Displaying 1-10 of 25 results found.
1, 7, 13, 97, 181, 1351, 2521, 18817, 35113, 262087, 489061, 3650401, 6811741, 50843527, 94875313, 708158977, 1321442641, 9863382151, 18405321661, 137379191137, 256353060613, 1913445293767, 3570537526921, 26650854921601, 49731172316281, 371198523608647
COMMENTS
See also A110294 (compare program code). a(2*n+1) = (a(2*n) + a(2*n+2))/2 and see A232765 for Diophantine equation that produces a sequence related to a(n). - Richard R. Forberg, Nov 30 2013
FORMULA
G.f.: (1+7*x-x^2-x^3) / ((1-4*x+x^2)*(1+4*x+x^2)).
a(n) = (3-(-1)^n)*((-3+2*sqrt(3))*(2-sqrt(3))^n + (3+2*sqrt(3))*(2+sqrt(3))^n )/(8*sqrt(3)).
a(n) = 14*a(n-2) - a(n-4) for n>3. (End)
MAPLE
seriestolist(series((1+7*x-x^2-x^3)/((1-4*x+x^2)*(1+4*x+x^2)), x=0, 25));
MATHEMATICA
CoefficientList[Series[(1+7x-x^2-x^3)/((1-4x+x^2)(1+4x+x^2)), {x, 0, 25}], x] (* Michael De Vlieger, Nov 01 2016 *)
PROG
(PARI) Vec((1+7*x-x^2-x^3)/((1-4*x+x^2)*(1+4*x+x^2)) + O(x^30)) \\ Colin Barker, Nov 01 2016 (Magma)
A001353:= func< n | Evaluate(ChebyshevSecond(n+1), 2) >;
(SageMath)
def A001353(n): return chebyshev_U(n, 2)
a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).
(Formerly M1769 N0700)
+10
104
1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647
COMMENTS
Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002 For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002 This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/ A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014] a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006 The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators= A005320, denominators= A001075. - Clark Kimberling, Aug 27 2008 The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators= A001075, denominators= A001353. - Clark Kimberling, Aug 27 2008 Pisano period lengths: 1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012 Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 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 + 48 = 0. - Colin Barker, Feb 10 2014 A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
1, 1, 0, 0, 0, 0, ...
2, 0, 3, 0, 0, 0, ...
2, 0, 0, 3, 0, 0, ...
2, 0, 0, 0, 3, 0, ...
2, 0, 0, 0, 0, 3, ...
...
The triangle generated from M is:
1,
1, 1,
3, 1, 3,
11, 3, 3, 9,
41, 11, 9, 9, 27,
...
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016 For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016 If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms. a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476. a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018 Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018 Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020 a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022
REFERENCES
Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.
FORMULA
G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120. a(n) = sqrt(1 + 3* A001353(n)) (cf. Richardson comment, Oct 10 2002). a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01, 2004 a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003 a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [ A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006 Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008 G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013 a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020 a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835. The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022 2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023
EXAMPLE
2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...
MAPLE
orthopoly[T](n, 2) ;
end proc:
MATHEMATICA
Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
LinearRecurrence[{4, -1}, {1, 2}, 30] (* Harvey P. Dale, Aug 22 2015 *) a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)
PROG
(PARI) {a(n) = subst(poltchebi(abs(n)), x, 2)};
(PARI) {a(n) = real((2 + quadgen(12))^abs(n))};
(PARI) {a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};
(PARI) my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017 (SageMath) [lucas_number2(n, 4, 1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009 (Haskell)
a001075 n = a001075_list !! n
a001075_list =
1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
(SageMath)
def a(n):
Q = QuadraticField(3, 't')
u = Q.units()[0]
(Magma) I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017
a(n) = 4*a(n-1) - a(n-2) with a(0) = 2, a(1) = 4.
(Formerly M1278)
+10
51
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
COMMENTS
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 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
REFERENCES
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.
