cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-8 of 8 results.

A005900 Octahedral numbers: a(n) = n*(2*n^2 + 1)/3.

Original entry on oeis.org

0, 1, 6, 19, 44, 85, 146, 231, 344, 489, 670, 891, 1156, 1469, 1834, 2255, 2736, 3281, 3894, 4579, 5340, 6181, 7106, 8119, 9224, 10425, 11726, 13131, 14644, 16269, 18010, 19871, 21856, 23969, 26214, 28595, 31116, 33781, 36594, 39559, 42680
Offset: 0

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Author

Keywords

Comments

Series reversion of g.f.: A(x) is Sum_{n>0} - A066357(n)(-x)^n.
Partial sums of centered square numbers A001844. - Paul Barry, Jun 26 2003
Also as a(n) = (1/6)*(4n^3 + 2n), n>0: structured tetragonal diamond numbers (vertex structure 5) (cf. A000447 - structured diamonds); and structured trigonal anti-prism numbers (vertex structure 5) (cf. A100185 - structured anti-prisms). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {3,4}.
If X is an n-set and Y and Z are disjoint 2-subsets of X then a(n-4) is equal to the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Starting with 1 = binomial transform of [1, 5, 8, 4, 0, 0, 0, ...] where (1, 5, 8, 4) = row 3 of the Chebyshev triangle A081277. - Gary W. Adamson, Jul 19 2008
a(n) = largest coefficient of (1 + ... + x^(n-1))^4. - R. H. Hardin, Jul 23 2009
Convolution square root of (1 + 6x + 19x^3 + ...) = (1 + 3x + 5x^2 + 7x^3 + ...) = A005408(x). - Gary W. Adamson, Jul 27 2009
Starting with offset 1 = the triangular series convolved with [1, 3, 4, 4, 4, ...]. - Gary W. Adamson, Jul 28 2009
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral, and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Let b be any product of four different primes. Then the divisor lattice of b^n is of width a(n+1). - Jean Drabbe, Oct 13 2010
Arises in Bezdek's proof on contact numbers for congruent sphere packings (see preprint). - Jonathan Vos Post, Feb 08 2011
Euler transform of length 2 sequence [6, -2]. - Michael Somos, Mar 27 2011
a(n+1) is the number of 2 X 2 matrices with all terms in {0,1,...,n} and (sum of terms) = 2n. - Clark Kimberling, Mar 19 2012
a(n) is the number of semistandard Young tableaux over all partitions of 3 with maximal element <= n. - Alois P. Heinz, Mar 22 2012
Self convolution of the odd numbers. - Reinhard Zumkeller, Apr 04 2012
a(n) is the number of (w,x,y,z) with all terms in {1,...,n} and w+x=y+z; also the number of (w,x,y,z) with all terms in {0,...,n} and |w-x|<=y. - Clark Kimberling, Jun 02 2012
The sequence is the third partial sum of (0, 1, 3, 4, 4, 4, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the number of join-irreducible elements in the Weyl group of type B_n with respect to the strong Bruhat order. - Rafael Mrden, Aug 26 2020
Number of unit octahedra contained in an n-scale octahedron composed of a tetrahedral-octahedral honeycomb. The number of unit tetrahedra in it is 8*A000292(n-1) = 4*(n^3 - n)/3. Also, the number of unit tetrahedra and unit octahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb is respectively A006527(n) = (n^3 + 2*n)/3 and A000292(n-1) = (n^3 - n)/6. - Jianing Song, Feb 24 2025

Examples

			G.f. = x + 6*x^2 + 19*x^3 + 44*x^4 + 85*x^5 + 146*x^6 + 231*x^7 + ...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Sums of 2 consecutive terms give A001845. Cf. A001844.
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. A022521.
Cf. A081277.
Row n=3 of A210391. - Alois P. Heinz, Mar 22 2012
Cf. A005408.
Cf. A002061.
Cf. A000292 (tetrahedral numbers), A000578 (cubes), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Similar sequence: A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193(m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

Programs

  • Haskell
    a005900 n = sum $ zipWith (*) odds $ reverse odds
                where odds = take n a005408_list
    a005900_list = scanl (+) 0 a001844_list
    -- Reinhard Zumkeller, Jun 16 2013, Apr 04 2012
    
  • Magma
    [n*(2*n^2+1)/3: n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015
    
  • Magma
    I:=[0,1,6,19]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Sep 12 2015
    
