A003154 -id:A003154 - cp's OEIS frontend

A015441 Generalized Fibonacci numbers.

0, 1, 1, 7, 13, 55, 133, 463, 1261, 4039, 11605, 35839, 105469, 320503, 953317, 2876335, 8596237, 25854247, 77431669, 232557151, 697147165, 2092490071, 6275373061, 18830313487, 56482551853, 169464432775, 508359743893, 1525146340543, 4575304803901, 13726182847159

Offset: 0

Olivier Gérard

a(n) is the coefficient of x^(n-1) in the bivariate Fibonacci polynomials F(n)(x,y) = xF(n-1)(x,y) + yF(n-2)(x,y), F(0)(x,y)=0, F(1)(x,y)=1, when y=6x^2. - Mario Catalani (mario.catalani(AT)unito.it), Dec 06 2002
For n>=1: number of length-(n-1) words with letters {0,1,2,3,4,5,6,7} where no two consecutive letters are nonzero, see fxtbook link below. - Joerg Arndt, Apr 08 2011
Starting with offset 1 and convolved with (1, 3, 3, 3, ...) = A003462: (1, 4, 13, 40, ...). - Gary W. Adamson, May 28 2009
a(n) is identical to its inverse binomial transform signed. Differences: A102901. - Paul Curtz, Feb 23 2010
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 7*a(n-2) equals the number of 7-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
Pisano period lengths: 1, 1, 1, 2, 20, 1, 6, 2, 3, 20, 5, 2, 12, 6, 20, 4, 16, 3, 18, 20, ... - R. J. Mathar, Aug 10 2012
A015441 and A015518 are the only integer sequences (from the family of homogeneous linear recurrence relation of order 2 with positive integer coefficients with initial values a(0)=0 and a(1)=1) whose ratio a(n+1)/a(n) converges to 3 as n approaches infinity. - Felix P. Muga II, Mar 14 2014
This is an autosequence of the first kind: the array of successive differences shows a main diagonal of zeros and the inverse binomial transform is identical to the sequence (with alternating signs). - Pointed out by Paul Curtz, Dec 05 2016
First two upper diagonals: A000400(n).
This is a variation on the Starhex honeycomb configuration A332243, see illustration in links. It is an alternating pattern of the 2nd iteration of the centered hexagonal numbers A003215 and centered 12-gonal 'Star' numbers A003154. - John Elias, Oct 06 2021

			G.f. = x + x^2 + 7*x^3 + 13*x^4 + 55*x^5 + 133*x^6 + 463*x^7 + 1261*x^8 + ...

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from T. D. Noe)
Joerg Arndt, Matters Computational (The Fxtbook), p. 317-318
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
John Elias, Illustration: A Starhex honeycomb variation
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
H. Li and T. MacHenry, Permanents and Determinants, Weighted Isobaric Polynomials, and Integer Sequences, J. Int. Seq. 16 (2013) #13.3.5, example 48.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
A. G. Shannon and J. V. Leyendekkers, The Golden Ratio family and the Binet equation, Notes on Number Theory and Discrete Mathematics, Vol. 21, No. 2, (2015), 35-42.
Index entries for linear recurrences with constant coefficients, signature (1,6).

Cf. A000079, A003462, A016153, A109466, A122117. Cf. A001656, A087451. Cf. A000400.

I:=[0,1]; [n le 2 select I[n] else Self(n-1) + 6*Self(n-2): n in [1..30]]; // G. C. Greubel, Jan 24 2018

A015441:=n->(1/5)*((3^n)-((-2)^n)); seq(A015441(n), n=0..30); # Wesley Ivan Hurt, Mar 14 2014

a[n_]:=(MatrixPower[{{1,4},{1,-2}},n].{{1},{1}})[[2,1]]; Table[Abs[a[n]], {n,-1,40}] (* Vladimir Joseph Stephan Orlovsky, Feb 19 2010 *)
LinearRecurrence[{1,6},{0,1},30] (* Harvey P. Dale, Apr 26 2011 *)
CoefficientList[Series[x/((1 + 2 x) (1 - 3 x)), {x, 0, 29}], x] (* Michael De Vlieger, Dec 05 2016 *)

