A135765 Distribute the odd numbers in columns based on the occurrence of "3" in each prime factorization; square array A(row, col) = 3^(row-1) * A007310(col), read by antidiagonals A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

1, 5, 3, 7, 15, 9, 11, 21, 45, 27, 13, 33, 63, 135, 81, 17, 39, 99, 189, 405, 243, 19, 51, 117, 297, 567, 1215, 729, 23, 57, 153, 351, 891, 1701, 3645, 2187, 25, 69, 171, 459, 1053, 2673, 5103, 10935, 6561, 29, 75, 207, 513, 1377, 3159, 8019, 15309, 32805

Offset: 1

Author

Alford Arnold, Nov 28 2007

Keywords

• easy
• nonn
• tabl

Comments

The Table can be constructed by multiplying sequence A000244 by A007310.

From Antti Karttunen, Jan 26 2015: (Start)

A permutation of odd numbers. Adding one to each term and then dividing by two gives a related table A254051, which for any odd number, located in this array as x = A(row,col), gives the result at A254051(row+1,col) after one combined Collatz step (3x+1)/2 -> x (A165355) has been applied.

Each odd number n occurs here in position A(A007949(n), A126760(n)).

Compare also to A135764.

(End)

Examples

			The top left corner of the array:
    1,    5,    7,   11,   13,   17,   19,   23,   25,   29,   31,   35, ...
    3,   15,   21,   33,   39,   51,   57,   69,   75,   87,   93,  105, ...
    9,   45,   63,   99,  117,  153,  171,  207,  225,  261,  279,  315, ...
   27,  135,  189,  297,  351,  459,  513,  621,  675,  783,  837,  945, ...
   81,  405,  567,  891, 1053, 1377, 1539, 1863, 2025, 2349, 2511, 2835, ...
  243, 1215, 1701, 2673, 3159, 4131, 4617, 5589, 6075, 7047, 7533, 8505, ...
etc.
For n = 6, we have [A002260(6), A004736(6)] = [3, 1] (that is 6 corresponds to location 3,1 (row,col) in above table) and A(3,1) = A000244(3-1) * A007310(1) = 3^2 * 1 = 9.
For n = 9, we have [A002260(9), A004736(9)] = [3, 2] (9 corresponds to location 3,2) and A(3,2) = A000244(3-1) * A007310(2) = 3^2 * 5 = 9*5 = 45.
For n = 13, we have [A002260(13), A004736(13)] = [3, 3] (13 corresponds to location 3,3) and A(3,3) = A000244(3-1) * A007310(3) = 3^2 * 7 = 9*7 = 63.
For n = 23, we have [A002260(23), A004736(23)] = [2, 6] (23 corresponds to location 2,6) and A(2,6) = A000244(2-1) * A007310(6) = 3^1 * 17 = 51.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10440; the first 144 antidiagonals of array
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Row 1: A007310.

Column 1: A000244.

Cf. A007949 (row index), A126760 (column index).

Cf. also A000265, A002260, A003961, A004736, A005408, A165355, A249745.

Related arrays: A135764, A254051, A254055, A254101, A254102.

Programs

Maple

N:= 20:
B:= [seq(op([6*n+1,6*n+5]),n=0..floor((N-1)/2))]:
[seq(seq(3^j*B[i-j],j=0..i-1),i=1..N)]; # Robert Israel, Jan 26 2015

Formula

From Antti Karttunen, Jan 26 2015: (Start)

With both row and col starting from 1:

A(row, col) = A000244(row-1) * A007310(col) = 3^(row-1) * A007310(col).

a(n) = (2*A254051(n))-1.

a(n) = A003961(A254053(n)).

Above in array form:

A(row,col) = A003961(A254053(row,col)) = A003961(A135764(row,A249745(col))).

(End)

Extensions

Name amended and examples edited by Antti Karttunen, Jan 26 2015