OFFSET
1,2
COMMENTS
From Ant King, Oct 19 2011: (Start)
lim_{n->infinity} a(2n+1)/a(2n) = (1/2)*(47+21*sqrt(5)).
lim_{n->infinity} a(2n)/a(2n-1) = (1/2)*(7+3*sqrt(5)).
(End)
From Raphie Frank, Nov 30 2012: (Start)
Where L_n is a Lucas number and F_n is Fibonacci number:
lim_{n->infinity} a(2n+1)/a(2n) = (1/2)*(L_8 + F_8*sqrt(5)),
lim_{n->infinity} a(2n)/a(2n-1) = (1/2)*(L_4 + F_4*sqrt(5)),
a(n) = L_1*a(n-1) + L_12*a(n-2) - L_12*a(n-3)- L_1*a(n-4) + L_1*a(n-5).
(End)
Values of n such that 2*n-1 and 10*n-1 are both perfect squares. - Colin Barker, Dec 03 2016
LINKS
Colin Barker, Table of n, a(n) for n = 1..798
Eric Weisstein's World of Mathematics, Heptagonal Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (1,322,-322,-1,1).
FORMULA
For n odd, a(n+2) = 322*a(n+1) - a(n) - 96; for n even, a(n+1) = 161*a(n) - 48 + 36*sqrt(20*a(n)^2 - 12*a(n)+1). - Richard Choulet, Sep 29 2007, Oct 09 2007
From Ant King, Oct 19 2011: (Start)
a(n) = 322*a(n-2) - a(n-4) - 96.
a(n) = a(n-1) + 322*a(n-2) - 322*a(n-3) - a(n-4) + a(n-5).
a(n) = (1/20)*((sqrt(5)-(-1)^n)*(sqrt(5)+2)^(2n-1) + (sqrt(5)+(-1)^n)*(sqrt(5)-2)^(2n-1)+6).
a(n) = ceiling((1/20)*(sqrt(5)-(-1)^n)*(2+sqrt(5))^(2n-1)).
G.f.: x*(1 + 4*x - 106*x^2 + 4*x^3 + x^4)/((1-x)*(1-18*x+x^2)*(1+18*x+x^2)).
(End)
MATHEMATICA
LinearRecurrence[{1, 322, -322, -1, 1}, {1, 5, 221, 1513, 71065}, 17] (* Ant King, Oct 19 2011 *)
Select[Range@240000000, IntegerQ@Sqrt[2 # - 1] && IntegerQ@Sqrt[10 # - 1] &] (* Vincenzo Librandi, Dec 04 2016 *)
PROG
(PARI) Vec(-x*(x^4+4*x^3-106*x^2+4*x+1)/((x-1)*(x^2-18*x+1)*(x^2+18*x+1)) + O(x^50)) \\ Colin Barker, Jun 23 2015
(Magma) [n: n in [1..2*10^8] |IsSquare(2*n-1) and IsSquare(10*n-1)]; // Vincenzo Librandi, Dec 04 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved