

A135764


Distribute the natural numbers in columns based on the occurrence of "2" in each prime factorization; square array A(row,col) = 2^(row1) * ((2*col)1), read by descending antidiagonals.


22



1, 3, 2, 5, 6, 4, 7, 10, 12, 8, 9, 14, 20, 24, 16, 11, 18, 28, 40, 48, 32, 13, 22, 36, 56, 80, 96, 64, 15, 26, 44, 72, 112, 160, 192, 128, 17, 30, 52, 88, 144, 224, 320, 384, 256, 19, 34, 60, 104, 176, 288, 448, 640, 768, 512, 21, 38, 68, 120, 208, 352, 576, 896, 1280, 1536, 1024, 23, 42, 76, 136, 240, 416, 704, 1152, 1792, 2560, 3072, 2048, 25, 46, 84, 152, 272, 480, 832, 1408, 2304, 3584, 5120, 6144, 4096, 27, 50, 92, 168, 304, 544, 960, 1664, 2816
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OFFSET

1,2


COMMENTS

The array in A135764 is identical to the array in A054582 [up to the transposition and different indexing.  Clark Kimberling, Dec 03 2010; comment amended by Antti Karttunen, Feb 03 2015; please see the illustration in Example section].
The array gives a bijection between the natural numbers N and N^2. A more usual bijection is to take the natural numbers A000027 and write them in the usual OEIS square array format. However this bijection has the advantage that it can be formed by iterating the usual bijection between N and 2N.  Joshua Zucker, Nov 04 2011
The array can be used to determine the configurations of kth Towers of Hanoi moves, by labeling odd row terms C,B,A,C,B,A,... and even row terms B,C,A,B,C,A,.... Then given k equal to or greater than term "a" in each nth row, but less than the next row term, record the label A, B, or C for term "a". This denotes the peg position for the disc corresponding to the nth row. For example, with k = 25, five discs are in motion since the binary for 25 = 11001, five bits. We find that 25 in row 5 is greater than 16 labeled C, but less than 48. Thus, disc 5 is on peg C. In the 4th row, 25 is greater than 24 (a C), but less than 40, so goes onto the C peg. Similarly, disc 3 is on A, 2 is on A, and disc 1 is on A. Thus, discs 2 and 3 are on peg A, while 1, 4, and 5 are on peg C.  Gary W. Adamson, Jun 22 2012
Shares with arrays A253551 and A254053 the property that A001511(n) = k for all terms n on row k and when going downward in each column, terms grow by doubling.  Antti Karttunen, Feb 03 2015
Let P be the infinite palindromic word having initial word 0 and midword sequence (1,2,3,4,...) = A000027. Row n of the array A135764 gives the positions of n1 in S. ("Infinite palindromic word" is defined at A260390.)  Clark Kimberling, Aug 13 2015
The probability distribution series 1 = 2/3 + 4/15 + 16/255 + 256/65535 + ... + A001146(n1)/A051179(n) governs the proportions of terms in A001511 from row n of the array. In A001511(1..15) there are ((2/3) * 15)) = ten terms from row one of the array, ((4/15) * 15)) = four terms from row two, and ((16/17) * 15)) = one (rounded), giving one term from row three (a 4).  Gary W. Adamson, Dec 16 2021
Subarrays representing the number of divisors of an integer can be mapped on the table. For 60, write the odd divisors on the top row: 1, 3, 5, 15. Since 60 has 12 divisors, let the left column equal 1, 2, 4, where 4 is the highest power of 2 dividing 60. Multiplying top row terms by left column terms, we get the result:
1 3 5 15
2 6 10 30
4 12 20 60. (End)


LINKS



FORMULA

A(row, col) = 2^(row1) * ((2*col)1) = A000079(row1) * A005408(col1).
A(row,col) = A(row+1,col)/2 [discarding the topmost row and halving the rest of terms gives the array back].
A(row,col) = A(row,col+1)  A000079(row) [discarding the leftmost column and subtracting 2^{row number} from the rest of terms gives the array back].
(End)
G.f.: ((2*x+1)*Sum_{i>=0} 2^i*x^(i*(i+1)/2) + 2*(12*x)*Sum_{i>=0} i*x^(i*(i+1)/2) + (16*x)*Sum_{i>=0} x^(i*(i+1)/2)  1  2*x)*x/(12*x)^2. These sums are related to Jacobi theta functions.  Robert Israel, Feb 03 2015


EXAMPLE

The table begins
1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, ...
2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, ...
4, 12, 20, 28, 36, 44, 52, 60, 68, 76, 84, 92, ...
8, 24, 40, 56, 72, 88, 104, 120, 136, 152, 168, 184, ...
16, 48, 80, 112, 144, 176, 208, 240, 272, 304, 336, 368, ...
32, 96, 160, 224, 288, 352, 416, 480, 544, 608, 672, 736, ...
etc.
For n = 6, we have [A002260(6), A004736(6)] = [3, 1] (i.e., 6 corresponds to location 3,1 (row,col) in above table) and A(3,1) = A000079(31) * A005408(11) = 2^2 * 1 = 4.
For n = 13, we have [A002260(13), A004736(13)] = [3, 3] (13 corresponds to location 3,3 (row,col) in above table) and A(3,3) = A000079(31) * A005408(31) = 2^2 * 5 = 20.
For n = 23, we have [A002260(23), A004736(23)] = [2, 6] (23 corresponds to location 2,6) and A(2,6) = A000079(21) * A005408(61) = 2^1 * 11 = 22.


MAPLE

seq(seq(2^(j1)*(2*(ij)+1), j=1..i), i=1..20); # Robert Israel, Feb 03 2015


MATHEMATICA

f[n_] := Block[{i, j}, {1}~Join~Flatten@ Last@ Reap@ For[j = 1, j <= n, For[i = j, i > 0, Sow[2^(j  i  1)*(2 i + 1)], i], j++]]; f@ 10 (* Michael De Vlieger, Feb 03 2015 *)


PROG

(Scheme)
(define (A135764bi row col) (* (A000079 ( row 1)) (+ 1 col col)))
(PARI) a(n) = {s = ceil((1 + sqrt(1 + 8*n)) / 2); r = n  binomial(s1, 2)  1; k = s  r  2; 2^r * (2 * k + 1) } \\ David A. Corneth, Feb 05 2015


CROSSREFS



KEYWORD



AUTHOR



EXTENSIONS

Name amended and the illustration of array in the example section transposed by Antti Karttunen, Feb 03 2015


STATUS

approved