PARI
```
{a(n) = (3^n - (-2)^n) / 5};
```

[lucas_number1(n,1,-6) for n in range(0, 27)] # Zerinvary Lajos, Apr 22 2009

G.f.: x/((1+2*x)*(1-3*x)).
a(n) = a(n-1) + 6*a(n-2).
a(n) = (1/5)*((3^n)-((-2)^n)). - henryk.wicke(AT)stud.uni-hannover.de
E.g.f.: (exp(3*x) - exp(-2*x))/5. - Paul Barry, Apr 20 2003
a(n+1) = Sum_{k=0..ceiling(n/2)} 6^k*binomial(n-k, k). - Benoit Cloitre, Mar 06 2004
a(n) = (A000244(n) - A001045(n+1)(-1)^n - A001045(n)(-1)^n)/5. - Paul Barry, Apr 27 2004
The binomial transform of [1,1,7,13,55,133,463,...] is A122117. - Philippe Deléham, Oct 19 2006
a(n+1) = Sum_{k=0..n} A109466(n,k)*(-6)^(n-k). - Philippe Deléham, Oct 26 2008
a(n) = 3a(n-1) + (-1)^(n+1)*A000079(n-1). - Paul Curtz, Feb 23 2010
G.f.: Q(0) -1, where Q(k) = 1 + 6*x^2 + (k+2)*x - x*(k+1 + 6*x)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 06 2013
a(n) = (Sum_{1<=k<=n, k odd} binomial(n,k)*5^(k-1))/2^(n-1). - Vladimir Shevelev, Feb 05 2014
a(-n) = -(-1)^n * a(n) / 6^n for all n in Z. - Michael Somos, Mar 18 2014
From Peter Bala, Apr 01 2015: (Start)
Sum_{n >= 0} a(n+1)*x^n = exp( Sum_{n >= 1} A087451(n)*x^n/n ).
For k = 0, 1, 2, ... and for n >= 1, (5^k)*a(n) | a((5^k)*n).
The expansion of exp( Sum_{n >= 1} a(5*n)/(5*a(n))*x^n/n ) has integral coefficients. Cf. A001656. (End)
From  Peter Bala, Jun 27 2025: (Start)
Sum_{n >= 1} (-6)^n/(a(n)*a(n+1)) = -2, since (-6)^n/(a(n)*a(n+1)) = (-2)^n/a(n) - (-2)^(n+1)/a(n+1) for n >= 1.
The following are examples of telescoping infinite products:
Product_{n >= 0} (1 + 6^n/a(2*n+2)) = 6, since (1 + 6^(2*n-1)/a(4*n))*(1 + 6^(2*n)/a(4*n+2)) = (6 - 4^(n+1)/b(n)) / (6 - 4^n/b(n-1)), where b(n) = (2*4^n + 3*9^n)/5 = A096951(n). Similarly,
Product_{n >= 1} (1 - 6^n/a(2*n+2)) = 3/13.
Product_{n >= 0} (1 + (-6)^n/a(2*n+2)) = 6/5.
Product_{n >= 1} (1 - (-6)^n/a(2*n+2)) = 15/13.
exp( Sum_{n >= 1} a(2*n)/a(n)*x^n/n ) = Sum_{n >= 0} a(n+1)*x^n. (End)

A032528 Concentric hexagonal numbers: floor(3*n^2/2).

Original entry on oeis.org

0, 1, 6, 13, 24, 37, 54, 73, 96, 121, 150, 181, 216, 253, 294, 337, 384, 433, 486, 541, 600, 661, 726, 793, 864, 937, 1014, 1093, 1176, 1261, 1350, 1441, 1536, 1633, 1734, 1837, 1944, 2053, 2166, 2281, 2400, 2521, 2646, 2773, 2904, 3037, 3174, 3313, 3456, 3601, 3750

