A003154 -id:A003154 - cp's OEIS frontend

A064200 a(n) = 12n(n-1).

0, 0, 24, 72, 144, 240, 360, 504, 672, 864, 1080, 1320, 1584, 1872, 2184, 2520, 2880, 3264, 3672, 4104, 4560, 5040, 5544, 6072, 6624, 7200, 7800, 8424, 9072, 9744, 10440, 11160, 11904, 12672, 13464, 14280, 15120, 15984, 16872, 17784, 18720, 19680, 20664, 21672

Offset: 0

Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr), Sep 22 2001

Luigi Berzolari, Allgemeine Theorie der Höheren Ebenen Algebraischen Kurven, Encyclopädie der Mathematischen Wissenschaften mit Einschluss ihrer Anwendungen, Band III_2, Heft 3, Leipzig: B. G. Teubner, 1906, p. 341.

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Leo Tavares, Illustration: Twin Stars.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A262221, A003154.

Table[12n(n-1),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,0,24},40] (* Harvey P. Dale, Jul 22 2015 *)
Join[{0},24*Accumulate[Range[0,60]]] (* Harvey P. Dale, Dec 17 2022 *)

a(n)=12*n*(n-1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 24*(n-1) + a(n-1) for n>0, with a(0)=0. - Vincenzo Librandi, Aug 07 2010
a(0)=0, a(1)=0, a(2)=24, a(n)=3*a(n-1)-3*a(n-2)+a(n-3). - Harvey P. Dale, Jul 22 2015
G.f.: -(24*x^2)/(x-1)^3. - Harvey P. Dale, Jul 22 2015
a(n) = 2*A003154(n) - 2. See Twin Stars illustration. - Leo Tavares, Aug 23 2021
From Amiram Eldar, Feb 22 2023: (Start)
Sum_{n>=2} 1/a(n) = 1/12.
Sum_{n>=2} (-1)^n/a(n) = (2*log(2) - 1)/12.
Product_{n>=2} (1 - 1/a(n)) = -(12/Pi)*cos(Pi/sqrt(3)).
Product_{n>=2} (1 + 1/a(n)) = (12/Pi)*cos(Pi/sqrt(6)). (End)

A103115 a(n) = 6n(n-1) - 1.

Original entry on oeis.org

-1, -1, 11, 35, 71, 119, 179, 251, 335, 431, 539, 659, 791, 935, 1091, 1259, 1439, 1631, 1835, 2051, 2279, 2519, 2771, 3035, 3311, 3599, 3899, 4211, 4535, 4871, 5219, 5579, 5951, 6335, 6731, 7139, 7559, 7991, 8435, 8891, 9359, 9839, 10331, 10835, 11351, 11879, 12419

Offset: 0

Jacob Landon (jacoblandon(AT)aol.com), May 09 2009

Star numbers A003154 minus 2.
What A163433 does for a triangle, this sequence is doing for a square but giving one-half the results. Take a square with vertices n, n+1, n+2, and n+3 and find the sum of the four products of each four vertices times the sum of the other three; at n you have n((n+1)+(n+2)+(n+3)) and so on for the other three vertices. The result of all four is 12*n^2 + 36*n + 22; half this is 6*n^2 + 18*n + 11 and gives the numbers in this sequence starting with n = 0. - J. M. Bergot, May 23 2012
Multiplying a(n) by 16 gives the sum of the convolution with itself of each of the 24 permutations of four consecutive numbers. - J. M. Bergot, May 15 2017

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

Cf. A000914, A003154, A163433.

[6*n*(n-1)-1: n in [0..50]]; // Vincenzo Librandi, May 16 2017

Table[6n(n-1)-1,{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{-1,-1,11},50] (* Harvey P. Dale, Nov 14 2011 *)
CoefficientList[Series[(1 - 2 x - 11 x^2) / (x - 1)^3, {x, 0, 50}], x] (* Vincenzo Librandi, May 16 2017 *)

a(n)=6*n*(n-1)-1 \\ Charles R Greathouse IV, Jun 16 2017

a(n) = A003154(n) - 2.
G.f.: (1-2*x-11*x^2)/(x-1)^3. - R. J. Mathar, May 11 2009 (adapted by Vincenzo Librandi, May 16 2017).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=-1, a(1)=-1, a(2)=11. - Harvey P. Dale, Nov 14 2011
a(n) = (n-2)*(n-1 + n + n+1) + (n-1)*(n + n+1) + n*(n+1), which is applying A000914 to four consecutive numbers. - J. M. Bergot, May 15 2017
Sum_{n>=1} 1/a(n) = tan(sqrt(5/3)*Pi/2)*Pi/(2*sqrt(15)). Amiram Eldar, Aug 20 2022
E.g.f.: exp(x)*(6*x^2 - 1). - Elmo R. Oliveira, Jan 16 2025

Edited and extended by R. J. Mathar, May 11 2009

Entries rechecked by N. J. A. Sloane, Jul 18 2009

A331952 a(n) = (-7 + (-1)^(1+n) + 6*n^2) / 8.

