0,2

Chebyshev polynomials of the first kind evaluated at 3.

a(n) solves for x in x^2 - 8*y^2 = 1, the corresponding y being A001109(n). For n>0, the ratios a(n)/A001090(n) may be obtained as convergents to sqrt(8): either successive convergents of [3; -6] or odd convergents of [2; 1, 4]. - Lekraj Beedassy (blekraj(AT)yahoo.com), Sep 09 2003

Formula: ((-1+sqrt(2))^n+(1+sqrt(2))^n+(1-sqrt(2))^n+(-1-sqrt(2))^n)/4 (with interpolated zeros) E.g.f. cosh(x)cosh(sqrt(2)x) (with interpolated zeros). - Paul Barry (pbarry(AT)wit.ie), Sep 18 2003

Also gives solutions to the equation x^2-1=floor(x*r*floor(x/r)) where r=sqrt(8) - Benoit Cloitre (benoit7848c(AT)orange.fr), Feb 14 2004

Appears to give all solutions >1 to the equation : x^2=ceiling(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 24, 2004

a(n+1) - A001542(n+1) = A090390(n+1) - A046729(n) = A001653(n); a(n+1) - 4*A079291(n+1) = (-1)^(n+1). Formula generated by the floretion - .5'i + .5'j - .5i' + .5j' - 'ii' + 'jj' - 2'kk' + 'ij' + .5'ik' + 'ji' + .5'jk' + .5'ki' + .5'kj' + e - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Nov 16 2004

This sequence give numbers n such that (n-1)*(n+1)/2 = j^2 = a square. Remark : (i-1)*(i+1)/2 = (i^2-1)/2 = -1 = i^2 with i=sqrt(-1) so i is also in the sequence. - Pierre CAMI (pierrecami(AT)tele2.fr), Apr 20 2005

a(n) is prime for n = {1, 2, 4, 8}. Prime a(n) are {3, 17, 577, 665857}, which belong to A001601(n). a(2k-1) is divisible by a(1) = 3. a(4k-2) is dovisible by a(2) = 17. a(8k-4) is divisible by a(4) = 577. a(16k-8) is divisible by a(8) = 665857. - Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 24 2006

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures}, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

I. Adler, Three diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.

H. Brocard, Notes e'le'mentaires sur le proble`me de Peel, Nouvelles Correspondance Math\'{e}matique, 4 (1878), 161-169.

D. H. Lehmer, A cotangent analogue of continued fractions, Duke Math. J., 4 (1935), 323-340.

D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.

Problem E1976, Amer. Math. Monthly, 75 (1968), 683-684.

J. W. L. Glaisher, On Eulerian numbers (formulae, residues, end-figures), with the values of the first twenty-seven, Quarterly Journal of Mathematics, vol. 45, 1914, pp. 1-51.

T. D. Noe, Table of n, a(n) for n=0..200

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures}, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

Tanya Khovanova, Recursive Sequences

Index entries for sequences related to Chebyshev polynomials.

a(n) = ((3+2*sqrt(2))^n + (3-2*sqrt(2))^n)/2.

a(n) = 3*A001109(n)-A001109(n-1), n >= 1. - Barry Williams and Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), May 05 2000.

G.f.: (1-3*x)/(1-6*x+x^2) - Barry Williams and Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), May 05 2000.

a(n) = sqrt{8*[(A001109(n))^2] + 1} = T(n, 3), with Chebyshev's T-polynomials A053120.

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

For all elements x of the sequence, 2*x^2 - 2 is a square. Lim. as n -> inf. of a(n)/a(n-1) = 3 + sqrt(2). - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 10 2002

E.g.f.: exp(3x)cosh(2sqrt(2)x). Binomial transform of A084128. - Paul Barry (pbarry(AT)wit.ie), May 16 2003

For n>=1, a(n) = A001652(n) - A001652(n-1) - Charlie Marion (charliem(AT)bestweb.net), Jul 01 2003

For n>0, a(n)^2 +1=2*A001653(n-1)*A001653(n); e.g. 17^2+1=290=2*5*29 - Charlie Marion (charliemath(AT)verizon.net), Dec 21 2003

a(n) = Sum_{k>=0} binomial(2*n, 2*k)*2^k = Sum_{k>=0} A086645(n, k)*2^k . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Feb 29 2004

a(n)*A002315(n+k)=A001652(2n+k)+A001652(k)+1; e.g. 3*1393=4069+119+1; for k>0, a(n+k)*A002315(n)=A001652(2n+k)-A001652(k-1); e.g. 577*7=4059-20 - Charlie Marion (charliemath(AT)verizon.net), Mar 17 2003

a(n)^2+a(n+1)^2=2*(A001653(2n+1)-A001652(2n)); e.g., 1^2+3^2=2(5-0); 17^2+99^2=2(5741-696) - Charlie Marion (charliemath(AT)verizon.net), Mar 17 2003

A053141(n+1) + A055997(n+1) = a(n+1) + A001109(n+1). - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Sep 16 2004

For n>k, a(n)*A001653(k)=A011900(n+k)+A053141(n-k-1); e.g. 99*5=495=493+2. For n<=k, a(n)*A001653(k)=A011900(n+k)+A053141(k-n); e.g. 3*29=87=85+2 - Charlie Marion (charliemath(AT)optonline.net), Oct 18 2004

a(n) = Sqrt[ A055997(2n) ]. - Alexander Adamchuk (alex(AT)kolmogorov.com), Nov 24 2006

a(2n) = A056771(n). a(2n+1) = 3*A077420(n). - Alexander Adamchuk (alex(AT)kolmogorov.com), Feb 01 2007

(A000129(n)^2)*4+(-1)^n - Vim Wenders (vim(AT)gmx.li), Mar 28 2007

2*a(k)*A001653(n)*A001653(n+k)=A001653(n)^2+A0001653(n+k)^2+A001542(k)^2; e.g., 2*3*5*29=5^2+29^2+2^2; 2*99*29*5741=2*99*29*5741=29^2+5741^2+70^2 - Charlie Marion (charliemath(AT)optonline.net), Oct 12 2007

a(n) = Cosh[2n*ArcSinh[1]] - Herbert Kociemba (kociemba(AT)t-online.de), Apr 24 2008

a[0]:=1: a[1]:=3: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 26 2006

A001541:=-(-1+3*z)/(1-6*z+z**2); [Conjectured by S. Plouffe in his 1992 dissertation.]

(PARI) a(n)=real((3+quadgen(32))^n)

(PARI) a(n)=subst(poltchebi(abs(n)), x, 3)

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

(PARI) a(n)=([1, 2, 2; 2, 1, 2; 2, 2, 3]^n)[3, 3] - Vim Wenders (vim(AT)gmx.li), Mar 28 2007

Bisection of A001333. A003499(n)=2a(n). Cf. A046090, A001109, A053142.

Cf. A084130.

Cf. A001109.

Cf. A055997 = numbers n such that n(n-1)/2 is a square. Cf. A001601.

Cf. A056771, A077420.

Cf. A005319.

Adjacent sequences: A001538 A001539 A001540 this_sequence A001542 A001543 A001544

Sequence in context: A086842 A010913 A056660 this_sequence A074565 A054365 A116886

nonn,easy,nice,new

njas