A001844 -id:A001844 - cp's OEIS frontend

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

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

A143003 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+5)a(n) - n^6a(n-1).

Original entry on oeis.org

0, 1, 21, 1091, 114520, 21298264, 6410456640, 2923097201856, 1920450126458880, 1747596822651334656, 2133806329230225408000, 3405545462439659704320000, 6950705677729940374290432000, 17807686090745585163974737920000

Offset: 0

Peter Bala, Jul 19 2008

This is the case m = 1 of the general recurrence a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+2*m^2+2*m+1)*a(n) - n^6*a(n-1) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series Sum_{k>=1} 1/k^3 for Apery's constant zeta(3). For other cases see A066989 (m=0), A143004 (m=2), A143005 (m=3) and A143006 (m=4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!^3*p_m(n)*Sum_{k = 1..n} 1/(k^3*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..n} C(2*k,k)^2*C(n+k,2*k)*C(x+k,2*k) is a polynomial in x of degree 2*m.
The first few are p_0(x) = 1, p_1(x) = 2*x^2 + 2*x + 1, p_2(x) = (3*x^4 + 6*x^3 + 9*x^2 + 6*x + 2)/2 and p_3(x) = (10*x^6 + 30*x^5 + 85*x^4 + 120*x^3 + 121*x^2 + 66*x + 18)/18. For fixed n, the sequence [p_n(k)]k>=0 is the crystal ball sequence for the product lattice A_n x A_n. See A143007 for the table of values [p_n(k)] n,k >= 0. Observe that [p_n(n)] n >= 0 is the sequence of Apery numbers A005259.
The reciprocity law p_m(n) = p_n(m) holds for nonnegative integers m and n. In particular we have p_m(1) = 2*m^2 + 2*m + 1 and p_m(2) = (3*m^4 + 6*m^3 + 9*m^2 + 6*m + 2)/2.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x+1)^3*f(x+1) + x^3*f(x-1) = (2*x+1)*(x^2+x+2*m^2+2*m+1)*f(x), normalized so that f(0) = 1. The reciprocity law now yields the Apery-like recursion m^3*p_m(x) + (m-1)^3*p_(m-2)(x) = (2*m-1)*(m^2-m+1+2*x^2+2*x)*p_(m-1)(x).
The polynomial functions p_m(x) have their zeros on the vertical line Re x = -1/2 in the complex plane; that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis (adapt the proof of the lemma on p. 4 of [BUMP et al.]).
The general recurrence in the first paragraph above has a second solution b(n) = n!^3*p_m(n) with initial conditions b(0) = 1, b(1) = 2*m^2+2*m+1. Hence the behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*p_m(k-1)*p_m(k)) = 1/((2*m^2+2*m+1) - 1^6/(3*(2*m^2+2*m+3) - 2^6/(5*(2*m^2+2*m+7) - 3^6/(7*(2*m^2+2*m+13) - ...)))) = Sum_{k>=1} 1/(m+k)^3. The final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii)].
For the corresponding results for the constant zeta(2) see A142995. For corresponding results for the constant log(2) see A142979 and A142992.

Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Seiichi Manyama, Table of n, a(n) for n = 0..181
D. Bump, K. Choi, P. Kurlberg and J. Vaaler, A local Riemann hypothesis, I, Math. Zeit. 233, (2000), 1-19.

Cf. A066989, A008288, A108625, A142979, A142992, A142995, A143004, A143005, A143006, A143007.

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

p := n -> 2*n^2+2*n+1: a := n -> n!^3*p(n)*sum (1/(k^3*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 0..14)

RecurrenceTable[{a[0]==0,a[1]==1,a[n+1]==(2n+1)(n^2+n+5)a[n]- n^6 a[n-1]}, a[n],{n,15}] (* Harvey P. Dale, Jun 20 2011 *)

a(n) = n!^3*p(n)*Sum_{k = 1..n} 1/(k^3*p(k-1)*p(k)), where p(n) = 2*n^2 + 2*n + 1 = A001844(n).
Recurrence: a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).
The sequence b(n):= n!^3*p(n) satisfies the same recurrence with the initial conditions b(0) = 1, b(1) = 5. Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - (n-1)^6/((2*n-1)*(n^2-n+5)))))), for n >= 2. The behavior of a(n) for large n is given by lim_{n -> infinity} a(n)/b(n) = Sum_{k>=1} 1/(k^3*(4*k^4 + 1)) = 1/(5 - 1^6/(21 - 2^6/(55 - 3^6/(119 - ... - n^6/((2*n+1)*(n^2+n+5) - ...))))) = zeta(3) - 1, where the final equality follows from a result of Ramanujan; see [Berndt, Chapter 12, Entry 32(iii) at x = 1].

A062786 Centered 10-gonal numbers.

Original entry on oeis.org

1, 11, 31, 61, 101, 151, 211, 281, 361, 451, 551, 661, 781, 911, 1051, 1201, 1361, 1531, 1711, 1901, 2101, 2311, 2531, 2761, 3001, 3251, 3511, 3781, 4061, 4351, 4651, 4961, 5281, 5611, 5951, 6301, 6661, 7031, 7411, 7801, 8201, 8611, 9031, 9461, 9901, 10351, 10811

