%I
%S 1,2,3,4,5,0,7,8,9,6,11,0,0,0,15,16,17,10,19,0,13,0,23,0,0,0,0,0,0,0,
%T 31,32,33,18,35,12,21,14,39,0,0,0,27,0,0,0,47,0,0,0,0,0,0,0,0,0,0,0,0,
%U 0,0,0,63,64,65,34,67,20,37,22,71,0,25,0,43,0,29,0,79,0,0,0,0,0,0,0,55,0,0
%N Tabular arrangement of the natural numbers: the row on which any nonzero term a(n) appears in is A053645(a(n))=A053645(n+1), and the column is A161511(a(n)). Table is presented by columns with 2^{k1} items in column k, unused positions are filled with 0's.
%C Note: 1 might be a more natural starting offset for this sequence, although the identities concerning A053645 and A161511 would have to be changed.  _Antti Karttunen_, Oct 12 2009.
%C This can be regarded as an arrangement of the partitions, indexed by position in A125106. The partitions in a given row all have the same remaining partition when the largest part is removed; specifically, the partition indexed by the row number in A125106 (with row 0 having the empty partition remaining).
%C The first value on row n is A004760(n+1). The second value on each row is A004760(n+1) plus A062383(n); subsequent values increase by ever enlarging powers of two. Or equivalently, each subsequent value on the row after the first nonzero value is given by A004754(previous value on the same row).
%C A055941(r) tells how many terms the row r (>= 0) has been shifted rightward from its "natural position", i.e. with how many zeros that row has been prepended.
%C The number of (nonzero) entries in column k is A000041(k).
%H A. Karttunen, <a href="/A126441/b126441.txt">Table of n, a(n) for n = 0..65534 (first 16 columns)</a>
%e The largest power of 2 <= 6 is 4, 6  4 = 2, so 6 is in row 2. By A125106, 6 corresponds to the partition [2^2], total 4, so 6 goes in column 4. Thus T(2,4) = 6.
%e The table begins:
%e 1.2.4..8.16.32.64.128.256.512.1024
%e ..3.5..9.17.33.65.129.257.513.1025
%e .......6.10.18.34..66.130.258..514
%e ....7.11.19.35.67.131.259.515.1027
%e ............12.20..36..68.132..260
%e .........13.21.37..69.133.261..517
%e ............14.22..38..70.134..262
%e ......15.23.39.71.135.263.519.1031
%e ...................24..40..72..136
%e ...............25..41..73.137..265
%e ...................26..42..74..138
%e ............27.43..75.139.267..523
%e .......................28..44...76
%e ...............29..45..77.141..269
%e ...................30..46..78..142
%e .........31.47.79.143.271.527.1039
%e ...........................48...80
%e .......................49..81..145
%e ...........................50...82
%e ...................51..83.147..275
%t columns = 7; row[n_] := n2^Floor[Log2[n]]; col[0] = 0; col[n_] := If[EvenQ[n], col[n/2] + DigitCount[n/2, 2, 1], col[(n1)/2]+1]; Clear[T]; T[_, _] = 0; Do[T[row[k], col[k]] = k, {k, 1, 2^columns}]; Table[T[n1, k], {k, 1, columns}, {n, 1, 2^(k1)}] // Flatten (* _JeanFrançois Alcover_, Sep 09 2017 *)
%o (GNU/MIT Scheme:)
%o (define (A126441 n) (A126441onebased (1+ n)))
%o (definec (A126441onebased n) (cond ((< n 2) n) (else (let ((prev (A126441onebased ( n (/ (A053644 n) 2))))) (if (or (= (A053644 n) (* 2 (A053644 (A053645 n)))) (zero? prev)) (let ((starter (A004760 (1+ (A053645 n))))) (if (> (A161511 starter) (1+ (A000523 n))) 0 starter)) (A004754 prev))))))
%Y Cf. A125106, A053645, A000041, A004760, A062383, A000079 (column lengths).
%Y A053645(a(A166274(n))) = A053645(1+A166274(n)) for all n>=1.
%Y Positions of zeros: A166275, this sequence without zeros: A161924. A161920(n) gives the position of the first nonzero term on the row n1.
%K nonn,tabf
%O 0,2
%A _Alford Arnold_, Jan 19 2007
%E Edited by _Franklin T. AdamsWatters_, Jan 23 2007
%E Further edited and Schemecode added by _Antti Karttunen_, Oct 12 2009
