A135503 -id:A135503 - cp's OEIS frontend

A006003 a(n) = n*(n^2 + 1)/2.

0, 1, 5, 15, 34, 65, 111, 175, 260, 369, 505, 671, 870, 1105, 1379, 1695, 2056, 2465, 2925, 3439, 4010, 4641, 5335, 6095, 6924, 7825, 8801, 9855, 10990, 12209, 13515, 14911, 16400, 17985, 19669, 21455, 23346, 25345, 27455, 29679, 32020, 34481, 37065, 39775

Offset: 0

N. J. A. Sloane, Simon Plouffe

Write the natural numbers in groups: 1; 2,3; 4,5,6; 7,8,9,10; ... and add the groups. In other words, "sum of the next n natural numbers". - Felice Russo
Number of rhombi in an n X n rhombus, if 'crossformed' rhombi are allowed. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Also the sum of the integers between T(n-1)+1 and T(n), the n-th triangular number (A000217). Sum of n-th row of A000027 regarded as a triangular array.
Unlike the cubes which have a similar definition, it is possible for 2 terms of this sequence to sum to a third. E.g., a(36) + a(37) = 23346 + 25345 = 48691 = a(46). Might be called 2nd-order triangular numbers, thus defining 3rd-order triangular numbers (A027441) as n(n^3+1)/2, etc. - Jon Perry, Jan 14 2004
Also as a(n)=(1/6)*(3*n^3+3*n), n > 0: structured trigonal diamond numbers (vertex structure 4) (cf. A000330 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The sequence M(n) of magic constants for n X n magic squares (numbered 1 through n^2) from n=3 begins M(n) = 15, 34, 65, 111, 175, 260, ... - Lekraj Beedassy, Apr 16 2005 [comment corrected by Colin Hall, Sep 11 2009]
The sequence Q(n) of magic constants for the n-queens problem in chess begins 0, 0, 0, 0, 34, 65, 111, 175, 260, ... - Paul Muljadi, Aug 23 2005
Alternate terms of A057587. - Jeremy Gardiner, Apr 10 2005
Also partial differences of A063488(n) = (2*n-1)*(n^2-n+2)/2. a(n) = A063488(n) - A063488(n-1) for n>1. - Alexander Adamchuk, Jun 03 2006
In an n X n grid of numbers from 1 to n^2, select -- in any manner -- one number from each row and column. Sum the selected numbers. The sum is independent of the choices and is equal to the n-th term of this sequence. - F.-J. Papp (fjpapp(AT)umich.edu), Jun 06 2006
Nonnegative X values of solutions to the equation (X-Y)^3 - (X+Y) = 0. To find Y values: b(n) = (n^3-n)/2. - Mohamed Bouhamida, May 16 2006
For the equation: m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 and m is an odd number the X values are given by the sequence defined by a(n) = (m*n^k+n)/2. The Y values are given by the sequence defined by b(n) = (m*n^k-n)/2. - Mohamed Bouhamida, May 16 2006
If X is an n-set and Y a fixed 3-subset of X then a(n-3) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
(m*(2n)^k+n, m*(2n)^k-n) solves the Diophantine equation: 2m*(X-Y)^k - (X+Y) = 0 with X >= Y, k >= 2 where m is a positive integer. - Mohamed Bouhamida, Oct 02 2007
Also c^(1/2) in a^(1/2) + b^(1/2) = c^(1/2) such that a^2 + b = c. - Cino Hilliard, Feb 09 2008
a(n) = n*A000217(n) - Sum_{i=0..n-1} A001477(i). - Bruno Berselli, Apr 25 2010
a(n) is the number of triples (w,x,y) having all terms in {0,...,n} such that at least one of these inequalities fails: x+y < w, y+w < x, w+x < y. - Clark Kimberling, Jun 14 2012
Sum of n-th row of the triangle in A209297. - Reinhard Zumkeller, Jan 19 2013
The sequence starting with "1" is the third partial sum of (1, 2, 3, 3, 3, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the largest eigenvalue of the matrix returned by the MATLAB command magic(n) for n > 0. - Altug Alkan, Nov 10 2015
a(n) is the number of triples (x,y,z) having all terms in {1,...,n} such that all these triangle inequalities are satisfied: x+y > z, y+z > x, z+x > y. - Heinz Dabrock, Jun 03 2016
Shares its digital root with the stella octangula numbers (A007588). See A267017. - Peter M. Chema, Aug 28 2016
Can be proved to be the number of nonnegative solutions of a system of three linear Diophantine equations for n >= 0 even: 2*a_{11} + a_{12} + a_{13} = n, 2*a_{22} + a_{12} + a_{23} = n and 2*a_{33} + a_{13} + a_{23} = n. The number of solutions is f(n) = (1/16)*(n+2)*(n^2 + 4n + 8) and a(n) = n*(n^2 + 1)/2 is obtained by remapping n -> 2*n-2. - Kamil Bradler, Oct 11 2016
For n > 0, a(n) coincides with the trace of the matrix formed by writing the numbers 1...n^2 back and forth along the antidiagonals (proved, see A078475 for the examples of matrix). - Stefano Spezia, Aug 07 2018
The trace of an n X n square matrix where the elements are entered on the ascending antidiagonals. The determinant is A069480. - Robert G. Wilson v, Aug 07 2018
Bisections are A317297 and A005917. - Omar E. Pol, Sep 01 2018
Number of achiral colorings of the vertices (or faces) of a regular tetrahedron with n available colors. An achiral coloring is identical to its reflection. - Robert A. Russell, Jan 22 2020
a(n) is the n-th centered triangular pyramidal number. - Lechoslaw Ratajczak, Nov 02 2021
a(n) is the number of words of length n defined on 4 letters {b,c,d,e} that contain one or no b's, one c or two d's, and any number of e's. For example, a(3) = 15 since the words are (number of permutations in parentheses): bce (6), bdd (3), cee (3), and dde (3). - Enrique Navarrete, Jun 21 2025

