A154685 -id:A154685 - cp's OEIS frontend

A151675 Row sums of A154685.

8, 27, 63, 122, 210, 333, 497, 708, 972, 1295, 1683, 2142, 2678, 3297, 4005, 4808, 5712, 6723, 7847, 9090, 10458, 11957, 13593, 15372, 17300, 19383, 21627, 24038, 26622, 29385, 32333, 35472, 38808, 42347, 46095, 50058, 54242, 58653, 63297

Offset: 1

N. J. A. Sloane, May 31 2009

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A154685, A162261.

I:=[8, 27, 63, 122]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 30 2012

CoefficientList[Series[(8-5*x+3*x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 30 2012 *)

def A151675(n): return n*(2*n**2 +5*n+9)//2
print([A151675(n) for n in range(1,51)]) # G. C. Greubel, Jan 21 2025

From R. J. Mathar, May 31 2009: (Start)
a(n) = n*(2*n^2 + 5*n + 9)/2.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: x*(8 - 5*x + 3*x^2)/(1-x)^4. (End)
a(n) = A162261(n) + 8*n. - L. Edson Jeffery, Oct 12 2012
E.g.f.: (1/2)*x*(16 + 11*x + 2*x^2)*exp(x). - G. C. Greubel, Jan 21 2025

Extended by R. J. Mathar, May 31 2009

A056220 a(n) = 2*n^2 - 1.

Original entry on oeis.org

-1, 1, 7, 17, 31, 49, 71, 97, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 647, 721, 799, 881, 967, 1057, 1151, 1249, 1351, 1457, 1567, 1681, 1799, 1921, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 3041, 3199, 3361, 3527, 3697, 3871, 4049, 4231, 4417, 4607, 4801

Offset: 0

N. J. A. Sloane, Aug 06 2000

Image of squares (A000290) under "little Hankel" transform that sends [c_0, c_1, ...] to [d_0, d_1, ...] where d_n = c_n^2 - c_{n+1}*c_{n-1}. - Henry Bottomley, Dec 12 2000
Surround numbers of an n X n square. - Jason Earls, Apr 16 2001
Numbers n such that 2*n + 2 is a perfect square. - Cino Hilliard, Dec 18 2003, Juri-Stepan Gerasimov, Apr 09 2016
The sums of the consecutive integer sequences 2n^2 to 2(n+1)^2-1 are cubes, as 2n^2 + ... + 2(n+1)^2-1 = (1/2)(2(n+1)^2 - 1 - 2n^2 + 1)(2(n+1)^2 - 1 + 2n^2) = (2n+1)^3. E.g., 2+3+4+5+6+7 = 27 = 3^3, then 8+9+10+...+17 = 125 = 5^3. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 29 2005
X values (other than 0) of solutions to the equation 2*X^3 + 2*X^2 = Y^2. To find Y values: b(n) = 2n*(2*n^2 - 1). - Mohamed Bouhamida, Nov 06 2007
Average of the squares of two consecutive terms is also a square. In fact: (2*n^2 - 1)^2 + (2*(n+1)^2 - 1)^2 = 2*(2*n^2 + 2*n + 1)^2. - Matias Saucedo (solomatias(AT)yahoo.com.ar), Aug 18 2008
Equals row sums of triangle A143593 and binomial transform of [1, 6, 4, 0, 0, 0, ...] with n > 1. - Gary W. Adamson, Aug 26 2008
Start a spiral of square tiles. Trivially the first tile fits in a 1 X 1 square. 7 tiles fit in a 3 X 3 square, 17 tiles fit in a 5 X 5 square and so on. - Juhani Heino, Dec 13 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jan 26 2010
For each n > 0, the recursive series, formula S(b) = 6*S(b-1) - S(b-2) - 2*a(n) with S(0) = 4n^2-4n+1 and S(1) = 2n^2, has the property that every even term is a perfect square and every odd term is twice a perfect square. - Kenneth J Ramsey, Jul 18 2010
Fourth diagonal of A154685 for n > 2. - Vincenzo Librandi, Aug 07 2010
First integer of (2*n)^2 consecutive integers, where the last integer is 3 times the first + 1. As example, n = 2: term = 7; (2*n)^2 = 16; 7, 8, 9, ..., 20, 21, 22: 7*3 + 1 = 22. - Denis Borris, Nov 18 2012
Chebyshev polynomial of the first kind T(2,n). - Vincenzo Librandi, May 30 2014
For n > 0, number of possible positions of a 1 X 2 rectangle in a (n+1) X (n+2) rectangular integer lattice. - Andres Cicuttin, Apr 07 2016
This sequence also represents the best solution for Ripà's n_1 X n_2 X n_3 dots problem, for any 0 < n_1 = n_2 < n_3 = floor((3/2)*(n_1 - 1)) + 1. - Marco Ripà, Jul 23 2018

