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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.
4

%I #12 Sep 03 2024 01:38:31

%S 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,

%T 15,22,23,29,34,39,42,11,13,14,16,17,19,25,33,37,41,49,103,107,125,13,

%U 14,16,17,19,20,23,27,28,32,37,40,46,53,82,83,15,18,21,26

%N 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.

%C 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.

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

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

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

%e 4: {5, 7}

%e 5: {7, 8}

%e 6: {9, 13, 17, 27}

%e 7: {8, 10, 11, 13, 16, 19}

%e 8: {10, 11, 13, 14, 19, 21, 25, 31, 59, 61}

%e 9: {12, 15, 22, 23, 29, 34, 39, 42}

%e 10: {11, 13, 14, 16, 17, 19, 25, 33, 37, 41, 49, 103, 107, 125}

%e 11: {13, 14, 16, 17, 19, 20, 23, 27, 28, 32, 37, 40, 46, 53, 82, 83}

%e 12: {15, 18, 21, 26, 29, 31, 34, 41, 43, 51, 54, 57, 61, 71, 159, 165, 209, 211}

%e ...

%e sqrt((2^2-1)*1^2 + 1*(1+1)^2) = sqrt(7) -> not an integer so not included.

%e sqrt((4^2-1)*1^2 + 1*(1+1)^2) = sqrt(19) -> not an integer so not included.

%e sqrt((4^2-3)*1^2 + 3*(1+1)^2) = 5 -> T(4,1).

%e sqrt((4^2-11)*1^2 + 11*(1+1)^2) = 7 -> T(4,2).

%e sqrt((5^2-8)*1^2 + 8*(1+1)^2) = 7 -> T(5,1).

%e sqrt((6^2-5)*2^2 + 5*(2+1)^2) = 13 -> T(6,2).

%o (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)

%o for(n=4, 12, print(n, " ", row(n)))

%Y Cf. A080782, A374602.

%K nonn,tabf,easy

%O 4,1

%A _Charles L. Hohn_, Aug 12 2024