A181510 -id:A181510 - cp's OEIS frontend

A059993 Pinwheel numbers: a(n) = 2n^2 + 6n + 1.

1, 9, 21, 37, 57, 81, 109, 141, 177, 217, 261, 309, 361, 417, 477, 541, 609, 681, 757, 837, 921, 1009, 1101, 1197, 1297, 1401, 1509, 1621, 1737, 1857, 1981, 2109, 2241, 2377, 2517, 2661, 2809, 2961, 3117, 3277, 3441, 3609, 3781, 3957, 4137, 4321, 4509, 4701, 4897

Offset: 0

Naohiro Nomoto, Mar 14 2001

Nonnegative integers m such that 2*m + 7 is a square. - Vincenzo Librandi, Mar 01 2013
Numbers of the form 4*(h+1)*(2*h-1) + 1, where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... . - Bruno Berselli, Feb 03 2017
a(n) is also the number of vertices of the Aztec diamond AZ(n) (see Lemma 2.1 of the Imran et al. paper). - Emeric Deutsch, Sep 23 2017

M. Imran and S. Hayat, On computation of topological indices of Aztec diamonds, Sci. Int. (Lahore), Vol. 26(4), 2014, pp. 1407-1412. - Emeric Deutsch, Sep 23 2017

Harry J. Smith, Table of n, a(n) for n = 0..1000
Author?, figure. [Wayback Machine link]
Ângela Mestre and José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. numbers n such that 2*n + 2*k + 1 is a square: A046092 (k=0), A142463 (k=1), A090288 (k=2), this sequence (k=3), A139570 (k=4), A222182 (k=5), A181510 (k=6).

[2*n^2+6*n+1: n in [0..50]]; // Vincenzo Librandi, Mar 01 2013

I:=[1,9]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2)+4: n in [1..50]]; // Vincenzo Librandi, Mar 01 2013

Table[2 n^2 + 6 n + 1, {n, 0, 46}] (* Zerinvary Lajos, Jul 10 2009 *)
LinearRecurrence[{3,-3,1},{1,9,21},50] (* Harvey P. Dale, Oct 01 2018 *)

a(n) = { 2*n^2 + 6*n + 1 } \\ Harry J. Smith, Jul 01 2009

a(n) = 4*n + a(n-1) + 4 for n > 0, a(0)=1. - Vincenzo Librandi, Aug 07 2010
G.f.: (1 + 6*x - 3*x^2)/(1-x)^3. - Arkadiusz Wesolowski, Dec 24 2011
a(n) = 2*a(n-1) - a(n-2) + 4. - Vincenzo Librandi, Mar 01 2013
a(n) = Hyper2F1([-2, n], [1], -2). - Peter Luschny, Aug 02 2014
Sum_{n>=0} 1/a(n) = 1/3 + Pi*tan(sqrt(7)*Pi/2)/(2*sqrt(7)). - Amiram Eldar, Dec 13 2022
From Elmo R. Oliveira, Nov 16 2024: (Start)
E.g.f.: exp(x)*(1 + 8*x + 2*x^2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A222182 Numbers m such that 2*m + 11 is a square.

Original entry on oeis.org

-5, -1, 7, 19, 35, 55, 79, 107, 139, 175, 215, 259, 307, 359, 415, 475, 539, 607, 679, 755, 835, 919, 1007, 1099, 1195, 1295, 1399, 1507, 1619, 1735, 1855, 1979, 2107, 2239, 2375, 2515, 2659, 2807, 2959, 3115, 3275, 3439, 3607, 3779, 3955, 4135, 4319, 4507, 4699

Offset: 1

Bruno Berselli, Mar 01 2013

Except the first term, main diagonal of A155546. - Vincenzo Librandi, Mar 04 2013

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

Cf. numbers n such that 2*n + 2*k + 1 is a square: A046092 (k=0), A142463 (k=1), A090288 (k=2), A059993 (k=3), A139570 (k=4), this sequence (k=5), A181510 (k=6).
Cf. A005408 (square roots of 2*a(n)+11), A155546.
After a(2), subsequence of A168489.