Original entry on oeis.org

-1, 0, 2, 6, 11, 18, 26, 36, 47, 60, 74, 90, 107, 126, 146, 168, 191, 216, 242, 270, 299, 330, 362, 396, 431, 468, 506, 546, 587, 630, 674, 720, 767, 816, 866, 918, 971, 1026, 1082, 1140, 1199, 1260, 1322, 1386, 1451, 1518, 1586, 1656, 1727, 1800, 1874, 1950, 2027

Offset: 0

Paul Curtz, Feb 02 2020

a(n+1) is once in the hexagonal spiral in A330707. a(n+2) is twice in the same spiral.
a(n) has one odd followed by three evens.
Difference table:
  -1, 0, 2, 6, 11, 18, 26, 36, ... = a(n)
   1, 2, 4, 5,  7,  8, 10, 11, ... = A001651(n+1)
   1, 2, 1, 2,  1,  2,  1,  2, ... = A000034.

			G.f. = -1 + 2*x^2 + 6*x^3 + 11*x^4 + 18*x^5 + 26*x^6 + 36*x^7 + 47*x^8 + ... - _Michael Somos_, Sep 08 2023

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).
Index entries for two-way infinite sequences

Cf. A000034, A001651, A158463, 6*A014105, 2*A003154, A152746(n+1), A008585(1+n).

Equals 2 less than A084684, 1 less than A077043, and 1 more than A276382(n-1). - Greg Dresden, Feb 22 2020

a:=[-1,0,2,6]; [n le 4 select a[n] else 2*Self(n-1)-2*Self(n-3)+Self(n-4): n in [1..45]]; // Marius A. Burtea, Feb 02 2020

LinearRecurrence[{2, 0, -2, 1}, {-1, 0, 2, 6}, 100] (* Amiram Eldar, Feb 02 2020 *)
a[n_] := Floor[(n^2 - 1)*3/4]; (* Michael Somos, Sep 08 2023 *)

Vec(-(1 - 2*x - 2*x^2) / ((1 - x)^3*(1 + x)) + O(x^40)) \\ Colin Barker, Feb 03 2020

{a(n) = (n^2 - 1)*3\4}; /* Michael Somos, Sep 08 2023 */

a(-n) = a(n).
a(20+n) - a(n) = 30*(10+n).
a(2+n) = a(n) + 3*(1+n), a(0)=-1 and a(1)=0.
a(4*n) = 12*n^2 - 1, a(1+4*n) = 6*n*(1+2*n), a(2+4*n) = 2 + 12*n*(1+n), a(3+4*n) = 6*(1+n)*(1+2*n) for n>= 0.
From Colin Barker, Feb 02 2020: (Start)
G.f.: -(1 - 2*x - 2*x^2) / ((1 - x)^3*(1 + x)).
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n>3.
a(n) = (-7 + (-1)^(1+n) + 6*n^2) / 8.
(End)
E.g.f.: (1/8)*(exp(x)*(6*x^2 + 6*x - 7) - exp(-x)). - Stefano Spezia, Feb 02 2020 after Colin Barker
a(n) = floor((n^2 - 1)*3/4). - Michael Somos, Sep 09 2023

a(42)-a(52) from Stefano Spezia, Feb 02 2020

A006060 Triangular star numbers.

Original entry on oeis.org

1, 253, 49141, 9533161, 1849384153, 358770992581, 69599723176621, 13501987525271953, 2619315980179582321, 508133798167313698381, 98575337528478677903653, 19123107346726696199610361

Offset: 1

N. J. A. Sloane

nonn

M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 20.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

B. Berselli, Table of n, a(n) for n = 1..400. [From _Bruno Berselli_, Jul 07 2010]
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Star Number
Index entries for linear recurrences with constant coefficients, signature (195, -195, 1).

Cf. A000567, A003154, A006061, A051673.