Offset: 1

Jason Earls, Jul 19 2001

Deleting the least significant digit yields the (n-1)-st triangular number: a(n) = 10*A000217(n-1) + 1. - Amarnath Murthy, Dec 11 2003
All divisors of a(n) are congruent to 1 or -1, modulo 10; that is, they end in the decimal digit 1 or 9. Proof: If p is an odd prime different from 5 then 5n^2 - 5n + 1 == 0 (mod p) implies 25(2n - 1)^2 == 5 (mod p), whence p == 1 or -1 (mod 10). - Nick Hobson, Nov 13 2006
Centered decagonal numbers. - Omar E. Pol, Oct 03 2011
The partial sums of this sequence give A004466. - Leo Tavares, Oct 04 2021
The continued fraction expansion of sqrt(5*a(n)) is [5n-3; {2, 2n-2, 2, 10n-6}]. For n=1, this collapses to [2; {4}]. - Magus K. Chu, Sep 12 2022
Numbers m such that 20*m + 5 is a square. Also values of the Fibonacci polynomial y^2 - x*y - x^2 for x = n and y = 3*n - 1. This is a subsequence of A089270. - Klaus Purath, Oct 30 2022
All terms can be written as a difference of two consecutive squares a(n) = A005891(n-1)^2 - A028895(n-1)^2, and they can be represented by the forms (x^2 + 2mxy + (m^2-1)y^2) and (3x^2 + (6m-2)xy + (3m^2-2m)y^2), both of discriminant 4. - Klaus Purath, Oct 17 2023

T. D. Noe, Table of n, a(n) for n = 1..1000
Leo Tavares, Illustration: Pentagonal Stars.
Leo Tavares, Illustration: Mid-section Stars.
Leo Tavares, Illustration: Mid-section Star Pillars.
Leo Tavares, Illustration: Trapezoidal Rays.
R. Yin, J. Mu, and T. Komatsu, The p-Frobenius Number for the Triple of the Generalized Star Numbers, Preprints 2024, 2024072280. See p. 2.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001263, A124080, A101321, A028387, A016861, A003154, A005891, A000217, A004466, A144390, A000326, A060544.

Cf. also A016754, A002378, A069099, A045943, A003215, A046092, A001844, A028896, A005448, A024966, A082970.

List([1..50], n-> 1+5*n*(n-1)); # G. C. Greubel, Mar 30 2019

[1+5*n*(n-1): n in [1..50]]; // G. C. Greubel, Mar 30 2019

FoldList[#1+#2 &, 1, 10Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
1+5*Pochhammer[Range[50]-1, 2] (* G. C. Greubel, Mar 30 2019 *)

j=[]; for(n=1,75,j=concat(j,(5*n*(n-1)+1))); j

for (n=1, 1000, write("b062786.txt", n, " ", 5*n*(n - 1) + 1) ) \\ Harry J. Smith, Aug 11 2009

def a(n): return(5*n**2-5*n+1) # Torlach Rush, May 10 2024

[1+5*rising_factorial(n-1, 2) for n in (1..50)] # G. C. Greubel, Mar 30 2019

a(n) = 5*n*(n-1) + 1.
From  Gary W. Adamson, Dec 29 2007: (Start)
Binomial transform of [1, 10, 10, 0, 0, 0, ...];
Narayana transform (A001263) of [1, 10, 0, 0, 0, ...]. (End)
G.f.: x*(1+8*x+x^2) / (1-x)^3. - R. J. Mathar, Feb 04 2011
a(n) = A124080(n-1) + 1. - Omar E. Pol, Oct 03 2011
a(n) = A101321(10,n-1). - R. J. Mathar, Jul 28 2016
a(n) = A028387(A016861(n-1))/5 for n > 0. - Art Baker, Mar 28 2019
E.g.f.: (1+5*x^2)*exp(x) - 1. - G. C. Greubel, Mar 30 2019
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(5))) / sqrt(5). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 20 2020: (Start)
Sum_{n>=1} a(n)/n! = 6*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 6/e - 1. (End)
a(n) = A005891(n-1) + 5*A000217(n-1). - Leo Tavares, Jul 14 2021
a(n) = A003154(n) - 2*A000217(n-1). See Mid-section Stars illustration. - Leo Tavares, Sep 06 2021
From Leo Tavares, Oct 06 2021: (Start)
a(n) = A144390(n-1) + 2*A028387(n-1). See Mid-section Star Pillars illustration.
a(n) = A000326(n) + A000217(n) + 3*A000217(n-1). See Trapezoidal Rays illustration.
a(n) = A060544(n) + A000217(n-1). (End)
From Leo Tavares, Oct 31 2021: (Start)
a(n) = A016754(n-1) + 2*A000217(n-1).
a(n) = A016754(n-1) + A002378(n-1).
a(n) = A069099(n) + 3*A000217(n-1).
a(n) = A069099(n) + A045943(n-1).
a(n) = A003215(n-1) + 4*A000217(n-1).
a(n) = A003215(n-1) + A046092(n-1).
a(n) = A001844(n-1) + 6*A000217(n-1).
a(n) = A001844(n-1) + A028896(n-1).
a(n) = A005448(n) + 7*A000217(n).
a(n) = A005448(n) + A024966(n). (End)
From Klaus Purath, Oct 30 2022: (Start)
a(n) = a(n-2) + 10*(2*n-3).
a(n) = 2*a(n-1) - a(n-2) + 10.
a(n) = A135705(n-1) + n.
a(n) = A190816(n) - n.
a(n) = 2*A005891(n-1) - 1. (End)

Better description from Terrel Trotter, Jr., Apr 06 2002

A084849 a(n) = 1 + n + 2*n^2.

Original entry on oeis.org

1, 4, 11, 22, 37, 56, 79, 106, 137, 172, 211, 254, 301, 352, 407, 466, 529, 596, 667, 742, 821, 904, 991, 1082, 1177, 1276, 1379, 1486, 1597, 1712, 1831, 1954, 2081, 2212, 2347, 2486, 2629, 2776, 2927, 3082, 3241, 3404, 3571, 3742, 3917, 4096, 4279, 4466

Offset: 0

Paul Barry, Jun 09 2003

Equals (1, 2, 3, ...) convolved with (1, 2, 4, 4, 4, ...). a(3) = 22 = (1, 2, 3, 4) dot (4, 4, 2, 1) = (4 + 8 + 6 + 4). - Gary W. Adamson, May 01 2009
a(n) is also the number of ways to place 2 nonattacking bishops on a 2 X (n+1) board. - Vaclav Kotesovec, Jan 29 2010
Partial sums are A174723. - Wesley Ivan Hurt, Apr 16 2016
Also the number of irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Aug 09 2017

