A253573 Rectangular array a(n,k) read by upwards antidiagonals: row A(n) is the result of applying the function defined in A098550 to the set comprising row n of A253572, for n >= 2.

1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 9, 1, 2, 3, 4, 9, 8, 1, 2, 3, 4, 9, 8, 27, 1, 2, 3, 4, 9, 8, 15, 16, 1, 2, 3, 4, 9, 8, 15, 16, 81, 1, 2, 3, 4, 9, 8, 15, 14, 5, 32, 1, 2, 3, 4, 9, 8, 15, 14, 5, 6, 243, 1, 2, 3, 4, 9, 8, 15, 14, 5, 6, 25, 64

Offset: 2

Author

L. Edson Jeffery, Jan 03 2015

Keywords

• nonn
• tabl

Comments

Let A(n) be the n-th row of this table and B(n) the n-th row of A253572. For n >= 2, A(n) is initialized with the first three entries of B(n). Let a(n,k) be the next entry of A(n) to be found. Then a(n,k) = m = the least number in B(n) such that m is not already in A(n), gcd(a(n,k-1),m) = 1 and gcd(a(n-2),m) > 1.

A(n) is not a permutation of B(n), for any n. Proof: Let a(n,k) be the k-th entry in A(n). By the definition of B(n) (see A253572), for k>1, there are distinct primes p,q in {prime(1),...,prime(n)} such that p divides a(n,k-1) and q divides a(n,k-2). It follows that we can never have a(n,k) = m*primorial(n), for any k>3 and any m>0. QED

Conjecture 1: Successive rows tend to A098550.

Conjecture 2: The main diagonal is A098550.

Conjecture 3: For each n, the primes prime(1), ..., prime(n) in row n appear in their natural order.

Conjecture 4: a(n,A251239(j)) = A000040(j), for j = 1,...,n inclusive.

Examples

			Array A starts:
{1, 2, 3, 4, 9, 8, 27, 16, 81, 32, 243, 64, 729, 128, 2187}
{1, 2, 3, 4, 9, 8, 15, 16,  5,  6,  25, 12, 125,  18,  625}
{1, 2, 3, 4, 9, 8, 15, 14,  5,  6,  25, 12,  35,  16,    7}
{1, 2, 3, 4, 9, 8, 15, 14,  5,  6,  25, 12,  35,  16,    7}
{1, 2, 3, 4, 9, 8, 15, 14,  5,  6,  25, 12,  35,  16,    7}
{1, 2, 3, 4, 9, 8, 15, 14,  5,  6,  25, 12,  35,  16,    7}
		

Crossrefs

Cf. A253572, A098550 (and many cross-references therein).

Programs

Mathematica

r = 13; max = 300; prev = Table[2^j, {j, 0, max}]; Do[y[n] = {}; g = {-1}; next = Take[Union[Flatten[Table[Prime[n]^j*prev, {j, 0, max}]]], max]; prev = next; Do[AppendTo[y[n], next[[1]]]; next = Delete[next, 1], {3}]; While[g != {0}, a = y[n][[-1]]; b = y[n][[-2]]; g = FirstPosition[next, v_ /; GCD[a, v] == 1 && GCD[b, v] > 1, 0]; If[g != {0}, y[n] = Flatten[Append[y[n], next[[g]]]]; next = Delete[next, g]]], {n, 2, r}]; Flatten[Table[y[n - k + 1][[k]], {n, 2, r}, {k, n - 1}]] (* Array antidiagonals flattened *)