  • Maple
    al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(3,n), n=0..100)];
    A005900:=(z+1)**2/(z-1)**4; # Simon Plouffe in his 1992 dissertation
    with(combinat): seq(fibonacci(4,2*n)/12, n=0..40); # Zerinvary Lajos, Apr 21 2008
  • Mathematica
    Table[(2n^3+n)/3, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,6,19},50] (* Harvey P. Dale, Oct 10 2013 *)
    CoefficientList[Series[x (1 + x)^2/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
  • Maxima
    makelist(n*(2*n^2+1)/3, n, 0, 20); /* Martin Ettl, Jan 07 2013 */
    
  • PARI
    {a(n) = n*(2*n^2+1)/3};
    
  • PARI
    concat([0],Vec(x*(1 + x)^2/(1 - x)^4 + O(x^50))) \\ Indranil Ghosh, Mar 16 2017
    
  • Python
    def a(n): return n*(2*n*n + 1)//3
    print([a(n) for n in range(41)]) # Michael S. Branicky, Sep 03 2021

Formula

a(n) = 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2. - Amarnath Murthy, May 28 2001
G.f.: x * (1 + x)^2 / (1 - x)^4. a(n) = -a(-n) = (2*n^3 + n) / 3.
a(n) = ( ((n+1)^5-n^5) - (n^5-(n-1)^5) )/30. - Xavier Acloque, Oct 17 2003
a(n) is the sum of the products pq, where p and q are both positive and odd and p + q = 2n, e.g., a(4) = 7*1 + 5*3 + 3*5 + 1*7 = 44. - Jon Perry, May 17 2005
a(n) = 4*binomial(n,3) + 4*binomial(n,2) + binomial(n,1). - Mitch Harris, Jul 06 2006
a(n) = binomial(n+2,3) + 2*binomial(n+1,3) + binomial(n,3), (this pair generalizes; see A014820, the 4-cross polytope numbers).
Sum_{n>=1} 1/a(n) = 3*gamma + 3*Psi((I*(1/2))*sqrt(2)) - (1/2)*(3*I)*Pi*coth((1/2)*Pi*sqrt(2)) - (1/2)*(3*I)*sqrt(2) = A175577, where I=sqrt(-1). - Stephen Crowley, Jul 14 2009
a(n) = A035597(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = A000578(n) - 2*A000292(n-1) for n>0. - J. M. Bergot, Apr 05 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Wesley Ivan Hurt, Sep 11 2015
E.g.f.: (1/3)*x*(3 + 6*x + 2*x^2)*exp(x). - Ilya Gutkovskiy, Mar 16 2017
a(n) = (A002061(A002061(n+1)) - A002061(A002061(n)))/6. - Daniel Poveda Parrilla, Jun 10 2017
a(n) = 6*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 6*log(2) - 4 = 1/(6 + 2/(6 + 6/(6 + ... + n*(n-1)/(6 + ...)))). See A142983. - Peter Bala, Mar 06 2024

A100145 Structured great rhombicosidodecahedral numbers.

Original entry on oeis.org

1, 120, 579, 1600, 3405, 6216, 10255, 15744, 22905, 31960, 43131, 56640, 72709, 91560, 113415, 138496, 167025, 199224, 235315, 275520, 320061, 369160, 423039, 481920, 546025, 615576, 690795, 771904, 859125, 952680, 1052791, 1159680
Offset: 1

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Author

James A. Record (james.record(AT)gmail.com), Nov 07 2004

Keywords

Comments

Structured polyhedral numbers are a type of figurate polyhedral numbers. Structurate polyhedra differ from regular figurate polyhedra by having appropriate figurate polygonal faces at any iteration, i.e., a regular truncated octahedron, n=2, would have 7 points on its hexagonal faces, whereas a structured truncated octahedron, n=2, would have 6 points - just as a hexagon, n=2, would have. Like regular figurate polygons, structured polyhedra seem to originate at a vertex and since many polyhedra have different vertices (a pentagonal diamond has 2 "polar" vertices with 5 adjacent vertices and 5 "equatorial" vertices with 4 adjacent vertices), these polyhedra have multiple structured number sequences, dependent on the "vertex structures" which are each equal to the one vertex itself plus its adjacent vertices. For polystructurate polyhedra the notation, structured polyhedra (vertex structure x) is used to differentiate between alternate vertices, where VS stands for vertex structure.

