A067902 -id:A067902 - cp's OEIS frontend

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

2, 4, 14, 52, 194, 724, 2702, 10084, 37634, 140452, 524174, 1956244, 7300802, 27246964, 101687054, 379501252, 1416317954, 5285770564, 19726764302, 73621286644, 274758382274, 1025412242452, 3826890587534, 14282150107684, 53301709843202, 198924689265124

Offset: 0

N. J. A. Sloane

a(n) gives values of x satisfying x^2 - 3*y^2 = 4; corresponding y values are given by 2*A001353(n).
If M is any given term of the sequence, then the next one is 2*M + sqrt(3*M^2 - 12). - Lekraj Beedassy, Feb 18 2002
For n > 0, the three numbers a(n) - 1, a(n), and a(n) + 1 form a Fleenor-Heronian triangle, i.e., a Heronian triangle with consecutive sides, whose area A(n) may be obtained from the relation [4*A(n)]^2 = 3([a(2n)]^2 - 4); or A(n) = 3*A001353(2*n)/2 and whose semiperimeter is 3*a[n]/2. The sequence is symmetrical about a[0], i.e., a[-n] = a[n].
For n > 0, a(n) + 2 is the number of dimer tilings of a 2*n X 2 Klein bottle (cf. A103999).
Tsumura shows that, for prime p, a(p) is composite (contrary to a conjecture of Juricevic). - Charles R Greathouse IV, Apr 13 2010
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 12 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 192 = 0. - Colin Barker, Feb 16 2014
A268281(n) - 1 is a member of this sequence iff A268281(n) is prime. - Frank M Jackson, Feb 27 2016
a(n) gives values of x satisfying 3*x^2 - 4*y^2 = 12; corresponding y values are given by A005320. - Sture Sjöstedt, Dec 19 2017
Middle side lengths of almost-equilateral Heronian triangles. - Wesley Ivan Hurt, May 20 2020
For all elements k of the sequence, 3*(k-2)*(k+2) is a square. - Davide Rotondo, Oct 25 2020

B. C. Berndt, Ramanujan's Notebooks Part IV, Springer-Verlag, see p. 82.
J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p.91.
Michael P. Cohen, Generating Heronian Triangles With Consecutive Integer Sides. Journal of Recreational Mathematics, vol. 30 no. 2 1999-2000 p. 123.
L. E. Dickson, History of The Theory of Numbers, Vol. 2 pp. 197;198;200;201. Chelsea NY.
Charles R. Fleenor, Heronian Triangles with Consecutive Integer Sides, Journal of Recreational Mathematics, Volume 28, no. 2 (1996-7) 113-115.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
V. D. To, "Finding All Fleenor-Heronian Triangles", Journal of Recreational Mathematics vol. 32 no.4 2003-4 pp. 298-301 Baywood NY.

T. D. Noe, Table of n, a(n) for n=0..200
P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
R. A. Beauregard and E. R. Suryanarayan, The Brahmagupta Triangles, The College Mathematics Journal 29(1) 13-7 1998 MAA.
Hacène Belbachir, Soumeya Merwa Tebtoub and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Daniel Birmajer, Juan B. Gil and Michael D. Weiner, Linear recurrence sequences with indices in arithmetic progression and their sums, arXiv preprint arXiv:1505.06339 [math.NT], 2015.
K. S. Brown's Mathpages, Some Properties of the Lucas Sequence(2, 4, 14, 52, 194, ...)
H. W. Gould, A triangle with integral sides and area, Fib. Quart., 11 (1973), 27-39.
Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2, ..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
Hideyuki Ohtskua, proposer, Problem B-1351, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 258.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Yu Tsumura, On compositeness of special types of integers, arXiv:1004.1244 [math.NT], 2010.
Eric Weisstein's World of Mathematics, Heronian Triangle
Wikipedia, Heronian triangle
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem, Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Cf. A001075, A001353, A001835.
Cf. A011945 (areas), A334277 (perimeters).
Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Cf. A001570, A002530, A005320, A006051, A048788, A174500, A268281.
Cf. A011943, A102344, A019673, A105531.

a003500 n = a003500_list !! n
a003500_list = 2 : 4 : zipWith (-)
   (map (* 4) $ tail a003500_list) a003500_list
-- Reinhard Zumkeller, Dec 17 2011

