A014634 -id:A014634 - cp's OEIS frontend

A143937 Triangle read by rows: T(n,k) is the number of unordered pairs of vertices at distance k in a benzenoid consisting of a linear chain of n hexagons (1 <= k <= 2n+1).

6, 6, 3, 11, 14, 12, 6, 2, 16, 22, 21, 14, 10, 6, 2, 21, 30, 30, 22, 18, 14, 10, 6, 2, 26, 38, 39, 30, 26, 22, 18, 14, 10, 6, 2, 31, 46, 48, 38, 34, 30, 26, 22, 18, 14, 10, 6, 2, 36, 54, 57, 46, 42, 38, 34, 30, 26, 22, 18, 14, 10, 6, 2, 41, 62, 66, 54, 50, 46, 42, 38, 34, 30, 26, 22

Offset: 1

Emeric Deutsch, Sep 06 2008

The entries in row n are the coefficients of the Wiener polynomial of the benzenoid consisting of a linear chain of n hexagons.
Sum of entries in row n is (2*n+1)*(4*n+1) = A014634(n).
Sum_{k=1..2n+1} k*T(n,k) = A143938(n) is the Wiener index of a benzenoid consisting of a linear chain of n hexagons.

			T(1,2)=6 because in a hexagon there are 6 distances equal to 2.
Triangle starts:
   6,  6,  3;
  11, 14, 12,  6,  2;
  16, 22, 21, 14, 10,  6,  2;
  21, 30, 30, 22, 18, 14, 10,  6,  2;

A. A. Dobrynin, I. Gutman, S. Klavzar, and P. Zigert, Wiener index of hexagonal systems, Acta Applicandae Mathematicae 72 (2002), pp. 247-294.
B. E. Sagan, Y-N. Yeh and P. Zhang, The Wiener Polynomial of a Graph, Internat. J. of Quantum Chem., 60, 1996, 959-969.

Cf. A014634, A143938.

T:=proc(n,k) if 2*n+1 < k then 0 elif k = 1 then 5*n+1 elif k = 3 then 9*n-6 elif `mod`(k, 2) = 0 then 8*n-4*k+6 else 8*n-4*k+6 end if end proc: for n to 8 do seq(T(n,k),k=1..2*n+1) end do; # yields sequence in triangular form

For 1 <= k <= 2n+1, T(n,k) is given by T(n,1) = 5*n+1, T(n,3) = 9*n - 6, T(n,2*p+1) = 8*n-8*p+2, T(n,2*p) = 8*n-8*p+6.

G.f.: q*z*(6+6*q-z+2*q*z+3*q^2+q^2*z^2-q^4*z)/((1-q^2*z)*(1-z)^2).

A139274 a(n) = n(8n-1).

Original entry on oeis.org

0, 7, 30, 69, 124, 195, 282, 385, 504, 639, 790, 957, 1140, 1339, 1554, 1785, 2032, 2295, 2574, 2869, 3180, 3507, 3850, 4209, 4584, 4975, 5382, 5805, 6244, 6699, 7170, 7657, 8160, 8679, 9214, 9765, 10332, 10915, 11514, 12129, 12760, 13407, 14070, 14749

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the triangular numbers A000217.

Polygonal number connection: 2*P_n + 5*S_n where P_n is the n-th pentagonal number and S_n is the n-th square. - William A. Tedeschi, Sep 12 2010

			a(1) = 16*1 + 0 - 9 = 7; a(2) = 16*2 + 7 - 9 = 30; a(3) = 16*3 + 30 - 9 = 69. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
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, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

[n*(8*n-1) : n in [0..50]]; // Wesley Ivan Hurt, Dec 04 2016

A139274:=n->n*(8*n-1): seq(A139274(n), n=0..100); # Wesley Ivan Hurt, Dec 04 2016

CoefficientList[Series[x (9 x + 7)/(1 - x)^3, {x, 0, 43}], x] (* Michael De Vlieger, Jan 11 2020 *)
Table[n(8n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,7,30},50] (* Harvey P. Dale, Apr 01 2024 *)

a(n)=n*(8*n-1) \\ Charles R Greathouse IV, Jun 17 2017

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) = 16*n + a(n-1) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(n) = (1/3) * Sum_{i=n..(7*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(9*x+7)/(1-x)^3.
E.g.f.: (8*x^2 + 7*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 4*log(2) + sqrt(2)*log(sqrt(2)+1) - (sqrt(2)+1)*Pi/2. - Amiram Eldar, Mar 18 2022

A139276 a(n) = n(8n+3).

