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A087986 a(n)=p[x] is the smallest prime such that 1+[2^n].p[x] is divisible by next-prime p[x+1]. 1
0, 2, 3, 2, 5, 2, 3, 2, 19, 2, 3, 2, 41, 2, 3, 2, 5, 2, 3, 2, 43, 2, 3, 2, 1217, 2, 3, 2, 5, 2, 3, 2, 67, 2, 3, 2, 61673, 2, 3, 2, 5, 2, 3, 2, 31, 2, 3, 2, 29, 2, 3, 2, 5, 2, 3, 2, 2087, 2, 3, 2, 691, 2, 3, 2, 5, 2, 3, 2, 29, 2, 3, 2, 449, 2, 3, 2, 5, 2, 3, 2, 31, 2, 3, 2, 229, 2, 3, 2, 5, 2, 3, 2, 89 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

LINKS

Table of n, a(n) for n=1..93.

FORMULA

a(n)=Min{p[x]; Mod[1+[2^n].p[x], p[x+1]]=0}; a(n)=A087985[2^n]

EXAMPLE

n=1: 1+(2^1).p=mq is not possible with {p,q} consecutive prime pair;

n=2s: 1+2^[2s].p=mq is solvable with {p,q}={2,3} primes so a[2s]=2;

n=4s+3: 1+2^[4s+3].p=mq is solvable with {p,q}={3,5} primes;

n=12s+5: 1+2^[12s+5].p=mq is solvable with {p,q}={5,7}.

If n=12s+1 opr n=12s+1 or n=12s+9 then larger nontrivial solutions

exist.

Eg:

n=37:2^37=137438953472

a(37)=61673=p[6206] because 1+137438953472.61673=8476272577478657=

61681.137421127697, 61681=p[6207].

Further set of solutions are derivable with special exponents of 2.

See e.g. n=60s+9 and n=60s+49 provide mostly a[n]=29 or more rarely

a[n]=19.

MATHEMATICA

{k=0, nu=0; sq={}}; Table[Print[{n-1, Min[Prime[sq]]}]; nu=0; sq={}; Do[s=Mod[(2^n)*Prime[x]+1, Prime[x+1]]; If[Equal[s, 0], nu=nu+1; sq=Append[sq, n]], {x, 1, 10000000}], {n, 1, 257}]

CROSSREFS

Cf. A087985-A087990.

Sequence in context: A265111 A101876 A260218 * A129088 A086418 A100761

Adjacent sequences:  A087983 A087984 A087985 * A087987 A087988 A087989

KEYWORD

nonn

AUTHOR

Labos Elemer, Oct 06 2003

STATUS

approved

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Last modified January 15 20:47 EST 2019. Contains 319184 sequences. (Running on oeis4.)