I:=[2,4]; [n le 2 select I[n] else 4*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2018

A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*procname(n-1)-procname(n-2); fi;
end proc;

a[0]=2; a[1]=4; a[n_]:= a[n]= 4a[n-1] -a[n-2]; Table[a[n], {n, 0, 23}]
LinearRecurrence[{4,-1},{2,4},30] (* Harvey P. Dale, Aug 20 2011 *)
Table[Round@LucasL[2n, Sqrt[2]], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)

x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ Altug Alkan, Apr 04 2016

[lucas_number2(n,4,1) for n in range(0, 24)] # Zerinvary Lajos, May 14 2009

a(n) = ( 2 + sqrt(3) )^n + ( 2 - sqrt(3) )^n.
a(n) = 2*A001075(n).
G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). Simon Plouffe in his 1992 dissertation.
a(n) = A001835(n) + A001835(n+1).
a(n) = trace of n-th power of the 2 X 2 matrix [1 2 / 1 3]. - Gary W. Adamson, Jun 30 2003 [corrected by Joerg Arndt, Jun 18 2020]
From the addition formula, a(n+m) = a(n)*a(m) - a(m-n), it is easy to derive multiplication formulas, such as: a(2*n) = (a(n))^2 - 2, a(3*n) = (a(n))^3 - 3*(a(n)), a(4*n) = (a(n))^4 - 4*(a(n))^2 + 2, a(5*n) = (a(n))^5 - 5*(a(n))^3 + 5*(a(n)), a(6*n) = (a(n))^6 - 6*(a(n))^4 + 9*(a(n))^2 - 2, etc. The absolute values of the coefficients in the expansions are given by the triangle A034807. - John Blythe Dobson, Nov 04 2007
a(n) = 2*A001353(n+1) - 4*A001353(n). - R. J. Mathar, Nov 16 2007
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n=0..infinity} (1 + x^(4*n + 1))/(1 + x^(4*n + 3)). Let alpha = 2 - sqrt(3). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.24561 99455 06551 88869 ... = 2 + 1/(4 + 1/(14 + 1/(52 + ...))). Cf. A174500.
Also F(-alpha) = 0.74544 81786 39692 68884 ... has the continued fraction representation 1 - 1/(4 - 1/(14 - 1/(52 - ...))) and the simple continued fraction expansion 1/(1 + 1/((4 - 2) + 1/(1 + 1/((14 - 2) + 1/(1 + 1/((52 - 2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((4^2 - 4) + 1/(1 + 1/((14^2 - 4) + 1/(1 + 1/((52^2 - 4) + 1/(1 + ...))))))).
(End)
a(2^n) = A003010(n). - John Blythe Dobson, Mar 10 2014
a(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015
E.g.f.: 2*exp(2*x)*cosh(sqrt(3)*x). - Ilya Gutkovskiy, Apr 27 2016
a(n) = Sum_{k=0..floor(n/2)} (-1)^k*n*(n - k - 1)!/(k!*(n - 2*k)!)*4^(n - 2*k) for n >= 1. - Peter Luschny, May 10 2016
From Peter Bala, Oct 15 2019: (Start)
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; -1, 4].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
2*Sum_{n >= 1} 1/( a(n) - 6/a(n) ) = 1.
6*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 2/a(n) ) = 1.
8*Sum_{n >= 1} 1/( a(n) + 24/(a(n) - 12/(a(n))) ) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/( a(n) + 8/(a(n) + 4/(a(n))) ) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/2 - 6*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 6)),
Sum_{n >= 1} 1/a(n) = 1/8 + 24*Sum_{n >= 1} 1/(a(n)*(a(n)^2 + 12)),
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/6 + 2*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 2)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 12)).
Sum_{n >= 1} 1/a(n) = ( theta_3(2-sqrt(3))^2 - 1 )/4 = 0.34770 07561 66992 06261 .... See Borwein and Borwein, Proposition 3.5 (i), p.91.
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - theta_3(sqrt(3)-2)^2 )/4. Cf. A003499 and A153415. (End)
a(n) = tan(Pi/12)^n + tan(5*Pi/12)^n. - Greg Dresden, Oct 01 2020
From Wolfdieter Lang, Sep 06 2021: (Start)
a(n) = S(n, 4) - S(n-2, 4) = 2*T(n, 2), for n >= 0, with S and T Chebyshev polynomials, with S(-1, x) = 0 and S(-2, x) = -1. S(n, 4) = A001353(n+1), for n >= -1, and T(n, 2) = A001075(n).
a(2*k) = A067902(k), a(2*k+1) = 4*A001570(k+1), for k >= 0. (End)
a(n) = sqrt(2 + 2*A011943(n+1)) = sqrt(2 + 2*A102344(n+1)), n>0. - Ralf Steiner, Sep 23 2021
Sum_{n>=1} arctan(3/a(n)^2) = Pi/6 - arctan(1/3) = A019673 - A105531 (Ohtskua, 2024). - Amiram Eldar, Aug 29 2024