Crossrefs

Cf. A051673, A100146 through A100156 - structured Archimedean solids; A100157 through A100175 - structured Catalan solids; A100147 - structured prisms; A000447 - structured diamonds; A100185 - structured anti-prisms; and A100188 - structured anti-diamonds.

Programs

Formula

a(n) = (1/6)*(222*n^3 - 312*n^2 + 96*n).
From Jaume Oliver Lafont, Sep 08 2009: (Start)
a(n) = (1+(n-1))*(1+22*(n-1)+37*(n-1)^2);
G.f.: x*(1+116*x+105*x^2)/(1-x)^4. (End)
E.g.f.: exp(x)*x*(1 + 59*x + 37*x^2). - Stefano Spezia, Jun 06 2025

Extensions

Corrected by T. D. Noe, Oct 25 2006

A100157 Structured rhombic dodecahedral numbers (vertex structure 9).

Original entry on oeis.org

1, 14, 55, 140, 285, 506, 819, 1240, 1785, 2470, 3311, 4324, 5525, 6930, 8555, 10416, 12529, 14910, 17575, 20540, 23821, 27434, 31395, 35720, 40425, 45526, 51039, 56980, 63365, 70210, 77531, 85344, 93665, 102510, 111895, 121836, 132349, 143450, 155155, 167480
Offset: 1

Views

Author

James A. Record (james.record(AT)gmail.com), Nov 07 2004

Keywords

Comments

Also structured triakis octahedral numbers (vertex structure 9) (Cf. A100171 = alternate vertex); and structured heptagonal anti-prism numbers (Cf. A100185 = structured anti-prisms).
If Y is a 2-subset of a 2n-set X then, for n>=2, a(n-1) is the number of 4-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007
Let M(2n-1) be a (2n-1)x(2n-1) matrix whose (i,j)-entry equals i^2/(i^2+sqrt(-1)) if i=j and equals 1 otherwise. Then a(n) equals (-1)^(n+1) times the real part of prod(k^2+sqrt(-1),k=1...2n-1) times the determinant of M(2n-1). - John M. Campbell, Sep 07 2011
Principal diagonal of the convolution array A213752. - Clark Kimberling, Jun 20 2012
The Fuss-Catalan numbers are Cat(d,k)= [1/(k*(d-1)+1)]*binomial(k*d,k) and enumerate the number of (d+1)-gon partitions of a (k*(d-1)+2)-gon (cf. Whieldon and Schuetz link). a(n)= Cat(n,4), so enumerates the number of (n+1)-gon partitions of a (4*(n-1)+2)-gon. Analogous series are A000326 (k=3) and A234043 (k=5). Also, a(n)= A006918(4n+1) = A008610(4n+1) = A053307(4n+1) with offset=0. - Tom Copeland, Oct 05 2014

Examples

			For n=4, sum( (4+i)^2, i=-3..3 ) = (4-3)^2+(4-2)^2+(4-1)^2+(4-0)^2+(4+1)^2+(4+2)^2+(4+3)^2 = 140 = a(4). - _Bruno Berselli_, Jul 24 2014
		

References

  • Jolley, Summation of Series, Dover (1961).

Crossrefs

Cf. A005915 = alternate vertex; A100145 for more on structured polyhedral numbers.

Programs

  • Magma
    [(1/6)*(16*n^3-12*n^2+2*n): n in [1..40]]; // Vincenzo Librandi, Jul 19 2011
    
  • Maple
    with(combstruct):ZL:=[st, {st=Prod(left, right), left=Set(U, card=r), right=Set(U, card=r), U=Sequence(Z, card>=1)}, unlabeled]: subs(r=1, stack): seq(count(subs(r=2, ZL), size=m*4), m=1..32) ; # Zerinvary Lajos, Jan 02 2008
  • PARI
    a(n)=(16*n^3-12*n^2+2*n)/6 \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = (16*n^3 - 12*n^2 + 2*n)/6.
a(n) = n*(2*n-1)*(4*n-1)/3 = A000330(2*n-1). - Reinhard Zumkeller, Jul 06 2009
Sum_{n>=1} 1/(24*a(n)) = Pi/8-log(2)/2 = 0.046125491418751... [Jolley eq. 251]
G.f.: x*(1+10*x+5*x^2)/(x-1)^4. - R. J. Mathar, Oct 03 2011
a(n) = binomial(2*n+1,3) + binomial(2*n,3). - John Molokach, Jul 10 2013
a(n) = Sum_{i=-(n-1)..(n-1)} (n+i)^2. - Bruno Berselli, Jul 24 2014
From Elmo R. Oliveira, Aug 04 2025: (Start)
E.g.f.: exp(x)*x*(8*x^2 + 18*x + 3)/3.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 4. (End)

A006484 a(n) = n*(n + 1)*(n^2 - 3*n + 5)/6.