LINKS
Hideyuki Ohtskua, proposer, Problem B-1351, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 62, No. 3 (2024), p. 258. 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.
FORMULA
a(n) = ( 2 + sqrt(3) )^n + ( 2 - sqrt(3) )^n.
G.f.: 2*(1 - 2*x)/(1 - 4*x + x^2). Simon Plouffe in his 1992 dissertation. 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 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(n) = [x^n] ( (1 + 4*x + sqrt(1 + 8*x + 12*x^2))/2 )^n for n >= 1. - Peter Bala, Jun 23 2015 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 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 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).
MAPLE
A003500 := proc(n) option remember; if n <= 1 then 2*n+2 else 4*procname(n-1)-procname(n-2); fi;
end proc;
MATHEMATICA
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 *)
PROG
(Sage) [lucas_number2(n, 4, 1) for n in range(0, 24)] # Zerinvary Lajos, May 14 2009 (Haskell)
a003500 n = a003500_list !! n
a003500_list = 2 : 4 : zipWith (-)
(map (* 4) $ tail a003500_list) a003500_list
(PARI) x='x+O('x^99); Vec(-2*(-1+2*x)/(1-4*x+x^2)) \\ Altug Alkan, Apr 04 2016 (Magma) 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
CROSSREFS
Cf. this sequence (middle side lengths), A016064 (smallest side lengths), A335025 (largest side lengths).
Standard deviation of A007654.
(Formerly M4948)
+10
45
0, 1, 14, 195, 2716, 37829, 526890, 7338631, 102213944, 1423656585, 19828978246, 276182038859, 3846719565780, 53577891882061, 746243766783074, 10393834843080975, 144767444036350576, 2016350381665827089, 28084137899285228670, 391161580208327374291, 5448177985017298011404
COMMENTS
a(n) corresponds also to one-sixth the area of Fleenor-Heronian triangle with middle side A003500(n). - Lekraj Beedassy, Jul 15 2002 a(n) give all (nontrivial, integer) solutions of Pell equation b(n+1)^2 - 48*a(n+1)^2 = +1 with b(n+1)= A011943(n), n>=0. For n>=3, a(n) equals the permanent of the (n-2) X (n-2) tridiagonal matrix with 14's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011 For n>1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,13}. - Milan Janjic, Jan 25 2015 6*a(n)^2 = 6*S(n-1, 14)^2 is the triangular number Tri((T(n, 7) - 1)/2) with Tri = A000217 and T = A053120. This is instance k = 3 of the general k-identity given in a comment to A001109. - Wolfdieter Lang, Feb 01 2016
REFERENCES
D. A. Benaron, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
E. K. Lloyd, The standard deviation of 1, 2, ..., n, Pell's equation and rational triangles, The Mathematical Gazette, Vol. 81, No. 491 (Jul., 1997), pp. 231-243.
FORMULA
a(n) = 14*a(n-1) - a(n-2).
G.f.: x^2/(1-14*x+x^2).
a(n+1) ~ 1/24*sqrt(3)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n+1) = S(n-1, 14), n>=0, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310. a(n+1) = ( (7+4*sqrt(3))^n - (7-4*sqrt(3))^n )/(8*sqrt(3)).
a(n+1) = sqrt(( A011943(n)^2 - 1)/48), n>=0. Chebyshev's polynomials U(n-2, x) evaluated at x=7.
a(n) = 13*( a(n-1) + a(n-2) ) - a(n-3).
a(n) = 15*( a(n-1) - a(n-2) ) + a(n-3). (End)
a(n) = b such that (-1)^n/4*Integral_{x=-Pi/2..Pi/2} (sin((2*n-2)*x))/(2-sin(x)) dx = c+b*log(3). - Francesco Daddi, Aug 02 2011 Product {n >= 1} (1 + 1/a(n)) = 1/3*(3 + 2*sqrt(3)). - Peter Bala, Dec 23 2012 Product {n >= 2} (1 - 1/a(n)) = 1/7*(3 + 2*sqrt(3)). - Peter Bala, Dec 23 2012 E.g.f.: 1 - exp(7*x)*(12*cosh(4*sqrt(3)*x) - 7*sqrt(3)*sinh(4*sqrt(3)*x))/12. - Stefano Spezia, Dec 11 2022
EXAMPLE
G.f. = x^2 + 14*x^3 + 195*x^4 + 2716*x^5 + 37829*x^6 + 526890*x^7 + ...
