A341970 Irregular triangle read by rows of the indices of the entries of 1's in the corresponding rows in A237048.

1, 1, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 2, 3, 1, 4, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 3, 5, 1, 1, 2, 1, 3, 4, 1, 2, 1, 5, 1, 2, 3, 6, 1, 4, 1, 2, 1, 3, 1, 2, 5, 1, 4, 1, 2, 3, 6, 1, 7, 1, 2, 1, 3, 4, 5, 1, 2, 1, 1, 2, 3, 6, 1, 4, 1, 2, 5, 7, 1, 3, 8, 1, 2, 1, 4, 1, 2, 3, 6, 1, 5

Offset: 1

Author

Hartmut F. W. Hoft, Feb 24 2021

Keywords

• nonn
• tabf

Comments

The number of entries in the n-th row of the table of this sequence is A001227(n), the number of odd divisors of n.

Let number n = 2^s * q with s >= 0 and q odd, let row(n) = floor( (sqrt(8*n+1) - 1)/2 ), let D_n = { d : d odd divisor of n and d <= row(n) }, let E_n = { e : e = 2^(s+1) * d, d in D_n and e <= row(n) } and let F_n be the union of D_n and E_n with its elements listed in increasing order. Then the numbers in F_n are exactly the positions of 1's in row n of A237048 and the numbers in row n of this sequence.

Examples

			a(8, 9) = { 1, 3 } is row 6 in this sequence with corresponding row 6  { 1,  0,  1 } in A237048.
a(26...29) = { 1, 2, 3, 5 } is row 15 in this sequence with corresponding row 15 { 1, 1, 1, 0, 1 } in A237048.
Table of the first 15 rows:
row    entries
1      1
2      1
3      1  1
4      1
5      1  2
6      1  3
7      1  2
8      1
9      1  2  3
10     1  4
11     1  2
12     1  3
13     1  2
14     1  4
15     1  2  3  5
		

Crossrefs

Cf. A001227, A060831, A237048, A237593.

Programs

Mathematica

row[n_] := Floor[(Sqrt[8 n+1]-1)/2]
oddD[n_] := Select[Divisors[n], OddQ[#]&&#<=row[n]&]
twoExp[n_] := Module[{f=FactorInteger[n]}, If[First[First[f]]==2, Last[First[f]], 0]]
dualD[n_] := Select[Map[2^(twoExp[n]+1)#&, oddD[n]], #<=row[n]&]
a341970[n_] := Union[oddD[n], dualD[n]]
Flatten[Map[a341970, Range[40]]] (* first 40 rows of table *)

Formula

Let 1 <= n, 1 <= d <= A001227(n) and k the index of the d-th 1 in row n of A237048.

a( A060831(n-1) + d ) = T(n, d) = k.