A224777 Triangle with integer geometric mean sqrt(n*m) for 1 <= m <= n, and 0 if sqrt(n*m) is not integer.
1, 0, 2, 0, 0, 3, 2, 0, 0, 4, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 7, 0, 4, 0, 0, 0, 0, 0, 8, 3, 0, 0, 6, 0, 0, 0, 0, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 0, 6, 0, 0, 0, 0, 0, 0, 0, 0, 12, 0, 0
Offset: 1
Examples
The triangle begins: n\m 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 ... 1: 1 2: 0 2 3: 0 0 3 4: 2 0 0 4 5: 0 0 0 0 5 6: 0 0 0 0 0 6 7: 0 0 0 0 0 0 7 8: 0 4 0 0 0 0 0 8 9: 3 0 0 6 0 0 0 0 9 10: 0 0 0 0 0 0 0 0 0 10 11: 0 0 0 0 0 0 0 0 0 0 11 12: 0 0 6 0 0 0 0 0 0 0 0 12 13: 0 0 0 0 0 0 0 0 0 0 0 0 13 14: 0 0 0 0 0 0 0 0 0 0 0 0 0 14 15: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 15 16: 4 0 0 8 0 0 0 0 12 0 0 0 0 0 0 16 ... a(8,2) = sqrt(16) = 4, a(8,8) = sqrt(64) = 8, h^2 == 0 (mod 8) has A000188(8) = 2 solutions from 1 <= h <= 8, namely h = 4 and h = 8.
Links
- Wolfdieter Lang, Rows n = 1..100 of triangle, flattened
Formula
a(n,m) = sqrt(n*m) > 0 if this is an integer and otherwise 0, for 1 <= m <= n. Due to commutativity this restriction is sufficient.
Comments