A061357 Number of 0

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

Offset: 1

Author

Amarnath Murthy, Apr 28 2001

Keywords

• nonn
• easy

Comments

Number of prime pairs (p,q) with p < n < q and q-n = n-p.

The same as the number of ways n can be expressed as the mean of two distinct primes.

Conjecture: for n>=4 a(n)>0. - Benoit Cloitre, Apr 29 2003

Conjectures from Rick L. Shepherd, Jun 24 2003: (Start)

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*(n-k-1)+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

a(n) = sum (A010051(2*n - p): p prime < n). [Reinhard Zumkeller, Oct 19 2011]

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

All 0Jamie Morken, Jun 02 2017

a(n) is the number of appearances of n in A143836. - Ya-Ping Lu, Mar 05 2023

Examples

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

Links

• Pierre CAMI, Table of n, a(n) for n = 1..60000

Crossrefs

Cf. A071681 (subsequence for prime n only).

Cf. A092953.

Bisection of A117929 (even part). - Alois P. Heinz, Nov 14 2012

Programs

Haskell

a061357 n = sum $
   zipWith (\u v -> a010051 u * a010051 v) [n+1..] $ reverse [1..n-1]
-- Reinhard Zumkeller, Nov 10 2012, Oct 19 2011

Mathematica

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

PARI

a(n)=my(s);forprime(p=2,n-1,s+=isprime(2*n-p));s \\ Charles R Greathouse IV, Mar 08 2013

Python

from sympy import primerange, isprime
def A061357(n): return sum(1 for p in primerange(n) if isprime((n<<1)-p)) # Chai Wah Wu, Sep 03 2024

Formula

a(n) = A045917(n) - A010051(n). - T. D. Noe, May 08 2007

a(n) = sum(A010051(n-k)*A010051(n+k): 1 <= k < n). - Reinhard Zumkeller, Nov 10 2012

a(n) = sum_{i=2..n-1} A010051(i)*A010051(2n-i). [Wesley Ivan Hurt, Aug 18 2013]

Extensions

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

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