MAPLE
0, seq(orthopoly[U](n, 7), n=0..30); # Robert Israel, Feb 04 2016
MATHEMATICA
Table[Sum[Binomial[n, 2 k - 1]*7^(n - 2 k + 1)*48^(k - 1), {k, 1, n}], {n, 0, 15}] (* Horst H. Manninger, Jan 16 2022 *)
PROG
(Sage) [lucas_number1(n, 14, 1) for n in range(0, 20)] # Zerinvary Lajos, Jun 25 2008 (Sage) [chebyshev_U(n, 7) for n in (-1..20)] # G. C. Greubel, Dec 23 2019 (Magma) [n le 2 select n-1 else 14*Self(n-1)-Self(n-2): n in [1..70]]; // Vincenzo Librandi, Feb 02 2016 (PARI) concat(0, Vec((x^2/(1-14*x+x^2) + O(x^30)))) \\ Michel Marcus, Feb 02 2016 (PARI) vector(21, n, polchebyshev(n-2, 2, 7) ) \\ G. C. Greubel, Dec 23 2019 (GAP) m:=7;; a:=[0, 1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019
CROSSREFS
Cf. A000217, A001570, A003500, A011922, A011943, A011945, A028230, A046184, A049310, A053120, A055793, A067900, A098301, A101950, A103974. Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), this sequence (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Triangle of coefficients of Chebyshev polynomials T_n(x).
+10
26
1, 1, -1, 2, -3, 4, 1, -8, 8, 5, -20, 16, -1, 18, -48, 32, -7, 56, -112, 64, 1, -32, 160, -256, 128, 9, -120, 432, -576, 256, -1, 50, -400, 1120, -1280, 512, -11, 220, -1232, 2816, -2816, 1024, 1, -72, 840, -3584, 6912, -6144, 2048, 13, -364, 2912, -9984, 16640, -13312, 4096
COMMENTS
The row length sequence of this irregular array is A008619(n), n >= 0. Even or odd powers appear in increasing order starting with 1 or x for even or odd row numbers n, respectively. This is the standard triangle A053120 with 0 deleted. - Wolfdieter Lang, Aug 02 2014 Let T* denote the triangle obtained by replacing each number in this triangle by its absolute value. Then T* gives the coefficients for cos(nx) as a polynomial in cos x. - Clark Kimberling, Aug 04 2024
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
E. A. Guilleman, Synthesis of Passive Networks, Wiley, 1957, p. 593.
Yaroslav Zolotaryuk, J. Chris Eilbeck, "Analytical approach to the Davydov-Scott theory with on-site potential", Physical Review B 63, p543402, Jan. 2001. The authors write, "Since the algebra of these is 'hyperbolic', contrary to the usual Chebyshev polynomials defined on the interval 0 <= x <= 1, we call the set of functions (21) the hyperbolic Chebyshev polynomials." (This refers to the triangle T* described in Comments.)
LINKS
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
FORMULA
a(n,m) = 2^(m-1) * n * (-1)^((n-m)/2) * ((n+m)/2-1)! / (((n-m)/2)! * m!) if n>0. - R. J. Mathar, Apr 20 2007 T_n(x) = 2*x*T_(n-1)(x) - T_(n-2)(x), T_0(x) = 1, T_1(x) = x.