Offset: 0

N. J. A. Sloane

From Omar E. Pol, Aug 20 2011: (Start)
Cellular automaton on the hexagonal net. The sequence gives the number of "ON" cells in the structure after n-th stage. A007310 gives the first differences. For a definition without words see the illustration of initial terms in the example section. Note that the cells become intermittent. A083577 gives the primes of this sequences.
A033581 and A003154 interleaved.
Row sums of an infinite square array T(n,k) in which column k lists 2*k-1 zeros followed by the numbers A008458 (see example).  (End)
Sequence found by reading the line from 0, in the direction 0, 1, ... and the same line from 0, in the direction 0, 6, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. Main axis perpendicular to A045943 in the same spiral. - Omar E. Pol, Sep 08 2011

			From _Omar E. Pol_, Aug 20 2011: (Start)
Using the numbers A008458 we can write:
  0, 1, 6, 12, 18, 24, 30, 36, 42,  48,  54, ...
  0, 0, 0,  1,  6, 12, 18, 24, 30,  36,  42, ...
  0, 0, 0,  0,  0,  1,  6, 12, 18,  24,  30, ...
  0, 0, 0,  0,  0,  0,  0,  1,  6,  12,  18, ...
  0, 0, 0,  0,  0,  0,  0,  0,  0,   1,   6, ...
And so on.
===========================================
The sums of the columns give this sequence:
0, 1, 6, 13, 24, 37, 54, 73, 96, 121, 150, ...
...
Illustration of initial terms as concentric hexagons:
.
.                                         o o o o o
.                         o o o o        o         o
.             o o o      o       o      o   o o o   o
.     o o    o     o    o   o o   o    o   o     o   o
. o  o   o  o   o   o  o   o   o   o  o   o   o   o   o
.     o o    o     o    o   o o   o    o   o     o   o
.             o o o      o       o      o   o o o   o
.                         o o o o        o         o
.                                         o o o o o
.
. 1    6        13           24               37
.
(End)

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

Cf. A003154, A007310, A008458, A033581, A083577, A000326, A001318, A005449, A045943, A032527, A195041. Column 6 of A195040.
Cf. A004524, A004525, A033436, A212959, A238410, A227017, A282513.
Cf. A033581, A003154, A011934, A184533.

a032528 n = a032528_list !! n
a032528_list = scanl (+) 0 a007310_list
-- Reinhard Zumkeller, Jan 07 2012

[Floor(3*n^2/2): n in [0..50]]; // Vincenzo Librandi, Aug 21 2011

f[n_, m_] := Sum[Floor[n^2/k], {k, 1, m}]; t = Table[f[n, 2], {n, 1, 90}] (* Clark Kimberling, Apr 20 2012 *)

a(n)=3*n^2\2 \\ Charles R Greathouse IV, Sep 24 2015

From Joerg Arndt, Aug 22 2011: (Start)
G.f.: (x+4*x^2+x^3)/(1-2*x+2*x^3-x^4) = x*(1+4*x+x^2)/((1+x)*(1-x)^3).
a(n) = +2*a(n-1) -2*a(n-3) +1*a(n-4). (End)
a(n) = (6*n^2+(-1)^n-1)/4. - Bruno Berselli, Aug 22 2011
a(n) = A184533(n), n >= 2. - Clark Kimberling, Apr 20 2012
First differences of A011934: a(n) = A011934(n) - A011934(n-1) for n>0. - Franz Vrabec, Feb 17 2013
From Paul Curtz, Mar 31 2019: (Start)
a(-n) = a(n).
a(n) = a(n-2) + 6*(n-1) for n > 1.
a(2*n) = A033581(n).
a(2*n+1) = A003154(n+1). (End)
E.g.f.: (3*x*(x + 1)*cosh(x) + (3*x^2 + 3*x - 1)*sinh(x))/2. - Stefano Spezia, Aug 19 2022
Sum_{n>=1} 1/a(n) = Pi^2/36 + tan(Pi/(2*sqrt(3)))*Pi/(2*sqrt(3)). - Amiram Eldar, Jan 16 2023

New name and more terms a(41)-a(50) from Omar E. Pol, Aug 20 2011

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

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.

