A391106 - OEIS

%I #17 Dec 11 2025 13:27:43

%S 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,

%T 26,30,32,33,35,36,23,42,40,34,39,45,44,48,38,29,50,31,49,54,52,55,51,

%U 46,56,60,64,37,57,63,66,65,72,41,70,68,75,43,58,80,81,77,78,69,62

%N 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.

%C 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.

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

%e                     __________________________________________

%e                    |    Infinite pool of prime factors        |

%e                    |  .    .    .    .     .     .     .      |

%e                    |  .    .    .    .     .     .     .      |

%e                    |  2    3    5    7    11    13    17  ... |

%e                    |__________________________________________|

%e                                       |

%e                                       v

%e 1st step:   2,                                 ->  prime factor 2 from pool

%e 2nd step:   2,   3                             ->  prime factor 3 from pool

%e 3rd step:   -    3,  2*2                       ->  prime factors 2 and 2 from set of

%e                                                    previous numbers and from pool

%e                                                      a(1) = 2 as 2 is used up

%e 4th step:   -    3,  2*2,  5                   ->  prime factor 5 from pool

%e 5th step:   -    -   2*-,  5,  2*3             ->  prime factors 2 and 3 from set of

%e                                                    previous numbers

%e                                                      a(2) = 3 as 3 is used up

%e And so on.

%e .

%e The creation of 15 = 3*5 involves using up the remnant prime factors of 9 and 10.

%e Thus 9 and 10 become two consecutive terms at the same time as a(8) = 9 and a(9) = 10:

%e                                          |    |                           |

%e                                          v    v                           v

%e            -, -, -*-, -, -*-. -, -*-*-, 3*-, -*5, 11, 2*2*3, 13, 2*7, (15 = 3*5)

%t 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 *)

%K nonn

%O 1,1

%A _Tamas Sandor Nagy_, Nov 29 2025