			G.f. = x + 5*x^2 + 15*x^3 + 34*x^4 + 65*x^5 + 111*x^6 + 175*x^7 + 260*x^8 + ...
For a(2)=5, the five tetrahedra have faces AAAA, AAAB, AABB, ABBB, and BBBB with colors A and B. - _Robert A. Russell_, Jan 31 2020

J.-M. De Koninck, Ces nombres qui nous fascinent, Entry 15, p. 5, Ellipses, Paris 2008.
F.-J. Papp, Colloquium Talk, Department of Mathematics, University of Michigan-Dearborn, March 6, 2005.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..1000
J. D. Bell, A translation of Leonhard Euler's "De Quadratis Magicis", E795, arXiv:math/0408230 [math.CO], 2004-2005.
James Grime and Brady Haran, Magic Hexagon, Numberphile video (2014).
Milan Janjic, Two Enumerative Functions.
Milan Janjic and Boris Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - _N. J. A. Sloane_, Feb 13 2013
Milan Janjic and Boris Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, Journal of Integer Sequences, 17 (2014), Article 14.3.5. - _Felix Fröhlich_, Oct 11 2016
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).
Ashish Kumar Pandey and Brajesh Kumar Sharma, A Note On Magic Squares And Magic Constants, Appl. Math. E-Notes (2023) Vol. 23, Art. No. 53, 577-582. See p. 577.
A. J. Turner and J. F. Miller, Recurrent Cartesian Genetic Programming Applied to Famous Mathematical Sequences, preprint, Proceedings of the Companion Publication of the 2015 Annual Conference on Genetic and Evolutionary Computation.
Eric Weisstein's World of Mathematics, Magic Constant.
Wikipedia, Floyd's triangle. - _Paul Muljadi_, Jan 25 2010
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
Index entries for sequences related to magic squares.
Index to sequences related to polygonal numbers.

