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A263879
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Length k of the longest chain of primes p_1, p_2, ..., p_k such that p_1 is the n-th prime and p_{i+1} equals 2*p_i + 1 or 2*p_i - 1 for all i < k, the +/- sign depending on i.
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5
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6, 5, 4, 2, 3, 1, 1, 3, 2, 2, 2, 2, 3, 1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 6, 2, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 5, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 1, 4, 1, 1, 1
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OFFSET
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1,1
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COMMENTS
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If the +/- signs are all + or all -, then p_1, p_2, ..., p_k is a Cunningham chain of the first or second kind, respectively.
If p_1 > 3, then the +/- signs must be all + or all -, because if e = +1 or -1, then one of p, 2*p + e, 2*(2*p + e) - e is divisible by 3; see Löh (1989), p. 751.
Cunningham chains of the first and second kinds of length > 1 cannot begin with the same prime p > 3, because one of the numbers p, 2*p-1, 2*p+1 is divisible by 3.
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REFERENCES
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R. K. Guy, Unsolved Problems in Number Theory, A7.
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LINKS
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FORMULA
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a(n) < prime(n) for n > 2; see Löh (1989), p. 751.
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EXAMPLE
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2, 3, 5, 11, 23, 47 is the longest such chain of primes starting with 2. Their indices are 1, 2, 3, 5, 9, 15, respectively, so a(1) = 6, a(2) = 5, a(3) = 4, a(5) = 3, a(9) = 2, and a(15) = 1.
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MAPLE
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f:= proc(n) option remember; local x;
if n mod 3 = 1 then x:= 2*n-1 else x:= 2*n+1 fi;
if isprime(x) then 1 + procname(x) else 1 fi;
end proc:
f(2):= 6: f(3):= 5:
map(f, [seq(ithprime(i), i=1..100)]); # Robert Israel, Jul 04 2023
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MATHEMATICA
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Table[p = Prime[n]; cnt = 1;
While[PrimeQ[2*p + 1] || PrimeQ[2*p - 1],
cnt++ && If[PrimeQ[2*p + 1], p = 2*p + 1, p = 2*p - 1 ]];
cnt, {n, 3, 100}]]
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PROG
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(Python)
from sympy import prime, isprime
if n <= 2: return 7-n
p, c = prime(n), 1
while isprime(p:=(p<<1)+(-1 if p%3==1 else 1)):
c += 1
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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