G. C. Greubel, Table of n, a(n) for n = 0..5000
W. Burrows and C. Tuffley, Maximising common fixtures in a round robin tournament with two divisions, arXiv:1502.06664 [math.CO], 2015.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
Aoife Hennessy, A Study of Riordan Arrays with Applications to Continued Fractions, Orthogonal Polynomials and Lattice Paths, Ph.D. Thesis, Waterford Institute of Technology, 2011.
Eric Weisstein's World of Mathematics, Cocktail Party Graph.
Eric Weisstein's World of Mathematics, Irredundant Set.
Wikipedia, Alexander polynomial and Seifert surface. [See _Peter Bala_'s comment.]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000027, A000217, A001844, A004767 (first differences), A014105, A058331, A060884, A100036, A100037, A100038, A100039, A100040, A100041, A131901, A134082, A174723, A177342.

[1+n+2*n^2 : n in [0..100]]; // Wesley Ivan Hurt, Apr 15 2016

A084849:=n->1+n+2*n^2: seq(A084849(n), n=0..100); # Wesley Ivan Hurt, Apr 15 2016

s = 1; lst = {s}; Do[s += n + 2; AppendTo[lst, s], {n, 1, 200, 4}]; lst (* Zerinvary Lajos, Jul 11 2009 *)
f[n_]:=(n*(2*n+1)+1);Table[f[n],{n,5!}] (* Vladimir Joseph Stephan Orlovsky, Feb 07 2010 *)
Table[1 + n + 2 n^2, {n, 0, 20}] (* Eric W. Weisstein, Aug 09 2017 *)
LinearRecurrence[{3, -3, 1}, {4, 11, 22}, {0, 20}] (* Eric W. Weisstein, Aug 09 2017 *)
CoefficientList[Series[(-1 - x - 2 x^2)/(-1 + x)^3, {x, 0, 20}], x] (* Eric W. Weisstein, Aug 09 2017 *)

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

a(n) = A058331(n) + A000027(n).
G.f.: (1 + x + 2*x^2)/(1 - x)^3.
a(n) = A014105(n) + 1; A100035(a(n)) = 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = ceiling((2*n + 1)^2/2) - n = A001844(n) - n. - Paul Barry, Jul 16 2006
From Gary W. Adamson, Oct 07 2007: (Start)
Row sums of triangle A131901.
(a(n): n >= 0) is the binomial transform of (1, 3, 4, 0, 0, 0, ...). (End)
Equals A134082 * [1,2,3,...]. -
a(n) = (1 + A000217(2*n-1) + A000217(2*n+1))/2. - Enrique Pérez Herrero, Apr 02 2010
a(n) = (A177342(n+1) - A177342(n))/2, with n > 0. - Bruno Berselli, May 19 2010
a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0, with n > 2. - Bruno Berselli, May 24 2010
a(n) = 4*n + a(n-1) - 1 (with a(0) = 1). - Vincenzo Librandi, Aug 08 2010
With an offset of 1, the polynomial a(t-1) = 2*t^2 - 3*t + 2 is the Alexander polynomial (with negative powers cleared) of the 3-twist knot. The associated Seifert matrix S is [[-1,-1], [0,-2]]. a(n-1) = det(transpose(S) - n*S). Cf. A060884. - Peter Bala, Mar 14 2012
E.g.f.: (1 + 3*x + 2*x^2)*exp(x). - Ilya Gutkovskiy, Apr 16 2016

A027861 Numbers k such that k^2 + (k+1)^2 is prime.

Original entry on oeis.org

1, 2, 4, 5, 7, 9, 12, 14, 17, 19, 22, 24, 25, 29, 30, 32, 34, 35, 39, 42, 47, 50, 60, 65, 69, 70, 72, 79, 82, 84, 85, 87, 90, 97, 99, 100, 102, 104, 109, 110, 115, 122, 130, 135, 137, 139, 144, 149, 154, 157, 160, 162, 164, 167, 172, 174, 185, 187, 189, 195, 199, 202

Offset: 1

Patrick De Geest

k > 1 never ends in 1, 3, 6 or 8 (that is, k*(k+1) does not end in 2). - Lekraj Beedassy, Jul 09 2004

k > 1 can never be congruent to (1 or 3) mod 5, because if it were, then k^2 + (k+1)^2 would be divisible by 5. In other words, for k > 1, this sequence cannot contain any values in A047219. This means that we can immediately discard 40% of all possible k. - Dmitry Kamenetsky, Sep 02 2008

T. D. Noe and Zak Seidov, Table of n, a(n) for n = 1..10000
Patrick De Geest, World!Of Numbers

Complement of A012132.
Cf. A002731 (2k+1 values), A027862 (resulting primes), A091277 (indices of resulting primes).
Cf. A047219 (k mod 5 = 1 or 3), A001844 (centered squares), A010051.

a027861 n = a027861_list !! (n-1)
a027861_list = filter ((== 1) . a010051 . a001844) [0..]
-- Reinhard Zumkeller, Jul 13 2014

[n: n in [0..1000] |IsPrime(n^2 + (n+1)^2)]; // Vincenzo Librandi, Nov 19 2010

Select[Range[250],PrimeQ[#^2+(#+1)^2]&] (* Harvey P. Dale, Dec 31 2017 *)

is(n)=isprime(n^2 + (n+1)^2) \\ Charles R Greathouse IV, Apr 28 2015

a(n) = (A002731(n)-1)/2.
a(n) = (sqrt(2*A027862(n)-1)-1)/2. - Zak Seidov, Jul 22 2013
A010051(A001844(a(n))) = 1. - Reinhard Zumkeller, Jul 13 2014
a(n) = floor(sqrt(A027862(n)/2)). - Rémi Guillaume, Apr 02 2025

A007588 Stella octangula numbers: a(n) = n(2n^2 - 1).