Cf. A000330, A000537, A066886, A057587, A027480, A002817 (partial sums).
Cf. A000578 (cubes).
Cf. A007742, A005449, A005448, A118465, A226449, A034262, A080992, A267017.
(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, this sequence, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Antidiagonal sums of array in A000027. Row sums of the triangular view of A000027.
Cf. A063488 (sum of two consecutive terms), A005917 (bisection), A317297 (bisection).
Cf. A105374 / 8.
Cf. A000292, A011863, A001620, A248177.
Tetrahedron colorings: A006008 (oriented), A000332(n+3) (unoriented), A000332 (chiral), A037270 (edges).
Other polyhedron colorings: A337898 (cube faces, octahedron vertices), A337897 (octahedron faces, cube vertices), A337962 (dodecahedron faces, icosahedron vertices), A337960 (icosahedron faces, dodecahedron vertices).
Row 3 of A325001 (simplex vertices and facets) and A337886 (simplex faces and peaks).

a_n:=List([0..nmax], n->n*(n^2 + 1)/2); # Stefano Spezia, Aug 12 2018

a006003 n = n * (n ^ 2 + 1) `div` 2
a006003_list = scanl (+) 0 a005448_list
-- Reinhard Zumkeller, Jun 20 2013

% Also works with FreeMat.
for(n=0:nmax); tm=n*(n^2 + 1)/2; fprintf('%d\t%0.f\n', n, tm); end
% Stefano Spezia, Aug 12 2018

[n*(n^2 + 1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015

[Binomial(n,3)+Binomial(n-1,3)+Binomial(n-2,3): n in [2..60]]; // Vincenzo Librandi, Sep 12 2015

Table[ n(n^2 + 1)/2, {n, 0, 45}]
LinearRecurrence[{4,-6,4,-1}, {0,1,5,15},50] (* Harvey P. Dale, May 16 2012 *)
CoefficientList[Series[x (1 + x + x^2)/(x - 1)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
With[{n=50},Total/@TakeList[Range[(n(n^2+1))/2],Range[0,n]]] (* Requires Mathematica version 11 or later *) (* Harvey P. Dale, Nov 28 2017 *)

a(n):=n*(n^2 + 1)/2$ makelist(a(n), n, 0, nmax); /* Stefano Spezia, Aug 12 2018 */

{a(n) = n * (n^2 + 1) / 2}; /* Michael Somos, Dec 24 2011 */

concat(0, Vec(x*(1+x+x^2)/(x-1)^4 + O(x^20))) \\ Felix Fröhlich, Oct 11 2016

def A006003(n): return n*(n**2+1)>>1 # Chai Wah Wu, Mar 25 2024

a(n) = binomial(n+2, 3) + binomial(n+1, 3) + binomial(n, 3). [corrected by Michel Marcus, Jan 22 2020]
G.f.: x*(1+x+x^2)/(x-1)^4. - Floor van Lamoen, Feb 11 2002
Partial sums of A005448. - Jonathan Vos Post, Mar 16 2006
Binomial transform of [1, 4, 6, 3, 0, 0, 0, ...] = (1, 5, 15, 34, 65, ...). - Gary W. Adamson, Aug 10 2007
a(n) = -a(-n) for all n in Z. - Michael Somos, Dec 24 2011
a(n) = Sum_{k = 1..n} A(k-1, k-1-n) where A(i, j) = i^2 + i*j + j^2 + i + j + 1. - Michael Somos, Jan 02 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=0, a(1)=1, a(2)=5, a(3)=15. - Harvey P. Dale, May 16 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 3. - Ant King, Jun 13 2012
a(n) = A000217(n) + n*A000217(n-1). - Bruno Berselli, Jun 07 2013
a(n) = A057145(n+3,n). - Luciano Ancora, Apr 10 2015
E.g.f.: (1/2)*(2*x + 3*x^2 + x^3)*exp(x). - G. C. Greubel, Dec 18 2015; corrected by Ilya Gutkovskiy, Oct 12 2016
a(n) = T(n) + T(n-1) + T(n-2), where T means the tetrahedral numbers, A000292. - Heinz Dabrock, Jun 03 2016
From Ilya Gutkovskiy, Oct 11 2016: (Start)
Convolution of A001477 and A008486.
Convolution of A000217 and A158799.
Sum_{n>=1} 1/a(n) = H(-i) + H(i) = 1.343731971048019675756781..., where H(k) is the harmonic number, i is the imaginary unit. (End)
a(n) = A000578(n) - A135503(n). - Miquel Cerda, Dec 25 2016
Euler transform of length 3 sequence [5, 0, -1]. - Michael Somos, Dec 25 2016
a(n) = A037270(n)/n for n > 0. - Kritsada Moomuang, Dec 15 2018
a(n) = 3*A000292(n-1) + n. - Bruce J. Nicholson, Nov 23 2019
a(n) = A011863(n) - A011863(n-2). - Bruce J. Nicholson, Dec 22 2019
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = C(n,1) + 3*C(n,2) + 3*C(n,3), where the coefficient of C(n,k) is the number of tetrahedron colorings using exactly k colors.
a(n) = C(n+3,4) - C(n,4).
a(n) = 2*A000332(n+3) - A006008(n) = A006008(n) - 2*A000332(n) = A000332(n+3) - A000332(n).
a(n) = A325001(3,n). (End)
From Amiram Eldar, Aug 21 2023: (Start)
Sum_{n>=1} 1/a(n) = 2 * (A248177 + A001620).
Product_{n>=2} (1 - 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi)/4.
Product_{n>=1} (1 + 1/a(n)) = cosh(sqrt(7)*Pi/2)*cosech(Pi). (End)

