A028230 Bisection of A001353. Indices of square numbers which are also octagonal.

1, 15, 209, 2911, 40545, 564719, 7865521, 109552575, 1525870529, 21252634831, 296011017105, 4122901604639, 57424611447841, 799821658665135, 11140078609864049, 155161278879431551, 2161117825702177665, 30100488280951055759, 419245718107612602961, 5839339565225625385695

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Chebyshev S-sequence with Diophantine property.

4*b(n)^2 - 3*a(n)^2 = 1 with b(n) = A001570(n), n>=0.

y satisfying the Pellian x^2 - 3*y^2 = 1, for even x given by A094347(n). - Lekraj Beedassy, Jun 03 2004

a(n) = L(n,-14)*(-1)^n, where L is defined as in A108299; see also A001570 for L(n,+14). - Reinhard Zumkeller, Jun 01 2005

Product x*y, where the pair (x, y) solves for x^2 - 3y^2 = -2, i.e., a(n) = A001834(n)*A001835(n). - Lekraj Beedassy, Jul 13 2006

Numbers n such that RootMeanSquare(1,3,...,2*A001570(k)-1) = n. - Ctibor O. Zizka, Sep 04 2008

As n increases, this sequence is approximately geometric with common ratio r = lim(n -> oo, a(n)/a(n-1)) = (2 + sqrt(3))^2 = 7 + 4 * sqrt(3). - Ant King, Nov 15 2011

References

• R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 329.
• J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..890
• K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
• Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
• T. N. E. Greville, Table for third-degree spline interpolations with equally spaced arguments, Math. Comp., 24 (1970), 179-183.
• W. D. Hoskins, Table for third-degree spline interpolation using equi-spaced knots, Math. Comp., 25 (1971), 797-801.
• Tanya Khovanova, Recursive Sequences
• E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 4, k=1, t=2.
• Dino Lorenzini, and Z. Xiang, Integral points on variable separated curves, Preprint 2016.
• F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
• Eric Weisstein's World of Mathematics, Octagonal Square Number.
• Index entries for linear recurrences with constant coefficients, signature (14,-1).
• Index entries for sequences related to Chebyshev polynomials.

Crossrefs

Cf. A001353, A001570, A001834, A001835, A001921, A007655, A036428, A046184, A049310, A094347.

Cf. A077416 with companion A077417.

Programs

GAP

a:=[1,15];; for n in [3..30] do a[n]:=14*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019

Magma

I:=[1,15]; [n le 2 select I[n] else 14*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 06 2019

Maple

seq(coeff(series((1+x)/(1-14*x+x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Dec 06 2019

Mathematica

LinearRecurrence[{14, - 1}, {1, 15}, 17] (* Ant King, Nov 15 2011 *)
CoefficientList[Series[(1+x)/(1-14x+x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Jun 17 2014 *)

PARI

Vec((1+x)/(1-14*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 16 2014

PARI

isok(n) = ispolygonal(n^2, 8); \\ Michel Marcus, Jul 09 2017

Sage

[(lucas_number2(n,14,1)-lucas_number2(n-1,14,1))/12 for n in range(1, 18)] # Zerinvary Lajos, Nov 10 2009

Formula

a(n) = 2*A001921(n)+1.

a(n) = 14*a(n-1) - a(n-2) for n>1.

a(n) = S(n, 14) + S(n-1, 14) = S(2*n, 4) with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(-1, x) = 0, S(n, 14) = A007655(n+1) and S(n, 4) = A001353(n+1).

G.f.: x*(1+x)/(1-14*x+x^2).

a(n) = (ap^(2*n+1) - am^(2*n+1))/(ap - am) with ap := 2+sqrt(3) and am := 2-sqrt(3).

a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*16^(n-k), n>=0.

a(n) = sqrt((4*A001570(n-1)^2 - 1)/3).

a(n) ~ 1/6*sqrt(3)*(2 + sqrt(3))^(2*n-1). - Joe Keane (jgk(AT)jgk.org), May 15 2002

4*a(n+1) = (A001834(n))^2 + 4*(A001835(n+1))^2 - (A001835(n))^2. E.g. 4*a(3) = 4*209 = 19^2 + 4*11^2 - 3^2 = (A001834(2))^2 + 4*(A001835(3))^2 - (A001835(2))^2. Generating floretion: 'i + 2'j + 3'k + i' + 2j' + 3k' + 4'ii' + 3'jj' + 4'kk' + 3'ij' + 3'ji' + 'jk' + 'kj' + 4e. - Creighton Dement, Dec 04 2004

a(n) = f(a(n-1),7) + f(a(n-2),7), where f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. - Marcos Carreira, Dec 27 2006

From Ant King, Nov 15 2011: (Start)

a(n) = 1/6 * sqrt(3) * ( (tan(5*Pi/12)) ^ (2n-1) - (tan(Pi/12)) ^ (2n-1) ).

a(n) = floor (1/6 * sqrt(3) * (tan(5*Pi/12)) ^ (2n-1)). (End)

a(n) = A001353(n)^2-A001353(n-1)^2. - Antonio Alberto Olivares, Apr 06 2020

E.g.f.: 1 - exp(7*x)*(3*cosh(4*sqrt(3)*x) - 2*sqrt(3)*sinh(4*sqrt(3)*x))/3. - Stefano Spezia, Dec 12 2022

a(n) = sqrt(A036428(n)). - Bernard Schott, Dec 19 2022

Extensions

Additional comments from Wolfdieter Lang, Nov 29 2002

Incorrect recurrence relation deleted by Ant King, Nov 15 2011

Minor edits by Vaclav Kotesovec, Jan 28 2015