			a(0) = 0^2-1*1 = -1, a(1) = 1^2 - 4*0 = 1, a(2) = 2^2 - 9*1 = 7, etc.
a(4) = 31 = (1, 3, 3, 1) dot (1, 6, 4, 0) = (1 + 18 + 12 + 0). - _Gary W. Adamson_, Aug 29 2008

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq., Vol. 13 (2010), Article # 10.7.8.
Mitch Phillipson, Manda Riehl, and Tristan Williams, Enumeration of Wilf classes in Sn ~ Cr for two patterns of length 3, PU. M. A., Vol. 21, No. 2 (2010), pp. 321-338.
Marco Ripà, The rectangular spiral or the n1 X n2 X ... X nk Points Problem , Notes on Number Theory and Discrete Mathematics, Vol. 20, No. 1 (2014), pp. 59-71.
Leo Tavares, Illustration: Twin Squares
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000105, A000217, A000290, A000384, A001082, A001653, A001844, A002378, A002593, A003215, A005563, A028347, A036666, A046092, A047875, A062717, A069074, A077585, A087475, A119258, A143593, A154685, A162610, A188653, A225227.
Cf. A066049 (indices of prime terms)
Column 2 of array A188644 (starting at offset 1).

List([0..50], n-> 2*n^2-1); # Muniru A Asiru, Jul 24 2018

[2*n^2-1 : n in [0..50]]; // Vincenzo Librandi, May 30 2014

A056220:=n->2*n^2-1; seq(A056220(n), n=0..50); # Wesley Ivan Hurt, Jun 16 2014

Array[2 #^2 - 1 &, 50, 0] (* Robert G. Wilson v, Jul 23 2018 *)
CoefficientList[Series[(x^2 +4x -1)/(1-x)^3, {x, 0, 50}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {-1, 1, 7}, 51] (* Robert G. Wilson v, Jul 24 2018 *)

