0,2

Sequence also gives values of x satisfying 3*y^2 - x^2 = 2, the corresponding y being given by A001835(n+1). Moreover, quadruples(p, q, r, s) satisfying p^2 + q^2 + r^2 = s^2, where p=q and r is either p+1 or p-1, are termed nearly isosceles Pythagorean and are given by p={x + (-1)^n}/3, r=p-(-1)^n, s=y for n>1. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 19 2002

a(n) = L(n,-4)*(-1)^n, where L is defined as in A108299; see also A001835 for L(n,+4). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 01 2005

a(n)= A002531(1+2*n) - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007

361 written in base A001835(n+1)-1 is the square of a(n). E.g. a(12)=2672279, A001835(13)-1=1542840. We have 361_(1542840)=3*1542840+6*1542840+1=2672279^2 - Richard Choulet (richardchoulet(AT)yahoo.fr), Oct 04 2007

The lower principal convergents to 3^(1/2), beginning with 1/1, 5/3, 19/11, 71/41, comprise a strictly increasing sequence; numerators=A001834, denominators=A001835. - Clark Kimberling (ck6(AT)evansville.edu), Aug 27 2008

General recurrence is a(n)=(a(1)-1)*a(n-1)-a(n-2), a(1)>=4, lim n->infinity a(n)= x*(k*x+1)^n, k =(a(1)-3), x=(1+sqrt((a(1)+1)/(a(1)-3)))/2. Examples in OEIS: a(1)=4 gives A002878, primes in it A121534. a(1)=5 gives A001834, primes in it A086386. a(1)=6 gives A030221, primes in it not in OEIS {29,139,3191,...}. a(1)=7 gives A002315, primes in it A088165. a(1)=8 gives A033890, primes in it not in OEIS (does there exist any ?). a(1)=9 gives A057080, primes in it not in OEIS {71,34649,16908641,...}. a(1)=10 gives A057081, primes in it not in OEIS {389806471,192097408520951,...}. [From Ctibor O. Zizka (ctibor.zizka(AT)seznam.cz), Sep 02 2008]

Inverse binomial transform of A030192. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Nov 19 2009]

L. Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 375.

Clark Kimberling, "Best lower and upper approximates to irrational numbers," Elemente der Mathematik, 52 (1997) 122-126.

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq. (44) rhs, m=6.

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'Interm\'{e}diaire des Math\'{e}maticiens, 10 (1903), 235-238.

F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.

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

Index entries for sequences related to linear recurrences with constant coefficients

L. Euler, Vollstaendige Anleitung zur Algebra, Zweiter Teil.

Tanya Khovanova, Recursive Sequences

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.

Index entries for sequences related to Chebyshev polynomials.

a(n) = ((1+sqrt(3))^(2*n+1)+(1-sqrt(3))^(2*n+1))/2^(n+1). - njas, Nov 10 2009

a(n) = (1/2) * ((1+sqrt(3))*(2+sqrt(3))^n + (1-sqrt(3))*(2-sqrt(3))^n). - Dean Hickerson (dean.hickerson(AT)yahoo.com), Dec 01 2002

With a=2+sqrt(3), b=2-sqrt(3): a(n)=(1/sqrt(2))(a^(n+1/2)-b^(n+1/2)). a(n)-a(n-1)=A003500(n). a(n)=sqrt(1+12*A061278(n)+12*A061278(n)^2). - Mario Catalani (mario.catalani(AT)unito.it), Apr 11 2003

a(n)=((1+sqrt[3])^(2*n+1)+(1-sqrt[3])^(2*n+1))/2^(n+1) - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007

G.f.: (1+x)/((1-4*x+x^2)). a(n)= S(2*n, sqrt(6)) = S(n, 4)+S(n-1, 4); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 4)= A001353(n).

For all members x of the sequence, 3*x^2 + 6 is a square. Lim. as n -> Inf. a(n)/a(n-1) = 2 + Sqrt(3). - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 10 2002

a(n)=2*A001571(n)+1 - Bruce Corrigan (scentman(AT)myfamily.com), Nov 04 2002

Let q(n, x)=sum(i=0, n, x^(n-i)*binomial(2*n-i, i)); then (-1)^n*q(n, -6)=a(n) - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 10 2002

a(n) = 2^(-n)*Sum{k>=0} binomial(2*n+1, 2*k)*3^k; see A091042 . - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Mar 01 2004

a(n) = floor(sqrt(3)*A001835(n+1)). - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Mar 03 2004

a(n+1) - 2*a(n) = 3*A001835(n+1). Using the known relation A001835(n+1) = sqrt((a(n)^2 + 2)/3) it follows that a(n+1) - 2*a(n) = sqrt(3*(a(n)^2+2)). Therefore a(n+1)^2 + a(n)^2 - 4*a(n+1)*a(n) - 6 = 0. - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Apr 18 2005

a(n)=Jacobi_P(n,1/2,-1/2,2)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry (pbarry(AT)wit.ie), Feb 03 2006

Equals binomial transform of A026150 starting (1, 4, 10, 28, 76,...) and double binomial transform of (1, 3, 3, 9, 9, 27, 27, 81, 81,...). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Nov 30 2007

Sequence satisfies 6 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos Sep 19 2008

a(-1-n) = -a(n). - Michael Somos Sep 19 2008

A001834:=(1+z)/(1-4*z+z**2); [S. Plouffe in his 1992 dissertation.]

f:=n->((1+sqrt(3))^(2*n+1)+(1-sqrt(3))^(2*n+1))/2^(n+1); [njas, Nov 10 2009]

a[0] = 1; a[1] = 5; a[n_] := a[n] = 4a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 25}] (from Robert G. Wilson v Apr 24 2004)

Table[Expand[((1+Sqrt[3])^(2*n+1)+(1+Sqrt[3])^(2*n+1))/2^(n+1)], {n, 0, 20}] - Anton Vrba (antonvrba(AT)yahoo.com), Feb 14 2007

q=24; s=0; lst={}; Do[s+=n; If[Sqrt[q*s+1]==Floor[Sqrt[q*s+1]], AppendTo[lst, Sqrt[q*s+1]]], {n, 0, 8!}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Apr 02 2009]

Floretion Algebra Multiplication Program, FAMP Code: A001834 = (4/3)vesseq[ - .25'i + 1.25'j - .25'k - .25i' + 1.25j' - .25k' + 1.25'ii' + .25'jj' - .75'kk' + .75'ij' + .25'ik' + .75'ji' - .25'jk' + .25'ki' - .25'kj' + .25e], apart from initial term

(PARI) {a(n) = real( (2 + quadgen(12))^n * (1 + quadgen(12)) )} /* Michael Somos Sep 19 2008 */

(PARI) {a(n) = subst( polchebyshev(n-1, 2) + polchebyshev(n, 2), x, 2)} /* Michael Somos Sep 19 2008 */

(Other) sage: [(lucas_number2(n, 4, 1)-lucas_number2(n-1, 4, 1))/2 for n in xrange(1, 27)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 10 2009]

A bisection of sequence A002531.

Cf. A001352, A001835.

Cf. A026150.

Sequence in context: A026590 A095073 A128349 this_sequence A099393 A083588 A149759

Adjacent sequences: A001831 A001832 A001833 this_sequence A001835 A001836 A001837

nonn,easy,nice,new

N. J. A. Sloane (njas(AT)research.att.com).

More terms from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Aug 07 2000