0,3

Solution to a Diophantine equation.

n such that n^2-1 is a triangular number - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 05 2002

For all elements "x" of the sequence, 8*x^2 - 7 is a square. Lim n-> inf. a(n)/a(n-2) = 3 + Sqrt(8). If n is odd, Lim n -> inf. a(n)/a(n-1) = (9 + 2*Sqrt(8))/7. If n is even, Lim n -> inf. a(n)/a(n-1) = (11 + 3*Sqrt(8))/7. - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 07 2002

A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Jul/Aug 1973 pp. 73-74.

J. O. Shallit, personal communication.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

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.

For n (even), a(n) = [ [(3 + Sqrt(8))^((n/2)+1) - (3 - Sqrt(8))^((n/2)+1)] - 2*[(3 + Sqrt(8))^((n/2)-1) - (3 - Sqrt(8))^((n/2)-1)] ] / (6*Sqrt(8)) For n (odd), a(n) = [ [(3 + Sqrt(8))^((n+1)/2) - (3 - Sqrt(8))^((n+1)/2)] - 2*[(3 + Sqrt(8))^((n-1)/2) - (3 - Sqrt(8))^((n-1)/2)] ] / (2*Sqrt(8)) - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 07 2002

A006452:=-(z-1)*(z**2+3*z+1)/(z**2+2*z-1)/(z**2-2*z-1); [Conjectured by S. Plouffe in his 1992 dissertation. Gives sequence except for one of the leading 1's.]

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

Cf. A006451, A006454.

Sequence in context: A026531 A038047 A061152 this_sequence A104430 A103669 A034485

Adjacent sequences: A006449 A006450 A006451 this_sequence A006453 A006454 A006455

nonn,easy

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

More terms from James A. Sellers (sellersj(AT)math.psu.edu), May 03 2000