PARI
```
a(n)=2*n^2-1;
```

[2*n^2-1 for n in (0..50)] # G. C. Greubel, Jul 07 2019

G.f.: (-1 + 4*x + x^2)/(1-x)^3. - Henry Bottomley, Dec 12 2000
a(n) = A119258(n+1,2) for n > 0. - Reinhard Zumkeller, May 11 2006
From Doug Bell, Mar 08 2009: (Start)
a(0) = -1,
a(n) = sqrt(A001844(n)^2 - A069074(n-1)),
a(n+1) = sqrt(A001844(n)^2 + A069074(n-1)) = sqrt(a(n)^2 + A069074(n-1)*2). (End)
a(n) + a(n+1) + 1 = (2n+1)^2. - Doug Bell, Mar 09 2009
a(n) = a(n-1) + 4*n - 2 (with a(0)=-1). - Vincenzo Librandi, Dec 25 2010
a(n) = A188653(2*n) for n > 0. - Reinhard Zumkeller, Apr 13 2011
a(n) = A162610(2*n-1,n) for n > 0. - Reinhard Zumkeller, Jan 19 2013
a(n) = ( Sum_{k=0..2} (C(n+k,3)-C(n+k-1,3))*(C(n+k,3)+C(n+k+1,3)) ) - (C(n+2,3)-C(n-1,3))*(C(n,3)+C(n+3,3)), for n > 3. - J. M. Bergot, Jun 16 2014
a(n) = j^2 + k^2 - 2 or 2*j*k if n >= 2 and j = n + sqrt(2)/2 and k = n - sqrt(2)/2. - Avi Friedlich, Mar 30 2015
a(n) = A002593(n)/n^2. - Bruce J. Nicholson, Apr 03 2017
a(n) = A000384(n) + n - 1. - Bruce J. Nicholson, Nov 12 2017
a(n)*a(n+k) + 2k^2 = m^2 (a perfect square), m = a(n) + (2n*k), for n>=1. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Aug 10 2020: (Start)
Sum_{n>=1} 1/a(n) = 1/2 - sqrt(2)*Pi*cot(Pi/sqrt(2))/4.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(2)*Pi*csc(Pi/sqrt(2))/4 - 1/2. (End)
From Amiram Eldar, Feb 04 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(2))*csc(Pi/sqrt(2)).
Product_{n>=2} (1 - 1/a(n)) = (Pi/(4*sqrt(2)))*csc(Pi/sqrt(2)). (End)
a(n) = A003215(n) - A000217(n-2)*2. - Leo Tavares, Jun 29 2021
Let T(n) = n*(n+1)/2. Then a(n)^2 = T(2n-2)*T(2n+1) + n^2. - Charlie Marion, Feb 12 2023
E.g.f.: exp(x)*(2*x^2 + 2*x - 1). - Stefano Spezia, Jul 08 2023

A051890 a(n) = 2*(n^2 - n + 1).

Original entry on oeis.org

2, 2, 6, 14, 26, 42, 62, 86, 114, 146, 182, 222, 266, 314, 366, 422, 482, 546, 614, 686, 762, 842, 926, 1014, 1106, 1202, 1302, 1406, 1514, 1626, 1742, 1862, 1986, 2114, 2246, 2382, 2522, 2666, 2814, 2966, 3122, 3282, 3446, 3614, 3786, 3962

Offset: 0

Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Apr 30 2000

Draw n ellipses in the plane (n > 0); sequence gives maximum number of regions into which the plane is divided (cf. A014206, A386480).
Least k such that Z(k,2) <= Z(n,3) where Z(m,s) = Sum_{i>=m} 1/i^s = zeta(s) - Sum_{i=1..m-1} 1/i^s. - Benoit Cloitre, Nov 29 2002
For n > 2, third diagonal of A154685. - Vincenzo Librandi, Aug 06 2010
a(k) is also the Moore lower bound A198300(k,6) on the order A054760(k,6) of a (k,6)-cage. Equality is achieved if and only if there exists a finite projective plane of order k - 1. A sufficient condition for this is that k - 1 be a prime power. - Jason Kimberley, Oct 17 2011 and Jan 01 2013
From Jess Tauber, May 20 2013: (Start)
For neutron shell filling in spherical atomic nuclei, this sequence shows numerical differences between filled spin-split suborbitals sharing all quantum numbers except the principal quantum number n, and here all n's must differ by 1. Only a small handful of exceptions exist.
This sequence consists of summed pairs of every other doubled triangular number. It also can be created by taking differences between nuclear magic numbers from the harmonic oscillator (HO)(doubled tetrahedral) set and the spin-orbit (SO) set (2,6,14,28,50,82,126,184,...), with either set being larger. So SO-HO: 2-0=2, 6-0=6, 14-0=14, 28-2=26, 50-8=42, 82-20=62, 126-40=86, 184-70=114, and HO-SO: 2-0=2, 8-2=6, 20-6=14, 40-14=26, 70-28=42, 112-50=62, 168-82=86, 240-126=114. From the perspective of idealized HO periodic structure, with suborbitals in order from largest to smallest spin, alternating by parity, the HO-SO set is spaced two period analogs PLUS one suborbital, while the SO-HO set is spaced two period analogs MINUS one suborbital. (End)
The known values of f(k,6) and F(k,6) in Brown (1967), Table 1, closely match this sequence. - N. J. A. Sloane, Jul 09 2015
Numbers k such that 2*k - 3 is a square. - Bruno Berselli, Nov 08 2017
Numbers written 222 in number base B, including binary with 'digit' 2: 222(2)=14, 222(3)=26, ... - Ron Knott, Nov 14 2017

