A139272 -id:A139272 - cp's OEIS frontend

A139593 A139276(n) followed by A139272(n+1).

0, 3, 11, 22, 38, 57, 81, 108, 140, 175, 215, 258, 306, 357, 413, 472, 536, 603, 675, 750, 830, 913, 1001, 1092, 1188, 1287, 1391, 1498, 1610, 1725, 1845, 1968, 2096, 2227, 2363, 2502, 2646, 2793, 2945, 3100, 3260, 3423, 3591, 3762

Offset: 0

Omar E. Pol, May 03 2008

Sequence found by reading the line from 0, in the direction 0, 3, ... and the same line from 0, in the direction 0, 11, ..., in the square spiral whose vertices are the triangular numbers A000217.

A139593 appears (both numerically and via back of an envelope algebra, but not a publishable proof) to be the cumulative sum of A047470. - Markus J. Q. Roberts, Jul 12 2009

			Array begins:
   0,   3;
  11,  22;
  38,  57;
  81, 108;

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A000217, A046092, A139272, A139276, A077221, A139591, A139592, A139594, A139595, A139596, A139597, A139598, A047470.

LinearRecurrence[{2,0,-2,1},{0,3,11,22},50] (* Harvey P. Dale, Feb 09 2019 *)

Array read by rows: row n gives 8*n^2 + 3n, 8*(n+1)^2 - 5(n+1).
From Colin Barker, Sep 15 2013: (Start)
a(n) = (-1 + (-1)^n + 6*n + 8*n^2)/4.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
G.f.: -x*(5*x+3) / ((x-1)^3*(x+1)). (End)

Edited by Omar E. Pol, Jul 13 2009

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

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

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, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

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) = 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) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A139278 a(n) = n(8n+7).

Original entry on oeis.org

0, 15, 46, 93, 156, 235, 330, 441, 568, 711, 870, 1045, 1236, 1443, 1666, 1905, 2160, 2431, 2718, 3021, 3340, 3675, 4026, 4393, 4776, 5175, 5590, 6021, 6468, 6931, 7410, 7905, 8416, 8943, 9486, 10045, 10620, 11211, 11818, 12441, 13080

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the segment (0, 15) together with the line from 15, in the direction 15, 46, ..., 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, A139275, A139276, A139277, A139279, A139280, A139281, A139282.

Table[n (8 n + 7), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 15, 46}, 50] (* Harvey P. Dale, Oct 07 2015 *)

a(n)=n*(8*n+7) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 8*n^2 + 7*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)-1 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(x+15)/(1-x)^3.
E.g.f.: (8*x^2 + 15*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/49 + (sqrt(2)+1)*Pi/14 - 4*log(2)/7 - sqrt(2)*log(sqrt(2)+1)/7. - Amiram Eldar, Mar 17 2022

A226492 a(n) = n(11n-5)/2.

Original entry on oeis.org

0, 3, 17, 42, 78, 125, 183, 252, 332, 423, 525, 638, 762, 897, 1043, 1200, 1368, 1547, 1737, 1938, 2150, 2373, 2607, 2852, 3108, 3375, 3653, 3942, 4242, 4553, 4875, 5208, 5552, 5907, 6273, 6650, 7038, 7437, 7847, 8268, 8700, 9143, 9597, 10062, 10538, 11025, 11523