Original entry on oeis.org

0, 11, 38, 81, 140, 215, 306, 413, 536, 675, 830, 1001, 1188, 1391, 1610, 1845, 2096, 2363, 2646, 2945, 3260, 3591, 3938, 4301, 4680, 5075, 5486, 5913, 6356, 6815, 7290, 7781, 8288, 8811, 9350, 9905, 10476, 11063, 11666, 12285, 12920

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 11,..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139272 in the same spiral.

			a(1)=16*1+0-5=11; a(2)=16*2+11-5=38; a(3)=16*3+38-5=81. - _Vincenzo Librandi_, Aug 03 2010

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Amelia Carolina Sparavigna, The groupoid of the Triangular Numbers and the generation of related integer sequences, Politecnico di Torino, Italy (2019).
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, A139275, A139277, A139278, A139279, A139280, A139281, A139282.

a[n_]:=n*(8*n+3); a[Range[0,60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3,-3,1},{0,11,38},50] (* Harvey P. Dale, Jul 02 2022 *)

a(n)=n*(8*n+3) \\ Charles R Greathouse IV, Jun 16 2017

a(n) = 8*n^2 + 3*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) = 16*n+a(n-1)-5 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(5*x + 11)/(1-x)^3.
E.g.f.: (8*x^2 + 11*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/9 - (sqrt(2)-1)*Pi/6 - 4*log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3. - Amiram Eldar, Mar 17 2022

A139277 a(n) = n(8n+5).

Original entry on oeis.org

0, 13, 42, 87, 148, 225, 318, 427, 552, 693, 850, 1023, 1212, 1417, 1638, 1875, 2128, 2397, 2682, 2983, 3300, 3633, 3982, 4347, 4728, 5125, 5538, 5967, 6412, 6873, 7350, 7843, 8352, 8877, 9418, 9975, 10548, 11137, 11742, 12363, 13000

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 13, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139273 in the same spiral.

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.

Cf. A139271, A139272, A139273, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Table[n (8 n + 5), {n, 0, 50}] (* Bruno Berselli, Aug 22 2018 *)
LinearRecurrence[{3,-3,1},{0,13,42},50] (* Harvey P. Dale, Dec 04 2018 *)

a(n)=n*(8*n+5) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 + 5*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) - 3 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
Sum_{n>=1} 1/a(n) = (sqrt(2)-1)*Pi/10 - 4*log(2)/5 + sqrt(2)*log(sqrt(2)+1)/5 + 8/25. - Amiram Eldar, Mar 18 2022
E.g.f.: exp(x)*x*(13 + 8*x). - Elmo R. Oliveira, Dec 15 2024

A194268 a(n) = 8n^2 + 7n + 1.

Original entry on oeis.org

1, 16, 47, 94, 157, 236, 331, 442, 569, 712, 871, 1046, 1237, 1444, 1667, 1906, 2161, 2432, 2719, 3022, 3341, 3676, 4027, 4394, 4777, 5176, 5591, 6022, 6469, 6932, 7411, 7906, 8417, 8944, 9487, 10046, 10621, 11212, 11819, 12442, 13081, 13736, 14407, 15094, 15797

Offset: 0

Omar E. Pol, Sep 05 2011

Sequence found by reading the line from 1, in the direction 1, 16,..., in the square spiral whose vertices are the triangular numbers A000217.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A051870, A069129, A139098, A194431.