Original entry on oeis.org

0, 1, 14, 51, 124, 245, 426, 679, 1016, 1449, 1990, 2651, 3444, 4381, 5474, 6735, 8176, 9809, 11646, 13699, 15980, 18501, 21274, 24311, 27624, 31225, 35126, 39339, 43876, 48749, 53970, 59551, 65504, 71841, 78574, 85715, 93276, 101269, 109706, 118599, 127960

Offset: 0

N. J. A. Sloane

Also as a(n)=(1/6)*(12*n^3-6*n), n>0: structured hexagonal anti-diamond numbers (vertex structure 13) (Cf. A005915 = alternate vertex; A100188 = structured anti-diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The only known square stella octangula number for n>1 is a(169) = 169*(2*169^2 - 1) = 9653449 = 3107^2. - Alexander Adamchuk, Jun 02 2008
Ljunggren proved that 9653449 = (13*239)^2 is the only square stella octangula number for n>1. See A229384 and the Wikipedia link. - Jonathan Sondow, Sep 30 2013
4*A007588 = A144138(ChebyshevU[3,n]). - Vladimir Joseph Stephan Orlovsky, Jun 30 2011
If A016813 is regarded as a regular triangle (with leading terms listed in A001844), a(n) provides the row sums of this triangle: 1, 5+9=14, 13+17+21=51 and so on. - J. M. Bergot, Jul 05 2013
Shares its digital root, A267017, with n*(n^2 + 1)/2 ("sum of the next n natural numbers" see A006003). - Peter M. Chema, Aug 28 2016

J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 51.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
W. Ljunggren, Zur Theorie der Gleichung x^2 + 1 = Dy^4, Avh. Norske Vid. Akad. Oslo. I. 1942 (5): 27.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Alexander Adamchuk and Vincenzo Librandi, Table of n, a(n) for n = 0..10000 [Alexander Adamchuk computed terms 0 - 169, Jun 02, 2008; Vincenzo Librandi computed the first 10000 terms, Aug 18 2011]
A. Bremner, R. Høibakk and D. Lukkassen, Crossed ladders and Euler’s quartic, Annales Mathematicae et Informaticae, 36 (2009) pp. 29-41. See p. 33.
John Elias, Nesting Cubes of the Surface Points of a Hexagonal Prism
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Amelia Carolina Sparavigna, Generalized Sum of Stella Octangula Numbers, Politecnico di Torino (Italy, 2021).
Amelia Carolina Sparavigna, Cardano Formula and Some Figurate Numbers, Politecnico di Torino (Italy, 2021).
Eric Weisstein's World of Mathematics, Stella Octangula Number
Wikipedia, Stella octangula number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Backwards differences give star numbers A003154: A003154(n)=a(n)-a(n-1).
1/12*t*(n^3-n)+ n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A001653 = Numbers n such that 2*n^2 - 1 is a square.
a(169) = (A229384(3)*A229384(4))^2.
Cf. A267017, A006003, A135503.
Cf. A002378, A005843, A061317, A062392.

[n*(2*n^2 - 1): n in [0..40]]; // Vincenzo Librandi, Aug 18 2011

A007588:=n->n*(2*n^2 - 1); seq(A007588(n), n=0..40); # Wesley Ivan Hurt, Mar 10 2014

Table[ n(2n^2-1), {n,0,169} ] (* Alexander Adamchuk, Jun 02 2008 *)
LinearRecurrence[{4,-6,4,-1},{0,1,14,51},50] (* Harvey P. Dale, Sep 16 2011 *)

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

def A007588(n): return n*(2*n**2-1) # Chai Wah Wu, Feb 18 2022

G.f.: x*(1+10*x+x^2)/(1-x)^4.
a(n) = n*A056220(n).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Harvey P. Dale, Sep 16 2011
From Ilya Gutkovskiy, Jul 02 2016: (Start)
E.g.f.: x*(1 + 6*x + 2*x^2)*exp(x).
Dirichlet g.f.: 2*zeta(s-3) - zeta(s-1). (End)
a(n) = A004188(n) + A135503(n). - Miquel Cerda, Dec 25 2016
a(n) = A061317(n) - A005843(n) = A062392(n) - A062392(n-1). - J.S. Seneschal, Jul 01 2025

In the formula given in the 1995 Encyclopedia of Integer Sequences, the second 2 should be an exponent.

A033282 Triangle read by rows: T(n, k) is the number of diagonal dissections of a convex n-gon into k+1 regions.

Original entry on oeis.org

1, 1, 2, 1, 5, 5, 1, 9, 21, 14, 1, 14, 56, 84, 42, 1, 20, 120, 300, 330, 132, 1, 27, 225, 825, 1485, 1287, 429, 1, 35, 385, 1925, 5005, 7007, 5005, 1430, 1, 44, 616, 4004, 14014, 28028, 32032, 19448, 4862, 1, 54, 936, 7644, 34398, 91728, 148512, 143208, 75582, 16796

