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 even k >= m + 2 and n >= 1, then the cofactor of F(m) is equal to F(m)/b = j*2^k + 1, where j is congruent to 7 mod 10 if n is odd, or j is congruent to 3 mod 10 if n is even. That is, the integer a(n) + j must be divisible by 10.
LINKS
Wikipedia, Fermat number
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) for n > 5.
a(n) = a(n-4) + 30.
G.f.: x*(3 + 4*x + 6*x^2 + 14*x^3 + 3*x^4)/((1 + x)*(1 + x^2)*(1 - x)^2).
EXAMPLE
37 belongs to this sequence and d = 37*2^16 + 1 is a divisor of F(9) = 2^(2^9) + 1, so 10 | (37 + (F(9)/d - 1)/2^16).
MATHEMATICA
LinearRecurrence[{1, 0, 0, 1, -1}, {3, 7, 13, 27, 33}, 60]
CoefficientList[ Series[(3 + 4x + 6x^2 + 14x^3 + 3x^4)/((-1 + x)^2 (1 + x + x^2 + x^3)), {x, 0, 54}], x] (* Robert G. Wilson v, Feb 08 2018 *)
PROG
(Magma) [n: n in [0..403] | n mod 30 in {3, 7, 13, 27}];
(PARI) Vec(x*(3 + 4*x + 6*x^2 + 14*x^3 + 3*x^4)/((1 + x)*(1 + x^2)*(1 - x)^2 + O(x^55)))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Arkadiusz Wesolowski, Feb 05 2018
STATUS
approved