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

Original entry on oeis.org

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401

Offset: 0

Ignacio Larrosa Cañestro, Feb 27 2000

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

			a(1)^2 = 121 is the 5th star number (A003154).

G. C. Greubel, Table of n, a(n) for n = 0..1000
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contrib. Discr. Math. 3 (2) (2008), pp. 76-114. See Section 13.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Star Number
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for linear recurrences with constant coefficients, signature (10,-1).
Index entries for sequences related to Chebyshev polynomials.

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A003154, A006061, A031138, A007667, A004189.
Cf. A138281. Cf. A100047.
Cf. A142238.

a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019

I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015

CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Numerator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A142238 *)
a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)

a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A049598 12 times triangular numbers.

Original entry on oeis.org

0, 12, 36, 72, 120, 180, 252, 336, 432, 540, 660, 792, 936, 1092, 1260, 1440, 1632, 1836, 2052, 2280, 2520, 2772, 3036, 3312, 3600, 3900, 4212, 4536, 4872, 5220, 5580, 5952, 6336, 6732, 7140, 7560, 7992, 8436, 8892, 9360, 9840, 10332, 10836, 11352

Offset: 0

Joe Keane (jgk(AT)jgk.org)

a(n-1) is the Wiener index of the helm graph H(n) (n>=3). The graph H(n) is obtained from an n-wheel graph (on n+1 nodes) by adjoining a pendant edge at each node of the cycle. The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph. The Wiener polynomial of H(n) is (1/2)*n*t*((n-3)t^3 + 2(n-2)t^2 + (n+3)t + 6). - Emeric Deutsch, Sep 28 2010
Also sequence found by reading the line from 0, in the direction 0, 12, ..., and the same line from 0, in the direction 0, 36, ..., in the square spiral whose vertices are the generalized tetradecagonal numbers A195818. Axis perpendicular to A195158 in the same spiral. - Omar E. Pol, Sep 29 2011
Also the Wiener index of the (n+1)-gear graph. - Eric W. Weisstein, Sep 08 2017

			a(1) = 12*1 + 0 = 12;
a(2) = 12*2 + 12 = 36;
a(3) = 12*3 + 36 = 72.

G. C. Greubel, Table of n, a(n) for n = 0..5000
B. E. Sagan, Y-N. Yeh and P. Zhang, The Wiener Polynomial of a Graph, Internat. J. of Quantum Chem., Vol. 60 (1996), pp. 959-969.
Leo Tavares, Illustration: Centroid Stars.
Eric Weisstein's World of Mathematics, Gear Graph.
Eric Weisstein's World of Mathematics, Helm Graph.
Eric Weisstein's World of Mathematics, Wiener Index.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A001844, A003154, A027468, A035008, A073577.

12 * Accumulate[Range[0, 50]] (* Harvey P. Dale, Feb 05 2013 *)
(* Start from Eric W. Weisstein, Sep 08 2017 *)
Table[6 n (n + 1), {n, 0, 20}]
12 PolygonalNumber[3, Range[0, 20]]
12 Binomial[Range[20], 2]
LinearRecurrence[{3, -3, 1}, {12, 36, 72}, {0, 20}]
(* End *)

a(n)=6*n*(n+1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 6*n*(n+1).
G.f.: 12*x/(1-x)^3.
a(n) = 12*A000217(n). - Omar E. Pol, Dec 11 2008
a(n) = 12*n + a(n-1) (with a(0)=0). - Vincenzo Librandi, Aug 06 2010
a(n) = A003154(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = A032528(2*n+1) - 1. - Adriano Caroli, Jul 19 2013
a(n) = A001844(n) + A073577(n). - Bruce J. Nicholson, Aug 06 2017
E.g.f.: 6*x*(x+2)*exp(x). - G. C. Greubel, Aug 23 2017
From Amiram Eldar, Feb 15 2022: (Start)
Sum_{n>=1} 1/a(n) = 1/6.
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/3 - 1/6. (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(6/Pi)*cos(sqrt(5/3)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (6/Pi)*cos(Pi/(2*sqrt(3))). (End)

A019557 Coordination sequence for G_2 lattice.