A006060:=-(1+58*z+z**2)/(z-1)/(z**2-194*z+1); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
a:= n-> (Matrix([[253,1,1]]). Matrix([[195,1,0], [ -195,0,1], [1,0,0]])^n)[1,3]: seq(a(n), n=1..20); # Alois P. Heinz, Aug 14 2008

a006060 = {}; Do[
If[Length[a006060] < 2, AppendTo[a006060, 1],
  AppendTo[a006060, 194*a006060[[-1]] + 60 - a006060[[-2]]]],  {n,
  20}]; TableForm[Transpose[List[Range[Length[a006060]], a006060]]] (* Michael De Vlieger *)
LinearRecurrence[{195,-195,1},{1,253,49141},20] (* Harvey P. Dale, Jan 12 2017 *)

G.f.: (1 + 58x + x^2)/((x-1)(1 - 194x + x^2)). - Ralf Stephan, Apr 23 2004
From Bruno Berselli, Jul 07 2010: (Start)
a(n) = 194*a(n-1) - a(n-2) + 60 (n>2).
a(n) = (3*((7 + 4*sqrt(3))^(2*n-1) + (7 - 4*sqrt(3))^(2*n-1)) - 10)/32 (n>0).
(End)

Extended by Eric W. Weisstein, Mar 01 2002

A063436 Write 1, 2, 3, 4, ... counterclockwise in a hexagonal spiral around 0 starting left down, then a(n) is the sequence found by reading from 0 in the vertical upward direction.

Original entry on oeis.org

0, 15, 54, 117, 204, 315, 450, 609, 792, 999, 1230, 1485, 1764, 2067, 2394, 2745, 3120, 3519, 3942, 4389, 4860, 5355, 5874, 6417, 6984, 7575, 8190, 8829, 9492, 10179, 10890, 11625, 12384, 13167, 13974, 14805, 15660, 16539, 17442, 18369, 19320, 20295, 21294, 22317

Offset: 0

Floor van Lamoen, Jul 21 2001

Related to parity of Beatty sequences for exp(-(1/2)/n). Let f(k,n)=-sum(i=1,n,sum(j=1,i,(-1)^floor(j*exp(-(1/2)/n)))), then a(n)=Max{f(k,n) : 1<=k<=4*a(n)-2} and for 0<=i<=4*a(n)-3, f(i,n)=f(4*a(n)-2-i,n). - Benoit Cloitre, May 26 2004

Or, sum of multiples of 2 and 3 from 0 to 6*n. - Zak Seidov, Aug 06 2016

			The spiral begins:
.
        16--15--14
        /         \
      17   5---4  13
      /   /     \   \
    18   6   0   3  12
    /   /   /   /   /
  19   7   1---2  11  26
    \   \         /   /
    20   8---9--10  25
      \             /
      21--22--23--24

Harry J. Smith, Table of n, a(n) for n = 0..1000
Leo Tavares, Illustration: Stellar Triangles.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000567, A003154, A007742, A062783, A272399.

a[n_] := 3*n*(4*n + 1); Array[a, 40, 0] (* Amiram Eldar, Mar 27 2022 *)

{ for (n=0, 1000, write("b063436.txt", n, " ", n*(12*n + 3)) ) } \\ Harry J. Smith, Aug 21 2009

a(n) = 3*n*(4*n+1) = 3*A007742(n).
a(n) = 24*n + a(n-1) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
From Colin Barker, Jul 07 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 3*x*(5 + 3*x)/(1-x)^3. (End)
a(n) = A272399(n+1) - A003154(n+1). - Leo Tavares, Mar 24 2022
From Amiram Eldar, Mar 27 2022: (Start)
Sum_{n>=1} 1/a(n) = 4/3 - Pi/6 - log(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(2)) + log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3 - 4/3. (End)
E.g.f.: 3*x*(5 + 4*x)*exp(x). - Elmo R. Oliveira, Oct 31 2024

A094421 a(n) = n * (6n^2 + 6n + 1).

Original entry on oeis.org

13, 74, 219, 484, 905, 1518, 2359, 3464, 4869, 6610, 8723, 11244, 14209, 17654, 21615, 26128, 31229, 36954, 43339, 50420, 58233, 66814, 76199, 86424, 97525, 109538, 122499, 136444, 151409, 167430, 184543, 202784, 222189, 242794

Offset: 1

Ralf Stephan, May 02 2004

Third column of array A094416.