Original entry on oeis.org

0, 1, 3, 10, 30, 75, 161, 308, 540, 885, 1375, 2046, 2938, 4095, 5565, 7400, 9656, 12393, 15675, 19570, 24150, 29491, 35673, 42780, 50900, 60125, 70551, 82278, 95410, 110055, 126325, 144336, 164208, 186065, 210035, 236250, 264846, 295963, 329745, 366340
Offset: 0

Views

Author

Dennis S. Kluk (mathemagician(AT)ameritech.net)

Keywords

Comments

Structured meta-pyramidal numbers, the n-th number from an n-gonal pyramidal number sequence. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The Gi4 triangle sums of A139600 are given by the terms of this sequence. For the definitions of the Gi4 and other triangle sums see A180662. - Johannes W. Meijer, Apr 29 2011

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. other meta sequences: A100177: prism; A000447: "polar" diamond; A059722: "equatorial diamond"; A100185: anti-prism; A100188: "polar" anti-diamond; and A100189: "equatorial" anti-diamond. Cf. A100145 for more on structured numbers.
Cf. A000332.

Programs

Formula

a(n) = (1/6)*(n^4 - 2*n^3 + 2*n^2 + 5*n). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
a(n) = binomial(n+3,4) - 2*binomial(n+2,4) + 5*binomial(n+1,4). - Johannes W. Meijer, Apr 29 2011

A096000 Cupolar numbers: a(n) = (n+1)*(5*n^2 + 7*n + 3)/3.

Original entry on oeis.org

1, 10, 37, 92, 185, 326, 525, 792, 1137, 1570, 2101, 2740, 3497, 4382, 5405, 6576, 7905, 9402, 11077, 12940, 15001, 17270, 19757, 22472, 25425, 28626, 32085, 35812, 39817, 44110, 48701, 53600, 58817, 64362, 70245, 76476, 83065, 90022, 97357, 105080, 113201, 121730
Offset: 0

Views

Author

N. J. A. Sloane, in memory of Harold Scott MacDonald Coxeter [Feb 09 1907 - Mar 31 2003], May 08 2004

Keywords

Comments

Number of equal balls that will fill a triangular cupola, formed by splitting a cuboctahedron along one of its four "equilateral" hexagons.
Also as a(n) = (1/6)*(10*n^3 - 6*n^2 + 10*n), n>0: structured pentagonal anti-prism numbers (Cf. A100185 = structured anti-prisms); and structured tetragonal anti-diamond numbers (vertex structure 7) (Cf. A000447 = alternate vertex; A100188 = structured anti-diamonds). Cf. A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004

References

  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.

Crossrefs

Programs

Formula

a(n) = (1/2)*(Q(n) + 3*n^2 + 3*n + 1), where Q(n) are the cuboctahedral numbers, A005902.
G.f.: (1+6*x+3*x^2)/(1-x)^4. - Paul Barry, Oct 28 2006
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Wesley Ivan Hurt, May 23 2015
E.g.f.: exp(x)*(3 + 27*x + 27*x^2 + 5*x^3)/3. - Elmo R. Oliveira, Aug 11 2025

A100182 Structured tetragonal anti-prism numbers.

Original entry on oeis.org

1, 8, 28, 68, 135, 236, 378, 568, 813, 1120, 1496, 1948, 2483, 3108, 3830, 4656, 5593, 6648, 7828, 9140, 10591, 12188, 13938, 15848, 17925, 20176, 22608, 25228, 28043, 31060, 34286, 37728, 41393, 45288, 49420, 53796, 58423, 63308, 68458, 73880, 79581, 85568, 91848, 98428, 105315, 112516
Offset: 1

Views

Author

James A. Record (james.record(AT)gmail.com), Nov 07 2004

Keywords

Comments

If offset is changed to 0, this is the number of magic labelings of the 5-node, 8-edge graph formed from a square with both diagonals drawn and a node at the center [Stanley]. - N. J. A. Sloane, Jul 07 2014

Crossrefs

Cf. A100185 - structured anti-prisms; A100145 for more on structured numbers.

