A120450 Number of ways to express a prime p as 2*p1 + 3*p2, where p1, p2 are primes or 1.

0, 0, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 4, 3, 3, 3, 5, 4, 4, 4, 4, 4, 4, 6, 3, 5, 4, 4, 4, 6, 5, 5, 5, 5, 7, 5, 6, 6, 6, 6, 7, 5, 6, 7, 6, 7, 9, 8, 8, 6, 7, 7, 8, 7, 9, 6, 10, 8, 6, 9, 7, 9, 8, 10, 9, 10, 9, 10, 11, 8, 7, 10, 8, 7, 10, 7, 11, 9, 8, 10, 10, 10

Offset: 1

Author

Vassilis Papadimitriou, Jul 20 2006

Keywords

• nonn

Comments

It seems that every prime p > 3 can be expressed as 2*p1 + 3*p2, where p1, p2 are primes or 1. I have tested it for the first 1500 primes (up to 12553) and it is true.

Examples

			a(11)=2 because we can write prime(11)=31 as 2*5 + 3*7 OR 2*11 + 3*3.
a(12)=3 because we can write prime(12)=37 as 2*2 + 3*11 OR 2*11 + 3*5 OR 2*17 + 3*1.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Programs

Python

from collections import Counter
from sympy import prime, primerange
def aupton(nn):
    primes, c = list(primerange(2, prime(nn)+1)), Counter()
    p2, p3 = [2] + [2*p for p in primes], [3] + [3*p for p in primes]
    for p in p2:
        if p > primes[-1]: break
        for q in p3:
            if p + q > primes[-1]: break
            c[p+q] += 1
    return [c[p] for p in primes]
print(aupton(83)) # Michael S. Branicky, Dec 21 2022

Extensions

a(59) and beyond from Michael S. Branicky, Dec 21 2022