[8*n^2 +7*n + 1: n in [0..50]]; // Vincenzo Librandi, Sep 07 2011

A194268:=n->8*n^2+7*n+1: seq(A194268(n), n=0..50); # Wesley Ivan Hurt, Jul 15 2014

Table[8n^2+7n+1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{1,16,47},50] (* Harvey P. Dale, Apr 06 2014 *)

a(n)=8*n^2+7*n+1 \\ Charles R Greathouse IV, Oct 07 2015

a(0)=1, a(1)=16, a(2)=47, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Apr 06 2014
From Elmo R. Oliveira, Oct 22 2024: (Start)
G.f.: (1 + 13*x + 2*x^2)/(1 - x)^3.
E.g.f.: (1 + 15*x + 8*x^2)*exp(x). (End)

A194431 a(n) = 8n^2 - 6n - 1.

Original entry on oeis.org

1, 19, 53, 103, 169, 251, 349, 463, 593, 739, 901, 1079, 1273, 1483, 1709, 1951, 2209, 2483, 2773, 3079, 3401, 3739, 4093, 4463, 4849, 5251, 5669, 6103, 6553, 7019, 7501, 7999, 8513, 9043, 9589, 10151, 10729, 11323, 11933, 12559, 13201, 13859, 14533, 15223, 15929

Offset: 1

Omar E. Pol, Sep 05 2011

Sequence found by reading the line from 1, in the direction 1, 19, ..., in the square spiral whose vertices are the triangular numbers A000217.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A051870, A069129, A139098, A194268.

[8*n^2 - 6*n - 1: n in [1..50]]; // Vincenzo Librandi, Sep 07 2011

Table[8n^2-6n-1,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{1,19,53},50] (* Harvey P. Dale, May 29 2021 *)

a(n)=8*n^2-6*n-1 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: x*(-1 - 16*x + x^2)/(x-1)^3. - R. J. Mathar, Sep 06 2011
From Elmo R. Oliveira, Jun 04 2025: (Start)
E.g.f.: 1 + (-1 + 2*x + 8*x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A185438 a(n) = 8n^2 - 2n + 1.

Original entry on oeis.org

1, 7, 29, 67, 121, 191, 277, 379, 497, 631, 781, 947, 1129, 1327, 1541, 1771, 2017, 2279, 2557, 2851, 3161, 3487, 3829, 4187, 4561, 4951, 5357, 5779, 6217, 6671, 7141, 7627, 8129, 8647, 9181, 9731, 10297, 10879, 11477, 12091, 12721, 13367, 14029, 14707, 15401, 16111, 16837, 17579

Offset: 0

Paul Curtz, Feb 03 2011

Odd numbers (A005408) written clockwise as a square spiral:
.
  41--43--45--47--49--51
   |                   |
  39  13--15--17--19  53
   |   |           |   |
  37  11   1---3  21  55
   |   |       |   |   |
  35   9---7---5  23  57
   |               |   |
  33--31--29--27--25  59
                       |
  71--69--67--65--63--61
.
Walking in straight lines away from the center:
  1, 17, 49, ... = A069129(n+1) =  1 - 8*n + 8*n^2,
  1,  3, 21, ... = A033567(n)   =  1 - 6*n + 8*n^2,
  1, 15, 45, ... = A014634(n)   =  1 + 6*n + 8*n^2,
  1,  5, 25, ... = A080856(n)   =  1 - 4*n + 8*n^2,
  1, 13, 41, ... = A102083(n)   =  1 + 4*n + 8*n^2,
  1,  7, 29, ... = a(n)         =  1 - 2*n + 8*n^2,
  1, 11, 37, ... = A188135(n)   =  1 + 2*n + 8*n^2,
  1,  9, 33, ... = A081585(n)   =  1       + 8*n^2,
  5, 29, 69, ... = A108928(n+1) = -3       + 8*n^2,
  7, 31, 71, ... = A157914(n+1) = -1       + 8*n^2,
  9, 35, 77, ... = A033566(n+1) = -1 + 2*n + 8*n^2.
All are quadrisections of sequences in A181407(n) (example: A014634(n) and A033567(n) in A064038(n+1)) or of this family (?): a(n) is a quadrisection of f(n) = 1,1,1,1,2,7,11,8,11,29,37,23,28,67,79,46,... f(n) is just before A064038(n+1) (fifth vertical) in A181407(n). The companion to a(n) is A188135(n), another quadrisection of f(n). Two last quadrisections of f(n) are A054552(n) and A033951(n).
For n >= 1, bisection of A193867. - Omar E. Pol, Aug 16 2011
Also the sequence may be obtained by starting with the segment (1, 7) followed by the line from 7 in the direction 7, 29, ... in the square spiral whose vertices are the generalized hexagonal numbers (A000217). - Omar E. Pol, Aug 01 2016