Better description from Albert Rich (Albert_Rich(AT)msn.com), Mar 1997

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.

A011886 a(n) = floor(n(n-1)(n-2)/4).

Original entry on oeis.org

0, 0, 0, 1, 6, 15, 30, 52, 84, 126, 180, 247, 330, 429, 546, 682, 840, 1020, 1224, 1453, 1710, 1995, 2310, 2656, 3036, 3450, 3900, 4387, 4914, 5481, 6090, 6742, 7440, 8184, 8976, 9817, 10710, 11655, 12654, 13708, 14820, 15990, 17220, 18511, 19866, 21285

Offset: 0

N. J. A. Sloane

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

Sequences of the form floor(n*(n-1)*(n-2)/m): A007531 (m=1), A135503 (m=2), A007290 (m=3), this sequence (m=4), A011887 (m=5), A000292 (m=6), A011889 (m=7), A011890 (m=8), A011891 (m=9), A011892 (m=10), A011893 (m=11), A011894 (m=12), A011895 (m=13), A011896 (m=14), A011897 (m=15), A011898 (m=16), A011899 (m=17), A011849 (m=18), A011901 (m=19), A011902 (m=20), A011903 (m=21), A011904 (m=22), A011905 (m=23), A011842 (m=24), A011907 (m=25), A011908 (m=26), A011909 (m=27), A011910 (m=28), A011911 (m=29), A011912 (m=30), A011912 (m=31), A011913 (m=32).

[Floor(n*(n-1)*(n-2)/4): n in [0..50]]; // Vincenzo Librandi, Jul 07 2012

Table[Floor[(n(n-1)(n-2))/4],{n,0,50}] (* or *) LinearRecurrence[{3,-3,1,1, -3,3,-1},{0,0,0,1,6,15,30}, 50] (* Harvey P. Dale, Feb 25 2012 *)
CoefficientList[Series[x^3*(1+3*x+2*x^3)/((1-x)^3*(1-x^4)),{x,0,50}],x] (* Vincenzo Librandi, Jul 07 2012 *)

[3*binomial(n,3)//2 for n in range(51)] # G. C. Greubel, Oct 06 2024

From R. J. Mathar, Apr 15 2010: (Start)
a(n) = +3*a(n-1) -3*a(n-2) +a(n-3) +a(n-4) -3*a(n-5) +3*a(n-6) -a(n-7).
G.f.: x^3*(1+3*x+2*x^3) / ( (1-x)^4*(1+x)*(1+x^2) ). (End)
a(n) = floor(Sum_{k=0..n} n*(k+1)/2) for n >= -2. - William A. Tedeschi, Sep 10 2010

More terms from William A. Tedeschi, Sep 10 2010

A004188 a(n) = n(3n^2 - 1)/2.