Offset: 3

N. J. A. Sloane

T(n+3, k) is also the number of compatible k-sets of cluster variables in Fomin and Zelevinsky's cluster algebra of finite type A_n. Take a row of this triangle regarded as a polynomial in x and rewrite as a polynomial in y := x+1. The coefficients of the polynomial in y give a row of the triangle of Narayana numbers A001263. For example, x^2 + 5*x + 5 = y^2 + 3*y + 1. - Paul Boddington, Mar 07 2003
Number of standard Young tableaux of shape (k+1,k+1,1^(n-k-3)), where 1^(n-k-3) denotes a sequence of n-k-3 1's (see the Stanley reference).
Number of k-dimensional 'faces' of the n-dimensional associahedron (see Simion, p. 168). - Mitch Harris, Jan 16 2007
Mirror image of triangle A126216. - Philippe Deléham, Oct 19 2007
For relation to Lagrange inversion or series reversion and the geometry of associahedra or Stasheff polytopes (and other combinatorial objects) see A133437. - Tom Copeland, Sep 29 2008
Row generating polynomials 1/(n+1)*Jacobi_P(n,1,1,2*x+1). Row n of this triangle is the f-vector of the simplicial complex dual to an associahedron of type A_n [Fomin & Reading, p. 60]. See A001263 for the corresponding array of h-vectors for associahedra of type A_n. See A063007 and A080721 for the f-vectors for associahedra of type B and type D respectively. - Peter Bala, Oct 28 2008
f-vectors of secondary polytopes for Grobner bases for optimization and integer programming (see De Loera et al. and Thomas). - Tom Copeland, Oct 11 2011
From Devadoss and O'Rourke's book: The Fulton-MacPherson compactification of the configuration space of n free particles on a line segment with a fixed particle at each end is the n-Dim Stasheff associahedron whose refined f-vector is given in A133437 which reduces to A033282. - Tom Copeland, Nov 29 2011
Diagonals of A132081 are rows of A033282. - Tom Copeland, May 08 2012
The general results on the convolution of the refined partition polynomials of A133437, with u_1 = 1 and u_n = -t otherwise, can be applied here to obtain results of convolutions of these polynomials. - Tom Copeland, Sep 20 2016
The signed triangle t(n, k) =(-1)^k* T(n+2, k-1), n >= 1, k = 1..n, seems to be obtainable from the partition array A111785 (in Abramowitz-Stegun order) by adding the entries corresponding to the partitions of n with the number of parts k. E.g., triangle t, row n=4: -1, (6+3) = 9, -21, 14. - Wolfdieter Lang, Mar 17 2017
The preceding conjecture by Lang is true. It is implicit in Copeland's 2011 comments in A086810 on the relations among a gf and its compositional inverse for that entry and inversion through A133437 (a differently normalized version of A111785), whose integer partitions are the same as those for A134685. (An inversion pair in Copeland's 2008 formulas below can also be used to prove the conjecture.) In addition, it follows from the relation between the inversion formula of A111785/A133437 and the enumeration of distinct faces of associahedra. See the MathOverflow link concernimg Loday and the Aguiar and Ardila reference in A133437 for proofs of the relations between the partition polynomials for inversion and enumeration of the distinct faces of the A_n associahedra, or Stasheff polytopes. - Tom Copeland, Dec 21 2017
The rows seem to give (up to sign) the coefficients in the expansion of the integer-valued polynomial (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/(n!*(n+1)!) in the basis made of the binomial(x+i,i). - F. Chapoton, Oct 07 2022
Chapoton's observation above is correct: the precise expansion is (x+1)*(x+2)^2*(x+3)^2*...*(x+n)^2*(x+n+1)/ (n!*(n+1)!) = Sum_{k = 0..n-1} (-1)^k*T(n+2,n-k-1)*binomial(x+2*n-k,2*n-k), as can be verified using the WZ algorithm. For example, n = 4 gives (x+1)*(x+2)^2*(x+3)^2*(x+4)^2*(x+5)/(4!*5!) = 14*binomial(x+8,8) - 21*binomial(x+7,7) + 9*binomial(x+6,6) - binomial(x+5,5). - Peter Bala, Jun 24 2023

			The triangle T(n, k) begins:
n\k  0  1   2    3     4     5      6      7     8     9
3:   1
4:   1  2
5:   1  5   5
6:   1  9  21   14
7:   1 14  56   84    42
8:   1 20 120  300   330   132
9:   1 27 225  825  1485  1287    429
10:  1 35 385 1925  5005  7007   5005   1430
11:  1 44 616 4004 14014 28028  32032  19448  4862
12:  1 54 936 7644 34398 91728 148512 143208 75582 16796
... reformatted. - _Wolfdieter Lang_, Mar 17 2017

S. Devadoss and J. O'Rourke, Discrete and Computational Geometry, Princeton Univ. Press, 2011 (See p. 241.)
Ronald L. Graham, Donald E. Knuth, Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994. Exercise 7.50, pages 379, 573.
T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.8.

