OFFSET
1,1
COMMENTS
It seems that D(n) = 4*a(n) gives precisely those even discriminants D from 4*A000037 of indefinite binary quadratic forms that have only improper solutions of the Pell equation x^2 - D*y^2 = +4. Conjecture tested for n = 1..66. Alternatively, the conjecture is that this sequence gives the r values for the Pell equation X^2 + r Y^2 = +1 whenever Y is even. See A261249 and A261250. - Wolfdieter Lang, Sep 16 2015
The proof of these two versions of the conjecture is given in the W. Lang link. - Wolfdieter Lang, Sep 19 2015 (revised Oct 03 2015)
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..200
J. H. Conway, On Happy Factorizations, J. Integer Sequences, Vol. 1, 1998, #1.
Wolfdieter Lang, Proof of a Conjecture Related to the 1-Happy Numbers.
FORMULA
a(n) is in the sequence if a(n) = C*B with integers B >= 1 and C >= 2, such that C*S^2 - B*R^2 = 1 has an integer solution (without loss of generality one may take S and R positive). See the Conway link. - Wolfdieter Lang, Sep 18 2015
EXAMPLE
From Wolfdieter Lang, Sep 18 2015: (Start)
a(1) = 5 = 5*1 and 5*1^2 - 1*2^2 = 1.
a(7) = 14 = 2*7 and 2*2^2 - 7*1^2 = 1. (End)
MATHEMATICA
r[b_, c_] := (red = Reduce[x>0 && y>0 && b*x^2 + 1 == c*y^2, {x, y}, Integers] /. C[1] -> 1 // Simplify; If[Head[red] === Or, First[red], red]); f[128] = {}(* to speed up *); f[n_] := f[n] = If[IntegerQ[Sqrt[n]], {}, Do[c = n/b; If[(r0 = r[b, c]) =!= False, {x0, y0} = {x, y} /. ToRules[r0]; Return[{b, c, x0, y0}]], {b, Divisors[n] // Most}]]; A007969 = Reap[Table[Print[n, " ", f[n]]; If[f[n] != {} && f[n] =!= Null, Sow[n]], {n, 1, 130}]][[2, 1]] (* Jean-François Alcover, Jun 26 2012, updated Sep 18 2015 *)
PROG
(Haskell)
a007969 n = a007969_list !! (n-1)
a007969_list = filter ((== 1) . a007968) [0..]
-- Reinhard Zumkeller, Oct 11 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved