A375336 For n>=4, irregular triangular array of successive integer solutions to sqrt((d-c)*b^2 + c*(b+1)^2) for square integers d = n^2, where b and c are positive integers and c < d, read by rows.

5, 7, 7, 8, 9, 13, 17, 27, 8, 10, 11, 13, 16, 19, 10, 11, 13, 14, 19, 21, 25, 31, 59, 61, 12, 15, 22, 23, 29, 34, 39, 42, 11, 13, 14, 16, 17, 19, 25, 33, 37, 41, 49, 103, 107, 125, 13, 14, 16, 17, 19, 20, 23, 27, 28, 32, 37, 40, 46, 53, 82, 83, 15, 18, 21, 26

Offset: 4

Author

Charles L. Hohn, Aug 12 2024

Keywords

• nonn
• tabf
• easy

Comments

Provable that every row n has a finite number of terms, with n < 4 producing no solutions, and T(n, k) never exceeding (n/2)^3.

This sequence excludes cases where c == 0, where all b produce integer solutions d*b.

Nonsquare d produce solutions following a different pattern, shown as A374602.

Examples

			4: {5, 7}
5: {7, 8}
6: {9, 13, 17, 27}
7: {8, 10, 11, 13, 16, 19}
8: {10, 11, 13, 14, 19, 21, 25, 31, 59, 61}
9: {12, 15, 22, 23, 29, 34, 39, 42}
10: {11, 13, 14, 16, 17, 19, 25, 33, 37, 41, 49, 103, 107, 125}
11: {13, 14, 16, 17, 19, 20, 23, 27, 28, 32, 37, 40, 46, 53, 82, 83}
12: {15, 18, 21, 26, 29, 31, 34, 41, 43, 51, 54, 57, 61, 71, 159, 165, 209, 211}
...
sqrt((2^2-1)*1^2 + 1*(1+1)^2) = sqrt(7) -> not an integer so not included.
sqrt((4^2-1)*1^2 + 1*(1+1)^2) = sqrt(19) -> not an integer so not included.
sqrt((4^2-3)*1^2 + 3*(1+1)^2) = 5 -> T(4,1).
sqrt((4^2-11)*1^2 + 11*(1+1)^2) = 7 -> T(4,2).
sqrt((5^2-8)*1^2 + 8*(1+1)^2) = 7 -> T(5,1).
sqrt((6^2-5)*2^2 + 5*(2+1)^2) = 13 -> T(6,2).
		

Crossrefs

Cf. A080782, A374602.

Programs

PARI

row(n)=my(d=n^2, t=n, v=List()); while(t
				

Formula

T(n, 1) = A080782(n+2).