

A061357


Number of 0<k<n such that nk and n+k are both primes.


21



0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 3, 2, 3, 4, 1, 3, 4, 3, 3, 5, 4, 3, 5, 3, 3, 6, 2, 5, 6, 2, 5, 6, 4, 5, 7, 4, 4, 8, 4, 4, 9, 4, 4, 7, 3, 6, 8, 5, 5, 8, 6, 7, 10, 6, 5, 12, 3, 5, 10, 3, 7, 9, 5, 5, 8, 7, 7, 11, 5, 5, 12, 4, 8, 11, 4, 8, 10, 5, 5, 13, 9, 6, 11, 7, 6, 14, 6, 8, 13, 5, 8, 11, 6, 9
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OFFSET

1,8


COMMENTS

Number of prime pairs (p,q) with p < n < q and qn = np.
The same as the number of ways n can be expressed as the mean of two distinct primes.
1) For each integer N>=1 there exists a positive integer m(N) such that for n>=m(N) a(n)>a(N). (After the first m(N)1 terms, a(N) does not reappear). In particular, for N=1 (or 2 or 3), m(N)=4 and a(N)=0, giving Benoit Cloitre's conjecture. (cont.)
(cont.) Conjectures based upon observing a(1),...,a(10000):
m(4)=m(5)=m(6)=m(7)=m(19)=20 for a(4)=a(5)=a(6)=a(7)=a(19)=1,
m(8)=...(7 others)...=m(34)=35 for a(8)=...(7 others)...=a(34)=2,
m(12)=...(10 others)...=m(64)=65 for a(12)=...(10 others)...=a(64)=3,
m(18)=...(10 others)...=m(79)=80 for a(18)=...(10 others)...=a(79)=4,
m(24)=...(14 others)...=m(94)=95 for a(24)=...(14 others)...=a(94)=5,
m(30)=...(17 others)...=m(199)=200 for a(30)=...(17 others)...=a(199)=6, etc.
2) Each nonnegative integer appears at least once in the current sequence.
3) Stronger than 2): A001477 (nonnegative integers) is a subsequence of the current sequence. (Supporting evidence: I've observed that 0,1,2,...,175 is a subsequence of a(1),...,a(10000)).
(End)
a(n) is also the number of k such that 2*k+1=p and 2*(nk1)+1=q are both odd primes with p < q with p*q = n^2  m^2. [Pierre CAMI, Sep 01 2008]
Also: Number of ways n^2 can be written as b^2+pq where 0<b<n1 and p,q are primes.  Erin Noel and George Panos (erin.m.noel(AT)rice.edu), Jun 27 2006
a(n) is also the number of partitions of 2*n into two distinct primes. See the first formula by T. D. Noe, and the Alois P. Heinz, Nov 14 2012, crossreference.  Wolfdieter Lang, May 13 2016


LINKS



FORMULA



EXAMPLE

a(10)= 2: there are two such pairs (3,17) and (7,13), as 10 = (3+17)/2 = (7+13)/2.


MAPLE

P:=proc(i) local a, b, c, n; print(0); print(0); print(0); for n from 4 by 1 to i do a:=0; b:=prevprime(n); while b>2 do c:=2*nb; if isprime(c) then a:=a+1; fi; b:=prevprime(b); od; print(a); od; end: P(100); # Paolo P. Lava, Dec 22 2008


MATHEMATICA

Table[Count[Range[n  1], k_ /; And[PrimeQ[n  k], PrimeQ[n + k]]], {n, 98}] (* Michael De Vlieger, May 14 2016 *)


PROG

(Haskell)
a061357 n = sum $
zipWith (\u v > a010051 u * a010051 v) [n+1..] $ reverse [1..n1]


CROSSREFS

Cf. A071681 (subsequence for prime n only).


KEYWORD

nonn,easy


AUTHOR



EXTENSIONS

More terms from Larry Reeves (larryr(AT)acm.org), May 15 2001


STATUS

approved