T_n(x) = ((x+sqrt(x^2-1))^n + (x-sqrt(x^2-1))^n)/2. (End)
T(n,x) = [z^n] ( z*x + sqrt(1 + z^2*(x^2 - 1)) )^n.
Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x).
exp( Sum_{n >= 1} T(n,x)*t^n/n ) = Sum_{n >= 0} P(n,x)*t^n, where P(n,x) denotes the n-th Legendre polynomial. (End)
EXAMPLE
Rows are: (1), (1), (-1,2), (-3,4), (1,-8,8), (5,-20,16) etc., since if c = cos(x): cos(0x) = 1, cos(1x) = 1c; cos(2x) = -1+2c^2; cos(3x) = -3c+4c^3, cos(4x) = 1-8c^2+8c^4, cos(5x) = 5c-20c^3+16c^5, etc.
This irregular triangle a(n,k) begins:
n\k 0 1 2 3 4 5 6 7 ...
0: 1
1: 1
2: -1 2
3: -3 4
4: 1 -8 8
5: 5 -20 16
6: -1 18 -48 32
7: -7 56 -112 64
8: 1 -32 160 -256 128
9: 9 -120 432 -576 256
10: -1 50 -400 1120 -1280 512
11: -11 220 -1232 2816 -2816 1024
12: 1 -72 840 -3584 6912 -6144 2048
13: 13 -364 2912 -9984 16640 -13312 4096
14: -1 98 -1568 9408 -26880 39424 -28672 8192
15: -15 560 -6048 28800 -70400 92160 -61440 16384
...
T(4,x) = 1 - 8*x^2 + 8*x^4, T(5,x) = 5*x - 20*x^3 +16*x^5.
(End)
MAPLE
A008310 := proc(n, m) local x ; coeftayl(simplify(ChebyshevT(n, x), 'ChebyshevT'), x=0, m) ; end: i := 0 : for n from 0 to 100 do for m from n mod 2 to n by 2 do printf("%d %d ", i, A008310(n, m)) ; i := i+1 ; od ; od ; # R. J. Mathar, Apr 20 2007
# second Maple program:
b:= proc(n) b(n):= `if`(n<2, 1, expand(2*b(n-1)-x*b(n-2))) end:
T:= n-> (p-> (d-> seq(coeff(p, x, d-i), i=0..d))(degree(p)))(b(n)):
MATHEMATICA
Flatten[{1, Table[CoefficientList[ChebyshevT[n, x], x], {n, 1, 13}]}]//DeleteCases[#, 0, Infinity]& (* or *) Flatten[{1, Table[Table[((-1)^k*2^(n-2 k-1)*n*Binomial[n-k, k])/(n-k), {k, Floor[n/2], 0, -1}], {n, 1, 13}]}] (* Eugeniy Sokol, Sep 04 2019 *)
CROSSREFS
Row sums are one. Polynomial evaluations include A001075 (x=2), A001541 (x=3), A001091, A001079, A023038, A011943, A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203.
Numbers k such that the standard deviation of 1,...,k is an integer.
(Formerly M3154)
+10
17
0, 3, 48, 675, 9408, 131043, 1825200, 25421763, 354079488, 4931691075, 68689595568, 956722646883, 13325427460800, 185599261804323, 2585064237799728, 36005300067391875, 501489136705686528, 6984842613812219523, 97286307456665386800, 1355023461779503195683
COMMENTS
Gives solutions k to the Diophantine equation m^2 = k*(k+1)/3. - Anton Lorenz Vrba (anton(AT)a-l-v.net), Jun 28 2005
If x=a(n), y=a(n+1), z=a(n+2) are three consecutive terms, then x^2 - 16*y*x + 14*x*z + 16*y^2 - 16*z*y + z^2 = 144. The formula is symmetric in x and z, so it is also valid for x=a(n+2), y=a(n+1), z=a(n). - Alexander Samokrutov, Jul 02 2015 Corresponding solutions m (of first comment) are in A011944. Equivalently, numbers k such that k/3 and k+1 are both perfect squares. (End)
REFERENCES
Guy Alarcon and Yves Duval, TS: Préparation au Concours Général, RMS, Collection Excellence, Paris, 2010, chapitre 13, Questions proposées aux élèves de Terminale S, Exercice 1, p. 220, p. 223.
