A166994 Triangle, read by rows, where T(n,k) = T(n,k-1)^2 - T(k-1,k-1)^2 for n>=k>1, with T(n,1) = n for n>=1.
1, 2, 3, 3, 8, 55, 4, 15, 216, 43631, 5, 24, 567, 318464, 99515655135, 6, 35, 1216, 1475631, 2175583184000, 4723258824886629604131775, 7, 48, 2295, 5264000, 27707792335839, 767711852760361479511965696
Offset: 1
Examples
Triangle begins: 1; 2, 3; 3, 8, 55; 4, 15, 216, 43631; 5, 24, 567, 318464, 99515655135; 6, 35, 1216, 1475631, 2175583184000, 4723258824886629604131775; 7, 48, 2295, 5264000, 27707792335839, 767711852760361479511965696, 589359179694820074404152604620573424809709490316113791; ... ILLUSTRATE THE RECURRENCE. For row 4, start with 4, then continue with the rule: "obtain the next term in the row by squaring the current term and subtracting the square of the first term in the current column": 4^2 - 1^2 = 15; 15^2 - 3^2 = 216; 216^2 - 55^2 = 43631. Likewise for row 5: 5^2 - 1^2 = 24; 24^2 - 3^2 = 567; 567^2 - 55^2 = 318464; 318464^2 - 43631^2 = 99515655135. Continuing in this way generates all rows of this triangle. ILLUSTRATE GENERATING METHOD USING NESTED RADICALS. Let a(n) = A083869(n), then row n equals the resulting integers at each stage in the successive nested radicals: sqrt(a(1)^2+sqrt(a(2)^2+sqrt(a(3)^2+(....+sqrt(a(n)^2)))...). For example, the terms in row n=3 are: 3 = sqrt(1^2 + sqrt(3^2 + sqrt(55^2))), 8 = sqrt(3^2 + sqrt(55^2)), 55 = sqrt(55^2). And the terms in row 4 are: 4 = sqrt(1^2 + sqrt(3^2 + sqrt(55^2 + sqrt(43631^2)))), 15 = sqrt(3^2 + sqrt(55^2 + sqrt(43631^2))), 216 = sqrt(55^2 + sqrt(43631^2)), 43631 = sqrt(43631^2).
Crossrefs
Cf. A083869.
Programs
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Mathematica
A[n_, 1] := n; A[n_, k_] := A[n, k - 1]^2 - A[n - 1, k - 1]^2; Flatten[Table[A[n, k], {n, 10}, {k, n}]] (* G. C. Greubel, May 30 2016 *) -
PARI
T(n,k)=if(k==1,n,T(n,k-1)^2-T(k-1,k-1)^2)
Formula
Main diagonal is A083869, which obeys an interesting recursion of nested radicals.