A135764 Distribute the natural numbers in columns based on the occurrence of "2" in each prime factorization; square array A(row,col) = 2^(row-1) * ((2*col)-1), read by descending antidiagonals.
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
Offset: 1
Examples
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(3-1) * A005408(1-1) = 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(3-1) * A005408(3-1) = 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(2-1) * A005408(6-1) = 2^1 * 11 = 22.
Links
Crossrefs
Programs
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Maple
seq(seq(2^(j-1)*(2*(i-j)+1),j=1..i),i=1..20); # Robert Israel, Feb 03 2015
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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 *)
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PARI
a(n) = {s = ceil((1 + sqrt(1 + 8*n)) / 2); r = n - binomial(s-1, 2) - 1;k = s - r - 2; 2^r * (2 * k + 1) } \\ David A. Corneth, Feb 05 2015
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Scheme
(define (A135764 n) (A135764bi (A002260 n) (A004736 n))) (define (A135764bi row col) (* (A000079 (- row 1)) (+ -1 col col))) ;; Antti Karttunen, Feb 03 2015
Formula
From Antti Karttunen, Feb 03 2015: (Start)
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*(1-2*x)*Sum_{i>=0} i*x^(i*(i+1)/2) + (1-6*x)*Sum_{i>=0} x^(i*(i+1)/2) - 1 - 2*x)*x/(1-2*x)^2. These sums are related to Jacobi theta functions. - Robert Israel, Feb 03 2015
Extensions
More terms from Sean A. Irvine, Nov 23 2010
Name amended and the illustration of array in the example section transposed by Antti Karttunen, Feb 03 2015
Comments