0,3

Indices of square numbers which are also generalized pentagonal numbers.

If t(n) denotes the n-th triangular number, t(A105038(n))=a(n)*a(n+1). - Robert Phillips (bobanne(AT)bellsouth.net), May 25 2008

The n:th term is a(n)=((5+Sqrt(24))^n-(5-Sqrt(24))^n)/(2*Sqrt(24)) [From Sture Sjoestedt (sture.sjostedt(AT)spray.se), May 31 2009]

Number of units of a(n) belongs to a periodic sequence: 0, 1, 0, 9.We conclude that a(n) and a(n+4) have the same number of units. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 05 2009]

A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case a=0,b=1; p=10, q=-1.

E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pps. 231-242.

W. Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38,5 (2000) 408-419; Eq.(44), lhs, m=12.

Index entries for sequences related to linear recurrences with constant coefficients

Tanya Khovanova, Recursive Sequences

Index entries for sequences related to Chebyshev polynomials.

a(n) = S(2*n-1, sqrt(12))/sqrt(12) = S(n-1, 10); S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(-1, x) := 0.

a(n)={[(5+2*sqrt(6))^n - (5-2*sqrt(6))^n]}/4*sqrt(6). G.f.(x)=x/(1-10*x+x^2). - Barry E. Williams, Aug 18 2000

G.f.: x/(1-10*x+x^2). a(-n)=-a(n). - Michael Somos Sep 05 2006

a(n) = 9*(a(n-1)+a(n-2))-a(n-3), a(n) = 11*(a(n-1)-a(n-2))+a(n-3). a(n)=10*a(n-1)-a(n-2). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), May 26 2007

a(2)=10 and (3(-8)^2-(-8))/2=10^2, a(3)=99 and (3(81)^2-(81))/2=99^2. - Michael Somos Sep 05 2006

lst={}; Do[AppendTo[lst, GegenbauerC[n, 1, 5]], {n, 0, 8^2}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Sep 11 2008]

(PARI) {a(n)=subst(poltchebi(n+1)-5*poltchebi(n), 'x, 5)/24} /* Michael Somos Sep 05 2006 */

sage: [lucas_number1(n, 10, 1) for n in range(22)] - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jun 25 2008

Cf. A000027, A001906, A001353, A004254, A001109, A004187, A001090, A018913.

A001079(n) = sqrt{24*[a(n)^2]+1}, that is a(n) = sqrt((A001079(n)^2-1)/24).

A046173(n)=a(2n-1).

Sequence in context: A135927 A129542 A081109 this_sequence A105694 A000456 A138365

Adjacent sequences: A004186 A004187 A004188 this_sequence A004190 A004191 A004192

easy,nonn

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

More terms from James A. Sellers (sellersj(AT)math.psu.edu), Sep 07 2000