OFFSET
0,5
COMMENTS
The solution of the linear Diophantine equation 9*x - 2^n*y = 1 with smallest positive x is x=a(n), n>= 0, and the corresponding y is given by y(n) = 5^(n+3) (mod 9) = A070366(n+3) with o.g.f. (8-4*x-2*x^2+7*x^3)/((1-x+x^2)*(1-x)*(1+x)) (derived from the one given in A070366). This is the periodic sequence with period [8, 4, 2, 1, 5, 7].
LINKS
Paolo Xausa, Table of n, a(n) for n = 0..1000
Wolfdieter Lang, On Collatz' Words, Sequences and Trees, arXiv preprint arXiv:1404.2710, 2014 and J. Int. Seq. 17 (2014) # 14.11.7.
Index entries for linear recurrences with constant coefficients, signature (3,-2,-8,24,-16)
FORMULA
a(n) = (1 + 2^n*(5^(n-3)(mod 9)))/3^2, n >= 3.
O.g.f.: (1-2*x+8*x^3-8*x^4)/((1-x)*(1-4*x^2)*(1-2*x+4*x^2)) (derived from the one for y(n) given above in a comment).
a(n) = 2*(a(n-1) - 4*a(n-3) + 8*a(n-4)) - 1, n >= 4, a(0)=a(1)=a(2)=a(3) = 1 (from the y(n) recurrence given in A070366).
EXAMPLE
n = 0: 9*1 - 1*8 = 1; n = 3: 9*1 - 8*1 = 1.
a(4) = (1 + 2^4*5)/9 = 9.
MATHEMATICA
LinearRecurrence[{3, -2, -8, 24, -16}, {1, 1, 1, 1, 9}, 50] (* Paolo Xausa, Nov 05 2024 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Feb 17 2014
STATUS
approved