G. C. Greubel, Table of n, a(n) for n = 0..5000
William G. Brown, On Hamiltonian regular graphs of girth six, J. London Math. Soc., 42 (1967): 514-520.
Steven Edwards and William Griffiths, On Generalized Delannoy Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.3.6.
William Q. Erickson and Jan Kretschmann, The structure and normalized volume of Monge polytopes, arXiv:2311.07522 [math.CO], 2023. See p. 10.
Parabola, Problem #Q607, vol. 20, no. 2, 1984, p. 27.
Eric Weisstein's World of Mathematics, Plane Division by Ellipses
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001844, A002061, A014206, A154685, A195600, A386480.

Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), A002522 (g=5), this sequence (g=6), A188377 (g=7).

List([0..50], n-> 2*(n^2-n+1)); # G. C. Greubel, Feb 21 2019

[2*(n^2-n+1): n in [0..50]]; // G. C. Greubel, Feb 21 2019

A051890 := n->2*(n^2-n+1); seq(A051890(n) = n=0..50);

Table[2*(n^2-n+1), {n, 0, 50}] (* G. C. Greubel, Jul 14 2017 *)
LinearRecurrence[{3,-3,1},{2,2,6},50] (* Harvey P. Dale, Jul 14 2025 *)

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

[2*(n^2-n+1) for n in (0..50)] # G. C. Greubel, Feb 21 2019

a(n) = 4*binomial(n, 2) + 2. - Francois Jooste (phukraut(AT)hotmail.com), Mar 05 2003
For n > 2, nearest integer to (Sum_{k>=n} 1/k^3)/(Sum_{k>=n} 1/k^5). - Benoit Cloitre, Jun 12 2003
a(n) = 2*A002061(n). - Jonathan Vos Post, Jun 19 2005
a(n) = 4*n + a(n-1) - 4 for n > 0, a(0)=2. - Vincenzo Librandi, Aug 06 2010
a(n) = 2*(n^2 - n +1) = 2*(n-1)^2 + 2(n-1) + 2 = 222 read in base n-1 (for n > 3). - Jason Kimberley, Oct 20 2011
G.f.: 2*(1 - 2*x + 3*x^2)/(1 - x)^3. - Colin Barker, Jan 10 2012
a(n) = A001844(n-1) + 1 = A046092(n-1) + 2. - Jaroslav Krizek, Dec 27 2013
E.g.f.: 2*(x^2 + 1)*exp(x). - G. C. Greubel, Jul 14 2017

A093328 a(n) = 2*n^2 + 3.

Original entry on oeis.org

3, 5, 11, 21, 35, 53, 75, 101, 131, 165, 203, 245, 291, 341, 395, 453, 515, 581, 651, 725, 803, 885, 971, 1061, 1155, 1253, 1355, 1461, 1571, 1685, 1803, 1925, 2051, 2181, 2315, 2453, 2595, 2741, 2891, 3045, 3203, 3365, 3531, 3701, 3875, 4053, 4235, 4421, 4611

Offset: 0

Ralf Stephan, Apr 25 2004

Number of 132-avoiding two-stack sortable permutations which also avoid 4321.
Conjecture: no perfect powers. - Zak Seidov, Sep 27 2015
Numbers k such that 2*k - 6 is a square. - Bruno Berselli, Nov 08 2017

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Steven Edwards and William Griffiths, Generalizations of Delannoy and cross polytope numbers, Fib. Q., Vol. 55, No. 4 (2017), pp. 356-366.
Steven Edwards and William Griffiths, On Generalized Delannoy Numbers, J. Int. Seq., Vol. 23 (2020), Article 20.3.6.
Eric S. Egge and Toufik Mansour, 132-avoiding two-stack sortable permutations, Fibonacci numbers, and Pell numbers, Discrete Applied Mathematics, Vol. 143, No. 1-3 (2004), pp. 72-83; arXiv preprint, arXiv:math/0205206 [math.CO], 2002.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001105, A005893, A058331, A154685.

[2*n^2+3: n in [0..50]]; // Vincenzo Librandi, Jul 08 2012

Table[2 n^2 + 3, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Feb 15 2011*)
CoefficientList[Series[(3 - 4 x + 5 x^2)/(1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jul 08 2012 *)
LinearRecurrence[{3, -3, 1}, {3, 5, 11}, 50] (* Harvey P. Dale, Apr 03 2016 *)

a(n)=2*n^2+3; \\ Zak Seidov, Sep 27 2015

a(n) = A005893(n)+1 = A058331(n)+2 = A001105(n)+3.
a(n+2) = A154685(n+1,n+2).
From Vincenzo Librandi, Jul 08 2012: (Start)
G.f.: (3 - 4*x + 5*x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
Sum_{n>=0} 1/a(n) = (1 + sqrt(3/2)*Pi*coth(sqrt(3/2)*Pi))/6. - Amiram Eldar, Nov 25 2020
E.g.f.: exp(x)*(3 + 2*x + 2*x^2). - Elmo R. Oliveira, Jan 17 2025

Simpler definition and new offset from Paul F. Brewbaker, Jun 23 2009

Edited by N. J. A. Sloane, Jun 27 2009

A137882 Number of (directed) Hamiltonian paths in the n-ladder graph.

Original entry on oeis.org

2, 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, 5728, 5944, 6164, 6388, 6616, 6848, 7084, 7324, 7568, 7816

Offset: 1

Eric W. Weisstein, Feb 20 2008

G. C. Greubel, Table of n, a(n) for n = 1..1000
Eric Weisstein's World of Mathematics, Hamiltonian Path.
Eric Weisstein's World of Mathematics, Ladder Graph.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001221, A001222, A003682, A154685.

A137882:=n->2*(n^2-n+2): 2,seq(A137882(n), n=2..100); # Wesley Ivan Hurt, Apr 25 2017

CoefficientList[Series[2*x*(1 + x - x^2 + x^3)/(1 - x)^3, {x,0,50}], x] (* G. C. Greubel, Apr 25 2017 *)
LinearRecurrence[{3,-3,1},{2,8,16,28},70] (* Harvey P. Dale, Nov 15 2018 *)

my(x='x+O('x^50)); Vec(2*x*(1 + x - x^2 + x^3)/(1 - x)^3) \\ G. C. Greubel, Apr 25 2017

For n>2, m = p^3*q (p,q = primes), a(n) = Sum_{d|m} (n-1)^(bigomega(d) - omega(d)) = Sum_{d|m} (n-1)^(A001222(d) - A001221(d)). - Jaroslav Krizek, Sep 24 2009
For n>1, a(n) = 2*(n^2 - n + 2); first diagonal of A154685. - Vincenzo Librandi, Nov 24 2010
G.f.: 2*x*(1+x-x^2+x^3)/(1-x)^3. - Colin Barker, Jan 20 2012
Sum_{n>=1} 1/a(n) = 1/4 + Pi*tanh(sqrt(7)*Pi/2)/(2*sqrt(7)). - Amiram Eldar, Dec 23 2022
From Elmo R. Oliveira, Jun 06 2025: (Start)
E.g.f.: 2*(exp(x)*(2 + x^2) - (2 + x)).
a(n) = 2*A003682(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 4. (End)

Extended and formula corrected by Max Alekseyev, Apr 11 2009

Corrected the formula which was confusing offsets - R. J. Mathar, Jun 04 2010

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

A153037 a(n) = 2n^2 + 16n + 23.

Original entry on oeis.org

23, 41, 63, 89, 119, 153, 191, 233, 279, 329, 383, 441, 503, 569, 639, 713, 791, 873, 959, 1049, 1143, 1241, 1343, 1449, 1559, 1673, 1791, 1913, 2039, 2169, 2303, 2441, 2583, 2729, 2879, 3033, 3191, 3353, 3519, 3689, 3863, 4041, 4223, 4409, 4599, 4793, 4991, 5193

Offset: 0

Vincenzo Librandi, Jan 25 2009

Sixth diagonal of triangle A154685.

