A046176 Indices of square numbers that are also hexagonal.
1, 35, 1189, 40391, 1372105, 46611179, 1583407981, 53789260175, 1827251437969, 62072759630771, 2108646576008245, 71631910824649559, 2433376321462076761, 82663163018885960315, 2808114166320660573949
Offset: 1
References
- M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..200
- 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=3.
- Serge Perrine, About the diophantine equation z^2 = 32y^2 - 16, SCIREA Journal of Mathematics (2019) Vol. 4, Issue 5, 126-139.
- Maciej Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008) , v_n in Theorem 5.6.
- P. H. van der Kamp, Global classification of two-component..., Found. Comput. Math. 9 (5) (2009) 559-597 near Eq. (4.7).
- Eric Weisstein's World of Mathematics, Hexagonal Square Number.
- Index entries for sequences related to Chebyshev polynomials.
- Index entries for linear recurrences with constant coefficients, signature (34,-1).
Programs
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GAP
List([0..20], n-> Lucas(2,-1, 4*n-2)[1]/2 ); # G. C. Greubel, Jan 13 2020
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Magma
I:=[1,35]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011
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Maple
seq( simplify(ChebyshevU(2*(n-1), 3)), n = 1..20); # G. C. Greubel, Jan 13 2020
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Mathematica
LinearRecurrence[{34, -1}, {1, 35}, 15] (* Ant King, Nov 08 2011 *) Fibonacci[4*Range[20] -2, 2]/2 (* G. C. Greubel, Jan 13 2020 *) -
PARI
vector(21, n, polchebyshev(2*(n-1), 2, 3) ) \\ G. C. Greubel, Jan 13 2020
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Sage
[lucas_number1(4*n-2, 2,-1)/2 for n in (1..20)] # G. C. Greubel, Jan 13 2020
Formula
a(n) = 34*a(n-1) - a(n-2); a(0)=-1, a(1)=1.
a(n+1) = S(2*n, 6) = S(n, 34) + S(n-1, 34), n >= 1, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(n, 34) = A029547(n).
G.f.: x*(1+x)/(1-34*x+x^2).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*6^(2*(n-k)), n >= 0.
a(n) = A001109(2n+1). - Lekraj Beedassy, Apr 23 2003
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),3),3). - Marcos Carreira, Dec 27 2006
From Antonio Alberto Olivares, Mar 22 2008: (Start)
a(n) = (sqrt(2)/8)*(3 + 2*sqrt(2))*(17 + 12*sqrt(2))^(n-1) - (sqrt(2)/8)*(3 - 2*sqrt(2))*(17 - 12*sqrt(2))^(n-1).
a(n) = (sqrt(2)/8)*( (17+12*sqrt(2))^(n-1/2) - (17-12*sqrt(2))^(n-1/2) ).
a(n) = (sqrt(2)/8)*( (3+2*sqrt(2))^(2n-1) - (3-2*sqrt(2))^(2n-1) ).
a(n) = (sqrt(2)/8)*( (1+sqrt(2))^(4n-2) - (1-sqrt(2))^(4n-2) ).
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). (End)
a(n+1) = 17*a(n) + 6*sqrt(8*a(n)^2+1) for n >= 0. - Richard Choulet, May 01 2009
a(n) = b such that (-1)^(n+1) * Integral_{x=-Pi/2..Pi/2} cos((2*n-1)*x)/(3-sin(x)) dx = c + b*log(2). - Francesco Daddi, Aug 01 2011
a(n) are the nonzero integer square roots of A227970. - Richard R. Forberg, Aug 01 2013
a(n) = y/5, where y are solutions to: y^2 = 2x^2 - x - 3. - Richard R. Forberg, Nov 24 2013
a(n) = Pell(4*n-2)/2. - G. C. Greubel, Jan 13 2020
Extensions
Chebyshev comments from Wolfdieter Lang, Nov 29 2002
Comments