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A341971 Irregular triangle read by rows in which row n lists the values in row n of A235791 whose indices are in row n of A341970; a number is signed when its index is even.

%I A341971 #39 Jun 20 2025 16:27:09
%S A341971 1,2,3,-1,4,5,-2,6,1,7,-3,8,9,-4,2,10,-1,11,-5,12,3,13,-6,14,-2,15,-7,
%T A341971 4,1,16,17,-8,18,5,-3,19,-9,20,2,21,-10,6,-1,22,-4,23,-11,24,7,25,-12,
%U A341971 3,26,-5,27,-13,8,-2,28,1,29,-14,30,9,-6,4,31,-15
%N A341971 Irregular triangle read by rows in which row n lists the values in row n of A235791 whose indices are in row n of A341970; a number is signed when its index is even.
%C A341971 Row n has length A001227(n), the number of odd divisors of n, and also the number of entries in row n of A341970.
%C A341971 Let 1 <= n, 1 <= d <= A001227(n) and k = A341970(A060831(n-1) + d). Expression s(n, k) = r*n - r*(r+1)*(r+2)/6 + k with r = floor((sqrt(8*n + 1) - 1)/2) translates position (row n, column k) in the triangle of A235791 to its position in sequence A235791.
%C A341971 The absolute values in row n are the smallest parts of the partitions of n into consecutive parts (with the partitions ordered by increasing number of parts). - _Omar E. Pol_, Dec 31 2024
%F A341971 a(A060831(n-1) + d) = T(n,d) = (-1)^(k+1)*A235791(s(n, k)).
%e A341971 Triangle begins:
%e A341971    1
%e A341971    2
%e A341971    3  -1
%e A341971    4
%e A341971    5  -2
%e A341971    6   1
%e A341971    7  -3
%e A341971    8
%e A341971    9  -4   2
%e A341971   10  -1
%e A341971   11  -5
%e A341971   12   3
%e A341971   13  -6
%e A341971   14  -2
%e A341971   15  -7   4   1
%e A341971   16
%e A341971   17  -8
%e A341971   18   5  -3
%e A341971   19  -9
%e A341971   20   2
%e A341971   21 -10   6  -1
%e A341971 ...
%e A341971 a(26)..a(29) = { 15, -7, 4, 1 } is row 15 in this sequence with corresponding row 15 { 1, 1, 1, 0, 1 } in A237048. The sum 15 - 7 = 8 represents the first (complete) part of the symmetric representation of sigma(15), while 4 gives to incomplete bottom subpart of the central part and 1 the complete subpart of the second layer on the diagonal of the central part. Because of the double counting on the diagonal the completed subparts of the symmetric central part have size 7 and 1 respectively, see A280851, so that the parts of the symmetric representation of sigma(15) are (8, 8, 8).
%e A341971 a(40)..a(43) = { 21, -10, 6, -1 } is row 21 in this sequence with corresponding row 21 { 1, 1, 1, 0, 0, 1 } in A237048. The sums 21 - 10 = 11 and 6 - 1 = 5 denote the sizes of the two (complete) parts of width 1 of the symmetric representation of sigma(21) up to the diagonal resulting in the four parts ( 11, 5, 5, 11 ).
%t A341971 (* Function a341970[ ] is defined in A341970 *)
%t A341971 widthL[n_] := Rest[FoldList[#1+If[OddQ[#2], +1, -1]&, 0, a341970[n]]]
%t A341971 sT[n_] := Map[(-1)^(#+1) Ceiling[(n+1)/# - (#+1)/2]&, a341970[n]]
%t A341971 a341971[n_] := Flatten[Map[sT, Range[n]]]
%t A341971 a341971[31] (* first 31 rows of table *)
%Y A341971 Cf. A001227, A060831, A235791, A237048, A237270, A237591, A237593, A249223, A280851 A341970.
%Y A341971 Cf. A286001, A299765.
%K A341971 sign,tabf
%O A341971 1,2
%A A341971 _Hartmut F. W. Hoft_, Feb 24 2021