

A045917


From Goldbach problem: number of decompositions of 2n into unordered sums of two primes.


79



0, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 3, 2, 3, 2, 4, 4, 2, 3, 4, 3, 4, 5, 4, 3, 5, 3, 4, 6, 3, 5, 6, 2, 5, 6, 5, 5, 7, 4, 5, 8, 5, 4, 9, 4, 5, 7, 3, 6, 8, 5, 6, 8, 6, 7, 10, 6, 6, 12, 4, 5, 10, 3, 7, 9, 6, 5, 8, 7, 8, 11, 6, 5, 12, 4, 8, 11, 5, 8, 10, 5, 6, 13, 9, 6, 11, 7, 7, 14, 6, 8, 13, 5, 8, 11, 7, 9
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,5


COMMENTS

Note that A002375 (which differs only at the n = 2 term) is the main entry for this sequence.
The graph of this sequence is called Goldbach's comet.  David W. Wilson, Mar 19 2012
This is the row length sequence of A182138, A184995 and A198292.  Jason Kimberley, Oct 03 2012
The Goldbach conjecture states that a(n) > 0 for n >= 2.  Wolfdieter Lang, May 14 2016
With the second Maple program, the command G(2n) yields all the unordered pairs of prime numbers having sum 2n; caveat: a pair {a,a} is listed as {a}. Example: G(26) yields {{13}, {3,23}, {7,19}}. The command G(100000) yields 810 pairs very fast.  Emeric Deutsch, Jan 03 2017
Conjecture: Let p denote any prime in any decomposition of 2n. 4 and 6 are the only numbers n such that 2n + p is prime for every p.  Ivan N. Ianakiev, Apr 06 2017
Conjecture: For all m >= 0, there exists at least one possible value of n such that a(n) = m.  Ahmad J. Masad, Jan 06 2018


REFERENCES

Calvin C. Clawson, "Mathematical Mysteries, the beauty and magic of numbers," Perseus Books, Cambridge, MA, 1996, Chapter 12, pages 236257.
H. Halberstam and H. E. Richert, 1974, "Sieve methods", Academic press, London, New York, San Francisco.


LINKS

H. J. Smith, Table of n, a(n) for n = 1..20000
M. Herkommer, Goldbach Conjecture Research
Eric Weisstein's World of Mathematics, Goldbach Partition
Wikipedia, Goldbach's conjecture
G. Xiao, WIMS server, Goldbach
Index entries for sequences related to Goldbach conjecture


FORMULA

From Halberstam and Richert: a(n) < (8+0(1))*c(n)*n/log(n)^2 where c(n) = Product_{p>2} (1  1/(p1)^2)*Product_{pn, p>2} (p1)/(p2). It is conjectured that the factor 8 can be replaced by 2.  Benoit Cloitre, May 16 2002
a(n) = ceiling(A035026(n) / 2) = (A035026(n) + A010051(n)) / 2.
a(n) = Sum_{i = 2..n} floor( 2/Omega(i * (2*ni) ) ).  Wesley Ivan Hurt, Jan 24 2013
a(n) = A224709(n) + (primepi(2n2)  primepi(n1)) + primepi(n) + 1  n.  Anthony Browne, May 03 2016
a(n) = A224708(2n)  A224708(2n+1) + A010051(n).  Anthony Browne, Jun 26 2016


MAPLE

A045917 := proc(n)
local a, i ;
a := 0 ;
for i from 1 to n do
if isprime(i) and isprime(2*ni) then
a := a+1 ;
end if;
end do:
a ;
end proc: # R. J. Mathar, Jul 01 2013
# second Maple program:
G := proc (n) local g, j: g := {}: for j from 2 to (1/2)*n do if isprime(j) and isprime(nj) then g := `union`(g, {{nj, j}}) end if end do: g end proc: seq(nops(G(2*n)), n = 1 .. 98); # Emeric Deutsch, Jan 03 2017


MATHEMATICA

f[n_] := Length[Select[2n  Prime[Range[PrimePi[n]]], PrimeQ]]; Table[ f[n], {n, 100}] (* Paul Abbott, Jan 11 2005 *)
nn = 10^2; ps = Boole[PrimeQ[Range[1, 2*nn, 2]]]; Join[{0, 1}, Table[Sum[ps[[i]] ps[[ni+1]], {i, Ceiling[n/2]}], {n, 3, nn}]] (* T. D. Noe, Apr 13 2011 *)


PROG

(PARI) a(n)=my(s); forprime(p=2, n, s+=isprime(2*np)); s \\ Charles R Greathouse IV, Mar 27 2012
(Haskell)
a045917 n = sum $ map (a010051 . (2 * n )) $ takeWhile (<= n) a000040_list
 Reinhard Zumkeller, Sep 02 2013
(Python)
from sympy import isprime
def A045917(n):
....x = 0
....for i in range(2, n+1):
........if isprime(i) and isprime(2*ni):
............x += 1
....return x # Chai Wah Wu, Feb 24 2015


CROSSREFS

Cf. A002375 (the main entry for this sequence (which differs only at the n=2 term)).
Cf. A023036 (first appearance of n), A000954 (last (assumed) appearance of n).
Cf. also A000040, A182138, A184995, A185297, A187129, A010051, A198292, A035026, A224708, A224709.
Sequence in context: A230443 A254610 A002375 * A240708 A235645 A294107
Adjacent sequences: A045914 A045915 A045916 * A045918 A045919 A045920


KEYWORD

easy,nice,nonn,look,changed


AUTHOR

Felice Russo


STATUS

approved