Original entry on oeis.org

1, 12, 30, 48, 66, 84, 102, 120, 138, 156, 174, 192, 210, 228, 246, 264, 282, 300, 318, 336, 354, 372, 390, 408, 426, 444, 462, 480, 498, 516, 534, 552, 570, 588, 606, 624, 642, 660, 678, 696, 714, 732, 750, 768, 786, 804, 822, 840, 858, 876, 894, 912, 930, 948, 966, 984, 1002, 1020, 1038, 1056

Offset: 0

Michael Baake (mbaake(AT)sunelc3.tphys.physik.uni-tuebingen.de)

Also, coordination sequence of Dual(3.12.12) tiling with respect to a 12-valent node. - N. J. A. Sloane, Jan 22 2018

For n > 1, also the number of minimum vertex colorings of the n-Andrásfai graph. - Eric W. Weisstein, Mar 03 2024

			From _Peter M. Chema_, Mar 20 2016: (Start)
Illustration of initial terms:
                                                       o
                                                      o o
                                    o                o   o
                                   o o        o o o o o o o o o o
                  o           o o o o o o o    o   o       o   o
               o o o o         o o     o o      o o         o o
     o          o   o           o       o        o           o
               o o o o         o o     o o      o o         o o
                  o           o o o o o o o    o   o       o   o
                                   o o        o o o o o o o o o o
                                    o                o   o
                                                      o o
                                                       o
     1           12                30                 48
Compare to A003154, A045946, and A270700. (End)

Michael Baake and Uwe Grimm, Coordination sequences for root lattices and related graphs, arXiv:cond-mat/9706122, Zeit. f. Kristallographie, 212 (1997), pp. 253-256
Roland Bacher, Pierre de la Harpe, and Boris Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Séries 1) (1997), pp. 1137-1142.
Roland Bacher, Pierre de la Harpe, and Boris Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, Annales de l'institut Fourier, 49 no. 3 (1999), pp. 727-762.
Tom Karzes, Tiling Coordination Sequences.
N. J. A. Sloane, Illustration of layers 0,1,2 in the graph of the Dual(3.12.12) tiling. Centered at a 12-valent node. Note that some of the blue edges are not part of the underlying graph.
N. J. A. Sloane, Overview of coordination sequences of Laves tilings. [Fig. 2.7.1 of Grünbaum-Shephard 1987 with A-numbers added and in some cases the name in the RCSR database]
Eric Weisstein's World of Mathematics, Andrásfai Graph.
Eric Weisstein's World of Mathematics, Minimum Vertex Coloring.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000290, A003154, A008706, A016789, A045946, A270700.
For partial sums see A082040.
List of coordination sequences for Laves tilings (or duals of uniform planar nets): [3,3,3,3,3.3] = A008486; [3.3.3.3.6] = A298014, A298015, A298016; [3.3.3.4.4] = A298022, A298024; [3.3.4.3.4] = A008574, A296368; [3.6.3.6] = A298026, A298028; [3.4.6.4] = A298029, A298031, A298033; [3.12.12] = A019557, A298035; [4.4.4.4] = A008574; [4.6.12] = A298036, A298038, A298040; [4.8.8] = A022144, A234275; [6.6.6] = A008458.

CoefficientList[Series[(1 + 10 x + 7 x^2)/(1 - x)^2, {x, 0, 59}], x] (* Michael De Vlieger, Mar 21 2016 *)

my(x='x+O('x^100)); Vec((1+10*x+7*x^2)/(1-x)^2) \\ Altug Alkan, Mar 20 2016

a(n) = 18*n - 6, n >= 1.
G.f.: (1 + 10*x + 7*x^2)/(1-x)^2.
From Elmo R. Oliveira, Apr 04 2025: (Start)
E.g.f.: 6*exp(x)*(3*x - 1) + 7.
a(n) = 6*A016789(n-1) for n >= 1.
a(n) = 2*a(n-1) - a(n-2) for n >= 3. (End)