D. A. Benaron, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
FORMULA
a(m) = 14*a(m-1) - a(m-2) + 6.
G.f.: -3*x^2*(1+x)/(-1+x)/(1-14*x+x^2) = -3 + (1/2)/(-1+x) + (1/2)*(-97*x+7)/(1-14*x+x^2). - R. J. Mathar, Nov 20 2007 a(n) = (-2 + (7-4*sqrt(3))^n*(7+4*sqrt(3)) + (7-4*sqrt(3))*(7+4*sqrt(3))^n)/4. - Colin Barker, Mar 05 2016
MATHEMATICA
RecurrenceTable[{a[m] == 14 a[m - 1] - a[m - 2] + 6, a[1] == 0, a[2] == 3}, a, {m, 1, 17}] (* Michael De Vlieger, Jul 02 2015 *) CoefficientList[Series[-3 x^2*(1 + x)/(-1 + x)/(1 - 14 x + x^2), {x, 0, 17}], x] (* Michael De Vlieger, Feb 02 2016 *)
PROG
(Magma) I:=[0, 3]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+6: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016
EXTENSIONS
Corrected by Keith Lloyd, Mar 15 1996
a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.
+10
13
1, 17, 577, 19601, 665857, 22619537, 768398401, 26102926097, 886731088897, 30122754096401, 1023286908188737, 34761632124320657, 1180872205318713601, 40114893348711941777, 1362725501650887306817, 46292552162781456490001
COMMENTS
This sequence {a(n)} gives all the nonnegative integer solutions of the Pell equation a(n)^2 - 32*(3* A091761(n))^2 = +1. - Wolfdieter Lang, Mar 09 2019
FORMULA
a(n) = (r^n + 1/r^n)/2 with r = 17 + sqrt(17^2-1).
a(n) = T(n, 17) = T(2*n, 3) with T(n, x) Chebyshev's polynomials of the first kind. See A053120. T(n, 3)= A001541(n). G.f.: (1-17*x)/(1-34*x+x^2).
G.f.: (1 - 17*x / (1 - 288*x / (17 - x))). - Michael Somos, Apr 05 2019 a(n) = (a^n + b^n)/2 where a = 17 + 12*sqrt(2) and b = 17 - 12*sqrt(2); sqrt(a(n)-1)/4 = A001109(n). - James R. Buddenhagen, Dec 09 2011 a(n) = sqrt(1 + 32*9* A091761(n)^2), n >= 0. See one of the Pell comments above. - Wolfdieter Lang, Mar 09 2019
EXAMPLE
G.f. = 1 + 17*x + 577*x^2 + 19601*x^3 + 665857*x^4 + 22619537*x^5 + ...
PROG
(Sage) [lucas_number2(n, 34, 1)/2 for n in range(0, 15)] # Zerinvary Lajos, Jun 27 2008 (Magma) I:=[1, 17]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011 (Maxima) makelist(expand(((17+sqrt(288))^n+(17-sqrt(288))^n))/2, n, 0, 15); // Vincenzo Librandi, Dec 18 2011 (PARI) {a(n) = polchebyshev( n, 1, 17)}; /* Michael Somos, Apr 05 2019 */
Member r=16 of the family of Chebyshev sequences S_r(n) defined in A092184.