Offset: 0

Bruno Berselli, Jun 11 2013

Sequences of numbers of the form n*(n*k - k + 6)/2:
. k from 0 to 10, respectively: A008585, A055998, A005563, A045943, A014105, A005475, A033428, A022264, A033991, A062741, A147874;
. k=11: a(n);
. k=12: A094159;
. k=13: 0, 3, 19, 48, 90, 145, 213, 294, 388, 495, 615, 748, 894, ...;
. k=14: 0, 3, 20, 51, 96, 155, 228, 315, 416, 531, 660, 803, 960, ...;
. k=15: A152773;
. k=16: A139272;
. k=17: 0, 3, 23, 60, 114, 185, 273, 378, 500, 639, 795, 968, ...;
. k=18: A152751;
. k=19: 0, 3, 25, 66, 126, 205, 303, 420, 556, 711, 885, 1078, ...;
. k=20: 0, 3, 26, 69, 132, 215, 318, 441, 584, 747, 930, 1133, ...;
. k=21: A152759;
. k=22: 0, 3, 28, 75, 144, 235, 348, 483, 640, 819, 1020, 1243, ...;
. k=23: 0, 3, 29, 78, 150, 245, 363, 504, 668, 855, 1065, 1298, ...;
. k=24: A152767;
. k=25: 0, 3, 31, 84, 162, 265, 393, 546, 724, 927, 1155, 1408, ...;
. k=26: 0, 3, 32, 87, 168, 275, 408, 567, 752, 963, 1200, 1463, ...;
. k=27: A153783;
. k=28: A195021;
. k=29: 0, 3, 35, 96, 186, 305, 453, 630, 836, 1071, 1335, 1628, ...;
. k=30: A153448;
. k=31: 0, 3, 37, 102, 198, 325, 483, 672, 892, 1143, 1425, 1738, ...;
. k=32: 0, 3, 38, 105, 204, 335, 498, 693, 920, 1179, 1470, 1793, ...;
. k=33: A153875.
Also:
a(n) - n         = A180223(n);
a(n) + n         = n*(11*n-3)/2 = 0, 4, 19, 45, 82, 130, 189, 259, ...;
a(n) - 2*n       = A051865(n);
a(n) + 2*n       = A022268(n);
a(n) - 3*n       = A152740(n-1);
a(n) + 3*n       = A022269(n);
a(n) - 4*n       = n*(11*n-13)/2 = 0, -1, 9, 30, 62, 105, 159, 224, ...;
a(n) + 4*n       = A254963(n);
a(n) - n*(n-1)/2 = A147874(n+1);
a(n) + n*(n-1)/2 = A094159(n) (case k=12);
a(n) - n*(n-1)   = A062741(n) (see above, this is the case k=9);
a(n) + n*(n-1)   = n*(13*n-7)/2 (case k=13);
a(n) - n*(n+1)/2 = A135706(n);
a(n) + n*(n+1)/2 = A033579(n);
a(n) - n*(n+1)   = A051682(n);
a(n) + n*(n+1)   = A186030(n);
a(n) - n^2       = A062708(n);
a(n) + n^2       = n*(13*n-5)/2 = 0, 4, 21, 51, 94, 150, 219, ..., etc.
Sum of reciprocals of a(n), for n > 0: 0.47118857003113149692081665034891...

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

Cf. sequences in Comments lines.
First differences are in A017425.
Cf. A033584, A180223.

Magma
```
[n*(11*n-5)/2: n in [0..50]];
```

I:=[0,3,17]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..46]]; // Vincenzo Librandi, Aug 18 2013

Table[n (11 n - 5)/2, {n, 0, 50}]
CoefficientList[Series[x (3 + 8 x) / (1 - x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{3,-3,1},{0,3,17},50] (* Harvey P. Dale, Jan 14 2019 *)

a(n)=n*(11*n-5)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(3+8*x)/(1-x)^3.
a(n) + a(-n) = A033584(n).
From Elmo R. Oliveira, Dec 27 2024: (Start)
E.g.f.: exp(x)*x*(6 + 11*x)/2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = n + A180223(n). (End)

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

A299989 Triangle read by rows: T(n,0) = 0 for n >= 0; T(n,2k+1) = A152842(2n,2(n-k)) and T(n,2k) = A152842(2n,2(n-k)+1) for n >= k > 0.

