A259301 Taken over all those prime-partitionable numbers m for which there exists a 2-partition 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.
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
Offset: 1
Keywords
Examples
The table below shows all p1a and p1b pairs for p1a <= 29 that demonstrate that m is prime-partitionable. . 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.
Links
- Christopher Hunt Gribble, Table of n, a(n) for n = 1..9592
Programs
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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 (p1a-3)/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):
Comments