A144555 -id:A144555 - cp's OEIS frontend

A195020 Vertex number of a square spiral in which the length of the first two edges are the legs of the primitive Pythagorean triple [3, 4, 5]. The edges of the spiral have length A195019.

0, 3, 7, 13, 21, 30, 42, 54, 70, 85, 105, 123, 147, 168, 196, 220, 252, 279, 315, 345, 385, 418, 462, 498, 546, 585, 637, 679, 735, 780, 840, 888, 952, 1003, 1071, 1125, 1197, 1254, 1330, 1390, 1470, 1533, 1617, 1683, 1771, 1840, 1932, 2004, 2100

Offset: 0

Omar E. Pol, Sep 07 2011 - Sep 12 2011

Zero together with the partial sums of A195019.
The spiral contains infinitely many Pythagorean triples in which the hypotenuses on the main diagonal are the positives A008587. The vertices on the main diagonal are the numbers A024966 = (3+4)*A000217 = 7*A000217, where both 3 and 4 are the first two edges in the spiral. The distance "a" between nearest edges that are perpendicular to the initial edge of the spiral is 3, while the distance "b" between nearest edges that are parallel to the initial edge is 4, so the distance "c" between nearest vertices on the same axis is 5 because from the Pythagorean theorem we can write c = (a^2+b^2)^(1/2) = sqrt(3^2+4^2) = sqrt(9+16) = sqrt(25) = 5.
Let an array have m(0,n)=m(n,0)=n*(n-1)/2 and m(n,n)=n*(n+1)/2. The first n+1 terms in row(n) are the numbers in the closed interval m(0,n) to m(n,n). The terms in column(n) are the same from m(n,0) to m(n,n). The first few antidiagonals are 0; 0,0; 1,1,1; 3,2,2,3; 6,4,3,4,6; 10,7,5,5,7,10. a(n) is the difference between the sum of the terms in the n+1 X n+1 matrices and those in the n X n matrices. - J. M. Bergot, Jul 05 2013 [The first five rows are: 0,0,1,3,6; 0,1,2,4,7; 1,2,3,5,8; 3,4,5,6,9; 6,7,8,9,10]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Ron Knott, Pythagorean triangles and triples
Eric Weisstein's World of Mathematics, Pythagorean Triple
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A024966, A008585, A008586, A001106, A022264, A024966, A033572, A144555, A158482, A158485, A195018, A195032, A195034, A195036.

[(2*n*(7*n+13)+(2*n-5)*(-1)^n+5)/16: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

With[{r = Range[50]}, Join[{0}, Accumulate[Riffle[3*r, 4*r]]]] (* or *)
LinearRecurrence[{1, 2, -2, -1, 1}, {0, 3, 7, 13, 21}, 100] (* Paolo Xausa, Feb 09 2024 *)

From Bruno Berselli, Oct 13 2011: (Start)
G.f.: x*(3+4*x)/((1+x)^2*(1-x)^3).
a(n) = (1/2)*A004526(n+2)*A047335(n+1) = (2*n*(7*n+13) + (2*n-5)*(-1)^n+5)/16.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) - a(n-2) = A047355(n+1). (End)

A195145 Concentric 14-gonal numbers.

Original entry on oeis.org

0, 1, 14, 29, 56, 85, 126, 169, 224, 281, 350, 421, 504, 589, 686, 785, 896, 1009, 1134, 1261, 1400, 1541, 1694, 1849, 2016, 2185, 2366, 2549, 2744, 2941, 3150, 3361, 3584, 3809, 4046, 4285, 4536, 4789, 5054, 5321, 5600, 5881, 6174, 6469, 6776, 7085, 7406

Offset: 0

Omar E. Pol, Sep 17 2011

Also concentric tetradecagonal numbers or concentric tetrakaidecagonal numbers. Also sequence found by reading the line from 0, in the direction 0, 14, ..., and the same line from 1, in the direction 1, 29, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Main axis, perpendicular to A024966 in the same spiral.