More terms from James Sellers, May 03 2000

Additional comments from Lekraj Beedassy, Feb 14 2002

A003499 a(n) = 6*a(n-1) - a(n-2), with a(0) = 2, a(1) = 6.

Original entry on oeis.org

2, 6, 34, 198, 1154, 6726, 39202, 228486, 1331714, 7761798, 45239074, 263672646, 1536796802, 8957108166, 52205852194, 304278004998, 1773462177794, 10336495061766, 60245508192802, 351136554095046, 2046573816377474, 11928306344169798, 69523264248641314

Offset: 0

N. J. A. Sloane

Two times Chebyshev polynomials of the first kind evaluated at 3.
Also 2(a(2*n)-2) and a(2*n+1)-2 are perfect squares. - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
Chebyshev polynomials of the first kind evaluated at 3, then multiplied by 2. - Michael Somos, Apr 07 2003
Also gives solutions > 2 to the equation x^2 - 3 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004
Output of Lu and Wu's formula for the number of perfect matchings of an m X n Klein bottle where m and n are both even specializes to this sequence for m=2. - Sarah-Marie Belcastro, Jul 04 2009
It appears that for prime P = 8*n +- 3, that a((P-1)/2) == -6 (mod P) and for all composites C = 8*n +- 3, there is at least one i < (C-1)/2 such that a(i) == -6 (mod P). Only a few of the primes P of the form 8*n +-3, e.g., 29, had such an i less than (P-1)/2. As for primes P = 8*n +- 1, it seems that the sum of the two adjacent terms, a((P-1)/2) and a((P+1)/2), is congruent to 8 (mod P). - Kenneth J Ramsey, Feb 14 2012 and Mar 05 2012
For n >= 1, a(n) is also the curvature of circles (rounded to the nearest integer) successively inscribed toward angle 90 degree of tangent lines, starting with a unit circle. The expansion factor is 5.828427... or 1/(3 - 2*sqrt(2)), which is also 3 + 2*sqrt(2) or A156035. See illustration in links. - Kival Ngaokrajang, Sep 04 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 6*x*y + y^2 + 32 = 0. - Colin Barker, Feb 08 2014

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 198.
Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; pp. 480-481.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, 2001, Wiley, pp. 77-79.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E. Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence K_3.
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Refik Keskin and Olcay Karaatli, Some New Properties of Balancing Numbers and Square Triangular Numbers, Journal of Integer Sequences, Vol. 15 (2012), #12.1.4.
Tanya Khovanova, Recursive Sequences
W. Lu and F. Y. Wu, Close-packed dimers on nonorientable surfaces, arXiv:cond-mat/0110035 [cond-mat.stat-mech], 2001-2002; Physics Letters A, 293(2002), 235-246. [From Sarah-Marie Belcastro, Jul 04 2009]
Kival Ngaokrajang, Illustration of initial terms
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211.
Jeffrey Shallit, An interesting continued fraction, Math. Mag., 48 (1975), 207-211. [Annotated scanned copy]
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).