Vincenzo Librandi, Table of n, a(n) for n = 3..2000
Paul Barry, On Integer-Sequence-Based Constructions of Generalized Pascal Triangles, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.4.
Paul Barry, Continued fractions and transformations of integer sequences, JIS 12 (2009) 09.7.6.
Paul Barry, On the Inverses of a Family of Pascal-Like Matrices Defined by Riordan Arrays, Journal of Integer Sequences, 16 (2013), #13.5.6.
Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
Paul Barry, On the f-Matrices of Pascal-like Triangles Defined by Riordan Arrays, arXiv:1805.02274 [math.CO], 2018.
Paul Barry, Generalized Catalan Numbers Associated with a Family of Pascal-like Triangles, J. Int. Seq., Vol. 22 (2019), Article 19.5.8.
Karin Baur and P. P. Martin, The fibres of the Scott map on polygon tilings are the flip equivalence classes, arXiv:1601.05080 [math.CO], 2016.
David Beckwith, Legendre polynomials and polygon dissections?, Amer. Math. Monthly, 105 (1998), 256-257.
William Butler, Andrew Kalotay and N. J. A. Sloane, Correspondence, 1974
Arthur Cayley, On the partitions of a polygon, Proc. London Math. Soc., 22 (1891), 237-262 = Collected Mathematical Papers. Vols. 1-13, Cambridge Univ. Press, London, 1889-1897, Vol. 13, pp. 93ff. (See p. 239.)
Adrian Celestino and Yannic Vargas, Schröder trees, antipode formulas and non-commutative probability, arXiv:2311.07824 [math.CO], 2023.
Frédéric Chapoton, Enumerative properties of generalized associahedra, Séminaire Lotharingien de Combinatoire, B51b (2004), 16 pp.
Johann Cigler, Some remarks and conjectures related to lattice paths in strips along the x-axis, arXiv:1501.04750 [math.CO], 2015.
Manosij Ghosh Dastidar and Michael Wallner, Bijections and congruences involving lattice paths and integer compositions, arXiv:2402.17849 [math.CO], 2024. See p. 16.
Jesús A. De Loera, Jörg Rambau, and Francisco Santos Leal, Triangulations of Point Sets [From Tom Copeland Oct 11 2011]
Satyan L. Devadoss, Combinatorial Equivalence of Real Moduli Spaces, Notices Amer. Math. Soc. 51 (2004), no. 6, 620-628.
Satyan L. Devadoss and Ronald C. Read, Cellular structures determined by polygons and trees, arXiv/0008145 [math.CO], 2000. [From Tom Copeland Nov 21 2017]
Anton Dochtermann, Face rings of cycles, associahedra, and standard Young tableaux, arXiv preprint arXiv:1503.06243 [math.CO], 2015.
Brian Drake, Ira M. Gessel, and Guoce Xin, Three Proofs and a Generalization of the Goulden-Litsyn-Shevelev Conjecture on a Sequence Arising in Algebraic Geometry, J. of Integer Sequences, Vol. 10 (2007), Article 07.3.7.
Cassandra Durell and Stefan Forcey, Level-1 Phylogenetic Networks and their Balanced Minimum Evolution Polytopes, arXiv:1905.09160 [math.CO], 2019.
P. Flajolet and M. Noy, Analytic Combinatorics of Non-crossing Configurations, Discrete Math., 204, 1999, 203-229.
Sergey Fomin and Nathan Reading, Root systems and generalized associahedra, Lecture notes for IAS/Park-City 2004, arXiv:math/0505518 [math.CO], 2005-2008. [From _Peter Bala_, Oct 28 2008]
Sergey Fomin and Andrei Zelevinsky, Cluster algebras I: Foundations, arXiv:math/0104151 [math.RT], 2001.
Sergey Fomin and Andrei Zelevinsky, Cluster algebras I: Foundations, J. Amer. Math. Soc. 15 (2002) no.2, 497-529.
Sergey Fomin and Andrei Zelevinsky, Y-systems and generalized associahedra, Ann. of Math. (2) 158 (2003), no. 3, 977-1018.
Ivan Geffner and Marc Noy, Counting Outerplanar Maps, Electronic Journal of Combinatorics 24(2) (2017), #P2.3.
Rijun Huang, Fei Teng, and Bo Feng, Permutation in the CHY-Formulation, arXiv:1801.08965 [hep-th], 2018.
Germain Kreweras, Sur les partitions non croisées d'un cycle, (French) Discrete Math. 1 (1972), no. 4, 333--350. MR0309747 (46 #8852).
Germain Kreweras, Sur les hiérarchies de segments, Cahiers Bureau Universitaire Recherche Opérationnelle, Cahier 20, Inst. Statistiques, Univ. Paris, 1973.
Germain Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)
Germain Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30.
Germain Kreweras, Les préordres totaux compatibles avec un ordre partiel, Math. Sci. Humaines No. 53 (1976), 5-30. (Annotated scanned copy)
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Sebastian Mizera, Combinatorics and topology of Kawai-Lewellen-Tye relations J. High Energy Phys. 2017, No. 8, Paper No. 97, 54 p. (2017).
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv:1403.5962 [math.CO], 2014.
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Rekha R. Thomas, Lectures in Geometric Combinatorics [_Tom Copeland_, Oct 11 2011]

Cf. diagonals: A000012, A000096, A033275, A033276, A033277, A033278, A033279; A000108, A002054, A002055, A002056, A007160, A033280, A033281; row sums: A001003 (Schroeder numbers, first term omitted). See A086810 for another version.
A007160 is a diagonal. Cf. A001263.
With leading zero: A086810.
Cf. A019538 'faces' of the permutohedron.
Cf. A063007 (f-vectors type B associahedra), A080721 (f-vectors type D associahedra), A126216 (mirror image).
Cf. A248727 for a relation to f-polynomials of simplices.
Cf. A111785 (contracted partition array, unsigned; see a comment above).
Antidiagonal sums give A005043. - Jordan Tirrell, Jun 01 2017

[[Binomial(n-3, k)*Binomial(n+k-1, k)/(k+1): k in [0..(n-3)]]: n in [3..12]];  // G. C. Greubel, Nov 19 2018

T:=(n,k)->binomial(n-3,k)*binomial(n+k-1,k)/(k+1): seq(seq(T(n,k),k=0..n-3),n=3..12); # Muniru A Asiru, Nov 24 2018

t[n_, k_] = Binomial[n-3, k]*Binomial[n+k-1, k]/(k+1);
Flatten[Table[t[n, k], {n, 3, 12}, {k, 0, n-3}]][[1 ;; 52]] (* Jean-François Alcover, Jun 16 2011 *)

Q=(1+z-(1-(4*w+2+O(w^20))*z+z^2+O(z^20))^(1/2))/(2*(1+w)*z);for(n=3,12,for(m=1,n-2,print1(polcoef(polcoef(Q,n-2,z),m,w),", "))) \\ Hugo Pfoertner, Nov 19 2018

for(n=3,12, for(k=0,n-3, print1(binomial(n-3,k)*binomial(n+k-1,k)/(k+1), ", "))) \\ G. C. Greubel, Nov 19 2018

[[ binomial(n-3,k)*binomial(n+k-1,k)/(k+1) for k in (0..(n-3))] for n in (3..12)] # G. C. Greubel, Nov 19 2018