Partial sums of A113801. - Reinhard Zumkeller, Jan 07 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

A144555 and A195314 interleaved.
Cf. A024966, A032527, A032528, A077221, A113801, A195045, A195046, A195142, A195143, A195146, A195147, A195148, A195149.
Column 14 of A195040. - Omar E. Pol, Sep 28 2011

a195145 n = a195145_list !! n
a195145_list = scanl (+) 0 a113801_list
-- Reinhard Zumkeller, Jan 07 2012

[(14*n^2+5*(-1)^n-5)/4: n in [0..50]]; // Vincenzo Librandi, Sep 27 2011

LinearRecurrence[{2, 0, -2, 1}, {0, 1, 14, 29}, 50] (* Amiram Eldar, Jan 16 2023 *)

G.f.: -x*(1+12*x+x^2) / ( (1+x)*(x-1)^3 ). - R. J. Mathar, Sep 18 2011
From Vincenzo Librandi, Sep 27 2011: (Start)
a(n) = (14*n^2 + 5*(-1)^n - 5)/4;
a(n) = a(-n) = -a(n-1) + 7*n^2 - 7*n + 1. (End)
Sum_{n>=1} 1/a(n) = Pi^2/84 + tan(sqrt(5/7)*Pi/2)*Pi/(2*sqrt(35)). - Amiram Eldar, Jan 16 2023
E.g.f.: (7*x*(x + 1)*cosh(x) + (7*x^2 + 7*x - 5)*sinh(x))/2. - Stefano Spezia, Nov 30 2024

A244630 a(n) = 17*n^2.

Original entry on oeis.org

0, 17, 68, 153, 272, 425, 612, 833, 1088, 1377, 1700, 2057, 2448, 2873, 3332, 3825, 4352, 4913, 5508, 6137, 6800, 7497, 8228, 8993, 9792, 10625, 11492, 12393, 13328, 14297, 15300, 16337, 17408, 18513, 19652, 20825, 22032, 23273, 24548, 25857, 27200, 28577, 29988

Offset: 0

Vincenzo Librandi, Jul 03 2014

First bisection of A195047. - Bruno Berselli, Jul 03 2014

Norms of purely imaginary numbers in Z[sqrt(-17)] (for example, 3*sqrt(-17) has norm 153). - Alonso del Arte, Jun 23 2018

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

Cf. A008599, A195047.

Cf. similar sequences of the type k*n^2: A000290 (k = 1), A001105 (k = 2), A033428 (k = 3), A016742 (k = 4), A033429 (k = 5), A033581 (k = 6), A033582 (k = 7), A139098 (k = 8), A016766 (k = 9), A033583 (k = 10), A033584 (k = 11), A135453 (k = 12), A152742 (k = 13), A144555 (k = 14), A064761 (k = 15), A016802 (k = 16), this sequence (k = 17), A195321 (k = 18), A244631 (k = 19), A195322 (k = 20), A064762 (k = 21), A195323 (k = 22), A244632 (k = 23), A195824 (k = 24), A016850 (k = 25), A244633 (k = 26), A244634 (k = 27), A064763 (k = 28), A244635 (k = 29), A244636 (k = 30).

List([0..45],n->17*n^2); # Muniru A Asiru, Jun 29 2018

Magma
```
[17*n^2: n in [0..40]];
```

seq(17*n^2,n=0..45); # Muniru A Asiru, Jun 29 2018

Mathematica
```
Table[17 n^2, {n, 0, 40}]
```

a(n)=17*n^2 \\ Charles R Greathouse IV, Oct 07 2015

for (i <- 0 to 50) yield 17 * (i * i) // Alonso del Arte, Jun 29 2018

G.f.: 17*x*(1 + x)/(1 - x)^3. [corrected by Bruno Berselli, Jul 03 2014]
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
a(n) = 17*A000290(n). - Omar E. Pol, Jul 03 2014
a(n) = a(-n). - Muniru A Asiru, Jun 29 2018
From Elmo R. Oliveira, Dec 02 2024: (Start)
E.g.f.: 17*x*(1 + x)*exp(x).
a(n) = n*A008599(n) = A195047(2*n). (End)

A193053 a(n) = (14n(n+3) + (2n-5)(-1)^n + 21)/16.

Original entry on oeis.org

1, 5, 10, 17, 26, 36, 49, 62, 79, 95, 116, 135, 160, 182, 211, 236, 269, 297, 334, 365, 406, 440, 485, 522, 571, 611, 664, 707, 764, 810, 871, 920, 985, 1037, 1106, 1161, 1234, 1292, 1369, 1430, 1511, 1575, 1660, 1727, 1816, 1886, 1979, 2052, 2149, 2225, 2326

Offset: 0

Bruno Berselli, Oct 20 2011 - based on remarks and sequences by Omar E. Pol

For an origin of this sequence, see the numerical spiral illustrated in the Links section.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A195020 (vertices of the numerical spiral in link).