A101321 Table T(n,m) = 1 + nm(m+1)/2 read by antidiagonals: centered polygonal numbers.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 7, 7, 4, 1, 1, 11, 13, 10, 5, 1, 1, 16, 21, 19, 13, 6, 1, 1, 22, 31, 31, 25, 16, 7, 1, 1, 29, 43, 46, 41, 31, 19, 8, 1, 1, 37, 57, 64, 61, 51, 37, 22, 9, 1, 1, 46, 73, 85, 85, 76, 61, 43, 25, 10, 1, 1, 56, 91, 109, 113, 106, 91, 71, 49, 28, 11, 1, 1, 67

Offset: 0

Eugene McDonnell (eemcd(AT)mac.com), Dec 24 2004

Row n gives the centered figurate numbers of the n-gon.

Antidiagonal sums are in A101338.

			The upper left corner of the infinite array T is
|0| 1   1   1   1   1   1   1   1   1   1 ... A000012
|1| 1   2   4   7  11  16  22  29  37  46 ... A000124
|2| 1   3   7  13  21  31  43  57  73  91 ... A002061
|3| 1   4  10  19  31  46  64  85 109 136 ... A005448
|4| 1   5  13  25  41  61  85 113 145 181 ... A001844
|5| 1   6  16  31  51  76 106 141 181 226 ... A005891
|6| 1   7  19  37  61  91 127 169 217 271 ... A003215
|7| 1   8  22  43  71 106 148 197 253 316 ... A069099
|8| 1   9  25  49  81 121 169 225 289 361 ... A016754
|9| 1  10  28  55  91 136 190 253 325 406 ... A060544

Cf. A000217, A001263, A101338.

A101321 := proc(n,k)
        n*k*(k+1)/2+1 ;
end proc: # R. J. Mathar, Jul 28 2016

T[n_, m_] := 1 + n m (m + 1)/2;
Table[T[n - m, m], {n, 0, 12}, {m, n, 0, -1}] // Flatten (* Jean-François Alcover, Mar 23 2020 *)

T(n,m) = 1 + n*m*(m+1)/2 \\ Charles R Greathouse IV, Jul 28 2016

T(n,2) = A016777(n). T(n,3) = A016921(n). T(n,4) = A017281(n).
T(10,m) = A062786(m+1).
T(11,m) = A069125(m+1).
T(12,m) = A003154(m+1).
T(13,m) = A069126(m+1).
T(14,m) = A069127(m+1).
T(15,m) = A069128(m+1).
T(16,m) = A069129(m+1).
T(17,m) = A069130(m+1).
T(18,m) = A069131(m+1).
T(19,m) = A069132(m+1).
T(20,m) = A069133(m+1).
T(n+1,m) = T(n,m) + m*(m+1)/2. - Gary W. Adamson and Michel Marcus, Oct 13 2015

Edited by R. J. Mathar, Oct 21 2009

A217843 Numbers which are the sum of one or more consecutive nonnegative cubes.

Original entry on oeis.org

0, 1, 8, 9, 27, 35, 36, 64, 91, 99, 100, 125, 189, 216, 224, 225, 341, 343, 405, 432, 440, 441, 512, 559, 684, 729, 748, 775, 783, 784, 855, 1000, 1071, 1196, 1241, 1260, 1287, 1295, 1296, 1331, 1584, 1728, 1729, 1800, 1925, 1989, 2016, 2024, 2025, 2197

Offset: 1

T. D. Noe, Oct 23 2012

nonn

Contains A000578 (cubes), A005898 (two consecutive cubes), A027602 (three consecutive cubes), A027603 (four consecutive cubes) etc. - R. J. Mathar, Nov 04 2012
See A265845 for sums of consecutive positive cubes in more than one way. - Reinhard Zumkeller, Dec 17 2015
From Lamine Ngom, Apr 15 2021: (Start)
a(n) can always be expressed as the difference of the squares of two triangular numbers (A000217).
A168566 is the subsequence A000217(n)^2 - 1.
a(n) is also the product of two nonnegative integers whose sum and difference are both promic.
See example and formula sections for details. (End)