G.f. G = G(t, z) satisfies (1+t)*G^2 - z*(1-z-2*t*z)*G + t*z^4 = 0.
T(n, k) = binomial(n-3, k)*binomial(n+k-1, k)/(k+1) for n >= 3, 0 <= k <= n-3.
From Tom Copeland, Nov 03 2008: (Start)
Two g.f.s (f1 and f2) for A033282 and their inverses (x1 and x2) can be derived from the Drake and Barry references.
1. a: f1(x,t) = y = {1 - (2t+1) x - sqrt[1 - (2t+1) 2x + x^2]}/[2x (t+1)] = t x + (t + 2 t^2) x^2 + (t + 5 t^2 + 5 t^3) x^3 + ...
b: x1 = y/[t + (2t+1)y + (t+1)y^2] = y {1/[t/(t+1) + y] - 1/(1+y)} = (y/t) - (1+2t)(y/t)^2 + (1+ 3t + 3t^2)(y/t)^3 +...
2. a: f2(x,t) = y = {1 - x - sqrt[(1-x)^2 - 4xt]}/[2(t+1)] = (t/(t+1)) x + t x^2 + (t + 2 t^2) x^3 + (t + 5 t^2 + 5 t^3) x^4 + ...
b: x2 = y(t+1) [1- y(t+1)]/[t + y(t+1)] = (t+1) (y/t) - (t+1)^3 (y/t)^2 + (t+1)^4 (y/t)^3 + ...
c: y/x2(y,t) = [t/(t+1) + y] / [1- y(t+1)] = t/(t+1) + (1+t) y + (1+t)^2 y^2 + (1+t)^3 y^3 + ...
x2(y,t) can be used along with the Lagrange inversion for an o.g.f. (A133437) to generate A033282 and show that A133437 is a refinement of A033282, i.e., a refinement of the f-polynomials of the associahedra, the Stasheff polytopes.
y/x2(y,t) can be used along with the indirect Lagrange inversion (A134264) to generate A033282 and show that A134264 is a refinement of A001263, i.e., a refinement of the h-polynomials of the associahedra.
f1[x,t](t+1) gives a generator for A088617.
f1[xt,1/t](t+1) gives a generator for A060693, with inverse y/[1 + t + (2+t) y + y^2].
f1[x(t-1),1/(t-1)]t gives a generator for A001263, with inverse y/[t + (1+t) y + y^2].
The unsigned coefficients of x1(y t,t) are A074909, reverse rows of A135278. (End)
G.f.: 1/(1-x*y-(x+x*y)/(1-x*y/(1-(x+x*y)/(1-x*y/(1-(x+x*y)/(1-x*y/(1-.... (continued fraction). - Paul Barry, Feb 06 2009
Let h(t) = (1-t)^2/(1+(u-1)*(1-t)^2) = 1/(u + 2*t + 3*t^2 + 4*t^3 + ...), then a signed (n-1)-th row polynomial of A033282 is given by u^(2n-1)*(1/n!)*((h(t)*d/dt)^n) t, evaluated at t=0, with initial n=2. The power series expansion of h(t) is related to A181289 (cf. A086810). - Tom Copeland, Sep 06 2011
With a different offset, the row polynomials equal 1/(1 + x)*Integral_{0..x} R(n,t) dt, where R(n,t) = Sum_{k = 0..n} binomial(n,k)*binomial(n+k,k)*t^k are the row polynomials of A063007. - Peter Bala, Jun 23 2016
n-th row polynomial = ( LegendreP(n-1,2*x + 1) - LegendreP(n-3,2*x + 1) )/((4*n - 6)*x*(x + 1)), n >= 3. - Peter Bala, Feb 22 2017
n*T(n+1, k) = (4n-6)*T(n, k-1) + (2n-3)*T(n, k) - (n-3)*T(n-1, k) for n >= 4. - Fang Lixing, May 07 2019

Missing factor of 2 for expansions of f1 and f2 added by Tom Copeland, Apr 12 2009

A053698 a(n) = n^3 + n^2 + n + 1.

Original entry on oeis.org

1, 4, 15, 40, 85, 156, 259, 400, 585, 820, 1111, 1464, 1885, 2380, 2955, 3616, 4369, 5220, 6175, 7240, 8421, 9724, 11155, 12720, 14425, 16276, 18279, 20440, 22765, 25260, 27931, 30784, 33825, 37060, 40495, 44136, 47989, 52060, 56355, 60880

Offset: 0

Henry Bottomley, Mar 23 2000

a(n) = 1111 in base n.

n^3 + n^2 + n + 1 = (n^2 + 1)*(n + 1), therefore a(n) is never prime. - Alonso del Arte, Apr 22 2014

			a(2) = 15 because 2^3 + 2^2 + 2 + 1 = 8 + 4 + 2 + 1 = 15.
a(3) = 40 because 3^3 + 3^2 + 3 + 1 = 27 + 9 + 3 + 1 = 40.
a(4) = 85 because 4^3 + 4^2 + 4 + 1 = 64 + 16 + 4 + 1 = 85.
From _Bruno Berselli_, Jan 02 2017: (Start)
The terms of the sequence are provided by the row sums of the following triangle (see the seventh formula in the previous section):
.   1;
.   3,   1;
.   9,   5,   1;
.  19,  13,   7,   1;
.  33,  25,  17,   9,   1;
.  51,  41,  31,  21,  11,   1;
.  73,  61,  49,  37,  25,  13,  1;
.  99,  85,  71,  57,  43,  29, 15,  1;
. 129, 113,  97,  81,  65,  49, 33, 17,  1;
. 163, 145, 127, 109,  91,  73, 55, 37, 19,  1;
. 201, 181, 161, 141, 121, 101, 81, 61, 41, 21, 1;
...
Columns from the first to the fifth, respectively: A058331, A001844, A056220 (after -1), A059993, A161532. Also, eighth column is A161549.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A237627 (subset of semiprimes).
Cf. A056106 (first differences).
Cf. A062158, A027444, A104878, A055129.

