A379746 - OEIS

%I #31 Jan 07 2025 08:42:50

%S 1,2,3,4,6,5,8,9,10,12,15,16,18,20,24,25,27,30,7,14,21,28,32,35,36,40,

%T 42,45,48,49,50,54,56,60,63,64,70,72,75,80,81,84,90,96,98,100,105,108,

%U 112,120,125,126,128,135,140,144,147,150,160,162,168,175,180

%N a(1)=1. For n>1 if a(n-1)=A002110(k), a(n)=prime(k+1). Otherwise a(n) is the smallest novel number whose prime factors have already occurred as previous terms.

%C Equivalent definition: Lexicographically earliest infinite sequence of distinct positive integers such that a(n) is the smallest novel number having prime power factorization Product_p_i^e_i where p_i is the least nondivisor prime of at most e_i distinct terms a(j); 1<=j<=n-1.

%C A permutation of the positive integers with prime powers q^k appearing in order (k>=1), and whose underlying sequence of least nondivisor primes is a permutation of A053669. Also, for distinct x, y; x<y, if rad(x)=rad(y) then x appears before y.

%C No multiple m*p (m>1) of a prime p can occur before p itself is a term.

%C From _Michael De Vlieger_, Jan 02 2025: (Start)

%C Efficient method of generating the sequence:

%C Define row k to be a(A363061(k)+1..A363061(k+1)).

%C Define R(i) to be { m <= i : rad(m) | i } = tensor product of prime power factor ranges of i that do not exceed i.

%C Then row k contains R(A002110(k+1)) \ R(A002110(k)).

%C Row 0 is R(1) = {1}.

%C Row 1 is R(2)\R(1) = {1, 2} \ {1} = {2},

%C   i.e., {row 2 of A162306} \ {row 1 of A162306}

%C       = {first A363061(1) terms of A000079} \ {1}.

%C Row 2 is R(6)\R(2) = {1, 2, 3, 4, 6} \ {1, 2} = {3, 4, 6},

%C   where R(6) = row 6 of A162306 = first A363061(2) terms of A003586.

%C Row 3 is R(30)\R(6)

%C   = {1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30} \ {1, 2, 3, 4, 6}

%C   = {5, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30},

%C   where R(30) = row 30 of A162306 = first A363061(3) terms of A051037, etc.

%C Therefore, for k > 1, within each row, terms strictly increase from prime(k) to primorial A002110(k).

%C Furthermore, a(1..A363061(k)) is a permutation of R(A002110(k)), hence the sequence is infinite and a permutation of natural numbers. (End)

%H Michael De Vlieger, <a href="/A379746/b379746.txt">Table of n, a(n) for n = 1..19985</a> (to include A002110(8).)

%H Michael De Vlieger, <a href="/A379746/a379746.txt">Efficient Wolfram program to generate this sequence</a>. (Note: pay attention to dataset size in A363061).

%H Michael De Vlieger, <a href="/A379746/a379746.png">Log log scatterplot of a(n)</a>, n = 1..19985.

%H Michael De Vlieger, <a href="/A379746/a379746_1.png">Plot p^m | a(n) at (x,y) = (n, pi(p))</a>, n = 1..2048, 12X vertical exaggeration, with a color function representing m = 1 in black, m = 2 in red, m = 3 in orange, ..., highest value of m in the dataset in magenta.

%F From _Michael De Vlieger_, Jan 02 2025: (Start)

%F a(A363061(k)) = A002110(k).

%F a(A363061(k)+1) = prime(k).

%F Seen as a table T(j,k), k = 1..A363061(j)-A363061(j-1) for j > 0, row 0 = {1},

%F   row j = {row A002110(j) of A162306} \ {row A002110(j-1) of A162306}. (End)

%e a(1) = 1 = A002110(0) therefore a(2) = A053669(1) = 2.

%e a(2) = 2 = A002110(1) therefore a(3) = A053669(2) = 3.

%e a(3) = 3 is not a primorial term so a(4)=4 = 2^2 is the smallest novel number whose prime factors do not exceed 3.

%e Using the second definition we have a(1,2,3,4)=1,2,3,4

%e                   with least nondivisor primes 2,3,2,3 respectively. Therefore a(5)=2^1*3^1=6, the smallest novel number whose prime factors (2,3) are nondivisor primes of the first 4 terms, and whose exponents do not exceed the number of times these primes have occurred in the underlying sequence of least nondivisor primes.

%t nn = 120; kk = 12;

%t c[_] := False; m[_] := 0; h = 0; q = j = 1; u = 2;

%t f[x_] := f[x] = FactorInteger[x][[All, 1]];

%t MapIndexed[Set[P[First[#2] - 1], #1] &, FoldList[Times, 1, Prime@ Range[kk]]];

%t {1}~Join~Reap[Do[

%t     If[j == P[h],

%t       If[h == kk, Break[]]; k = Prime[h + 1]; h++; q = Prime[h],

%t       k = u; While[Or[c[k], ! AllTrue[f[k], # <= q &]], k++]];

%t     j = Sow[k]; c[k] = True; If[k == u, While[c[u], u++] ],

%t {n, 2, nn}] ][[-1, 1]] (* _Michael De Vlieger_, Jan 02 2025 *)

%Y Cf. A000027, A000040, A002110, A007947, A053669, A162306, A351495, A363061, A368108, A377566.

%K nonn,easy,new

%O 1,2

%A _David James Sycamore_, Jan 01 2025