			From _Lamine Ngom_, Apr 15 2021: (Start)
Arrange the positive terms in a triangle as follows:
n\k |   1    2    3    4    5    6    7
----+-----------------------------------
  0 |   1;
  1 |   8,   9;
  2 |  27,  35,  36;
  3 |  64,  91,  99, 100;
  4 | 125, 189, 216, 224, 225;
  5 | 216, 341, 405, 432, 440, 441;
  6 | 343, 559, 684, 748, 775, 783, 784;
Column 1: cubes = A000217(n+1)^2 - A000217(n)^2.
The difference of the squares of two consecutive triangular numbers (A000217) is a cube (A000578).
Column 2: sums of 2 consecutive cubes (A027602).
Column 3: sums of 3 consecutive cubes (A027603).
etc.
Column k: sums of k consecutive cubes.
Row n: A000217(n)^2 - A000217(m)^2, m < n.
T(n,n) = A000217(n)^2 (main diagonal).
T(n,n-1) = A000217(n)^2 - 1 (A168566) (2nd diagonal).
Now rectangularize this triangle as follows:
n\k |   1    2     3     4    5     6   ...
----+--------------------------------------
  0 |   1,   9,   36,  100,  225,  441, ...
  1 |   8,  35,   99,  224,  440,  783, ...
  2 |  27,  91,  216,  432,  775, 1287, ...
  3 |  64, 189,  405,  748, 1260, 1989, ...
  4 | 125, 341,  684, 1196, 1925, 2925, ...
  5 | 216, 559, 1071, 1800, 2800, 4131, ...
  6 | 343, 855, 1584, 2584, 3915, 5643, ...
The general form of terms is:
T(n,k) = [n^4 + A016825(k)*n^3 + A003154(k)*n^2 + A300758(k)*n]/4, sum of n consecutive cubes after k^3.
This expression can be factorized into [n*(n + A005408(k))*(n*(n + A005408(k)) + 4*A000217(k))]/4.
For k = 1, the sequence provides all cubes: T(n,1) = A000578(k).
For k = 2, T(n,2) = A005898(k), centered cube numbers, sum of two consecutive cubes.
For k = 3, T(n,3) = A027602(k), sum of three consecutive cubes.
For k = 4, T(n,4) = A027603(k), sum of four consecutive cubes.
For k = 5, T(n,5) = A027604(k), sum of five consecutive cubes.
T(n,n) = A116149(n), sum of n consecutive cubes after n^3 (main diagonal).
For n = 0, we obtain the subsequence T(0,k) = A000217(n)^2, product of two numbers whose difference is 0*1 (promic) and sum is promic too.
For n = 1, we obtain the subsequence T(1,k) = A168566(x), product of two numbers whose difference is 1*2 (promic) and sum is promic too.
For n = 2, we obtain the subsequence T(2,k) = product of two numbers whose difference is 2*3 (promic) and sum is promic too.
etc.
For n = x, we obtain the subsequence formed by products of two numbers whose difference is the promic x*(x+1) and sum is promic too.
Consequently, if m is in the sequence, then m can be expressed as the product of two nonnegative integers whose sum and difference are both promic. (End)

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A034705, A217844-A217850, A062682, A131643, A240137.
Cf. A000578, A005898, A027602, A027603, A027604.
Cf. A265845 (subsequence).
Cf. A000217 (triangular numbers), A046092 (4*A000217).
Cf. A168566 (A000217^2 - 1).
Cf. A002378 (promics), A016825 (singly even numbers), A003154 (stars numbers).
Cf. A000330 (square pyramidal numbers), A300758 (12*A000330).
Cf. A005408 (odd numbers).