Cf. A001106, A022264, A033572, A144555, A152760, A158482, A158485, A185019, A195021, A195023-A195030, A195320, A198017 [incomplete list].

[(14*n*(n+3)+(2*n-5)*(-1)^n+21)/16: n in [0..50]];

Table[(14*n*(n + 3) + (2*n - 5)*(-1)^n + 21)/16, {n, 0, 50}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{1,5,10,17,26},60] (* Harvey P. Dale, Jun 19 2020 *)

for(n=0, 50, print1((14*n*(n+3)+(2*n-5)*(-1)^n+21)/16", "));

O.g.f.: (1 + 4*x + 3*x^2 - x^3)/((1 + x)^2*(1 - x)^3).
E.g.f.: (1/16)*((21 + 56*x + 14*x^2)*exp(x) - (5 + 2*x)*exp(-x)). - G. C. Greubel, Aug 19 2017
a(n) = A195020(n) + n + 1.
a(n)   - a(-n-1)  = A047336(n+1).
a(n+1) - a(-n)    = A113804(n+1).
a(n+2) - a(n)     = A047385(n+3).
a(n+4) - a(n)     = A113803(n+4).
a(2*n) + a(2*n-1) = A069127(n+1).
a(2*n) - a(2*n-1) = A016813(n).
a(2*n+1) - a(2*n) = A016777(n+1).
a(n+2) + 2*a(n+1) + a(n) = A024966(n+2).

A195321 a(n) = 18*n^2.

Original entry on oeis.org

0, 18, 72, 162, 288, 450, 648, 882, 1152, 1458, 1800, 2178, 2592, 3042, 3528, 4050, 4608, 5202, 5832, 6498, 7200, 7938, 8712, 9522, 10368, 11250, 12168, 13122, 14112, 15138, 16200, 17298, 18432, 19602, 20808, 22050, 23328, 24642, 25992, 27378, 28800, 30258, 31752

Offset: 0

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 0, in the direction 0, 18, ..., in the square spiral whose vertices are the generalized hendecagonal numbers A195160. Semi-axis opposite to A195316 in the same spiral.
Area of a square with diagonal 6n. - Wesley Ivan Hurt, Jun 19 2014
Number of identical tessellation tiles that are composed of 48 equilateral edge joined triangles that can be formed into a order n hexagon. The example tiles shown in the link below are tessellated with eight sphinx tiles. See A291582. - Craig Knecht, Sep 02 2017

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

Bisection of A195147.
Cf. A016802, A033581, A033583, A135453, A139098, A144555, A195322, A195323.
Cf. A000290, A001105, A008600, A016766, A033428, A195160, A195316, A291582.

[18*n^2:n in [0..40]]; // Vincenzo Librandi, Sep 20 2011

A195321:=n->18*n^2; seq(A195321(n), n=0..50); # Wesley Ivan Hurt, Jun 19 2014

18 Range[0, 50]^2 (* or *) CoefficientList[Series[18 x*(1 + x)/(1 - x)^3, {x, 0, 30}], x] (* Wesley Ivan Hurt, Jun 20 2014 *)
LinearRecurrence[{3,-3,1},{0,18,72},50] (* Harvey P. Dale, Mar 26 2023 *)

a(n)=18*n^2 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 18*A000290(n) = 9*A001105(n) = 6*A033428(n) = 3*A033581(n) = 2*A016766(n).
G.f.: 18*x*(1+x)/(1-x)^3. - Wesley Ivan Hurt, Jun 20 2014
From Elmo R. Oliveira, Dec 01 2024: (Start)
E.g.f.: 18*x*(1 + x)*exp(x).
a(n) = n*A008600(n) = A195147(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A195322 a(n) = 20*n^2.

Original entry on oeis.org

0, 20, 80, 180, 320, 500, 720, 980, 1280, 1620, 2000, 2420, 2880, 3380, 3920, 4500, 5120, 5780, 6480, 7220, 8000, 8820, 9680, 10580, 11520, 12500, 13520, 14580, 15680, 16820, 18000, 19220, 20480, 21780, 23120, 24500, 25920, 27380, 28880, 30420, 32000, 33620, 35280