[m: m in [-5..5000] | IsSquare(2*m+11)];

I:=[-5,-1,7]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Mar 04 2013

Mathematica
```
Table[2 n^2 - 2 n - 5, {n, 50}]
```

makelist(coeff(taylor(-(5-14*x+5*x^2)/(1-x)^3, x, 0, n), x, n), n, 0, 50);

a(n)=2*n^2-2*n-5 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: -x*(5 - 14*x + 5*x^2)/(1-x)^3.
a(n) = a(-n+1) = 2*n^2 - 2*n - 5.
a(n) = A046092(n-1) - 5.
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(11)*Pi/2)/(2*sqrt(11)). - Amiram Eldar, Dec 23 2022
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(2*x^2 - 5) + 5.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A271625 a(n) = = 2*(n+1)^2 - 5.

Original entry on oeis.org

3, 13, 27, 45, 67, 93, 123, 157, 195, 237, 283, 333, 387, 445, 507, 573, 643, 717, 795, 877, 963, 1053, 1147, 1245, 1347, 1453, 1563, 1677, 1795, 1917, 2043, 2173, 2307, 2445, 2587, 2733, 2883, 3037, 3195, 3357, 3523, 3693, 3867, 4045, 4227, 4413, 4603, 4797, 4995, 5197, 5403, 5613, 5827

Offset: 1

Juri-Stepan Gerasimov, Apr 11 2016

Numbers n such that 2*n + 10 is a perfect square.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000290, A201713.

Numbers h such that 2*h + k is a perfect square: A294774 (k=-9), A255843 (k=-8), A271649 (k=-7), A093328 (k=-6), A097080 (k=-5), A271624 (k=-4), A051890 (k=-3), A058331 (k=-2), A001844 (k=-1), A001105 (k=0), A046092 (k=1), A056222 (k=2), A142463 (k=3), A054000 (k=4), A090288 (k=5), A268581 (k=6), A059993 (k=7), (-1)*A147973 (k=8), A139570 (k=9), this sequence (k=10), A222182 (k=11), A152811 (k=12), A181510 (k=13), A161532 (k=14), no sequence (k=15).

Magma
```
[ 2*n^2 + 4*n - 3: n in [1..60]];
```

[ n: n in [1..6000] | IsSquare(2*n+10)];

Table[2 n^2 + 4 n - 3, {n, 53}] (* Michael De Vlieger, Apr 11 2016 *)
LinearRecurrence[{3,-3,1},{3,13,27},60] (* Harvey P. Dale, Jun 08 2023 *)
2*Range[2,60]^2 -5 (* G. C. Greubel, Jan 21 2025 *)

x='x+O('x^99); Vec(x*(3+4*x-3*x^2)/(1-x)^3) \\ Altug Alkan, Apr 11 2016

def A271625(n): return 2*pow(n+1,2) - 5
print([A271625(n) for n in range(1,61)]) # G. C. Greubel, Jan 21 2025

G.f.: x*(3 + 4*x - 3*x^2)/(1 - x)^3. - Ilya Gutkovskiy, Apr 11 2016
Sum_{n>=1} 1/a(n) = 13/30 - Pi*cot(sqrt(5/2)*Pi)/(2*sqrt(10)) = 0.5627678459924... . - Vaclav Kotesovec, Apr 11 2016
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(2*x^2 + 6*x - 3) + 3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)
a(n) = 2*A000290(n+1) - 5. - G. C. Greubel, Jan 21 2025

Name simplified by G. C. Greubel, Jan 21 2025

A155724 Triangle read by rows: T(n, k) = 2nk + n + k - 4.

Original entry on oeis.org

0, 3, 8, 6, 13, 20, 9, 18, 27, 36, 12, 23, 34, 45, 56, 15, 28, 41, 54, 67, 80, 18, 33, 48, 63, 78, 93, 108, 21, 38, 55, 72, 89, 106, 123, 140, 24, 43, 62, 81, 100, 119, 138, 157, 176, 27, 48, 69, 90, 111, 132, 153, 174, 195, 216, 30, 53, 76, 99, 122, 145, 168, 191, 214, 237, 260

Offset: 1

Vincenzo Librandi, Jan 25 2009

			Triangle begins:
   0;
   3,  8;
   6, 13, 20;
   9, 18, 27, 36;
  12, 23, 34, 45,  56;
  15, 28, 41, 54,  67,  80;
  18, 33, 48, 63,  78,  93, 108;
  21, 38, 55, 72,  89, 106, 123, 140;
  24, 43, 62, 81, 100, 119, 138, 157, 176;
  27, 48, 69, 90, 111, 132, 153, 174, 195, 216;

Vincenzo Librandi, Rows n = 1..100, flattened

All terms are in A155723.
Cf. A008585, A016885, A017053, A020705, A154685, A155722.
Cf. A162261 (row sums).
Columns k: A008585 (k=1), A016885 (k=2), A017053 (k=3), 9*A020705 (k=4).
Diagonals include: A139570, A181510, A271625.