import Data.Set (singleton, deleteFindMin, insert, Set)
a217843 n = a217843_list !! (n-1)
a217843_list = f (singleton (0, (0,0))) (-1) where
   f s z = if y /= z then y : f s'' y else f s'' y
              where s'' = (insert (y', (i, j')) $
                           insert (y' - i ^ 3 , (i + 1, j')) s')
                    y' = y + j' ^ 3; j' = j + 1
                    ((y, (i, j)), s') = deleteFindMin s
-- Reinhard Zumkeller, Dec 17 2015, May 12 2015

nMax = 3000; t = {0}; Do[k = n; s = 0; While[s = s + k^3; s <= nMax, AppendTo[t, s]; k++], {n, nMax^(1/3)}]; t = Union[t]

lista(nn) = {my(list = List([0])); for (i=1, nn, my(s = 0); forstep(j=i, 1, -1, s += j^3; if (s > nn^3, break); listput(list, s););); Set(list);} \\ Michel Marcus, Nov 13 2020

a(n) >> n^2. Probably a(n) ~ kn^2 for some k but I cannot prove this. - Charles R Greathouse IV, Aug 07 2013

a(n) is of the form [x*(x+2*k+1)*(x*(x+2*k+1)+2*k*(k+1))]/4, sum of n consecutive cubes starting from (k+1)^3. - Lamine Ngom, Apr 15 2021

Name edited by N. J. A. Sloane, May 24 2021

A014634 a(n) = (2n+1)(4*n+1).

Original entry on oeis.org

1, 15, 45, 91, 153, 231, 325, 435, 561, 703, 861, 1035, 1225, 1431, 1653, 1891, 2145, 2415, 2701, 3003, 3321, 3655, 4005, 4371, 4753, 5151, 5565, 5995, 6441, 6903, 7381, 7875, 8385, 8911, 9453, 10011, 10585, 11175, 11781, 12403, 13041, 13695, 14365, 15051

Offset: 0

Mohammad K. Azarian, N. J. A. Sloane

Odd hexagonal numbers. Bisection of A000384. - Omar E. Pol, Apr 06 2008
Sequence found by reading the line from 1, in the direction 1, 15, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 03 2011
a(n) is also the sum of natural numbers which can be placed in a center box and expanded ones on 4 arms on N, S, E, W or NE, NW, SW, SE directions. See illustration in links. - Kival Ngaokrajang, Jul 08 2014

G. C. Greubel, Table of n, a(n) for n = 0..5000
Kival Ngaokrajang, Illustration of initial terms.
Leo Tavares, Illustration: Dual Square Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A000384, A005408, A016813, A033954, A157870, A194268.

Cf. A003154, A000290.

[(2*n+1)*(4*n+1) : n in [0..50]]; // Wesley Ivan Hurt, Jul 09 2014

A014634:=n->(2*n+1)*(4*n+1); seq(A014634(k), k=0..100); # Wesley Ivan Hurt, Nov 04 2013

lst={};Do[a=(2*n+1)*(4*n+1);AppendTo[lst,a],{n,0,5!}];lst (* Vladimir Joseph Stephan Orlovsky, Mar 10 2009 *)
Table[(2 n + 1) (4 n + 1), {n, 0, 50}] (* Wesley Ivan Hurt, Jul 09 2014 *)
LinearRecurrence[{3,-3,1},{1,15,45},50] (* Harvey P. Dale, Aug 30 2021 *)

a(n)=(2*n+1)*(4*n+1) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = A157870(n)/2. - Vladimir Joseph Stephan Orlovsky, Mar 10 2009
a(n) = a(n-1) + 16*n-2 (with a(0)=1). - Vincenzo Librandi, Nov 20 2010
G.f.: (1+12*x+3*x^2)/(1-x)^3. - Colin Barker, Jan 08 2012
a(n) = A005408(n) * A016813(n). - Omar E. Pol, Nov 05 2013
a(n) = 2*A033954(n) + 1 = A194268(n) - n. - Wesley Ivan Hurt, Jul 14 2014
E.g.f.: (8*x^2 +14*x + 1)*exp(x). - G. C. Greubel, Jul 18 2017
From Amiram Eldar, Feb 28 2022: (Start)
Sum_{n>=0} 1/a(n) = Pi/4 + log(2)/2.
Sum_{n>=0} (-1)^n/a(n) = Pi*(sqrt(2)-1)/4 + log(sqrt(2)+1)/sqrt(2). (End)
a(n) = A003154(n+1) + 2*A000290(n). - Leo Tavares, Mar 26 2022

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A014634 a(n) = (2n+1)(4*n+1).