Offset: 0

Omar E. Pol, Sep 16 2011

Sequence found by reading the line from 0, in the direction 0, 20, ..., in the square spiral whose vertices are the generalized dodecagonal numbers A195162. Semiaxis opposite to A195317 in the same spiral.
a(n) is the sum of all the integers less than 10*n which are not multiple of 2 or 5. a(2) = (1 + 3 + 7 + 9) + (11 + 13 + 17 + 19) = 20 + 60 = 80 = 20 * 2^2. (Link Crux Mathematicorum). - Bernard Schott, May 15 2017
Number of terms less than 10^k (k=0, 1, 2, ...): 1, 1, 3, 8, 23, 71, 224, 708, 2237, 7072, 22361, 70711, ... - Muniru A Asiru, Feb 01 2018

			From _Muniru A Asiru_, Feb 01 2018: (Start)
n=0, a(0) = 20*0^2 = 0.
n=1, a(1) = 20*1^2 = 20.
n=1, a(2) = 20*2^2 = 80.
n=1, a(3) = 20*3^2 = 180.
n=1, a(4) = 20*4^2 = 320.
...
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Léo Sauvé, Problem 53, Crux Mathematicorum, Vol. 1, Nov. 1975, page 88.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A195148.
Cf. A016802, A033581, A033583, A135453, A139098, A144555, A195321, A195323.
Cf. A000290, A001105, A005899, A010014, A016742, A033429, A195162, A195317.
Cf. A008602.

List([0..10^3],n->20*n^2); # Muniru A Asiru, Feb 01 2018

[20*n^2: n in [0..40]]; // Vincenzo Librandi, Sep 20 2011

a := n -> 20*n^2; seq(a(n), n=0..10^3); # Muniru A Asiru, Feb 01 2018

20 Range[0, 40]^2 (* or *) LinearRecurrence[{3, -3, 1}, {0, 20, 80}, 50] (* Harvey P. Dale, Jan 18 2013 *)

a(n) = 20*n^2 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 20*A000290(n) = 10*A001105(n) = 5*A016742(n) = 4*A033429(n) = 2*A033583(n).
a(0)=0, a(1)=20, a(2)=80; for n > 2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jan 18 2013
a(n) = A010014(n) - A005899(n) for n > 0. - R. J. Cano, Sep 29 2015
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 20*x*(1 + x)/(1-x)^3.
E.g.f.: 20*x*(1 + x)*exp(x).
a(n) = n*A008602(n) = A195148(2*n). (End)

A198017 a(n) = n(7n + 11)/2 + 1.

Original entry on oeis.org

1, 10, 26, 49, 79, 116, 160, 211, 269, 334, 406, 485, 571, 664, 764, 871, 985, 1106, 1234, 1369, 1511, 1660, 1816, 1979, 2149, 2326, 2510, 2701, 2899, 3104, 3316, 3535, 3761, 3994, 4234, 4481, 4735, 4996, 5264, 5539, 5821, 6110, 6406, 6709, 7019, 7336, 7660, 7991

Offset: 0

Bruno Berselli, Oct 21 2011 - based on remarks and sequences by Omar E. Pol

First bisection of A193053 (see also the numerical spiral illustrated in the Links section).

The inverse binomial transform yields 1, 9, 7, 0, 0 (0 continued).

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

Cf. A195020 (vertices of the numerical spiral in link).
Cf. A017005 (first differences).
Cf. A069099, A001106.
Cf. A001106, A022264, A033572, A144555, A152760, A158482, A158485, A185019, A193053, A195021, A195023-A195030, A195320 [incomplete list].

Magma
```
[n*(7*n+11)/2+1: n in [0..47]];
```

Table[(n(7n+11))/2+1,{n,0,60}] (* or *) LinearRecurrence[{3,-3,1},{1,10,26},60] (* Harvey P. Dale, Mar 03 2013 *)

for(n=0, 47, print1(n*(7*n+11)/2+1", "));

G.f.: (1 + 7*x - x^2)/(1-x)^3.
a(n) = A195020(2*n) + 2*n + 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) = 2*a(n-1) - a(n-2) + 7.
From Elmo R. Oliveira, Dec 24 2024: (Start)
E.g.f.: exp(x)*(2 + 18*x + 7*x^2)/2.
a(n) = n + A001106(n+1). (End)

A195021 a(n) = n(14n - 11).

Original entry on oeis.org

0, 3, 34, 93, 180, 295, 438, 609, 808, 1035, 1290, 1573, 1884, 2223, 2590, 2985, 3408, 3859, 4338, 4845, 5380, 5943, 6534, 7153, 7800, 8475, 9178, 9909, 10668, 11455, 12270, 13113, 13984, 14883, 15810, 16765, 17748, 18759, 19798, 20865, 21960, 23083, 24234, 25413

Offset: 0

Omar E. Pol, Sep 07 2011

Sequence found by reading the first two vertices [0, 3] together with the line from 34, in the direction 34, 93, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020, which is related to the primitive Pythagorean triple [3, 4, 5]. For another version see A195030.