+10
12
0, 1, 16, 225, 3136, 43681, 608400, 8473921, 118026496, 1643897025, 22896531856, 318907548961, 4441809153600, 61866420601441, 861688079266576, 12001766689130625, 167163045568562176, 2328280871270739841, 32428769152221795600, 451674487259834398561
COMMENTS
Also m such that (3*m^2 + m)/4 = m*(3*m + 1)/4 is a perfect square. - Ctibor O. Zizka, Oct 15 2010
FORMULA
a(n) = (T(n, 7)-1)/6 with Chebyshev's polynomials of the first kind evaluated at x=7: T(n, 7) = A011943(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n)/2; therefore: a(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n - 2)/12. a(n) = A001353(n)^2 = S(n-1, 4)^2 with Chebyshev's polynomials of the second kind evaluated at x=4, S(n, 4):=U(n, 2). a(n) = 14*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), n >= 3.
G.f.: x*(1+x)/((1-x)*(1 - 14*x + x^2)) = x*(1+x)/(1 - 15*x + 15*x^2 - x^3) (from the Stephan link, see A092184).
MATHEMATICA
LinearRecurrence[{# - 1, -# + 1, 1}, {0, 1, #}, 20] &[16] (* Michael De Vlieger, Feb 23 2021 *)
PROG
(PARI) concat(0, Vec(x*(1+x)/((1-x)*(1-14*x+x^2)) + O(x^50))) \\ Colin Barker, Jun 15 2015
Array of (k^n + k^(-n))/2 where k = (sqrt(x^2-1) + x)^2 for integers x >= 1.
+10
10
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
COMMENTS
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
FORMULA
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
EXAMPLE
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, ...
...
MATHEMATICA
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 *)
CROSSREFS
Cf. A188645 (f(x, y) as above with y=1).
Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is Chebyshev polynomial of the first kind T_{n}(x), evaluated at x=k.
+10
10
1, 1, 0, 1, 1, -1, 1, 2, 1, 0, 1, 3, 7, 1, 1, 1, 4, 17, 26, 1, 0, 1, 5, 31, 99, 97, 1, -1, 1, 6, 49, 244, 577, 362, 1, 0, 1, 7, 71, 485, 1921, 3363, 1351, 1, 1, 1, 8, 97, 846, 4801, 15124, 19601, 5042, 1, 0, 1, 9, 127, 1351, 10081, 47525, 119071, 114243, 18817, 1, -1
FORMULA
A(0,k) = 1, A(1,k) = k and A(n,k) = 2 * k * A(n-1,k) - A(n-2,k) for n > 1.
A(n,k) = n * Sum_{j=0..n} (2*k-2)^j * binomial(n+j,2*j)/(n+j) for n > 0. - Seiichi Manyama, Mar 05 2021
EXAMPLE
Square array begins:
1, 1, 1, 1, 1, 1, 1, ...
0, 1, 2, 3, 4, 5, 6, ...
-1, 1, 7, 17, 31, 49, 71, ...
0, 1, 26, 99, 244, 485, 846, ...
1, 1, 97, 577, 1921, 4801, 10081, ...
0, 1, 362, 3363, 15124, 47525, 120126, ...
-1, 1, 1351, 19601, 119071, 470449, 1431431, ...
MATHEMATICA
Table[ChebyshevT[n-k, k], {n, 0, 10}, {k, n, 0, -1}] // Flatten (* Amiram Eldar, Dec 28 2018 *)
PROG
(PARI) T(n, k) = polchebyshev(n, 1, k);
(PARI) T(n, k) = if(n==0, 1, n*sum(j=0, n, (2*k-2)^j*binomial(n+j, 2*j)/(n+j))); \\ Seiichi Manyama, Mar 05 2021
CROSSREFS
Columns 0-20 give A056594, A000012, A001075, A001541, A001091, A001079, A023038, A011943(n+1), A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203, A322888, A056771, A322889, A078986, A322890. Rows 0-10 give A000012, A001477, A056220, A144129, A144130, A243131, A243132, A243133, A243134, A243135, A243136. Cf. A323182 (Chebyshev polynomial of the second kind).
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