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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), ...
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%I #25 Feb 03 2015 16:09:49

%S 1,5,3,7,15,9,11,21,45,27,13,33,63,135,81,17,39,99,189,405,243,19,51,

%T 117,297,567,1215,729,23,57,153,351,891,1701,3645,2187,25,69,171,459,

%U 1053,2673,5103,10935,6561,29,75,207,513,1377,3159,8019,15309,32805

%N 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), ...

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

%C From _Antti Karttunen_, Jan 26 2015: (Start)

%C 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.

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

%C Compare also to A135764.

%C (End)

%H Antti Karttunen, <a href="/A135765/b135765.txt">Table of n, a(n) for n = 1..10440; the first 144 antidiagonals of array</a>

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>

%F From _Antti Karttunen_, Jan 26 2015: (Start)

%F With both row and col starting from 1:

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

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

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

%F Above in array form:

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

%F (End)

%e The top left corner of the array:

%e 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, ...

%e 3, 15, 21, 33, 39, 51, 57, 69, 75, 87, 93, 105, ...

%e 9, 45, 63, 99, 117, 153, 171, 207, 225, 261, 279, 315, ...

%e 27, 135, 189, 297, 351, 459, 513, 621, 675, 783, 837, 945, ...

%e 81, 405, 567, 891, 1053, 1377, 1539, 1863, 2025, 2349, 2511, 2835, ...

%e 243, 1215, 1701, 2673, 3159, 4131, 4617, 5589, 6075, 7047, 7533, 8505, ...

%e etc.

%e 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.

%e 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.

%e 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.

%e 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.

%p N:= 20:

%p B:= [seq(op([6*n+1,6*n+5]),n=0..floor((N-1)/2))]:

%p [seq(seq(3^j*B[i-j],j=0..i-1),i=1..N)]; # _Robert Israel_, Jan 26 2015

%o (Scheme, two versions)

%o (define (A135765 n) (A135765bi (A002260 n) (A004736 n)))

%o (define (A135765bi row col) (* (A000244 (- row 1)) (A007310 col)))

%o (define (A135765 n) (+ -1 (* 2 (A254051 n))))

%Y Row 1: A007310.

%Y Column 1: A000244.

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

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

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

%K easy,nonn,tabl

%O 1,2

%A _Alford Arnold_, Nov 28 2007

%E Name amended and examples edited by _Antti Karttunen_, Jan 26 2015