A209274 Table T(n,k) = n*(n+2^k-1)/2, n, k > 0 read by antidiagonals.

1, 2, 3, 4, 5, 6, 8, 9, 9, 10, 16, 17, 15, 14, 15, 32, 33, 27, 22, 20, 21, 64, 65, 51, 38, 30, 27, 28, 128, 129, 99, 70, 50, 39, 35, 36, 256, 257, 195, 134, 90, 63, 49, 44, 45, 512, 513, 387, 262, 170, 111, 77, 60, 54, 55, 1024, 1025, 771, 518, 330, 207, 133, 92, 72, 65, 66

Offset: 1

Author

Boris Putievskiy, Jan 15 2013

Keywords

• nonn
• tabl

Comments

Column number 1 A000217 n*(n+1)/2,

column number 2 A000096 n*(n+3)/2,

column number 3 A055999 n*(n+7)/2,

column number 4 A056121 n*(n+15)/2,

column number 5 A132758 n*(n+31)/2.

Row number 1 A000079 2^k,

row number 2 A000051 2^k + 1.

Examples

			The start of the sequence as table:
  1....2...4...8...16...32...64...
  3....5...9..17...33...65..129...
  6....9..15..27...51...99..195...
  10..14..22..38...70..134..262...
  15..20..30..50...90..170..330...
  21..27..39..63..111..207..399...
  28..35..49..77..133..245..469...
  . . .
The start of the sequence as triangle array read by rows:
  1;
  2,3;
  4,5,6;
  8,9,9,10;
  16,17,15,14,15;
  32,33,27,22,20,21;
  64,65,51,38,30,27,28;
  . . .
Row number r contains r numbers.
		

Links

• Boris Putievskiy, Rows n = 1..140 of triangle, flattened
• Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.

Crossrefs

Cf. A000217, A000096, A055999, A056121, A132758, A000079, A000051.

Programs

Mathematica

b[n_] := n - d[n]*(d[n] + 1)/2; c[n_] := (d[n]^2 + 3*d[n] + 4)/2 - n; d[n_] := Floor[(-1 + Sqrt[8*n - 7])/2]; a[n_] := b[n]*(b[n] + 2^c[n] - 1)/2; Table[a[n], {n, 1, 50}] (* G. C. Greubel, Jan 04 2018 *)

PARI

a(n, k) = n*(n+2^k-1)/2
array(rows, cols) = for(x=1, rows, for(y=1, cols, print1(a(x, y), ", ")); print(""))
/* Print initial 7 rows and 8 columns of table as follows */
array(7, 8) \\ Felix Fröhlich, Jan 05 2018

Python

t=int((math.sqrt(8*n-7) - 1)/ 2)
i=n-t*(t+1)/2
j=(t*t+3*t+4)/2-n
result = i*(i+2**j-1)/2

Formula

a(n) = A002260(n)*(A002260(n)+2^A004736(n)-1)/2.

a(n) = i*(i+2^j-1)/2,

where i = n-t*(t+1)/2, j = (t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2).