A075300 Array A read by antidiagonals upwards: A(n, k) = array A054582(n,k) - 1 = 2^n*(2*k+1) - 1 with n,k >= 0.
0, 1, 2, 3, 5, 4, 7, 11, 9, 6, 15, 23, 19, 13, 8, 31, 47, 39, 27, 17, 10, 63, 95, 79, 55, 35, 21, 12, 127, 191, 159, 111, 71, 43, 25, 14, 255, 383, 319, 223, 143, 87, 51, 29, 16, 511, 767, 639, 447, 287, 175, 103, 59, 33, 18, 1023, 1535, 1279, 895, 575, 351, 207, 119
Offset: 0
Examples
The array A begins: 0 2 4 6 8 10 12 14 16 18 ... 1 5 9 13 17 21 25 29 33 37 ... 3 11 19 27 35 43 51 59 67 75 ... 7 23 39 55 71 87 103 119 135 151 ... 15 47 79 111 143 175 207 239 271 303 ... 31 95 159 223 287 351 415 479 543 607 ... ... - _Philippe Deléham_, Feb 19 2014 From _Wolfdieter Lang_, Jan 31 2019: (Start) The triangle T begins: n\k 0 1 2 3 4 5 6 7 8 9 10 ... 0: 0 1: 1 2 2: 3 5 4 3: 7 11 9 6 4: 15 23 19 13 8 5 31 47 39 27 17 10 6: 63 95 79 55 35 21 12 7: 127 191 159 111 71 43 25 14 8: 255 383 319 223 143 87 51 29 16 9: 511 767 639 447 287 175 103 59 33 18 10: 1023 1535 1279 895 575 351 207 119 67 37 20 ... T(3, 1) = 2^2*(2*1+1) - 1 = 12 - 1 = 11. (End)
Crossrefs
Programs
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Maple
A075300bi := (x,y) -> (2^x * (2*y + 1))-1; A075300 := n -> A075300bi(A025581(n), A002262(n)); A002262 := n -> n - binomial(floor((1/2)+sqrt(2*(1+n))),2); A025581 := n -> binomial(1+floor((1/2)+sqrt(2*(1+n))),2) - (n+1);
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Mathematica
Table[(2^# (2 k + 1)) - 1 &[m - k], {m, 0, 10}, {k, 0, m}] (* Michael De Vlieger, Jun 05 2016 *)
Formula
From Wolfdieter Lang, Jan 31 2019: (Start)
Array A(n, k) = 2^n*(2*k+1) - 1, for n >= 0 and m >= 0.
The triangle is T(n, k) = A(n-k, k) = 2^(n-k)*(2*k+1) - 1, n >= 0, k=0..n.
See also A054582 after subtracting 1. (End)
From Ruud H.G. van Tol, Mar 17 2025: (Start)
A(0, k) is even. For n > 0, A(n, k) is odd and (3 * A(n, k) + 1) / 2 = A(n-1, 3*k+1).
A(n, k) = 2^n - 1 (mod 2^(n+1)) (equivalent to the comment about trailing 1-bits). (End)
Comments