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

4,1

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.

Links

Table of n, a(n) for n=4..69.

Formula

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

Example

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).

Prog

(PARI) row(n)=my(d=n^2, t=n, v=List()); while(t<n^3, my(b=floor(sqrt(t^2/d)), r=t^2-d*b^2); if (r && r%(b*2+1)==0, listput(v, t)); t++); concat(v)
for(n=4, 12, print(n, " ", row(n)))

Crossrefs

Cf. A080782, A374602.

Sequence in context: A001989 A086056 A154475 * A264527 A137755 A258316

Adjacent sequences: A375333 A375334 A375335 * A375337 A375338 A375339

Keyword

nonn,tabf,easy

Author

Charles L. Hohn, Aug 12 2024

Status

approved