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A005408, A014634, A014635, A033566, A033567, A033951, A054552, A064038.

Cf. A069129, A080856, A081585, A102083, A108928, A157914, A181407, A188135, A193867.

[1-2*n+8*n^2: n in [0..50]]; // Vincenzo Librandi, Feb 03 2011

Table[1 - 2n + 8n^2, {n, 0, 39}] (* Alonso del Arte, Feb 03 2011 *)
CoefficientList[Series[(-1 - 4 x - 11 x^2)/(x - 1)^3, {x, 0, 47}], x] (* Michael De Vlieger, Aug 01 2016 *)

a(n)=8*n^2-2*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = a(n-1) + 16*n - 10 (n > 0).
a(n) = 2*a(n-1) - a(n-2) + 16 (n > 1).
a(n) = 3*(n-1) - 3*a(n-2) + a(n-3) (n > 2).
G.f.: (-1 - 4*x - 11*x^2)/(x-1)^3. - R. J. Mathar, Feb 03 2011
a(n) = A014635(n) + 1. - Bruno Berselli, Apr 09 2011
E.g.f.: exp(x)*(1 + 6*x + 8*x^2). - Elmo R. Oliveira, Nov 17 2024

A143218 Triangle read by rows, A127775 * A000012 * A127775; 1<=k<=n.

Original entry on oeis.org

1, 3, 9, 5, 15, 25, 7, 21, 35, 49, 9, 27, 45, 63, 81, 11, 33, 55, 77, 99, 121, 13, 39, 65, 91, 117, 143, 169, 15, 45, 75, 105, 135, 165, 195, 225, 17, 51, 85, 119, 153, 187, 221, 255, 289, 19, 57, 95, 133, 171, 209, 247, 285, 323, 361, 21, 63, 105, 147, 189, 231, 273, 315, 357, 399, 441

Offset: 1

Gary W. Adamson, Jul 30 2008

			First few rows of the triangle =
   1;
   3,  9;
   5, 15, 25;
   7, 21, 35, 49;
   9, 27, 45, 63,  81;
  11, 33, 55, 77,  99, 121;
  13, 39, 65, 91, 117, 143, 169;
  ...
T(5,3) = 45 = 9*5 = (2*5 - 1) * (2*3 - 1).

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Cf. A000012, A005408, A014634, A015237 (row sums).

Cf. A016754, A024598, A033567, A127775, A131507.

[(2*n-1)*(2*k-1): k in [1..n], n in [1..12]]; // G. C. Greubel, Jul 12 2022

Table[(2*k-1)*(2*n-1), {n,12}, {k,n}]//Flatten (* G. C. Greubel, Jul 12 2022 *)

flatten([[(2*n-1)*(2*k-1) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Jul 12 2022

Triangle read by rows, A127775 * A000012 * A127775.
T(n, k) = (2*n - 1) * (2*k - 1), 1<=k<=n.
Sum_{k=1..n} T(n, k) = A015237(n) = n^2 * (2*n-1).
From G. C. Greubel, Jul 12 2022: (Start)
T(n, k) = A131507(n,k) * A127775(n,k).
T(n, n) = A016754(n-1) = (2*n-1)^2, n >= 1.
T(2*n-1, n) = A014634(n-1), n >= 1.
T(2*n-2, n-1) = A033567(n-1), n >= 2.
Sum_{k=1..floor((n+1)/2)} T(n-k+1, k) = A024598(n), n >= 1. (End)

A157870 a(n) = (4n+1)(4n+2) = (4n+2)!/(4*n)!.