Original entry on oeis.org

0, 1, 11, 39, 94, 185, 321, 511, 764, 1089, 1495, 1991, 2586, 3289, 4109, 5055, 6136, 7361, 8739, 10279, 11990, 13881, 15961, 18239, 20724, 23425, 26351, 29511, 32914, 36569, 40485, 44671, 49136, 53889, 58939, 64295, 69966, 75961

Offset: 0

Albert D. Rich (Albert_Rich(AT)msn.com)

3-dimensional analog of centered polygonal numbers.

(1), (4+7), (10+13+16), (19+22+25+28), ... - Jon Perry, Sep 10 2004

E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
T. P. Martin, Shells of atoms, Phys. Reports, 273 (1996), 199-241, eq. (11).

Vincenzo Librandi, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-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. A002412, A016061, A051682.
Cf. A236770 (partial sums).

[n*(3*n^2-1)/2: n in [0..50]]; //Vincenzo Librandi, May 15 2011

seq(binomial(2*n+1,3)+binomial(n+1,3), n=0..37); # Zerinvary Lajos, Jan 21 2007

Table[n(3n^2-1)/2,{n,0,80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,11,39},40] (* Harvey P. Dale, Jul 19 2019 *)

PARI
```
vector(40, n, n*(3*n^2-1)/2)
```

Partial sums of n-1 3-spaced triangular numbers, e.g., a(4) = t(1) + t(4) + t(7) = 1 + 10 + 28 = 39. - Jon Perry, Jul 23 2003
a(n) = C(2*n+1,3) + C(n+1,3), n >= 0. - Zerinvary Lajos, Jan 21 2007
a(n) = A000447(n) + A000292(n). - Zerinvary Lajos, Jan 21 2007
G.f.: x*(1+7*x+x^2) / (x-1)^4. - R. J. Mathar, Oct 08 2011
From Miquel Cerda, Dec 25 2016: (Start)
a(n) = A000578(n) + A135503(n).
a(n) = A007588(n) - A135503(n). (End)
E.g.f.: (x/2)*(2 + 9*x + 3*x^2)*exp(x). - G. C. Greubel, Sep 01 2017

A342758 Array read by ascending antidiagonals: T(k, n) is the maximum value of the magic constant in a perimeter-magic k-gon of order n.

Original entry on oeis.org

12, 15, 23, 19, 30, 37, 22, 37, 48, 54, 26, 44, 60, 71, 74, 29, 51, 71, 88, 97, 97, 33, 58, 83, 105, 121, 128, 123, 36, 65, 94, 122, 144, 159, 162, 152, 40, 72, 106, 139, 168, 190, 202, 201, 184, 43, 79, 117, 156, 191, 221, 241, 250, 243, 219, 47, 86, 129, 173, 215, 252, 281, 299, 303, 290, 257

Offset: 3

Stefano Spezia, Mar 21 2021

			The array begins:
k\n|  3   4   5    6    7 ...
---+---------------------
3  | 12  23  37   54   74 ...
4  | 15  30  48   71   97 ...
5  | 19  37  60   88  121 ...
6  | 22  44  71  105  144 ...
7  | 26  51  83  122  168 ...
...

Terrel Trotter, Perimeter-Magic Polygons, Journal of Recreational Mathematics Vol. 7, No. 1, 1974, pp. 14-20 (see equations 11 and 13).

Cf. A017005 (n = 4), A135503 (diagonal), A341740 (k = 3), A342719, A342757 (minimum).

T[k_,n_]:= (n+k(n^2-2)+(Mod[k,2]-1)Mod[n,2])/2; Table[T[k+3-n,n],{k,3,13},{n,3,k}]//Flatten

G.f.: (- x^2*(2*y^2 + y - 1) - x*(y^2 + 2*y - 1) + (y - 1)*y^2)/((x - 1)^2*(x + 1)*(y - 1)^3*(y + 1)).
T(k, n) = (n^2/2 - 1)*k + n/2 if n is even or both n and k are odd.
T(k, n) = (n^2/2 - 1)*k + (n - 1)/2 if n is odd and k is even.
T(k, n) = (n + k*(n^2 - 2) + ((k mod 2) - 1)*(n mod 2))/2.