Numbers of the form 2*k^2 - 9. - Bruno Berselli, Oct 30 2012

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

Cf. A153039, A154685.

I:=[23, 41, 63]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 22 2012

Table[2*n^2 + 16*n + 23, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Feb 03 2012 *)
LinearRecurrence[{3, -3, 1}, {23, 41, 63}, 50] (* Vincenzo Librandi, Feb 22 2012 *)
CoefficientList[Series[(23 - 28*x  +9*x^2)/(1 -x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Jan 04 2013 *)

a(n)=2*n^2+16*n+23 \\ Charles R Greathouse IV, Jan 11 2012

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Vincenzo Librandi, Feb 22 2012
G.f.: (23 - 28*x + 9*x^2)/(1-x)^3. - Vincenzo Librandi, Jan 04 2013
From Amiram Eldar, Feb 25 2023: (Start)
Sum_{n>=0} 1/a(n) = 137/126 - cot(3*Pi/sqrt(2))*Pi/(6*sqrt(2)).
Sum_{n>=0} (-1)^n/a(n) = 43/42 - cosec(3*Pi/sqrt(2))*Pi/(6*sqrt(2)). (End)
E.g.f.: exp(x)*(23 + 18*x + 2*x^2). - Elmo R. Oliveira, Feb 08 2025

Erroneously duplicated terms removed by Vincenzo Librandi, Feb 22 2012

A163652 Triangle read by rows where T(n,m)=2mn + m + n + 6.

Original entry on oeis.org

10, 13, 18, 16, 23, 30, 19, 28, 37, 46, 22, 33, 44, 55, 66, 25, 38, 51, 64, 77, 90, 28, 43, 58, 73, 88, 103, 118, 31, 48, 65, 82, 99, 116, 133, 150, 34, 53, 72, 91, 110, 129, 148, 167, 186, 37, 58, 79, 100, 121, 142, 163, 184, 205, 226, 40, 63, 86, 109, 132, 155, 178

Offset: 1

Vincenzo Librandi, Aug 02 2009

The numbers 2*T(n,m)-11 = (2*n+1)*(2*m+1) are not prime, and 2*T(n,n) = (2n+1)^2.

First column: A112414, second column: A016885, third column: A017005, fourth column: A017173. - Vincenzo Librandi, Nov 20 2012

			Triangle begins:
  10;
  13, 18;
  16, 23, 30;
  19, 28, 37, 46;
  22, 33, 44, 55,  66;
  25, 38, 51, 64,  77,  90;
  28, 43, 58, 73,  88,  103, 118;
  31, 48, 65, 82,  99,  116, 133, 150;
  34, 53, 72, 91,  110, 129, 148, 167, 186;
  37, 58, 79, 100, 121, 142, 163, 184, 205, 226;
  40, 63, 86, 109, 132, 155, 178, 201, 224, 247, 270;
  etc.

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

Cf. A153045, A154685, A163657, A112414, A016885, A017005, A017173.

[2*n*k + n + k + 6: k in [1..n], n in [1..11]]; // Vincenzo Librandi, Nov 20 2012

t[n_,k_]:=2 n*k + n + k +  6; Table[t[n, k], {n, 15}, {k, n}]//Flatten (* Vincenzo Librandi, Nov 20 2012 *)

T(n,m) = A154685(n,m)+2 = A163657(n,m)-2. [R. J. Mathar, Oct 22 2009]

Comment clarified by R. J. Mathar, Oct 22 2009

cp's OEIS Frontend

A151675 Row sums of A154685.

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A056220 a(n) = 2*n^2 - 1.

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A051890 a(n) = 2*(n^2 - n + 1).

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A093328 a(n) = 2*n^2 + 3.

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A137882 Number of (directed) Hamiltonian paths in the n-ladder graph.

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

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

A153037 a(n) = 2n^2 + 16n + 23.

A163652 Triangle read by rows where T(n,m)=2mn + m + n + 6.