A152578 a(n) = 5^(2^(n-1)) + 2.

7, 27, 627, 390627, 152587890627, 23283064365386962890627, 542101086242752217003726400434970855712890627

Offset

1,1

Comments

Except for the first term, these numbers are divisible by 3. This follows from the identity I: a^n-b^n = (a+b)(a^(n-1) - a^(n-2)b + ... + b^(n-1)) for odd values of n. In this example, by expanding the binomial (3+2)^(2^n)+2, we get 3h + 2^(2^n)+2 for some h. Now 2^(2^n)+2 = 2*(2^(2^n-1)+1). Since 2^n-1 is odd, by identity I, 3 divides 2^(2^n)+2 + 3h. Therefore 3 divides 5^(2^n)+2 for n > 0.

Links

Table of n, a(n) for n=1..7.

Prog

(PARI) a(n) = 5^(2^(n-1)) + 2

Crossrefs

Sequence in context: A033910 A196647 A274579 * A300529 A299468 A200974

Adjacent sequences: A152575 A152576 A152577 * A152579 A152580 A152581

Keyword

nonn,easy

Author

Cino Hilliard, Dec 08 2008

Status

approved