

A214280


Total number of primes which can be reached via Cunningham chains, starting with prime(n), not counting this starting prime.


1



7, 6, 3, 1, 2, 0, 0, 2, 1, 1, 1, 1, 2, 0, 0, 1, 0, 0, 0, 0, 0, 2, 1, 5, 1, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 4, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 0, 2, 1, 0, 0, 0, 3, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 2, 1
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OFFSET

1,1


COMMENTS

Other definition: Count of the binary descendants of the nth prime. Prime q is a binary descendant of prime p if 2*p1 <= q <= 2*p+1.
a(n) is the total count of direct binary descendants of the nth prime plus their binary descendants, and so on.
q=(p1)*2 + b*2 + 1, where b is either 0 or 1. Thus if p>2 then in base 2: q is p with a digit inserted before the least significant digit.
It is conjectured there are arbitrarily big terms (see the MathWorld link).
If p==2 (mod 3), then only 2p+1 (==2 (mod 3) again) may be prime; 2p1 will be divisible by 3. For p==1 (mod 3), only 2p1 (==1 (mod 3) again) can be prime. Therefore, there can be no alternance in the +1/1 choice (except for starting primes 2 and 3) when looking at all possible descendants. This leads to the formula by T. D. Noe.  M. F. Hasler, Jul 13 2012


LINKS

Table of n, a(n) for n=1..86.
Jens Kruse Andersen and Eric W. Weisstein, MathWorld: Cunningham Chain


FORMULA

a(n) = max(A181697(n), A181715(n))  1 for n > 2.  T. D. Noe, Jul 12 2012


EXAMPLE

prime(3)=5 has one binary descendant 11, which has one b.d. 23, which has one b.d. 47. Because 47 has no binary descendants, a(3)=3.
prime(4)=7 has one binary descendant 13, which has no binary descendants, so a(4)=1.
As explained in the comment, for n>2 this equals the maximum length of either of the Cunningham chains, i.e., iterations of x>2x+1 resp. x>2x1 before getting a composite. For prime(2)=3, the first map yields (3)>7>(15) and the second map yields (3)>5>(9), so there are 2 primes, but one has to add the a(4)+a(3)=3+1 descendants of these 2 primes, whence a(2)=2+4=6.
Starting with prime(1)=2, the "2x1" map yields 3, to which one must add its a(2)=6 descendants. They already include the prime 5 = 2*2+1 and its descendants. Thus, a(1)=1+6=7.


MATHEMATICA

des[n_] := {If[PrimeQ[2*n1], s = AppendTo[s, 2*n1]; des[2*n1]]; If[PrimeQ[2*n+1], s = AppendTo[s, 2*n+1]; des[2*n+1]]}; Table[s = {}; des[Prime[n]]; Length[Union[s]], {n, 100}] (* T. D. Noe, Jul 11 2012 *)


CROSSREFS

Cf. A000040.
Cf. A005383  primes of the form prime*21.
Cf. A005385  primes of the form prime*2+1.
Cf. A214342  count of the decimal descendants.
Cf. A181697, A181715 (two kinds of Cunningham chains).
Sequence in context: A244920 A073011 A086312 * A291081 A030797 A019908
Adjacent sequences: A214277 A214278 A214279 * A214281 A214282 A214283


KEYWORD

nonn,base,easy


AUTHOR

Alex Ratushnyak, Jul 09 2012


STATUS

approved