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

List([1..40], n-> n*(6*n^2+6*n+1)); # G. C. Greubel, Oct 30 2019

[n*(6*n^2+6*n+1): n in [1.40]]; // G. C. Greubel, Oct 30 2019

A094421:=n->n * (6*n^2 + 6*n + 1); seq(A094421(n), n=1..40); # Wesley Ivan Hurt, Feb 12 2014

Table[n(6n^2+6n+1), {n, 40}] (* Wesley Ivan Hurt, Feb 12 2014 *)
LinearRecurrence[{4,-6,4,-1},{13,74,219,484},40] (* Harvey P. Dale, Jan 04 2016 *)

vector(40, n, n*(6*n^2+6*n+1)) \\ G. C. Greubel, Oct 30 2019

[n*(6*n^2+6*n+1) for n in (1..40)] # G. C. Greubel, Oct 30 2019

Equals n * A003154(n) (star numbers).
G.f.: x*(13 + 22*x + x^2)/(1-x)^4. - Colin Barker, Aug 02 2012
E.g.f.: x*(13 + 24*x + 6*x^2)*exp(x). - G. C. Greubel, Oct 30 2019

A162245 Triangle T(n,m) = 6mn + 3m + 3n + 1 read by rows.

Original entry on oeis.org

13, 22, 37, 31, 52, 73, 40, 67, 94, 121, 49, 82, 115, 148, 181, 58, 97, 136, 175, 214, 253, 67, 112, 157, 202, 247, 292, 337, 76, 127, 178, 229, 280, 331, 382, 433, 85, 142, 199, 256, 313, 370, 427, 484, 541, 94, 157, 220, 283, 346, 409, 472, 535, 598, 661

Offset: 1

Vincenzo Librandi, Jun 28 2009

If h belongs to the main diagonal of the triangle then 6*h+3 is a square since T(n,n) = (3/2)*(2*n+1)^2-1/2 and 6*T(n,n)+3 = 9*(2*n+1)^2. Also, the first column is A017209 (after 4). - Vincenzo Librandi, Nov 20 2012

			Triangle begins:
13;
22, 37;
31, 52,  73;
40, 67,  94,  121;
49, 82,  115, 148, 181;
58, 97,  136, 175, 214, 253;
67, 112, 157, 202, 247, 292, 337;
76, 127, 178, 229, 280, 331, 382, 433; etc.

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

Cf. A003154, A017209, A083487, A163433.

[6*n*k + 3*n + 3*k + 1:  k in [1..n],  n in [1..11]]; // Vincenzo Librandi, Nov 20 2012

Flatten@Table[6*m*n + 3*m + 3*n + 1, {n, 20}, {m, n}] (* Vincenzo Librandi, Mar 03 2012 *)

Row sums: Sum_{m=1..n} T(n,m) = n*(5+6*n^2+15*n)/2. - R. J. Mathar, Jul 26 2009
T(n,m) = 3*A083487(n,m)+1. - R. J. Mathar, Jul 26 2009
T(k,k) = A003154(k+1) and T(k+1,k) = A163433(k+2). - Avi Friedlich, May 22 2015

Edited by R. J. Mathar, Jul 26 2009

A181475 a(n) = 3n^4 + 6n^3 - 3*n + 1.

Original entry on oeis.org

1, 7, 91, 397, 1141, 2611, 5167, 9241, 15337, 24031, 35971, 51877, 72541, 98827, 131671, 172081, 221137, 279991, 349867, 432061, 527941, 638947, 766591, 912457, 1078201, 1265551, 1476307, 1712341, 1975597, 2268091, 2591911, 2949217, 3342241, 3773287, 4244731

Offset: 0

Bruno Berselli, Oct 25 2010 - Oct 29 2010

If gcd(n,7) = gcd(n+1,7) = gcd(2*n+1,7) = 1 then a(n) == 0 (mod 7) (E. Picutti, see References).

Ettore Picutti, Sul numero e la sua storia, Feltrinelli Economica, 1977, p. 208.

Bruno Berselli, Table of n, a(n) for n = 0..10000.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Subsequence of A003215.