Programs

  • Magma
    [(1/6)*(7*n^3-3*n^2+2*n): n in [1..40]]; // Vincenzo Librandi, Aug 18 2011
    
  • Mathematica
    Table[(7*n^3 - 3*n^2 + 2*n)/6, {n,1,40}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {1, 8, 28, 68}, 40] (* G. C. Greubel, Nov 08 2018 *)
  • PARI
    vector(40, n, (7*n^3 -3*n^2 +2*n)/6) \\ G. C. Greubel, Nov 08 2018

Formula

a(n) = (1/6)*(7*n^3 - 3*n^2 + 2*n). [Corrected by Luca Colucci, Mar 01 2011]
G.f.: x*(1 + 4*x + 2*x^2)/(1-x)^4. - Colin Barker, Jun 08 2012
E.g.f.: (6*x +18*x^2 +7*x^3)*exp(x)/6. - G. C. Greubel, Nov 08 2018
a(n) = binomial(n,3) + n^3. - Pedro Caceres, Jul 28 2019

A100183 Structured hexagonal anti-prism numbers.

Original entry on oeis.org

1, 12, 46, 116, 235, 416, 672, 1016, 1461, 2020, 2706, 3532, 4511, 5656, 6980, 8496, 10217, 12156, 14326, 16740, 19411, 22352, 25576, 29096, 32925, 37076, 41562, 46396, 51591, 57160, 63116, 69472
Offset: 1

Views

Author

James A. Record (james.record(AT)gmail.com), Nov 07 2004

Keywords

Crossrefs

Cf. A100185 - structured anti-prisms; A100145 for more on structured numbers.

Programs

  • Magma
    [(1/6)*(13*n^3-9*n^2+2*n): n in [1..40]]; // Vincenzo Librandi, Aug 18 2011
    
  • Mathematica
    Table[(13*n^3 - 9*n^2 + 2*n)/6, {n,1,40}] (* or *) LinearRecurrence[{4, -6, 4, -1}, {1, 12, 46, 116}, 40] (* G. C. Greubel, Nov 08 2018 *)
  • PARI
    vector(40, n, (13*n^3 - 9*n^2 + 2*n)/6) \\ G. C. Greubel, Nov 08 2018

Formula

a(n) = (1/6)*(13*n^3 - 9*n^2 + 2*n). [Corrected by Luca Colucci, Mar 01 2011]
G.f.: x*(1 + 8*x + 4*x^2)/(1-x)^4. - Colin Barker, Jun 08 2012
E.g.f.: (6*x +30*x^2 +13*x^3)*exp(x)/6. - G. C. Greubel, Nov 08 2018

A100184 Structured octagonal anti-prism numbers.

Original entry on oeis.org

1, 16, 64, 164, 335, 596, 966, 1464, 2109, 2920, 3916, 5116, 6539, 8204, 10130, 12336, 14841, 17664, 20824, 24340, 28231, 32516, 37214, 42344, 47925, 53976, 60516, 67564, 75139, 83260, 91946, 101216, 111089, 121584, 132720, 144516, 156991, 170164, 184054, 198680
Offset: 1

Views

Author

James A. Record (james.record(AT)gmail.com), Nov 07 2004

Keywords

Crossrefs

Cf. A100185 (structured anti-prisms), A100145 (for more on structured numbers).

Programs

  • GAP
    List([1..33], n -> (1/6)*(19*n^3-15*n^2+2*n)); # Muniru A Asiru, Feb 14 2018
  • Magma
    [(1/6)*(19*n^3-15*n^2+2*n): n in [1..40]]; // Vincenzo Librandi, Aug 18 2011
    
  • Maple
    a:=n->(1/6)*(19*n^3-15*n^2+2*n): seq(a(n),n=1..33); # Muniru A Asiru, Feb 14 2018
  • Mathematica
    Rest@ CoefficientList[Series[x (1 + 12 x + 6 x^2)/(1 - x)^4, {x, 0, 32}], x] (* Michael De Vlieger, Feb 15 2018 *)

Formula

a(n) = (1/6)*(19*n^3-15*n^2+2*n). [Corrected by Luca Colucci, Mar 01 2011]
G.f.: x*(1 + 12*x + 6*x^2)/(1 - x)^4. - Colin Barker, Jun 08 2012
a(n) = Sum_{i = 0..n-1} (n + i)*(n + 2*i). - Bruno Berselli, Feb 14 2018
E.g.f.: exp(x)*x*(6 + 42*x + 19*x^2)/6. - Stefano Spezia, Oct 11 2023
Showing 1-8 of 8 results.