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A007588 Stella octangula numbers: a(n) = n*(2*n^2 - 1).
(Formerly M4932)
38

%I M4932 #150 Sep 09 2022 15:58:29

%S 0,1,14,51,124,245,426,679,1016,1449,1990,2651,3444,4381,5474,6735,

%T 8176,9809,11646,13699,15980,18501,21274,24311,27624,31225,35126,

%U 39339,43876,48749,53970,59551,65504,71841,78574,85715,93276,101269,109706,118599,127960

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

%C 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

%C 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

%C 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

%C 4*A007588 = A144138(ChebyshevU[3,n]). - _Vladimir Joseph Stephan Orlovsky_, Jun 30 2011

%C 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

%C 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

%D J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, p. 51.

%D E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.

%D W. Ljunggren, Zur Theorie der Gleichung x^2 + 1 = Dy^4, Avh. Norske Vid. Akad. Oslo. I. 1942 (5): 27.

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

%H Alexander Adamchuk and Vincenzo Librandi, <a href="/A007588/b007588.txt">Table of n, a(n) for n = 0..10000</a> [Alexander Adamchuk computed terms 0 - 169, Jun 02, 2008; Vincenzo Librandi computed the first 10000 terms, Aug 18 2011]

%H A. Bremner, R. Høibakk and D. Lukkassen, <a href="http://ami.ektf.hu/uploads/papers/finalpdf/AMI_36_from29to41.pdf">Crossed ladders and Euler’s quartic</a>, Annales Mathematicae et Informaticae, 36 (2009) pp. 29-41. See p. 33.

%H John Elias, <a href="/A007588/a007588.png">Nesting Cubes of the Surface Points of a Hexagonal Prism</a>

%H T. P. Martin, <a href="http://dx.doi.org/10.1016/0370-1573(95)00083-6">Shells of atoms</a>, Phys. Reports, 273 (1996), 199-241, eq. (11).

%H Amelia Carolina Sparavigna, <a href="https://doi.org/10.5281/zenodo.4662348">Generalized Sum of Stella Octangula Numbers</a>, Politecnico di Torino (Italy, 2021).

%H Amelia Carolina Sparavigna, <a href="https://doi.org/10.5281/zenodo.4663050">Cardano Formula and Some Figurate Numbers</a>, Politecnico di Torino (Italy, 2021).

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/StellaOctangulaNumber.html">Stella Octangula Number</a>

%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Stella_octangula_number">Stella octangula number</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F G.f.: x*(1+10*x+x^2)/(1-x)^4.

%F a(n) = n*A056220(n).

%F 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

%F From _Ilya Gutkovskiy_, Jul 02 2016: (Start)

%F E.g.f.: x*(1 + 6*x + 2*x^2)*exp(x).

%F Dirichlet g.f.: 2*zeta(s-3) - zeta(s-1). (End)

%F a(n) = A004188(n) + A135503(n). - _Miquel Cerda_, Dec 25 2016

%p A007588:=n->n*(2*n^2 - 1); seq(A007588(n), n=0..40); # _Wesley Ivan Hurt_, Mar 10 2014

%t Table[ n(2n^2-1), {n,0,169} ] (* _Alexander Adamchuk_, Jun 02 2008 *)

%t LinearRecurrence[{4,-6,4,-1},{0,1,14,51},50] (* _Harvey P. Dale_, Sep 16 2011 *)

%o (PARI) a(n)=n*(2*n^2-1)

%o (Magma) [n*(2*n^2 - 1): n in [0..40]]; // _Vincenzo Librandi_, Aug 18 2011

%o (Python)

%o def A007588(n): return n*(2*n**2-1) # _Chai Wah Wu_, Feb 18 2022

%Y Backwards differences give star numbers A003154: A003154(n)=a(n)-a(n-1).

%Y 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.

%Y Cf. A001653 = Numbers n such that 2*n^2 - 1 is a square.

%Y a(169) = (A229384(3)*A229384(4))^2.

%Y Cf. A267017, A006003, A135503.

%K nonn,easy,nice

%O 0,3

%A _N. J. A. Sloane_

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

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Last modified April 24 18:15 EDT 2024. Contains 371962 sequences. (Running on oeis4.)