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.

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, 4, 1, 16, 17, -8, 18, 5, -3, 19, -9, 20, 2, 21, -10, 6, -1, 22, -4, 23, -11, 24, 7, 25, -12, 3, 26, -5, 27, -13, 8, -2, 28, 1, 29, -14, 30, 9, -6, 4, 31, -15

Offset: 1

Author

Hartmut F. W. Hoft, Feb 24 2021

Keywords

• sign
• tabf

Comments

Row n has length A001227(n), the number of odd divisors of n, and also the number of entries in row n of A341970.

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.

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

Examples

			Triangle begins:
   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   4   1
  16
  17  -8
  18   5  -3
  19  -9
  20   2
  21 -10   6  -1
...
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).
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 ).
		

Crossrefs

Cf. A001227, A060831, A235791, A237048, A237270, A237591, A237593, A249223, A280851 A341970.

Cf. A286001, A299765.

Programs

Mathematica

(* Function a341970[ ] is defined in A341970 *)
widthL[n_] := Rest[FoldList[#1+If[OddQ[#2], +1, -1]&, 0, a341970[n]]]
sT[n_] := Map[(-1)^(#+1) Ceiling[(n+1)/# - (#+1)/2]&, a341970[n]]
a341971[n_] := Flatten[Map[sT, Range[n]]]
a341971[31] (* first 31 rows of table *)

Formula

a(A060831(n-1) + d) = T(n,d) = (-1)^(k+1)*A235791(s(n, k)).