/* Triangle: */ [[2*m*n+m+n-4: m in [1..n]]: n in [1..10]]; // Bruno Berselli, Aug 31 2012

Flatten[Table[2 n m + m + n - 4, {n, 10}, {m, n}]] (* Vincenzo Librandi, Mar 01 2012 *)

def A155724(n,k): return 2*n*k+n+k-4
print(flatten([[A155724(n,k) for k in range(1,n+1)] for n in range(1,16)])) # G. C. Greubel, Jan 21 2025

T(n, k) = A154685(n, k) - 8. - L. Edson Jeffery, Oct 12 2012
2*T(n, k) + 9 = (2*k+1)*(2*n+1). - Vincenzo Librandi, Nov 18 2012
From G. C. Greubel, Jan 21 2025: (Start)
T(2*n-1, n) = 4*n^2 + n - 5 (main diagonal).
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (1/4)*( 4*(-1)^(n+1)*n^2 + 2*(2-3*(-1)^n)*n - 7*(1-(-1)^n)).
G.f.: x*y*(3*x + 3*y - 4*x*y)/((1-x)*(1-y))^2. (End)

Edited by N. J. A. Sloane, Jun 23 2010

A271624 a(n) = 2n^2 - 4n + 4.

Original entry on oeis.org

2, 4, 10, 20, 34, 52, 74, 100, 130, 164, 202, 244, 290, 340, 394, 452, 514, 580, 650, 724, 802, 884, 970, 1060, 1154, 1252, 1354, 1460, 1570, 1684, 1802, 1924, 2050, 2180, 2314, 2452, 2594, 2740, 2890, 3044, 3202, 3364, 3530, 3700, 3874, 4052, 4234, 4420, 4610, 4804, 5002, 5204, 5410, 5620

Offset: 1

Juri-Stepan Gerasimov, Apr 11 2016

Numbers n such that 2*n - 4 is a perfect square.

For n > 2, the number of square a(n)-gonal numbers is finite. - Muniru A Asiru, Oct 16 2016

			a(1) = 2*1^2 - 4*1 + 4 = 2.

Muniru A Asiru, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, numbers n such that 2*n + k is a perfect square: no sequence (k = -9), A255843 (k = -8), A271649 (k = -7), A093328 (k = -6), A097080 (k = -5), this sequence (k = -4), A051890 (k = -3), A058331 (k = -2), A001844 (k = -1), A001105 (k = 0), A046092 (k = 1), A056222 (k = 2), A142463 (k = 3), A054000 (k = 4), A090288 (k = 5), A268581 (k = 6), A059993 (k = 7), (-1)*A147973 (k = 8), A139570 (k = 9), A271625 (k = 10), A222182 (k = 11), A152811 (k = 12), A181510 (k = 13), A161532 (k = 14), no sequence (k = 15).

Cf. A005893, A160457.

Magma
```
[ 2*n^2 - 4*n + 4: n in [1..60]];
```

[ n: n in [1..6000] | IsSquare(2*n-4)];

Table[2 n^2 - 4 n + 4, {n, 54}] (* Michael De Vlieger, Apr 11 2016 *)
LinearRecurrence[{3,-3,1},{2,4,10},60] (* Harvey P. Dale, Jul 18 2023 *)

x='x+O('x^99); Vec(2*x*(1-x+2*x^2)/(1-x)^3) \\ Altug Alkan, Apr 11 2016

a(n)=2*n^2-4*n+4 \\ Charles R Greathouse IV, Apr 11 2016

a(n) = 2*A002522(n-1).
G.f.: 2*x*(1 - x + 2*x^2)/(1 - x)^3. - Ilya Gutkovskiy, Apr 11 2016
Sum_{n>=1} 1/a(n) = (1 + Pi*coth(Pi))/4 = 1.038337023734290587067... . - Vaclav Kotesovec, Apr 11 2016
a(n) = A005893(n-1), n > 1. - R. J. Mathar, Apr 12 2016
a(n) = 2 + 2*(n-1)^2. - Tyler Skywalker, Jul 21 2016
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: 2*(exp(x)*(x^2 - x + 2) - 2).
a(n) = 2*A160457(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A271649 a(n) = 2*(n^2 - n + 2).