Original entry on oeis.org

0, 1, 0, 3, 4, 1, 0, 9, 24, 22, 8, 1, 0, 27, 108, 171, 136, 57, 12, 1, 0, 81, 432, 972, 1200, 886, 400, 108, 16, 1, 0, 243, 1620, 4725, 7920, 8430, 5944, 2810, 880, 175, 20, 1, 0, 729, 5832, 20898, 44280, 61695, 59472, 40636, 19824, 6855, 1640, 258, 24, 1

Offset: 0

Franck Maminirina Ramaharo, Feb 26 2018

T(n,k) is the number of state diagrams having k components of n connected summed trefoil knots.

Row sums gives A001018.

			The triangle T(n, k) begins:
n\k 0     1      2      3       4       5       6      7        8       9
0:  0     1
1:  0     3      4      1
2:  0     9     24     22       8       1
3:  0    27    108    171     136      57      12       1
4:  0    81    432    972    1200     886     400     108      16       1

V. I. Arnold, Topological Invariants of Plane Curves and Caustics, American Math. Soc., 1994.

Michael De Vlieger, Table of n, a(n) for n = 0..10301 (rows 0 <= n <= 100, flattened.)
Ryo Hanaki, Pseudo diagrams of knots, links and spatial graphs, Osaka Journal of Mathematics, Vol. 47 (2010), 863-883.
Louis H. Kauffman, State models and the Jones polynomial, Topology, Vol. 26 (1987), 395-407.
Carolina Medina, Jorge Ramírez-Alfonsín and Gelasio Salazar, On the number of unknot diagrams, arXiv:1710.06470 [math.CO], 2017.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the pretzel knot, arXiv:1805.10680 [math.CO], 2018.
Franck Ramaharo, A bracket polynomial for 2-tangle shadows, arXiv:2002.06672 [math.CO], 2020.

Row 2: row 5 of A158454.
Row 3: row 2 of A220665.
Row 4: row 5 of A219234.

row[n_] := CoefficientList[x*(x^2 + 4*x + 3)^n, x]; Array[row, 7, 0] // Flatten (* Jean-François Alcover, Mar 16 2018 *)

g(x, y) := taylor(x/(1 - y*(x^2 + 4*x + 3)), y, 0, 10)$
a : makelist(ratcoef(g(x, y), y, n), n, 0, 10)$
T : []$
for i:1 thru 11 do
  T : append(T, makelist(ratcoef(a[i], x, n), n, 0, 2*i - 1))$
T;

T(n, k) = polcoeff(x*(x^2 + 4*x + 3)^n, k);
tabf(nn) = for (n=0, nn, for (k=0, 2*n+1, print1(T(n, k), ", ")); print); \\ Michel Marcus, Mar 03 2018

T(n,k) = coefficients of x*(x^2 + 4*x + 3)^n.
T(n,k) = T(n-1,k-2) + 4*T(n-1,k-1) + 3*T(n-1,k), with T(n,0) = 0, T(n,1) = 3^n and  T(n,2) =  4*n*3^(n-1).
T(n,n+k+1) = A152842(2*n,n+k) and T(n,n-k) = A152842(2*n,n+k+1), for n >= k >= 0.
T(n,1) =  A000244(n).
T(n,2) =  A120908(n).
T(n,n+1) = A069835(n).
T(n,2*n-1) = A139272(n).
T(n,2*n) = A008586(n).
T(n,2*n-2) = A140138(4*n) = A185872(2n,2) for n >= 1.
G.f.: x/(1 - y*(x^2 + 4*x + 3)).

Typo in row 6 corrected by Jean-François Alcover, Mar 16 2018

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

cp's OEIS Frontend

A139593 A139276(n) followed by A139272(n+1).

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

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

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A139273 a(n) = n*(8*n - 3).

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A139278 a(n) = n*(8*n+7).

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A226492 a(n) = n*(11*n-5)/2.

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

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

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

A139273 a(n) = n(8n - 3).

A139278 a(n) = n(8n+7).

A226492 a(n) = n(11n-5)/2.

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

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

A299989 Triangle read by rows: T(n,0) = 0 for n >= 0; T(n,2k+1) = A152842(2n,2(n-k)) and T(n,2k) = A152842(2n,2(n-k)+1) for n >= k > 0.

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