A040081 -id:A040081 - cp's OEIS frontend

A244609 Least prime divisor of 659*2^n-1.

2, 3, 5, 3, 13, 3, 5, 3, 73, 3, 5, 3, 7, 3, 5, 3, 13, 3, 5, 3, 977, 3, 5, 3, 7, 3, 5, 3, 13, 3, 5, 3, 31, 3, 5, 3, 7, 3, 5, 3, 13, 3, 5, 3, 73, 3, 5, 3, 7, 3, 5, 3, 13, 3, 5, 3, 13477, 3, 5, 3, 7, 3, 5, 3, 13, 3, 5, 3, 48430237, 3, 5, 3, 7, 3, 5, 3, 13

Offset: 0

Robert Israel, Jul 01 2014

nonn

a(n) = 3 if n is odd.
a(n) = 5 if n == 2 (mod 4).
From Bruno Berselli, Jul 02 2014: (Start)
a(n) = 7 if n == 0 (mod 12) for n>0.
a(n) = 13 if n == 4 (mod 12).
a(n) == 3 or 7 (mod 12) for n>1. (End)
A040081(659) = 800516, so 800516 is the first n for which a(n) = 659*2^n-1 (found by David W Linton in 2004). - Jens Kruse Andersen, Jul 02 2014

			For n=4, 659*2^4-1 = 10543 = 13 * 811 so a(4) = 13.

Robert Israel, Table of n, a(n) for n = 0..355
The Prime Pages, 659*2^800516-1

Cf. A020639, A038699, A057026.

[PrimeDivisors(659*2^n-1)[1]: n in [0..100]]; // Bruno Berselli, Jul 02 2014

f:= proc(m) local F;
   F:= map(t -> t[1],ifactors(659*2^m-1,easy)[2]);
   F:= select(type,F,integer);
   if nops(F) = 0 then
     F:= map(t -> t[1],ifactors(659*2^m-1)[2]);
     min(F);
   else min(F)
   fi
end proc;
seq(f(n), n= 0 .. 100);

A253176 Least k>=0 such that both 3n2^k+1 and 3n2^k-1 are primes, or -1 if no such k exists.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 3, 2, 0, 6, 1, 3, 0, 2, 2, 1, 1, 2, 0, 14, 5, 1, 0, 1, 2, 5, 5, 2, 1, 4, 1, 1, 0, 2, 0

Offset: 1

Eric Chen, Mar 16 2015

If n*2^k+1 and n*2^k-1 are both primes, then n must be divisible by 3.
a(79) = -1, since 237*2^k+1 or 237*2^k-1 must divisible by 5, 7, 13, 17, or 241. Similarly, a(269) = -1 (cover: {5, 7, 13, 19, 37, 73}), a(1527) = -1 (cover: {5, 7, 13, 17, 241}).
Conjecture: if n < 79, then a(n) >= 0.
a(38) - a(40) = {1, 4, 1}, a(42) - a(50) = {13, 3, 4, 1, 0, 1, 3, 44, 0}, a(52) = 1, a(54) - a(56) = {4, 2, 4}, a(58) - a(60) = {1, 12, 0}, a(n) is currently unknown for n = {37, 41, 51, 53, 57, ...}
a(37), if it exists, is > 160000.

			a(11) = 6 since 33*2^n+1 and 33*2^n-1 are not both primes for all 0 <= n <= 5, but they are both primes for n = 6.

Ray Ballinger and Wilfrid Keller, Primes of the form k*2^n+1 for k up to 300
Ray Ballinger and Wilfrid Keller, Primes of the form k*2^n-1 for k up to 300
Eric Chen, Table of n, a(n) for n = 1..1000 (-1 if this term is unknown or does not exist)
Carlos Rivera, Problem 49. Sierpinski-like numbers

Cf. A040076, A040081, A194591.

Table[k = 0; While[! PrimeQ[3*n*2^k + 1] || ! PrimeQ[3*n*2^k - 1], k++]; k, {n, 60}]

a(n) = for(k=0, 2^24, if(ispseudoprime(3*n*2^k+1) && ispseudoprime(3*n*2^k-1), return(k)))

If a(n) > 0, then a(2n) = a(n) - 1.

A343914 Riesel problem in base 3: a(n) is the smallest k >= 0 such that (2n)3^k-1 is prime, or -1 if no such k exists.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 2, 1, 0, 0, 1, 1, 0, 2, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 0, 1, 1, 0, 3, 0, 0, 1, 1, 0, 3, 0, 1, 1, 0, 2, 1, 2, 0, 3, 0, 0, 1, 0, 0, 3, 0, 1, 1, 1, 2, 3, 9, 0, 1, 0, 1, 2, 0, 0, 2, 1, 6, 1, 0, 0, 1, 1, 0, 1, 3, 0, 2, 0, 1, 3, 0

Offset: 1

Felix Fröhlich, May 04 2021

sign

31532322469 (A273987(3)/2) is the smallest n such that a(n) = -1.

			For n = 11: (2*11)*3^k-1 is prime for k = 2, with 2 being the smallest such k, so a(11) = 2.

Wikipedia, Riesel number

Cf. A040081 (base 2), A273987.

a(n) = for(k=0, oo, if(ispseudoprime((2*n)*3^k-1), return(k)))

A345403 Riesel problem in base 5: a(n) is the smallest k >= 0 such that (2n)5^k - 1 is prime, or -1 if no such k exists.

Original entry on oeis.org

4, 0, 0, 0, 3, 0, 0, 1, 0, 0, 1, 0, 4, 1, 0, 0, 163, 1, 0, 1, 0, 0, 1, 0, 2, 5, 0, 2, 7, 0, 0, 5, 5, 0, 1, 0, 0, 1, 1, 0, 1, 0, 2058, 1, 0, 26, 5, 1, 0, 1, 0, 0, 3, 0, 0, 3, 0, 32, 17, 1, 2, 1, 3, 0, 3, 0, 8, 21, 0, 0, 1, 1, 4, 1, 0, 0, 1, 4, 0, 7, 1, 0, 1, 0

Offset: 1

Felix Fröhlich, Jun 18 2021

sign

a(346802/2) = a(173401) = -1 (see A273987).

			For n = 5: 10*5^k - 1 is composite for k = 0, 1, 2 and prime for k = 3. Since 3 is the smallest such k, a(5) = 3.

Joe O, Project Description, Mersenne forum.
Reggie, Welcome to the Sierpinski/Riesel Base 5 Project, PrimeGrid forum.
Wikipedia, Riesel number

Cf. A040081 (base 2), A343914 (base 3), A250205 (base 6).

Cf. A273987.

a(n) = for(k=0, oo, if(ispseudoprime((2*n)*5^k-1), return(k)))

cp's OEIS Frontend

A244609 Least prime divisor of 659*2^n-1.

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A253176 Least k>=0 such that both 3n*2^k+1 and 3n*2^k-1 are primes, or -1 if no such k exists.

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A343914 Riesel problem in base 3: a(n) is the smallest k >= 0 such that (2*n)*3^k-1 is prime, or -1 if no such k exists.

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A345403 Riesel problem in base 5: a(n) is the smallest k >= 0 such that (2*n)*5^k - 1 is prime, or -1 if no such k exists.

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A253176 Least k>=0 such that both 3n2^k+1 and 3n2^k-1 are primes, or -1 if no such k exists.

A343914 Riesel problem in base 3: a(n) is the smallest k >= 0 such that (2n)3^k-1 is prime, or -1 if no such k exists.

A345403 Riesel problem in base 5: a(n) is the smallest k >= 0 such that (2n)5^k - 1 is prime, or -1 if no such k exists.