Original entry on oeis.org

4, 8, 16, 28, 44, 64, 88, 116, 148, 184, 224, 268, 316, 368, 424, 484, 548, 616, 688, 764, 844, 928, 1016, 1108, 1204, 1304, 1408, 1516, 1628, 1744, 1864, 1988, 2116, 2248, 2384, 2524, 2668, 2816, 2968, 3124, 3284, 3448, 3616, 3788, 3964, 4144, 4328, 4516, 4708, 4904, 5104, 5308, 5516

Offset: 1

Juri-Stepan Gerasimov, Apr 11 2016

Numbers n such that 2*n - 7 is a perfect square.

Galois numbers for three-dimensional vector space, defined as the total number of subspaces in a three-dimensional vector space over GF(n-1), when n-1 is a power of a prime. - Artur Jasinski, Aug 31 2016, corrected by Robert Israel, Sep 23 2016

			a(1) = 2*(1^2 - 1 + 2) = 4.

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

Cf. A000124, A014206, A137882.

Numbers h such that 2*h + k is a perfect square: no sequence (k=-9), A255843 (k=-8), this sequence (k=-7), A093328 (k=-6), A097080 (k=-5), A271624 (k=-4), A051890 (k=-3), A058331 (k=-2), A001844 (k=-1), A001105 (k=0), A046092 (k=1), A056222 (k=2), A142463 (k=3), A054000 (k=4), A090288 (k=5), A268581 (k=6), A059993 (k=7), (-1)*A147973 (k=8), A139570 (k=9), A271625 (k=10), A222182 (k=11), A152811 (k=12), A181510 (k=13), A161532 (k=14), no sequence (k=15).

Magma
```
[ 2*n^2 - 2*n + 4: n in [1..60]];
```

[ n: n in [1..6000] | IsSquare(2*n-7)];

A271649:=n->2*(n^2-n+2): seq(A271649(n), n=1..60); # Wesley Ivan Hurt, Aug 31 2016

Table[2 (n^2 - n + 2), {n, 53}] (* or *)
Select[Range@ 5516, IntegerQ@ Sqrt[2 # - 7] &] (* or *)
Table[SeriesCoefficient[(-4 (1 - x + x^2))/(-1 + x)^3, {x, 0, n}], {n, 0, 52}] (* Michael De Vlieger, Apr 11 2016 *)
LinearRecurrence[{3,-3,1},{4,8,16},60] (* Harvey P. Dale, Jun 14 2022 *)

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

a(n) = 4*A000124(n).
a(n) = 2*A014206(n).
a(n) = A137882(n), n > 1. - R. J. Mathar, Apr 12 2016
Sum_{n>=1} 1/a(n) = tanh(sqrt(7)*Pi/2)*Pi/(2*sqrt(7)). - Amiram Eldar, Jul 30 2024
From Elmo R. Oliveira, Nov 18 2024: (Start)
G.f.: 4*x*(1 - x + x^2)/(1 - x)^3.
E.g.f.: 2*(exp(x)*(x^2 + 2) - 2).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A213197 T(n,k) = (2(n+k)^2 - 2(n+k) - 4k + 6 + (2k-2)(-1)^n + (2k-1)(-1)^k + (-2n+1)*(-1)^(n+k))/4; n, k > 0, read by antidiagonals.