Original entry on oeis.org

2, 30, 90, 182, 306, 462, 650, 870, 1122, 1406, 1722, 2070, 2450, 2862, 3306, 3782, 4290, 4830, 5402, 6006, 6642, 7310, 8010, 8742, 9506, 10302, 11130, 11990, 12882, 13806, 14762, 15750, 16770, 17822, 18906, 20022, 21170, 22350, 23562, 24806, 26082, 27390, 28730, 30102, 31506, 32942

Offset: 0

SUNKU Sai Swaroop (sai2020(AT)gmail.com), Mar 08 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002378, A014634, A016813, A016825, A157870.

(4*n+1)*(4*n+2); // Vincenzo Librandi Jul 10 2012

Table[(4n+1)*(4n+2),{n,0,50}] (* Vincenzo Librandi, Jul 10 2012 *)

a(n)=(4*n+1)*(4*n+2) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = A002378(4*n+1) = 2*A014634(n). - R. J. Mathar, Mar 11 2009
From Vincenzo Librandi, Jul 10 2012: (Start)
G.f.: 2*(1+12*x+3*x^2)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
From Amiram Eldar, Mar 01 2022: (Start)
Sum_{n>=0} 1/a(n) = Pi/8 + log(2)/4.
Sum_{n>=0} (-1)^n/a(n) = ((sqrt(2)-1)*Pi + sqrt(2)*log((2+sqrt(2))/(2-sqrt(2))))/8. (End)
From Elmo R. Oliveira, Oct 30 2024: (Start)
E.g.f.: 2*exp(x)*(1 + 14*x + 8*x^2).
a(n) = A016813(n)*A016825(n). (End)

Definition corrected and sequence extended by R. J. Mathar, Mar 11 2009

A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 15, 6, 14, 10, 45, 15, 33, 21, 91, 28, 60, 36, 153, 45, 95, 55, 231, 66, 138, 78, 325, 91, 189, 105, 435, 120, 248, 136, 561, 153, 315, 171, 703, 190, 390, 210, 861, 231, 473, 253, 1035, 276, 564, 300, 1225, 325

Offset: 0

Paul Curtz, Mar 12 2012

a(n) is divisible by the n-th term of the sequence 3, 3, 1, 1, 3, 3 (periodically repeated with period 6).
a(n) is divisible by b(floor((n-1)/3)), where b(n) = 1, 3, 2, 3, 7, 3, 5, 3, 13, 3, 8, 3, 19, 3,... , n>=0, is defined by inserting a 3 after each entry of A165355.
(n+1)*(n+2)*(n+3)/2=3*A000292(n+1) is divisible by a(n+2), so there is an integer sequence c(n)= 3*A000292(n+1)/a(n+2) = 3, 12, 10, 20, 7, 28, 18,... with c(2*n)=A123167(n+1) and c(n)/A109613(n+2)=A176895(n).
The sequence of denominators of a(n+2)/n has period length 8: 1, 2, 1, 4, 1, 1, 1, 4.
A table T(k,c) = a(1+c*(1+2k)) of (2*k+1)-sections starts as follows:
0  1  1   3   3   15...
0  3  6  45  21   60...
0 15 15  60  55  325...
0 14 28 231 105  315...
0 45 45 189 171 1035...
The table of T'(k,c) = T(k,c)/(2k+1), columns c>=0, looks as follows, construction similar to A165943:
0 1 1  3  3  15  6  14   k=0
0 1 2 15  7  20 15  77   k=1
0 3 3 12 11  65 24  63   k=2
0 2 4 33 15  45 33 175   k=3
0 5 5 21 19 115 42 112   k=4
0 3 6 51 23  70 51 273   k=5
The entries T'(k,c) are divisible by A060819(c).
Differences are T'(2,c)-T'(0,c) = T'(4,c)-T'(2,c) = 0, 2, 2, 9, 8, 50, 18, 49, 32, ... which is A168077(c) multiplied by the c-th term of the period-4 sequence 2, 2, 2, 1.
Differences are T'(3,c)- T'(1,c) = T'(5,c)-T'(3,c) = 0, 1, 2, 18, 8, 25, 18, 98, 32,... which is  A168077(c) multiplied by the period-4 sequence 2, 1, 2, 2.
The reduced fractions T'(0,c)/T'(1,c) = 1, 1/2, 1/5, 3/7, 3/4, 2/5, 2/11, 5/13, 5/7, 3/8, 3/17, 7/19, .., c>=1, have a numerator sequence A026741(floor(c/2)+1). The denominator sequence is f(c) = 1, 2, 5, 7, 4, 5,.. = A001651(c+1)/A130658(c+1), with f(2*c+1) +f(2*c+2) = 3, 12, 9, 24 .. =3*A022998(c).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000217, A014634, A026741, A033991, A064038, A060819, A165355, A208950.

