A209492 a(0)=1; for n >= 1, let k = floor((1 + sqrt(8*n-7))/2), m = n - (k^2 - k+2)/2. Then a(n) = 2^k + 2^(m+1) - 1.

1, 3, 5, 7, 9, 11, 15, 17, 19, 23, 31, 33, 35, 39, 47, 63, 65, 67, 71, 79, 95, 127, 129, 131, 135, 143, 159, 191, 255, 257, 259, 263, 271, 287, 319, 383, 511, 513, 515, 519, 527, 543, 575, 639, 767, 1023, 1025, 1027, 1031, 1039, 1055, 1087, 1151, 1279, 1535, 2047, 2049, 2051, 2055, 2063, 2079, 2111, 2175, 2303, 2559, 3071

Offset: 0

Author

Vladimir Shevelev, Mar 09 2012

Keywords

• nonn
• easy
• tabl

Comments

The sequence is concatenation of rows of triangle which begins

i\j | 0 1 2 3 4 5 6 7 8

======+====================================================

0 | 1

1 | 3 5

2 | 7 9 11

3 | 15 17 19 23

4 | 31 33 35 39 47

5 | 63 65 67 71 79 95

6 | 127 129 131 135 143 159 191

7 | 255 257 259 263 271 287 319 383

8 | 511 513 515 519 527 543 575 639 767

Examples

			Consider n=19. Then k = floor((1 + sqrt(145))/2) = 6 and m = 19 - 16 = 3. Thus a(19) = 2^6 + 2^4 - 1 = 79.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A000225, A224195 (binary reversal).

Programs

Mathematica

k = Floor[(1 + Sqrt[8*n - 7])/2]; m = n - (k^2 - k + 2)/2; a[n_] = If[n == 0, 1, 2^k + 2^(m + 1) - 1]; Table[a[n], {n, 0, 100}]

Formula

For i=0,1,..., the i-th row is 2^(i+1)-1, if j=0, and 2^(i+1)+2^j-1, if j=1,...,i.