Original entry on oeis.org

1, 3, 4, 2, 6, 5, 8, 9, 11, 12, 7, 15, 10, 14, 13, 17, 18, 20, 21, 23, 24, 16, 28, 19, 27, 22, 26, 25, 30, 31, 33, 34, 36, 37, 39, 40, 29, 45, 32, 44, 35, 43, 38, 42, 41, 47, 48, 50, 51, 53, 54, 56, 57, 59, 60, 46, 66, 49, 65, 52, 64, 55, 63, 58, 62, 61, 68

Offset: 1

Boris Putievskiy, Mar 01 2013

Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Enumeration table T(n,k). Let m be natural number. The order of the list:
T(1,1)=1;
T(3,1), T(2,2), T(1,3);
T(2,1), T(1,2);
...
T(1,2*m+1), T(1,2*m), T(2, 2*m-1), T(3, 2*m-1),... T(2*m,1), T(2*m+1,1);
T(2*m,2), T(2*m-2,4), ...T(2,2*m);
...
Movement along two adjacent antidiagonals. The first row consists of phases: step to the west, step to the southwest, step to the south. The second row consists of phases: 2 steps to the north, 2 steps to the east. The length of each step is 1.

			The start of the sequence as a table:
   1,  3,  2,  8,  7, 17, 16, ...
   4,  6,  9, 15, 18, 28, 31, ...
   5, 11, 10, 20, 19, 33, 32, ...
  12, 14, 21, 27, 34, 44, 51, ...
  13, 23, 22, 36, 35, 53, 52, ...
  24, 26, 37, 43, 54, 64, 75, ...
  25, 39, 38, 56, 55, 77, 76, ...
  ...
The start of the sequence as a triangular array read by rows:
   1;
   3,  4;
   2,  6,  5;
   8,  9, 11, 12;
   7, 15, 10, 14, 13;
  17, 18, 20, 21, 23, 24;
  16, 28, 19, 27, 22, 26, 25;
  ...
The start of the sequence as an array read by rows, the length of row r is 4*r-3.
First 2*r-2 numbers are from row 2*r-2 of the triangular array above.
Last  2*r-1 numbers are from row 2*r-1 of the triangular array above.
   1;
   3,  4,  2,  6,  5;
   8,  9, 11, 12,  7, 15, 10, 14, 13;
  17, 18, 20, 21, 23, 24, 16, 28, 19, 27, 22, 26, 25;
  ...
Row r contains permutation of 4*r-3 numbers from 2*r*r-5*r+4 to 2*r*r-r:
2*r*r-5*r+5, 2*r*r-5*r+6, ..., 2*r*r-2*r+2, 2*r*r-2*r+1.

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Eric Weisstein's World of Mathematics, Pairing functions
Index entries for sequences that are permutations of the natural numbers

Cf. A002260, A004736, A003056, A003057.

Table T(n,k) contains: in rows A130883, A033816, A100037, A000384, A100038, A014106, A091823; in columns A001844, A142463, A090288, A139570, A046092, A051890, A059993, A097080, A181510, A137882, A152813.

T:=(n,k)->(2*(n+k)^2-2*(n+k)-4*k+6+(2*k-2)*(-1)^n+(2*k-1)*(-1)^k+(1-+2*n)*(-1)^(n+k))/4: seq(seq(T(k,n-k),k=1..n-1),n=1..13); # Muniru A Asiru, Dec 06 2018

T[n_, k_] := (2(n+k)^2 - 2(n+k) - 4k + 6 + (2k-2)(-1)^n + (2k-1)(-1)^k + (-2n+1)(-1)^(n+k))/4;
Table[T[n-k+1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Dec 06 2018 *)

t=int((math.sqrt(8*n-7) - 1)/ 2)
i=n-t*(t+1)/2
j=(t*t+3*t+4)/2-n
result=(2*(t+2)**2-2*(t+2)-4*j+6 +(2*j-2)*(-1)**i+(2*j-1)*(-1)**j+(-2*i+1)*(-1)**t)/4

As a table:
T(n,k) = (2*(n+k)^2 - 2*(n+k) - 4*k + 6 + (2*k-2)*(-1)^n + (2*k-1)*(-1)^k + (-2*n+1)*(-1)^(n+k))/4.
As a linear sequence:
a(n) = (2*A003057(n)^2 - 2*A003057(n) - 4*A004736(n) + 6 + (2*A004736(n)-2)*(-1)^A002260(n) + (2*A004736(n)-1)*(-1)^A004736(n) + (-2*A002260(n)+1)*(-1)^A003056(n))/4;
a(n) = (2*(t+2)^2 - 2*(t+2) - 4*j + 6 + (2*j-2)*(-1)^i + (2*j-1)*(-1)^j + (-2*i+1)*(-1)^t)/4, where i = n - t*(t+1)/2, j = (t*t + 3*t + 4)/2 - n, t = floor((-1+sqrt(8*n-7))/2).