A081555(n) = 1 + a(n).
Bisection of A002203.
First row of array A103999.
Row 1 * 2 of array A188645. A174501.

a:=[2,6];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 16 2020

I:=[2,6]; [n le 2 select I[n] else 6*Self(n-1) -Self(n-2): n in [1..25]]; // G. C. Greubel, Jan 16 2020

R:=PowerSeriesRing(Integers(), 25); Coefficients(R!( (2-6*x)/(1 - 6*x + x^2) )); // Marius A. Burtea, Jan 16 2020

A003499:=-2*(-1+3*z)/(1-6*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation

a[0]=2; a[1]=6; a[n_]:= 6a[n-1] -a[n-2]; Table[a[n], {n,0,25}] (* Robert G. Wilson v, Jan 30 2004 *)
Table[Tr[MatrixPower[{{6, -1}, {1, 0}}, n]], {n, 25}] (* Artur Jasinski, Apr 22 2008 *)
LinearRecurrence[{6, -1}, {2, 6}, 25] (* Vladimir Joseph Stephan Orlovsky, Feb 26 2012 *)
CoefficientList[Series[(2-6x)/(1-6x+x^2), {x,0,25}], x] (* Vincenzo Librandi, Jun 07 2013 *)
(* From Eric W. Weisstein, Apr 17 2018 *)
Table[(3-2Sqrt[2])^n + (3+2Sqrt[2])^n, {n,0,25}]//Expand
Table[(1+Sqrt[2])^(2n) + (1-Sqrt[2])^(2n), {n,0,25}]//FullSimplify
Join[{2}, Table[Fibonacci[4n, 2]/Fibonacci[2n, 2], {n, 25}]]
2*ChebyshevT[Range[0, 25], 3] (* End *)

PARI
```
a(n)=2*real((3+quadgen(32))^n)
```
PARI
```
a(n)=2*subst(poltchebi(abs(n)),x,3)
```

a(n)=if(n<0,a(-n),polsym(1-6*x+x^2,n)[n+1])

[lucas_number2(n,6,1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

G.f.: (2-6*x)/(1 - 6*x + x^2).
a(n) = (3+2*sqrt(2))^n + (3-2*sqrt(2))^n = 2*A001541(n).
For all sequence elements n, 2*n^2 - 8 is a perfect square. Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002
a(2*n)+2 is a perfect square, 2(a(2*n+1)+2) is a perfect square. a(n), a(n-1) and A077445(n), n > 0, satisfy the Diophantine equation x^2 + y^2 - 3*z^2 = -8. - Mario Catalani (mario.catalani(AT)unito.it), Mar 24 2003
a(n+1) is the trace of n-th power of matrix {{6, -1}, {1, 0}}. - Artur Jasinski, Apr 22 2008
a(n) = Product_{r=1..n} (4*sin^2((4*r-1)*Pi/(4*n)) + 4). [Lu/Wu] - Sarah-Marie Belcastro, Jul 04 2009
a(n) = (1 + sqrt(2))^(2*n) + (1 + sqrt(2))^(-2*n). - Gerson Washiski Barbosa, Sep 19 2010
For n > 0, a(n) = A001653(n) + A001653(n+1). - Charlie Marion, Dec 27 2011
For n > 0, a(n) = b(4*n)/b(2*n) where b(n) is the Pell sequence, A000129. - Kenneth J Ramsey, Feb 14 2012
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 3 - 2*sqrt(2). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.16585 37786 96882 80543 ... = 2 + 1/(6 + 1/(34 + 1/(198 + ...))). Cf. A174501.
Also F(-alpha) = 0.83251219269380007634 ... has the continued fraction representation 1 - 1/(6 - 1/(34 - 1/(198 - ...))) and the simple continued fraction expansion 1/(1 + 1/((6-2) + 1/(1 + 1/((34-2) + 1/(1 + 1/((198-2) + 1/(1 + ...))))))). Cf. A174501 and A003500.
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((6^2-4) + 1/(1 + 1/((34^2-4) + 1/(1 + 1/((198^2-4) + 1/(1 + ...))))))).
(End)
G.f.: G(0), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
Inverse binomial transform of A228568 [Bhadouria]. - R. J. Mathar, Nov 10 2013
From Peter Bala, Oct 16 2019: (Start)
4*Sum_{n >= 1} 1/(a(n) - 8/a(n)) = 1.
8*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 4/a(n)) = 1.
Series acceleration formulas for sum of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/4 - 8*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 8)) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/8 + 4*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 4)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(3-2*sqrt(2)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(2*sqrt(2)-3))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A067902.
(End)
E.g.f.: 2*exp(3*x)*cosh(2*sqrt(2)*x). - Stefano Spezia, Oct 18 2019
a(2*n)+2 = a(n)^2. - Greg Dresden and Shraya Pal, Jun 29 2021

A188644 Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.