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

Cf. A144555, A185019, A193053, A195019, A195020, A195023, A195024, A195025, A195030, A195320, A198017.

Cf. numbers of the form n*(n*k - k + 6)/2, this sequence is the case k=28: see Comments lines of A226492.

[14*n^2 - 11*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[n*(14*n - 11), {n, 0, 50}] (* Paolo Xausa, Jan 15 2025 *)

a(n)=n*(14*n-11) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 14*n^2 - 11*n.
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(3+25*x)/(1-x)^3. (End)
E.g.f.: exp(x)*x*(3 + 14*x). - Elmo R. Oliveira, Dec 30 2024

Edited by Bruno Berselli, Oct 18 2011

A195023 a(n) = 14n^2 - 4n.

Original entry on oeis.org

0, 10, 48, 114, 208, 330, 480, 658, 864, 1098, 1360, 1650, 1968, 2314, 2688, 3090, 3520, 3978, 4464, 4978, 5520, 6090, 6688, 7314, 7968, 8650, 9360, 10098, 10864, 11658, 12480, 13330, 14208, 15114, 16048, 17010, 18000, 19018, 20064, 21138, 22240, 23370, 24528, 25714

Offset: 0

Omar E. Pol, Oct 13 2011

Sequence found by reading the line from 0, in the direction 0, 10, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-axis of the square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

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

Cf. A135703, A144555, A152760, A193053, A195019, A195020, A195024, A195320, A198017.

[14*n^2 - 4*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[14n^2-4n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,10,48},50] (* Harvey P. Dale, Sep 05 2012 *)

a(n)=14*n^2-4*n \\ Charles R Greathouse IV, Apr 10 2012

a(n) = 2*A135703(n). - Bruno Berselli, Oct 13 2011
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 2*x*(5+9*x)/(1-x)^3. (End)
E.g.f.: 2*exp(x)*x*(5 + 7*x). - Elmo R. Oliveira, Dec 30 2024

Corrected by Vincenzo Librandi, Oct 14 2011

A195024 a(n) = n(14n - 1).

Original entry on oeis.org

0, 13, 54, 123, 220, 345, 498, 679, 888, 1125, 1390, 1683, 2004, 2353, 2730, 3135, 3568, 4029, 4518, 5035, 5580, 6153, 6754, 7383, 8040, 8725, 9438, 10179, 10948, 11745, 12570, 13423, 14304, 15213, 16150, 17115, 18108, 19129, 20178, 21255, 22360, 23493, 24654, 25843

Offset: 0

Omar E. Pol, Oct 13 2011

Related to the primitive Pythagorean triple [3, 4, 5].
Sequence found by reading the line from 0, in the direction 0, 13, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-diagonals of the square spiral.
Also sequence found by reading the line from 0, in the direction 0, 13, ..., in the square spiral whose vertices are the generalized 9-gonal numbers A118277. - Omar E. Pol, Jul 28 2012

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

Cf. A118277, A144555, A152760, A185019, A193053, A195019, A195020, A195320, A198017.

[14*n^2 - n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[n(14n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,13,54},50] (* Harvey P. Dale, Jul 28 2012 *)

a(n)=n*(14*n-1) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 14*n^2 - n.
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(13+15*x)/(1-x)^3. (End)
E.g.f.: exp(x)*x*(13 + 14*x). - Elmo R. Oliveira, Jan 12 2025

cp's OEIS Frontend

A195020 Vertex number of a square spiral in which the length of the first two edges are the legs of the primitive Pythagorean triple [3, 4, 5]. The edges of the spiral have length A195019.

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A195145 Concentric 14-gonal numbers.

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A244630 a(n) = 17*n^2.

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A193053 a(n) = (14*n*(n+3) + (2*n-5)*(-1)^n + 21)/16.

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A195321 a(n) = 18*n^2.

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A195322 a(n) = 20*n^2.

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A198017 a(n) = n*(7*n + 11)/2 + 1.

A193053 a(n) = (14n(n+3) + (2n-5)(-1)^n + 21)/16.

A198017 a(n) = n(7n + 11)/2 + 1.

A195021 a(n) = n(14n - 11).

A195023 a(n) = 14n^2 - 4n.

A195024 a(n) = n(14n - 1).