

A259301


Taken over all those primepartitionable numbers m for which there exists a 2partition of the set of primes < m that has one subset containing two primes only, a(n) is the frequency with which the smaller prime occurs, where n is the prime index.


1



0, 0, 1, 1, 3, 3, 3, 2, 4, 4, 3, 4, 5, 7, 8, 5, 8, 7, 8, 9, 10, 10, 11, 12, 12, 14, 13, 13, 12, 15, 14, 14, 17, 14, 19, 17, 12, 18, 13, 19, 20, 22, 20, 23, 21, 15, 21, 21, 23, 25, 26, 23, 26, 26, 19, 23, 27, 24, 29, 27, 26, 28, 31, 29, 30, 25, 30, 29, 34, 30
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OFFSET

1,5


COMMENTS

A number n is called a prime partitionable number if there is a partition {P1,P2} of the primes less than n such that for any composition n1+n2=n, either there is a prime p in P1 such that p  n1 or there is a prime p in P2 such that p  n2.
To demonstrate that a positive integer m is primepartitionable, a suitable 2partition {P1, P2} of the set of primes < m must be found. In this sequence we are interested in primepartitionable numbers such that P1 contains 2 odd primes.
Conjecture: If P1 = {p1a, p1b} with p1a and p1b odd primes, p1a < p1b and p1b = 2*k*p1a + 1 for some positive integer k such that 2*k <= p1a  3 and if m = p1a + p1b then m is primepartitionable.


LINKS



EXAMPLE

The table below shows all p1a and p1b pairs for p1a <= 29 that demonstrate that m is primepartitionable.
. n p1a p1b 2k m
. 3 5 11 2 16
. 4 7 29 4 36
. 5 11 23 2 34
. 11 67 6 78
. 11 89 8 100
. 6 13 53 4 66
. 13 79 6 92
. 13 131 10 144
. 7 17 103 6 120
. 17 137 8 154
. 17 239 14 256
. 8 19 191 10 210
. 19 229 12 248
. 9 23 47 2 70
. 23 139 6 162
. 23 277 12 300
. 23 461 20 484
.10 29 59 2 88
. 29 233 8 262
. 29 349 12 378
. 29 523 18 552
By examining the p1a column it can be seen that
a(1) = 0, a(2) = 0, a(3) = 1, a(4) = 1, a(5) = 3, a(6) = 3,
a(7) = 3, a(8) = 2, a(9) = 4, a(10) = 4.


MAPLE

# Makes use of conjecture in COMMENTS section.
ppgen := proc (ub)
local freq_p1a, i, j, k, nprimes, p1a, p1b, pless;
# Construct set of primes < ub in pless.
pless := {};
for i from 3 to ub do
if isprime(i) then
pless := `union`(pless, {i});
end if
end do;
nprimes := numelems(pless);
# Determine frequency of each p1a.
printf("0, "); # For prime 2.
for j to nprimes do
p1a := pless[j];
freq_p1a := 0;
for k to (p1a3)/2 do
p1b := 2*k*p1a+1;
if isprime(p1b) then
freq_p1a := freq_p1a+1;
end if;
end do;
printf("%d, ", freq_p1a);
end do;
end proc:
ub := 1000:
ppgen(ub):


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



