A387016 Permutation of the odd integers >= 3 formed by ordering them first by odd k >= 3 and then by integer m >= 1 in their unique representation (k - 2^m)*2^m + 1.

3, 7, 5, 11, 13, 15, 21, 9, 19, 29, 25, 23, 37, 41, 27, 45, 57, 31, 53, 73, 17, 35, 61, 89, 49, 39, 69, 105, 81, 43, 77, 121, 113, 47, 85, 137, 145, 51, 93, 153, 177, 55, 101, 169, 209, 59, 109, 185, 241, 63, 117, 201, 273, 33, 67, 125, 217, 305, 97

Offset

1,1

Comments

A term t must have m = A007814(t-1), and k follows from that so that the representation is unique.

For a given odd k >= 3, terms have 1 < 2^m < k which is m is in the range 1 <= m <= floor(log_2(k)).

The k,m pairs are k = 2*i + 1 and m = n - A001855(i) where i = A030530(n).

The terms where m is the maximum for each k are A369901 (in that order) and are a permutation of the Proth numbers A080075.

Links

Amiram Eldar, Table of n, a(n) for n = 1..11703 (terms corresponding to k < 2500)

Formula

a(n) = 2*A386386(n-1) + 3.

a(n) = (k - 2^m)*2^m + 1 = k*2^m - 4^m + 1 where n = A001855((k-1)/2) + m with 1 < 2^m < k odd.

a(n) = A089654(n)*(2*A030530(n) - A089654(n) + 1) + 1.

Example

Initial terms and their k and m begin
  n    = 1   2   3   4   5   6   7   8   9  10  11  ...
  a(n) = 3,  7,  5, 11, 13, 15, 21,  9, 19, 29, 25, ...
  k    = 3   5   5   7   7   9   9   9  11  11  11
  m    = 1   1   2   1   2   1   2   3   1   2   3
  runs       \---/   \---/   \-------/   \-------/
For odd k=7 and m=2, we have 1<2^2<7, so a(n)=(7-2^2)*2^2+1=13 where n=A001855((7-1)/2)+2=3+2=5.

Mathematica

Table[(k - 2^m)*2^m + 1, {k, 3, 35, 2}, {m, 1, Log2[k-1]}] // Flatten (* Amiram Eldar, Aug 13 2025 *)

Crossrefs

Cf. A001855, A007814 (2-adic valuation), A030530, A080075, A089654, A369901, A386386.

Sequence in context: A328190 A090978 A113830 * A237055 A074368 A074588

Adjacent sequences: A387013 A387014 A387015 * A387017 A387018 A387019

Keyword

nonn,easy,look

Author

Thomas Ordowski, Aug 13 2025

Extensions

More terms from Amiram Eldar, Aug 13 2025

Status

approved