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A166994
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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.
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1
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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
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OFFSET
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1,2
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LINKS
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FORMULA
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Main diagonal is A083869, which obeys an interesting recursion of nested radicals.
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EXAMPLE
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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).
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MATHEMATICA
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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 *)
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PROG
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(PARI) T(n, k)=if(k==1, n, T(n, k-1)^2-T(k-1, k-1)^2)
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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