A391106 When constructing the sequence of the positive integers > 1 from prime factors, we reuse the prime factors of the smallest earlier constructed numbers where we can. The sequence is the numbers whose prime factors were first reused entirely in the process.

2, 3, 4, 6, 5, 8, 7, 9, 10, 12, 15, 14, 11, 16, 18, 20, 13, 21, 24, 22, 17, 25, 28, 27, 19, 26, 30, 32, 33, 35, 36, 23, 42, 40, 34, 39, 45, 44, 48, 38, 29, 50, 31, 49, 54, 52, 55, 51, 46, 56, 60, 64, 37, 57, 63, 66, 65, 72, 41, 70, 68, 75, 43, 58, 80, 81, 77, 78, 69, 62

Offset

1,1

Comments

For the process, we source the prime factors from an infinite set of prime factors containing every prime, each with infinite multiplicity. However, whenever we can, we also recycle, so to speak, any needed prime factors from the earlier made numbers. When the prime factors of such numbers are fully depleted, then those numbers become terms. When more than one number is used up at once, they take place in the sequence in order of their magnitude.

With 1 prepended, the sequence is a permutation of the positive integers.

Links

Table of n, a(n) for n=1..70.

Example

                    __________________________________________
                   |    Infinite pool of prime factors        |
                   |  .    .    .    .     .     .     .      |
                   |  .    .    .    .     .     .     .      |
                   |  2    3    5    7    11    13    17  ... |
                   |__________________________________________|
                                      |
                                      v
1st step:   2,                                 ->  prime factor 2 from pool
2nd step:   2,   3                             ->  prime factor 3 from pool
3rd step:   -    3,  2*2                       ->  prime factors 2 and 2 from set of
                                                   previous numbers and from pool
                                                     a(1) = 2 as 2 is used up
4th step:   -    3,  2*2,  5                   ->  prime factor 5 from pool
5th step:   -    -   2*-,  5,  2*3             ->  prime factors 2 and 3 from set of
                                                   previous numbers
                                                     a(2) = 3 as 3 is used up
And so on.
.
The creation of 15 = 3*5 involves using up the remnant prime factors of 9 and 10.
Thus 9 and 10 become two consecutive terms at the same time as a(8) = 9 and a(9) = 10:
                                         |    |                           |
                                         v    v                           v
           -, -, -*-, -, -*-. -, -*-*-, 3*-, -*5, 11, 2*2*3, 13, 2*7, (15 = 3*5)

Mathematica

seq[len_] := Module[{s = {}, t = {}, n = 1, f, available, p, k, pos, target, newterms, primes, j}, primes[k_] := Flatten[Table[#[[1]], {#[[2]]}] & /@ FactorInteger[k]]; While[Length[s] < len, n++; f = primes[n]; available = Sort[t[[All, 1]]]; Do[p = f[[i]]; Do[k = available[[m]]; j = 1; While[j <= Length[t] && t[[j, 1]] != k, j++]; pos = j; target = t[[pos, 2]]; If[MemberQ[target, p], t[[pos, 2]] = DeleteCases[target, p, 1, 1]; Break[]], {m, Length[available]}], {i, Length[f]}]; newterms = Sort[First /@ Select[t, Last[#] === {} &]]; s = Join[s, newterms]; t = Select[t, ! MemberQ[newterms, First[#]] &]; t = Append[t, {n, f}]]; s]; seq[70] (* Amiram Eldar, Dec 05 2025 *)

Crossrefs

Sequence in context: A104650 A397055 A083179 * A123503 A123717 A123718

Adjacent sequences: A391103 A391104 A391105 * A391107 A391108 A391109

Keyword

nonn

Author

Tamas Sandor Nagy, Nov 29 2025

Status

approved