Magma
```
[3*n^4+6*n^3-3*n+1: n in [0..31]];
```

Table[3 n^4 + 6 n^3 - 3 n + 1, {n, 0, 40}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{1,7,91,397,1141},40] (* Harvey P. Dale, Jul 12 2022 *)

G.f.: (1 + 2*x + 66*x^2 + 2*x^3 + x^4)/(1-x)^5.
a(n) = a(-n-1) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 6*12.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 6*A008594(n-1).
a(n) = 2*a(n-1) - a(n-2) + 6*A003154(n).
a(n) = a(n-1) + 6*A007588(n).
a(n) = 1 + 6*A062392(n).
a(n) = 7*A000540(n)/A000330(n) = A154105(A000096(n-1)) for n > 0.
Sum_{i=0..n} a(i) = (3*n^5 + 15*n^4 + 20*n^3 - 3*n + 5)/5.
a(n) = 7*(3*n^2 + 3*n - 1)*(Sum_{k=1..n} k^6)/(5*Sum_{k=1..n} k^4), n > 0. - Gary Detlefs, Oct 18 2011

Formula, program and crossref added by Bruno Berselli, Aug 22 2011

A184899 n such that the n-th centered 12-gonal number is prime. Indices of prime star numbers.

Original entry on oeis.org

2, 3, 4, 6, 8, 9, 10, 11, 13, 14, 19, 20, 21, 23, 24, 31, 32, 33, 34, 36, 37, 39, 42, 43, 44, 46, 47, 48, 52, 56, 59, 61, 66, 68, 74, 75, 78, 79, 85, 87, 89, 94, 96, 98, 101, 102, 105, 107, 108, 110, 113, 118, 121, 124, 125, 127, 130, 131, 133, 135, 136, 138

Offset: 1

Jonathan Vos Post, Feb 01 2011

For n > 3, A003154(a(n)) = A057199(2*a(n)-2). - Chai Wah Wu, Sep 17 2019

			98 is in the sequence because 6*98^2 - 6*98 + 1 = 57037 is prime.

Zak Seidov, Table of n, a(n) for n = 1..1000

Cf. A083577, A000040, A003154, A057199.

Select[Range[140],PrimeQ[6#^2-6#+1]&]  (* Harvey P. Dale, Feb 16 2011 *)

{n: 6*n^2 - 6*n + 1 is prime} = {n: A003154(n) is in A000040}.

A277978 a(n) = 3n(n+3).

Original entry on oeis.org

0, 12, 30, 54, 84, 120, 162, 210, 264, 324, 390, 462, 540, 624, 714, 810, 912, 1020, 1134, 1254, 1380, 1512, 1650, 1794, 1944, 2100, 2262, 2430, 2604, 2784, 2970, 3162, 3360, 3564, 3774, 3990, 4212, 4440, 4674, 4914, 5160, 5412, 5670, 5934, 6204, 6480

Offset: 0

Emeric Deutsch, Nov 08 2016

For n>= 3, a(n) is the second Zagreb index of the wheel graph with n+1 vertices. The second Zagreb index of a simple connected graph is the sum of the degree products d(i)d(j) over all edges ij of g.

			a(3) = 54. Indeed, the wheel graph with 4 vertices consists of 6 edges, each connecting two vertices of degree 3. Then, the second Zagreb index is 6*3*3 = 54.

Leo Tavares, Illustration: Hexagonal Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000096, A028569, A140091.

Maple
```
seq(3*n*(n+3), n = 0 .. 45);
```

A277978[n_] := 3 n (n + 3); Array[A277978, 45] (* JungHwan Min, Nov 08 2016 *)

a(n)=3*n*(n+3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 2 * A140091(n) = 3 * A028552(n) = 6 * A000096(n).
G.f.: 6*x*(2-x)/(1-x)^3
a(n) = A003154(n+1) - A003215(n-1). See Hexagonal Stars illustration. - Leo Tavares, Aug 20 2021

cp's OEIS Frontend

A064200 a(n) = 12*n*(n-1).

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A103115 a(n) = 6*n*(n-1) - 1.

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A331952 a(n) = (-7 + (-1)^(1+n) + 6*n^2) / 8.

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A006060 Triangular star numbers.

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A063436 Write 1, 2, 3, 4, ... counterclockwise in a hexagonal spiral around 0 starting left down, then a(n) is the sequence found by reading from 0 in the vertical upward direction.

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A094421 a(n) = n * (6*n^2 + 6*n + 1).

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A162245 Triangle T(n,m) = 6*m*n + 3*m + 3*n + 1 read by rows.

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A064200 a(n) = 12n(n-1).

A103115 a(n) = 6n(n-1) - 1.

A094421 a(n) = n * (6n^2 + 6n + 1).

A162245 Triangle T(n,m) = 6mn + 3m + 3n + 1 read by rows.

A181475 a(n) = 3n^4 + 6n^3 - 3*n + 1.

A277978 a(n) = 3n(n+3).