OFFSET
1,1
COMMENTS
For any m >= 0, if F(m) = 2^(2^m) + 1 has a factor of the form b = a(n)*2^k + 1 with k >= m + 2 and n >= 1, then the integer F(m)/b is congruent to 13 or 19 mod 30.
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).
FORMULA
a(n) = a(n-1) + a(n-4) - a(n-5), n >= 6.
a(n) = a(n-4) + 30.
G.f.: x*(7 + 4*x + 2*x^2 + 16*x^3 + x^4)/((1 + x)*(1 + x^2)*(1 - x)^2).
a(n) = (-15 + 5*(-1)^n + (3+9*i)*(-i)^n + (3-9*i)*i^n + 30*n) / 4 where i=sqrt(-1). - Colin Barker, Dec 19 2017
E.g.f.: (5*(e^(-x) + (6*x - 3)*e^x) + 6*cos(x) + 18*sin(x))/4. - Iain Fox, Dec 19 2017
MATHEMATICA
LinearRecurrence[{1, 0, 0, 1, -1}, {7, 11, 13, 29, 37}, 60]
PROG
(Magma) [n: n in [0..403] | n mod 30 in {7, 11, 13, 29}];
(PARI) Vec(x*(7 + 4*x + 2*x^2 + 16*x^3 + x^4)/((1 + x)*(1 + x^2)*(1 - x)^2 + O(x^55)))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Arkadiusz Wesolowski, Dec 19 2017
STATUS
approved