A249369 Rectangular array by descending antidiagonals: t(n,k) is the position of prime(n+1)*k^2 when the numbers prime(j+1)*h^2 are jointly ordered, for j>=1 and h>=1.

1, 5, 2, 11, 9, 3, 21, 19, 12, 4, 31, 34, 26, 18, 6, 43, 50, 45, 39, 22, 7, 55, 70, 67, 68, 48, 28, 8, 74, 91, 93, 101, 79, 59, 32, 10, 89, 116, 122, 138, 117, 100, 64, 37, 13, 109, 142, 156, 181, 164, 148, 110, 78, 47, 14, 128, 172, 189, 233, 211, 205, 165

Offset: 1

Author

Clark Kimberling, Oct 26 2014

Keywords

• nonn
• tabl
• easy

Comments

Equivalently, let S be the set of positive integer multiples of the square roots of the odd primes. Then t(n,k) is the position of k*sqrt(prime(n+1)) in the ordered union of S.

Every positive integer occurs exactly once in the array {t(n,k)}.

Examples

			Northwest corner:
1   5    11   21   31    43    55
2   9    19   34   50    70    91
3   12   26   45   67    93    122
4   18   39   68   101   138   181
6   22   48   79   117   164   211
The numbers 3*k^2 are (3,12,27,48,75,...);
the numbers 5*k^2 are (5,20,45,80,125,...);
the numbers 7*k^2 are (7,28,63,112,175,...);
the joint ranking of all such numbers is (3,5,7,11,12,13,...) = A249370,
in which numbers of the form 3*k^2 occupy positions 1,5,11,21,... which is row 1 of the present array.  Similarly, the numbers 5*k^2 occupy positions 2,9,19,34,...
		

Crossrefs

Cf. A249368, A249370.

Programs

Mathematica

z = 20000; e[h_] := e[h] = Select[Range[2000], Prime[h]*(#^2) < z &];
t = Table[Prime[n]*e[n]^2, {n, 2, 2000}]; s = Sort[Flatten[t]];
u[n_, k_] := Position[s, Prime[n]*k^2];
TableForm[Table[u[n, k], {n, 2, 15}, {k, 1, 15}]]   (* A249369 array *)
Table[u[k, n - k + 1], {n, 15}, {k, 1, n}] // Flatten  (* A249369 sequence *)