OFFSET
1,9
COMMENTS
For even n > 2, a(n) = A061358(n-2). - Reinhard Zumkeller, Aug 08 2009
Vinogradov proved that every sufficiently large odd number is the sum of three primes. - T. D. Noe, Mar 27 2013
The two Helfgott papers show that every odd number greater than 5 is the sum of three primes (this is the Odd Goldbach Conjecture). - T. D. Noe, May 14 2013, N. J. A. Sloane, May 18 2013
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 1..36000 (first 10000 terms from T. D. Noe)
H. A. Helfgott, Minor arcs for Goldbach's problem, arXiv:1205.5252 [math.NT], 2012.
H. A. Helfgott, Major arcs for Goldbach's theorem, arXiv:1305.2897 [math.NT], 2013.
H. A. Helfgott, The ternary Goldbach conjecture is true, arxiv:1312.7748 [math.NT], 2013.
H. A. Helfgott, The ternary Goldbach problem, arXiv:1404.2224 [math.NT], 2014.
Yannick Saouter, Checking the odd Goldbach conjecture up to 10^20, Math. Comp. 67 (222) (1998) 863-866.
Eric Weisstein's World of Mathematics, Vinogradov's Theorem
Wikipedia, Goldbach's conjecture.
FORMULA
a(n) = Sum_{k=1..floor(n/3)} Sum_{i=k..floor((n-k)/2)} A010051(i) * A010051(k) * A010051(n-i-k). - Wesley Ivan Hurt, Mar 26 2019
a(n) = [x^n y^3] Product_{k>=1} 1/(1 - y*x^prime(k)). - Ilya Gutkovskiy, Apr 18 2019
EXAMPLE
a(6) = 1 because 6 = 2+2+2,
a(9) = 2 because 9 = 2+2+5 = 3+3+3,
a(15) = 3 because 15 = 2+2+11 = 3+5+7 = 5+5+5,
a(17) = 4 because 17 = 2+2+13 = 3+3+11 = 3+7+7 = 5+5+7.
- Zak Seidov, Jun 29 2017
MATHEMATICA
f[n_] := Block[{c = 0, lmt = PrimePi@ Floor[n/2], p, q}, Do[p = Prime@ i; q = Prime@ j; r = n - p - q; If[ PrimeQ@ r && r >= p, c++ ], {i, lmt}, {j, i}]; c]; Array[f, 91] (* Robert G. Wilson v, Apr 13 2008 *)
Table[Count[IntegerPartitions[n, {3}], _?(AllTrue[#, PrimeQ]&)], {n, 50}] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Sep 10 2019 *)
PROG
(PARI) a(n)=my(s); forprime(p=(n+2)\3, n-4, forprime(q=(n-p+1)\2, min(n-p-2, p), if(isprime(n-p-q), s++))); s \\ Charles R Greathouse IV, Jun 29 2017
(Python)
from sympy import isprime, primerange, floor
def a(n):
s=0
for p in primerange(((n + 2)//3), n - 3):
for q in primerange(((n - p + 1)//2), min(n - p - 2, p) + 1):
if isprime(n - p - q): s+=1
return s
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 01 2017, after PARI code by Charles R Greathouse IV
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Naohiro Nomoto, Feb 24 2002
EXTENSIONS
More terms from Vladeta Jovovic, Mar 10 2002
STATUS
approved