_n := a(n+1) - a(n) = 4 + Dv_n <= 4+16*n.
These upper bounds can be achieved by taking n interwoven concave quadrilaterals (for n=1,2,3 see the attached Keyang Li links), and we achieve a(n) = 8n^2 - 8n + 2 (and v_n = 8n^2 - 4n, e_n = 4n*(4n-3)) for n > 0. QED (End)
For n > 0: A071974(a(n)) = 2*n+1, A071975(a(n)) = 2. - Reinhard Zumkeller, Jul 10 2011
a(n) = 1 + A069129(n), if n >= 1. - Omar E. Pol, Sep 05 2011
a(n) = 2*A016754(n-1), if n >= 1. - Omar E. Pol, Sep 05 2011
G.f.: (1 - x + 15*x^2 + x^3)/(1-x)^3. - Colin Barker, Feb 23 2012
E.g.f.: (8*x^2 + 2)*exp(x) - 1. - G. C. Greubel, Jul 15 2017
From Amiram Eldar, Jan 29 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/16.
Sum_{n>=1} (-1)^(n+1)/a(n) = G/2, where G is Catalan constant (A006752).
Product_{n>=1} (1 + 1/a(n)) = cosh(Pi/sqrt(8)).
Product_{n>=1} (1 - 1/a(n)) = cos(Pi/sqrt(8)). (End)

A139275 a(n) = n(8n+1).

Original entry on oeis.org

0, 9, 34, 75, 132, 205, 294, 399, 520, 657, 810, 979, 1164, 1365, 1582, 1815, 2064, 2329, 2610, 2907, 3220, 3549, 3894, 4255, 4632, 5025, 5434, 5859, 6300, 6757, 7230, 7719, 8224, 8745, 9282, 9835, 10404, 10989, 11590, 12207, 12840

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 9,..., in the square spiral whose vertices are the triangular numbers A000217.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139273, A139274, A139276, A139278, A139279, A139280, A139281, A139282.

Table[n (8 n + 1), {n, 0, 40}] (* Bruno Berselli, Sep 21 2016 *)
LinearRecurrence[{3,-3,1},{0,9,34},50] (* Harvey P. Dale, Apr 21 2020 *)

a(n) = n*(8*n+1); \\ Altug Alkan, Sep 21 2016

a(n) = 8*n^2 + n.
Sequences of the form a(n) = 8*n^2+c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 7 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
a(n) = A000217(5*n) - A000217(3*n). - Bruno Berselli, Sep 21 2016
Sum_{n>=1} 1/a(n) = 8 - (1+sqrt(2))*Pi/2 - 4*log(2) - sqrt(2) * log(1+sqrt(2)) = 0.1887230016056779928... . - Vaclav Kotesovec, Sep 21 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(7*x + 9)/(1-x)^3.
E.g.f.: (8*x^2 + 9*x)*exp(x). (End)

A139271 a(n) = 2n(4*n-3).

Original entry on oeis.org

0, 2, 20, 54, 104, 170, 252, 350, 464, 594, 740, 902, 1080, 1274, 1484, 1710, 1952, 2210, 2484, 2774, 3080, 3402, 3740, 4094, 4464, 4850, 5252, 5670, 6104, 6554, 7020, 7502, 8000, 8514, 9044, 9590, 10152, 10730, 11324, 11934, 12560, 13202, 13860, 14534, 15224

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 2, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A033585 in the same spiral.
Twice decagonal numbers (or twice 10-gonal numbers). - Omar E. Pol, May 15 2008
a(n) is the number of walks in a cubic lattice of n dimensions that reach the point of origin for the first time after 4 steps. - Shel Kaphan, Mar 20 2023

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.
Cf. A001107.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=16). - Bruno Berselli, Jun 10 2013
Row n=2 of A361397.

Table[8n^2-6n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,2,20},50] (* Harvey P. Dale, Sep 26 2016 *)

a(n)=2*n*(4*n-3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 - 6*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3a(n-1) - 3a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = A001107(n)*2. - Omar E. Pol, May 15 2008
a(n) = 16*n + a(n-1) - 14 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: (2*x)*(7*x+1)/(1-x)^3.
E.g.f.: (8*x^2 + 2*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = Pi/12 + log(2)/2. - Amiram Eldar, Mar 28 2023

Corrected by Harvey P. Dale, Sep 26 2016

cp's OEIS Frontend

A275543 A081585 and A069129 interleaved.

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A035008 Total number of possible knight moves on an (n+2) X (n+2) chessboard, if the knight is placed anywhere.

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A077221 a(0) = 0 and then alternately even and odd numbers in increasing order such that the sum of any two successive terms is a square.

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A139098 a(n) = 8*n^2.

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A051870 18-gonal (or octadecagonal) numbers: a(n) = n*(8*n-7).

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A101321 Table T(n,m) = 1 + n*m*(m+1)/2 read by antidiagonals: centered polygonal numbers.

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A051870 18-gonal (or octadecagonal) numbers: a(n) = n(8n-7).

A101321 Table T(n,m) = 1 + nm(m+1)/2 read by antidiagonals: centered polygonal numbers.

A139275 a(n) = n(8n+1).

A139271 a(n) = 2n(4*n-3).