cp's OEIS Frontend

A379770 Irregular triangle T(j,k) read by rows: split the natural numbers m > 1 into groups of length 2^s and then sort each group in the order defined in A019565.

%I A379770 #11 Jan 04 2025 22:46:29
%S A379770 1,2,3,4,5,6,7,8,9,10,12,11,13,14,15,16,17,18,20,19,24,21,25,22,26,23,
%T A379770 28,27,29,30,31,32,33,34,36,35,40,37,48,41,38,42,49,39,50,44,43,52,51,
%U A379770 45,56,46,53,57,54,47,58,55,60,59,61,62,63,64,65,66,68,67
%N A379770 Irregular triangle T(j,k) read by rows: split the natural numbers m > 1 into groups of length 2^s and then sort each group in the order defined in A019565.
%C A379770 In other words, function f(m) = Product_{i=0..Floor(log_2(m))} prime(i), with i the index of 1's in the binary expansion of m.
%C A379770 A permutation of the natural numbers.
%H A379770 Michael De Vlieger, <a href="/A379770/b379770.txt">Table of n, a(n) for n = 0..16384</a> (rows j = 0..13, flattened)
%H A379770 Michael De Vlieger, <a href="/A379770/a379770.png">Scatterplot of a(n)</a>, n = 0..131071.
%H A379770 Michael De Vlieger, <a href="/A379770/a379770_1.png">Log log scatterplot of a(n)</a>, n = 0..131071
%H A379770 Michael De Vlieger, <a href="/A379770/a379770_2.png">Fan style binary tree of T(j,k)</a>, j = 0..10, k = 1..2^(j-1), i.e., a(n) = 0..2047, with a color function where k = 1 is green, k = 2^(j-1) in red, indicating displacement of a(n) from its order in the natural numbers.
%H A379770 <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the natural numbers</a>
%F A379770 Length of row j is 2^(j-1) = A000079(j-1).
%F A379770 T(0,1) = a(0) = 1.
%F A379770 T(j,1) = a(2^(j-1)) = 2^(j-1) + 1.
%F A379770 T(j, 2^(j-1)) = a(2^j - 1) = 2^j.
%e A379770 Table below shows rows j = 0..5:
%e A379770 j\k   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16
%e A379770 -------------------------------------------------------------------
%e A379770 0:    1;
%e A379770 1:    2;
%e A379770 2:    3,  4;
%e A379770 3:    5,  6,  7,  8;
%e A379770 4:    9, 10, 12, 11, 13, 14, 15, 16;
%e A379770 5:   17, 18, 20, 19, 24, 21, 25, 22, 26, 23, 28, 27, 29, 30, 31, 32;
%e A379770 .
%e A379770 These correspond with values f(T(j,k)) as shown below:
%e A379770 j\k   1   2   3   4   5   6    7    8    9   10   11   12   13   14    15    16
%e A379770 -------------------------------------------------------------------------------
%e A379770 0:    1;
%e A379770 1:    2;
%e A379770 2:    3,  6;
%e A379770 3:    5, 10, 15, 30;
%e A379770 4:    7, 14, 21, 35, 42, 70, 105, 210;
%e A379770 5:   11, 22, 33, 55, 66, 77, 110, 154, 165, 231, 330, 385, 462, 770, 1155, 2310;
%e A379770 .
%e A379770 T(4,4) = a(11) = 35, while A019565(11) = 42, since 11_2 = "1011", f(11) = 2*3*7 = 42, but A019565(12) = 35 since 12_2 = "1100", f(12) = 5*7 is smaller than 42, therefore a(11) = 35, and a(12) = 42.
%t A379770 Flatten@ Table[
%t A379770   SortBy[Range[2^n, 2^(n + 1) - 1],
%t A379770     Times @@ Flatten@
%t A379770       MapIndexed[Prime[#2]^#1 &,
%t A379770         Reverse@ IntegerDigits[#, 2]] &], {n, 0, 8}]
%Y A379770 Cf. A019565.
%K A379770 nonn,tabf,easy
%O A379770 0,2
%A A379770 _Michael De Vlieger_, Jan 02 2025