A228405 Pellian Array, A(n, k) with numbers m such that 2*m^2 +- 2^k is a square, and their corresponding square roots, read downward by diagonals.
0, 1, 1, 0, 1, 2, 2, 2, 3, 5, 0, 2, 4, 7, 12, 4, 4, 6, 10, 17, 29, 0, 4, 8, 14, 24, 41, 70, 8, 8, 12, 20, 34, 58, 99, 169, 0, 8, 16, 28, 48, 82, 140, 239, 408, 16, 16, 24, 40, 68, 116, 198, 338, 577, 985, 0, 16, 32, 56, 96, 164, 280, 478, 816, 1393, 2378
Offset: 0
Examples
With row # as n. and column # as k, and n, k =>0, the array begins: 0, 1, 0, 2, 0, 4, 0, 8, ... 1, 1, 2, 2, 4, 4, 8, 8, ... 2, 3, 4, 6, 8, 12, 16, 24, ... 5, 7, 10, 14, 20, 28, 40, 56, ... 12, 17, 24, 34, 48, 68, 96, 136, ... 29, 41, 58, 82, 116, 164, 232, 328, ... 70, 99, 140, 198, 280, 396, 560, 792, ... 169, 239, 338, 478, 676, 956, 1352, 1912, ... 408, 577, 816, 1154, 1632, 2308, 3264, 4616, ...
Links
- Seiichi Manyama, Antidiagonals n = 0..139, flattened
- MacTutor, D'Arcy Thompson on Greek irrationals
- D'Arcy Thompson, Excess and Defect: Or the Little More and the Little Less, Mind, New Series, Vol. 38, No. 149 (Jan., 1929), pp. 43-55 (13 pages). See page 50.
Crossrefs
Formula
If using the left column and top row to initialize: A(n,k) = A(n,k-1) + A(n-1,k-1).
If using only the top row to initialize, then each column for k = i is the binomial transform of A(0,k) restricted to k=> i as input to the transform with an appropriate down shift of index. The inverse binomial transform with a similar condition can produce each row from A000129.
If using only the first two rows to initialize then the Pell equation produces each column, as: A(n,k) = 2*A(n-1, k) + A(n-2, k).
If using only the left column (A000219(n) = Pell Numbers) to initialize then the following two equations will produce each row:
A(n,k) = sqrt(2*A(n,k-1) + (-2)^(k-1)) for even rows
A(n,k) = sqrt(2*A(n,k-1) - (-2)^(k-1)) for odd rows.
Interestingly, any portion of the array can also be filled "backwards" given the top row and any column k, using only: A(n,k-1) = A(n-1,k-1) + A(n-1, k), or if given any column and its column number by rearranging the sqrt recursions above.
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