[n^3+n^2+n+1: n in [0..50]]; // Vincenzo Librandi, May 01 2011

A053698:=n->n^3 + n^2 + n + 1; seq(A053698(n), n=0..50); # Wesley Ivan Hurt, Apr 22 2014

Table[n^3 + n^2 + n + 1, {n, 0, 39}] (* Alonso del Arte, Apr 22 2014 *)
FromDigits["1111", Range[0, 50]] (* Paolo Xausa, May 11 2024 *)

Vec((1 + 5*x^2) / (1 - x)^4 + O(x^50)) \\ Colin Barker, Jan 02 2017

def a(n): return (n**3+n**2+n+1) # Torlach Rush, May 08 2024

For n >= 2, a(n) = (n^4-1)/(n-1) = A024002(n)/A024000(n) = A002522(n)*(n+1) = A002061(n+1) + A000578(n).
G.f.: (1+5*x^2) / (1-x)^4. - Colin Barker, Jan 06 2012
a(n) = -A062158(-n). - Bruno Berselli, Jan 26 2016
a(n) = Sum_{i=0..n} 2*n*(n-i)+1. - Bruno Berselli, Jan 02 2017
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 3. - Colin Barker, Jan 02 2017
a(n) = A104878(n+3,n) = A055129(4,n) for n > 0. - Mathew Englander, Jan 06 2021
E.g.f.: exp(x)*(x^3+4*x^2+3*x+1). - Nikolaos Pantelidis, Feb 06 2023

A069129 Centered 16-gonal numbers.

Original entry on oeis.org

1, 17, 49, 97, 161, 241, 337, 449, 577, 721, 881, 1057, 1249, 1457, 1681, 1921, 2177, 2449, 2737, 3041, 3361, 3697, 4049, 4417, 4801, 5201, 5617, 6049, 6497, 6961, 7441, 7937, 8449, 8977, 9521, 10081, 10657, 11249, 11857, 12481, 13121, 13777, 14449, 15137, 15841

Offset: 1

Terrel Trotter, Jr., Apr 07 2002

Also, sequence found by reading the line from 1, in the direction 1, 17, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139098 in the same spiral. - Omar E. Pol, Apr 26 2008
The subsequence of primes begins: 17, 97, 241, 337, 449, 577, 881, 1249, 3041, 3361, 3697, 4049, 4801, 6961, 7937, 9521,  10657, 13121, 14449. See A184899: n such that the n-th centered 12-gonal number is prime. Indices of prime star numbers. - Jonathan Vos Post, Feb 27 2011
Binomial transform of [1, 16, 16, 0, 0, 0, ...] and Narayana transform (A001263) of [1, 16, 0, 0, 0, ...]. - Gary W. Adamson, Jul 28 2011
Centered hexadecagonal numbers or centered hexakaidecagonal numbers. - Omar E. Pol, Oct 03 2011
a(n) = m(n,n) for an array constructed by using the terms in A016813 as the antidiagonals; the first few antidiagonals are 1; 5,9; 13,17,21; 25,29,33,37. - J. M. Bergot, Jul 05 2013
[The first five rows begin: 1,9,21,37,57; 5,17,33,53,77; 13,29,49,73,101; 25,45,69,97,129; 41,65,93,125,161.]

			a(5) = 161 because 8*5^2 - 8*5 + 1 = 200 - 40 + 1 = 161.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Leo Tavares, Illustration: Octagonal Stars.
Eric Weisstein's World of Mathematics, Centered Polygonal Numbers.
R. Yin, J. Mu, and T. Komatsu, The p-Frobenius Number for the Triple of the Generalized Star Numbers, Preprints 2024, 2024072280. See p. 2.
Index entries for sequences related to centered polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005448, A001844, A005891, A003215, A069099, A000217, A035008, A139098.

Bisection of A077221.

[8*n^2-8*n+1: n in [0..50]]; // Vincenzo Librandi, Feb 05 2013

FoldList[#1 + #2 &, 1, 16 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
Rest[CoefficientList[Series[-x(1+14x+x^2)/(x-1)^3,{x,0,50}],x]]  (* Harvey P. Dale, Apr 22 2011 *)

a(n)=8*n^2-8*n+1 \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 8*n + 1.
a(n) = A035008(n-1) + 1. - Omar E. Pol, Apr 26 2008
a(n) = 16*n + a(n-1) - 16 with n > 1, a(1)=1. - Vincenzo Librandi, Aug 08 2010
G.f.: -x*(1+14*x+x^2) / (x-1)^3. - R. J. Mathar, Feb 04 2011
E.g.f.: (8*x^2 + 1)*exp(x). - G. C. Greubel, Jul 18 2017
a(n) = A056220(2n-1). - Bruce J. Nicholson, Aug 31 2017
Sum_{n>=1} 1/a(n) = Pi * tan(Pi/(2*sqrt(2))) / (4*sqrt(2)). - Vaclav Kotesovec, Jul 23 2019
From Amiram Eldar, Jun 21 2020: (Start)
Sum_{n>=1} a(n)/n! = 9*e - 1.
Sum_{n>=1} (-1)^n * a(n)/n! = 9/e - 1. (End)
Product_{n>=2} (a(n) - 1) / (a(n) + 1) = Pi/4. - Dimitris Valianatos, Jun 27 2020
a(n) = A016754(n-1) + 8*A000217(n-1). - Leo Tavares, Jul 19 2021

cp's OEIS Frontend

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

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A143003 a(0) = 0, a(1) = 1, a(n+1) = (2*n+1)*(n^2+n+5)*a(n) - n^6*a(n-1).

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A062786 Centered 10-gonal numbers.

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A084849 a(n) = 1 + n + 2*n^2.

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A027861 Numbers k such that k^2 + (k+1)^2 is prime.

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A007588 Stella octangula numbers: a(n) = n*(2*n^2 - 1).

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A143003 a(0) = 0, a(1) = 1, a(n+1) = (2n+1)(n^2+n+5)a(n) - n^6a(n-1).

A007588 Stella octangula numbers: a(n) = n(2n^2 - 1).