OFFSET
1,1
COMMENTS
Let c(m) be the m-th composite and p(n) be the n-th prime. The prime-composite array, B, is defined such that each element B(m,n) is the highest power of p(n) that is contained within c(m). Diagonals can also be specified, where the m-th diagonal consists of the infinite number of elements B(m,1), B(m+1,2), B(m+2,3), ... The m-th antidiagonal of the array consists of the m elements B(m,1), B(m-1,2), B(m-2,3), ..., B(1,m).
The Third Borve Conjecture states that there are infinitely many integers m for which the m-th diagonal and m-th antidiagonal are both zero-only.
The prime-composite array begins:
1 2 3 4 5 6 7 8 (n)
(2) (3) (5) (7) (11) (13) (17) (19) (p_n)
1 (4) 2 0 0 0 0 0 0 0 ...
2 (6) 1 1 0 0 0 0 0 0 ...
3 (8) 3 0 0 0 0 0 0 0 ...
4 (9) 0 2 0 0 0 0 0 0 ...
5 (10) 1 0 1 0 0 0 0 0 ...
6 (12) 2 1 0 0 0 0 0 0 ...
7 (14) 1 0 0 1 0 0 0 0 ...
8 (15) 0 1 1 0 0 0 0 0 ...
9 (16) 4 0 0 0 0 0 0 0 ...
LINKS
EXAMPLE
Thus each composite has its own row, consisting of the indices of its prime factors. For example, the 10th composite is 18 and 18 = 2^1 * 3^2 * 5^0 * 7^0 * 11^0 * ..., so the 10th row reads: 1, 2, 0, 0, 0, .... Similarly, B(6,2) = 1 because c(6) = 12, p(2) = 3 and the highest power of 3 contained within 12 is 3^1 = 3. And B(34,3) = 2 because c(34) = 50, p(3) = 5 and the highest power of 5 contained within 50 is 5^2 = 25.
MATHEMATICA
Composite[n_Integer] := FixedPoint[n + PrimePi[ # ] + 1 &, n + PrimePi[n] + 1]; m = 750; a = Table[0, {m}, {m}]; Do[b = Transpose[ FactorInteger[ Composite[n]]]; a[[n, PrimePi[First[b]]]] = Last[b], {n, 1, m}]; Do[ If[ Union[ Join[ Table[a[[n - i + 1, i]], {i, 1, n}], Table[a[[n + i - 1, i]], {i, 1, m - n + 1}]]] == {0}, Print[n]], {n, 1, m}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert G. Wilson v, Feb 04 2002
STATUS
approved