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A152578 Numbers of the form 5^(2^n) + 2. 1
7, 27, 627, 390627, 152587890627, 23283064365386962890627, 542101086242752217003726400434970855712890627 (list; graph; refs; listen; history; text; internal format)
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) g(a, n) = if(a%2, b=2, b=1); for(x=0, n, y=a^(2^x)+b; print1(y", "))

CROSSREFS

Sequence in context: A033910 A196647 A274579 * A300529 A299468 A200974

Adjacent sequences:  A152575 A152576 A152577 * A152579 A152580 A152581

KEYWORD

nonn

AUTHOR

Cino Hilliard, Dec 08 2008

STATUS

approved

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Last modified January 29 14:07 EST 2020. Contains 331338 sequences. (Running on oeis4.)