A185138 := proc(n)
        if n mod 4 = 0 then
                return n/4*(n-1) ;
        elif n mod 2 = 1 then
                return (n-1)*(n+1)/8 ;
        else
                return (n-1)*n/2 ;
        end if;
end proc: # R. J. Mathar, Apr 05 2012

Clear[b];b[1] = 0; b[2] = 0; b[3] = 1; b[4] = 1; b[5] = 3; b[6] = 3; b[7] = 15;b[8] = 6;b[n_Integer] := b[n] = ((-2 + n) (-4 (-4 + n) (-3 + n) (-2 + n) (8 + n (-9 + 2 n)) b[-3 + n] + (-5 + n) ((-3 +n) ((-4 + n) (211 + 2 n (-215 + n (147 + n (-41 + 4 n)))) - 4 (-1 + n) (19 + n (-13 + 2 n)) b[-2 + n]) - 4 (-4 + n)^2 (8 + n (-9 + 2 n)) b[-1 + n])))/(4 (-5 + n) (-4 + n) (-3 + n)^2 (19 + n (-13 + 2 n)))
a = Table[b[n], {n, 1, 52}] (* Roger L. Bagula, Mar 14 2012 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,0,1,1,3,3,15,6,14,10,45,15},60] (* Harvey P. Dale, Nov 23 2015 *)

x='x+O('x^50); concat([0,0], Vec(-x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3))) \\ G. C. Greubel, Jun 23 2017

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*n) = A064038(2*n), a(2*n+1) = A000217(n).
a(n) = 3*A208950(n)/A109613(n).
a(n+1) = A060819(n) * A026741(n+2)(floor(n/2)).
G.f.: -x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3). - R. J. Mathar, Mar 22 2012
a(n) = (4*n^2-3*n-1+(2*n^2-3*n+1)*(-1)^n + n*(n-1)*(1+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4))/16. - Luce ETIENNE, May 13 2016
Sum_{n>=2} 1/a(n) = 2 - Pi/4 + 7*log(2)/2. - Amiram Eldar, Aug 12 2022

cp's OEIS Frontend

A143937 Triangle read by rows: T(n,k) is the number of unordered pairs of vertices at distance k in a benzenoid consisting of a linear chain of n hexagons (1 <= k <= 2n+1).

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A139274 a(n) = n*(8*n-1).

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A139276 a(n) = n*(8*n+3).

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A139277 a(n) = n*(8*n+5).

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A194268 a(n) = 8*n^2 + 7*n + 1.

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A194431 a(n) = 8*n^2 - 6*n - 1.

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A185438 a(n) = 8*n^2 - 2*n + 1.

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A139274 a(n) = n(8n-1).

A139276 a(n) = n(8n+3).

A139277 a(n) = n(8n+5).

A194268 a(n) = 8n^2 + 7n + 1.

A194431 a(n) = 8n^2 - 6n - 1.

A185438 a(n) = 8n^2 - 2n + 1.

A157870 a(n) = (4n+1)(4n+2) = (4n+2)!/(4*n)!.

A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).