Original entry on oeis.org

1, 1, 1, 1, 7, 1, 1, 97, 17, 1, 1, 1351, 577, 31, 1, 1, 18817, 19601, 1921, 49, 1, 1, 262087, 665857, 119071, 4801, 71, 1, 1, 3650401, 22619537, 7380481, 470449, 10081, 97, 1, 1, 50843527, 768398401, 457470751, 46099201, 1431431, 18817, 127, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given the function f(x,y) = (sqrt(x^2+y) + x)^2 and constant k=f(x,y), then for all integers x >= 1 and y=[+-]1, k may be irrational, but (k^n + k^(-n))/2 always produces integer sequences; y=-1 results shown here; y=1 results are A188645.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{2*k}(x), evaluated at x=n. - Seiichi Manyama, Dec 30 2018

			Row 2 gives {( (2+sqrt(3))^(2*n) + (2-sqrt(3))^(2*n) )/2}.
Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,   7,     97,      1351,        18817, ...
   3 | 1,  17,    577,     19601,       665857, ...
   4 | 1,  31,   1921,    119071,      7380481, ...
   5 | 1,  49,   4801,    470449,     46099201, ...
   6 | 1,  71,  10081,   1431431,    203253121, ...
   7 | 1,  97,  18817,   3650401,    708158977, ...
   8 | 1, 127,  32257,   8193151,   2081028097, ...
   9 | 1, 161,  51841,  16692641,   5374978561, ...
  10 | 1, 199,  79201,  31521799,  12545596801, ...
  11 | 1, 241, 116161,  55989361,  26986755841, ...
  12 | 1, 287, 164737,  94558751,  54276558337, ...
  13 | 1, 337, 227137, 153090001, 103182433537, ...
  14 | 1, 391, 305761, 239104711, 186979578241, ...
  15 | 1, 449, 403201, 362074049, 325142092801, ...
  ...

Row 2 is A011943, row 3 is A056771, row 8 is A175633, (row 2)*2 is A067902, (row 9)*2 is A089775.
Column 0-5 give A000012, A056220, A144130, A243132, A243134, A243136.
(column 1)*2 is A060626.
Cf. A188645 (f(x, y) as above with y=1).
Diagonals give A173129, A322899.
Cf. A188646, A322836.

max = 9; y = -1; t = Table[k = ((x^2 + y)^(1/2) + x)^2; ((k^n) + (k^(-n)))/2 // FullSimplify, {n, 0, max - 1}, {x, 1, max}]; Table[ t[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jul 17 2013 *)

A(n,k) = (A188646(n,k-1) + A188646(n,k))/2.

A(n,k) = Sum_{j=0..k} binomial(2*k,2*j)*(n^2-1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited by Seiichi Manyama, Dec 30 2018

More terms from Seiichi Manyama, Jan 01 2019

A103999 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2M x 2N Klein bottle.

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 16, 34, 1, 1, 54, 196, 198, 1, 1, 196, 1666, 2704, 1154, 1, 1, 726, 16384, 64152, 37636, 6726, 1, 1, 2704, 171394, 1844164, 2549186, 524176, 39202, 1, 1, 10086, 1844164, 57523158, 220581904, 101757654, 7300804, 228486, 1

Offset: 0

Ralf Stephan, Feb 26 2005

			Array begins:
  1,   1,     1,        1,           1,             1,                1, ...
  1,   6,    34,      198,        1154,          6726,            39202, ...
  1,  16,   196,     2704,       37636,        524176,          7300804, ...
  1,  54,  1666,    64152,     2549186,     101757654,       4064620168, ...
  1, 196, 16384,  1844164,   220581904,   26743369156,    3252222705664, ...
  1, 726,171394, 57523158, 21050622914, 7902001927776, 2988827208115522, ...

Cliff, Danny and Zoe Stoll, About Klein bottles
W. T. Lu and F. Y. Wu, Dimer statistics on the Moebius strip and the Klein bottle, arXiv:cond-mat/9906154 [cond-mat.stat-mech], 1999.

Rows include A003499, A067902+2. Columns include A003500+2.
Main diagonal gives A340557.
Cf. A099390, A103997.

