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Search: id:A087986
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| A087986 |
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a(n)=p[x] is the smallest prime such that 1+[2^n].p[x] is divisible by next-prime p[x+1]. |
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2
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| 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; listen)
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OFFSET
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1,2
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FORMULA
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a(n)=Min{p[x]; Mod[1+[2^n].p[x], p[x+1]]=0}; a(n)=A087985[2^n]
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EXAMPLE
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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.
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MATHEMATICA
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{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}]
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CROSSREFS
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Cf. A087985-A087990.
Sequence in context: A056927 A094290 A101876 this_sequence A129088 A086418 A100761
Adjacent sequences: A087983 A087984 A087985 this_sequence A087987 A087988 A087989
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KEYWORD
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nonn
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AUTHOR
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Labos E. (labos(AT)ana.sote.hu), Oct 06 2003
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