A135509 Nonnegative integers c such that there are nonnegative integers a and b that satisfy a^(1/2) + b^(1/2) = c^(1/2) and a^2 + b = c.

Original entry on oeis.org

0, 1, 25, 225, 1156, 4225, 12321, 30625, 67600, 136161, 255025, 450241, 756900, 1221025, 1901641, 2873025, 4227136, 6076225, 8555625, 11826721, 16080100, 21538881, 28462225, 37149025, 47941776, 61230625, 77457601, 97121025

Offset: 0

Cino Hilliard, Feb 09 2008

Define FLTR as Fermat's Last Theorem with rational exponents. Consider x + y = x + y. Then (x^m)^(1/m) + (y^m)^(1/m) = ((x+y)^m)^(1/m) for integer m >= 1.
For m = 2, we have (x^2)^(1/2) + (y^2)^(1/2) = ((x+y)^2)^(1/2). Thus, a = x^2, b = y^2 and c = (x+y)^2. Then a^2 + b = c => x^4 + y^2 = (x+y)^2 => x^4 + y^2 = x^2 + 2*x*y + y^2 => y = (x^3-x)/2. It follows that c = (x+y)^2 = (x^3 + x)^2/4 is the generating function for this sequence for x = 0, 1, 2, 3, ...
For m = 2, there are infinitely many nonnegative integer solutions for the FLTR proposition. The same holds for m = 3, i.e., there are also infinitely many nonnegative integer solutions to a^(1/3) + b^(1/3) = c^(1/3). E.g., 8^(1/3) + 27^(1/3) = 125^(1/3). Moreover, there are infinitely many solutions to FLTR for a general positive integer m.
However, in conjunction with a^2 + b = c, I could not find any nontrivial solutions when m >= 3. Perhaps there is another formula that will yield solutions. [Edited by Petros Hadjicostas, Dec 21 2019]

			For a = 9, b = 144, and c = 225, we obtain 9^(1/2) + 144^(1/2) = 225^(1/2) and 9^2 + 144 = 225. Thus, c = 225 is an entry in this sequence.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A006003, A135503.

Table[(n + n^3)^2/4, {n, 0, 25}] (* G. C. Greubel, Oct 16 2016 *)

flt2(n) = {local(a, b); for(a=0, n, b = (a^3+a)/2; print1(b^2", "))} /* edited by Petros Hadjicostas, Dec 21 2019 */

From Colin Barker, May 02 2012: (Start)
a(n) = (n + n^3)^2/4 = A006003(n)^2.
G.f.: x*(1 + x)*(1 + 4*x + x^2)*(1 + 13*x + x^2)/(1 - x)^7. (End)
E.g.f.: (1/4)*x*(4 + 46*x + 102*x^2 + 67*x^3 + 15*x^4 + x^5)*exp(x). - G. C. Greubel, Oct 16 2016

cp's OEIS Frontend

A006003 a(n) = n*(n^2 + 1)/2.

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

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A011886 a(n) = floor(n*(n-1)*(n-2)/4).

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A004188 a(n) = n*(3*n^2 - 1)/2.

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A342758 Array read by ascending antidiagonals: T(k, n) is the maximum value of the magic constant in a perimeter-magic k-gon of order n.

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A135509 Nonnegative integers c such that there are nonnegative integers a and b that satisfy a^(1/2) + b^(1/2) = c^(1/2) and a^2 + b = c.

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

A011886 a(n) = floor(n(n-1)(n-2)/4).

A004188 a(n) = n(3n^2 - 1)/2.

A360282 Triangle read by rows. T(n, k) = (1/2) * binomial(2(n - k + 1), n - k + 1) binomial(2*n - k, k - 1) for n > 0, T(0, 0) = 1.