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A227244 Smallest prime factor of (2^(2^n) + 1)*2^(2^(2^n)) - 1. 0
11, 79, 1114111, 29758566933990262223857743147232792318290386059069624958140599090033674317463551, 16267, 11, 563, 139, 47, 11, 131107211, 47, 163430017, 11, 563 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

From Bernard Schott, Oct 23 2019: (Start)

G_n = (2^(2^n) + 1) * 2^(2^(2^n)) - 1 = F_n * 2^(F_n -1) - 1 with F_n is the n-th Fermat number.

For the first four terms, a(n) = F_n  * 2^(F_n -1) - 1 because the numbers G_n are then primes. It is in 1968 that Williams and Zarnke showed that G_3 = a(3) with 80 digits was a prime number. [Ribenboim]

Proof that the 1st conjecture of Chai Wah Wu is right.

G_n = F_n * 2^(F_n -1) - 1.

1) a(n) <> 2

G_n is clearly odd, so 2 does not divide G_n and a(n) > 2.

2) a(n) <> 3

For n >= 2, F_n == 5 (mod 6) [Proof by induction with F_(n+1) = (F_n - 1)^2 + 1], so F_n - 1 == 4 (mod 6). Yet, if q even, 2^q  == 4 (mod 6), so G_n == 5 * 4 -1 == 19 == 1 (mod 6) ==> G_n == 1 (mod 3). G_n is not divisible by 3 ==> a(n) > 3.

3) a(n) <> 5

For n >= 2, the final two digits of F_n are periodically repeated with period 4: {17, 57, 37, 97}, hence F_n = 7 (mod 10) and F_n - 1 ends with {16, 56, 36, 96}.

Fn - 1 == 0 (mod 4) but when q == 0 (mod 4) , 2^q == 6 (mod 10).

G_n == 7 * 6 - 1 = 41 = 1 (mod 10) and G_n is not divisible by 5 ==> a(n) > 5.

4) a(n) <> 7

If n is even, F_n == 3 (mod 7) and if n is odd, F_n = 5 (mod 7). We have 2^q == 1, 2 or 4 (mod 7).

If n even, G_n == 3*1 - 1 == 2 (mod 7) or G_n == 3*2 - 1 === 5 (mod 7) or G(n) = 3*4 - 1 == 6 (mod 7), and,

if n odd, G_n == 5*1 - 1 == 4 (mod 7) or G_n == 5*2 - 1 === 9 == 2 (mod 7) or G(n) = 5*4 - 1 == 19 == 5 (mod 7).

G_n is not divisible by 7 ==> a(n) > 7.

5) a(0) = a(5) = 11 and the conjecture is proved.

(End)

REFERENCES

Paulo Ribenboim, The New Book of Prime Number Records, 3rd edition, Springer-Verlag, New York, 1995, p. 89.

LINKS

Table of n, a(n) for n=0..14.

FORMULA

From Chai Wah Wu, Oct 21 2019: (Start)

a(n) >= 11 for n >= 0 (conjectured).

a(4n+5) = 11 for n >= 0 (conjectured).

(End)

PROG

(PARI) a(n) = factor((2^(2^n) + 1)*2^(2^(2^n)) - 1)[1, 1]; \\ Michel Marcus, Oct 22 2013

(PARI) a(n)=my(N=2^n); forprime(p=3, , if((Mod(2, p)^N+1)*Mod(2, p)^lift(Mod(2, p-1)^N)==1, return(p))) \\ Charles R Greathouse IV, Oct 22 2013

CROSSREFS

Cf. A000215.

Sequence in context: A155619 A126506 A159663 * A026897 A021024 A127021

Adjacent sequences:  A227241 A227242 A227243 * A227245 A227246 A227247

KEYWORD

nonn,hard

AUTHOR

Arkadiusz Wesolowski, Oct 20 2013

EXTENSIONS

a(12)-a(14) from Charles R Greathouse IV, Oct 22 2013

STATUS

approved

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Last modified November 16 19:17 EST 2019. Contains 329201 sequences. (Running on oeis4.)