T[m_, n_] := Product[4 Sin[(4k-1) Pi/(4n)]^2 + 4 Cos[j Pi/(2m+1)]^2, {j, 1, m}, {k, 1, n}] // Round;
Table[T[m-n, n], {m, 0, 9}, {n, 0, m}] // Flatten (* Jean-François Alcover, Aug 20 2018 *)

default(realprecision, 120);
{T(n, k) = round(prod(a=1, n, prod(b=1, k, 4*sin((4*a-1)*Pi/(4*n))^2+4*sin((2*b-1)*Pi/(2*k))^2)))} \\ Seiichi Manyama, Jan 11 2021

T(M, N) = Product_{m=1..M} Product_{n=1..N} ( 4sin(Pi*(4n-1)/(4N))^2 + 4sin(Pi*(2m-1)/(2M))^2 ).

A067900 a(n) = 14*a(n-1) - a(n-2); a(0) = 0, a(1) = 8.

Original entry on oeis.org

0, 8, 112, 1560, 21728, 302632, 4215120, 58709048, 817711552, 11389252680, 158631825968, 2209456310872, 30773756526240, 428623135056488, 5969950134264592, 83150678744647800, 1158139552290804608, 16130803053326616712, 224673103194281829360, 3129292641666618994328

Offset: 0

Lekraj Beedassy, May 13 2003

Solves for y in x^2 - 3*y^2 = 4. Quadruples (a=b-y, b, c=b+y, d), with b=y^2 + 1 and d=x*y, where (x, y) solves x^2 - 3*y^2 = 4, satisfy the triangle relation (a^2 + b^2 + c^2 + d^2)^2 = 3*(a^4 + b^4 + c^4 + d^4). Thus d corresponds to the distance sum of the Fermat (or Torriccelli) point from its vertices in a triangle whose sides are in A.P. with middle side b and common difference y.

G. C. Greubel, Table of n, a(n) for n = 0..870
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Cf. A067902.
First differences of A045899.
Equals 8 * A007655(n+1).

m:=7;; a:=[0,8];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

m:=7; I:=[0,8]; [n le 2 select I[n] else 2*m*Self(n-1) -Self(n-2): n in [1..20]]; // G. C. Greubel, Dec 23 2019

a := proc(n) option remember: if n=0 then RETURN(0) fi: if n=1 then RETURN(8) fi: 14*a(n-1)-a(n-2): end: for n from 0 to 30 do printf(`%d,`,a(n)) od:
seq( simplify(8*ChebyshevU(n-1, 7)), n=0..20); # G. C. Greubel, Dec 23 2019

LinearRecurrence[{14,-1}, {0,8}, 17] (* Jean-François Alcover, Sep 19 2017 *)
8*ChebyshevU[Range[21] -2, 7] (* G. C. Greubel, Dec 23 2019 *)

vector(21, n, 8*polchebyshev(n-2, 2, 7) ) \\ G. C. Greubel, Dec 23 2019

[8*chebyshev_U(n-1,7) for n in (0..20)] # G. C. Greubel, Dec 23 2019

G.f.: 8*x/(1-14*x+x^2). - Philippe Deléham, Nov 17 2008

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

A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

Original entry on oeis.org

2, 2, 2, 2, 3, 2, 2, 4, 7, 2, 2, 5, 14, 18, 2, 2, 6, 23, 52, 47, 2, 2, 7, 34, 110, 194, 123, 2, 2, 8, 47, 198, 527, 724, 322, 2, 2, 9, 62, 322, 1154, 2525, 2702, 843, 2, 2, 10, 79, 488, 2207, 6726, 12098, 10084, 2207, 2, 2, 11, 98, 702, 3842, 15127, 39202, 57965, 37634, 5778, 2

Offset: 0

William W. Collier, Feb 18 2018

Note the similarity in form of the recursive steps in the array definition above and the polynomial definition under FORMULA.

