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

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

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

Comments

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

References

• 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).

Links

• 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).

Crossrefs

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.

Programs

Magma

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

Maple

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

Mathematica

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)

Python

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

Formula

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

Extensions

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