A294774 a(n) = 2n^2 + 2n + 5.

Original entry on oeis.org

5, 9, 17, 29, 45, 65, 89, 117, 149, 185, 225, 269, 317, 369, 425, 485, 549, 617, 689, 765, 845, 929, 1017, 1109, 1205, 1305, 1409, 1517, 1629, 1745, 1865, 1989, 2117, 2249, 2385, 2525, 2669, 2817, 2969, 3125, 3285, 3449, 3617, 3789, 3965, 4145, 4329, 4517, 4709, 4905

Offset: 0

Bruno Berselli, Nov 08 2017

This is the case k = 9 of 2*n^2 + (1-(-1)^k)*n + (2*k-(-1)^k+1)/4 (similar sequences are listed in Crossrefs section). Note that:

2*( 2*n^2 + (1-(-1)^k)*n + (2*k-(-1)^k+1)/4 ) - k = ( 2*n + (1-(-1)^k)/2 )^2. From this follows an alternative definition for the sequence: Numbers h such that 2*h - 9 is a square. Therefore, if a(n) is a square then its base is a term of A075841.

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

1st diagonal of A154631, 3rd diagonal of A055096, 4th diagonal of A070216.
Second column of Mathar's array in A016813 (Comments section).
Subsequence of A001481, A001983, A004766, A020668, A046711 and A057653 (because a(n) = (n+2)^2 + (n-1)^2); A097268 (because it is also a(n) = (n^2+n+3)^2 - (n^2+n+2)^2); A047270; A243182 (for y=1).
Similar sequences (see the first comment): A161532 (k=-14), A181510 (k=-13), A152811 (k=-12), A222182 (k=-11), A271625 (k=-10), A139570 (k=-9), (-1)*A147973 (k=-8), A059993 (k=-7), A268581 (k=-6), A090288 (k=-5), A054000 (k=-4), A142463 or A132209 (k=-3), A056220 (k=-2), A046092 (k=-1), A001105 (k=0), A001844 (k=1), A058331 (k=2), A051890 (k=3), A271624 (k=4), A097080 (k=5), A093328 (k=6), A271649 (k=7), A255843 (k=8), this sequence (k=9).

seq(2*n^2 + 2*n + 5, n=0..100); # Robert Israel, Nov 10 2017

Table[2n^2+2n+5,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{5,9,17},50] (* Harvey P. Dale, Sep 18 2023 *)

Vec((5 - 6*x + 5*x^2) / (1 - x)^3 + O(x^50)) \\ Colin Barker, Nov 13 2017

O.g.f.: (5 - 6*x + 5*x^2)/(1 - x)^3.
E.g.f.: (5 + 4*x + 2*x^2)*exp(x).
a(n) = a(-1-n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 5*A000217(n+1) - 6*A000217(n) + 5*A000217(n-1).
n*a(n) - Sum_{j=0..n-1} a(j) = A002492(n) for n>0.
a(n) = Integral_{x=0..2n+4} |3-x| dx. - Pedro Caceres, Dec 29 2020

cp's OEIS Frontend

A059993 Pinwheel numbers: a(n) = 2*n^2 + 6*n + 1.

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A222182 Numbers m such that 2*m + 11 is a square.

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A271625 a(n) = = 2*(n+1)^2 - 5.

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A155724 Triangle read by rows: T(n, k) = 2*n*k + n + k - 4.

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A271624 a(n) = 2*n^2 - 4*n + 4.

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A271649 a(n) = 2*(n^2 - n + 2).

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A059993 Pinwheel numbers: a(n) = 2n^2 + 6n + 1.

A155724 Triangle read by rows: T(n, k) = 2nk + n + k - 4.

A271624 a(n) = 2n^2 - 4n + 4.

A213197 T(n,k) = (2(n+k)^2 - 2(n+k) - 4k + 6 + (2k-2)(-1)^n + (2k-1)(-1)^k + (-2n+1)*(-1)^(n+k))/4; n, k > 0, read by antidiagonals.

A294774 a(n) = 2n^2 + 2n + 5.