			i\j |0  1   2    3      4       5        6          7           8            9
----+-------------------------------------------------------------------------
   0|2  2   2    2      2       2        2          2           2            2
   1|2  3   7   18     47     123      322        843        2207         5778
   2|2  4  14   52    194     724     2702      10084       37634       140452
   3|2  5  23  110    527    2525    12098      57965      277727      1330670
   4|2  6  34  198   1154    6726    39202     228486     1331714      7761798
   5|2  7  47  322   2207   15127   103682     710647     4870847     33385282
   6|2  8  62  488   3842   30248   238142    1874888    14760962    116212808
   7|2  9  79  702   6239   55449   492802    4379769    38925119    345946302
   8|2 10  98  970   9602   95050   940898    9313930    92198402    912670090
   9|2 11 119 1298  14159  154451  1684802   18378371   200477279   2186871698
  10|2 12 142 1692  20162  240252  2862862   34114092   406506242   4843960812
  11|2 13 167 2158  27887  360373  4656962   60180133   777684767  10049721838
  12|2 14 194 2702  37634  524174  7300802  101687054  1416317954  19726764302
  13|2 15 223 3330  49727  742575 11088898  165590895  2472774527  36926027010
  14|2 16 254 4048  64514 1028176 16386302  261152656  4162056194  66331746448
  15|2 17 287 4862  82367 1395377 23639042  400468337  6784322687 114933017342
  16|2 18 322 5778 103682 1860498 33385282  599074578 10749957122 192900153618
  17|2 19 359 6802 128879 2441899 46267202  876634939 16609796639 314709501202
  18|2 20 398 7940 158402 3160100 63043598 1257711860 25091193602 500566160180
  19|2 21 439 9198 192719 4037901 84603202 1772629341 37140612959 778180242798

William W. Collier, a(i,j) = f(i+2,j)
William W. Collier, Experimental Mathematics on Wisteria Tables, Talk to Poughkeepsie ACM Chapter.
OEIS Wiki, The (1,2) Pascal Triangle.

The array first appeared in A298675.
Rows 1 through 29 of the array appear in these OEIS entries: A005248, A003500, A003501, A003499, A056854, A086903, A056918, A087799, A057076, A087800, A078363, A067902, A078365, A090727, A078367, A087215, A078369, A090728, A090729, A090730, A090731, A090732, A090733, A090247, A090248, A090249, A090251. Also entries occur for rows 45, 121, and 320:  A087265, A065705, A089775. Each of these entries asserts that a(i,j)=f(i+2,j) is true for that row.
A few of the columns appear in the OEIS: A008865 (for column 2), A058794 and A007754 (for column 3), and A230586 (for column 5).
Main diagonal gives A343261.

A:= proc(i, j) option remember; `if`(min(i, j)=0, 2,
      `if`(j=1, i+2, (i+2)*A(i, j-1)-A(i, j-2)))
    end:
seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Mar 05 2019

a[, 0] = a[0, ] = 2; a[i_, 1] := i + 2;
a[i_, j_] := a[i, j] =(i + 2) a[i, j - 1] - a[i, j - 2];
Table[a[i - j, j], {i, 0, 10}, {j, 0, i}] // Flatten (* Jean-François Alcover, Dec 07 2019 *)

Let k be an integer, and let r1 and r2 be the roots of x + 1/x = k. Then f(k,n) = r1^n + r2^n is an integer, for integer n >= 0. Theorem: a(i,j) = f(i+2,j), for i,j >= 0. Proof: See the Collier link.
Define polynomials recursively by:
    p[0](n) = 2, for n >= 0 ( [ and ] demark subscripts).
    p[1](n) = n + 2, for n >= 0.
    p[j](n) = p[j-1](n) * p[1](n) - p[j-2](n), for j > 1, n >= 0. The coefficients of these polynomials occur as the even numbered, upward diagonals in the OEIS Wiki link. Conjecture: a(i,j) = p[j](i), i,j >= 0.

Edited by N. J. A. Sloane, Apr 04 2018

cp's OEIS Frontend

A003500 a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.

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A003499 a(n) = 6*a(n-1) - a(n-2), with a(0) = 2, a(1) = 6.

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A188644 Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.

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A103999 Square array T(M,N) read by antidiagonals: number of dimer tilings of a 2M x 2N Klein bottle.

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A067900 a(n) = 14*a(n-1) - a(n-2); a(0) = 0, a(1) = 8.

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A299741 Array read by antidiagonals upwards: a(i,0) = 2, i >= 0; a(i,1) = i+2, i >= 0; a(i,j) = (i+2) * a(i,j-1) - a(i,j-2), for i >= 0, j > 1.

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