cp's OEIS Frontend

A361534 Let h,i,j be the latest 3 terms in the sequence, starting with a(1)=1, a(2)=2, a(3)=3. Let R = rad(h*i*j), where rad is A007947, and let S be the smallest number of terms in U = {h,i,j} which are divisible by any prime p dividing R. Then, a(n) is the least novel multiple of the greatest such prime p.

%I A361534 #19 Jun 24 2025 11:55:32
%S A361534 1,2,3,6,9,4,12,15,5,8,18,10,20,21,7,25,30,14,28,35,40,42,24,49,56,27,
%T A361534 33,11,22,36,39,13,16,45,26,52,50,55,65,78,44,66,91,104,77,88,117,130,
%U A361534 99,110,143,156,60,121,169,182,132,154,195,208,165,176,221,17,234,48,34,51,68,54,85,70,102
%N A361534 Let h,i,j be the latest 3 terms in the sequence, starting with a(1)=1, a(2)=2, a(3)=3. Let R = rad(h*i*j), where rad is A007947, and let S be the smallest number of terms in U = {h,i,j} which are divisible by any prime p dividing R. Then, a(n) is the least novel multiple of the greatest such prime p.
%C A361534 In the first 2^20 terms there are only 3 occasions of adjacent terms with same rad, (e.g., a(12,13) = 10, 20); only two occasions of rad(h) = rad(j) (e.g., a(3,4,5) = 3,6,9) and there is no occasion of rad(h) = rad(i) = rad(j) (possible iff S = 3), so it's reasonable to assume that it never happens. In any case the definition would ensure continuation of the sequence since a(n) would be the least novel multiple of the greatest prime divisor of R.
%C A361534 The sequence is conjectured to be a permutation of the positive integers with the primes appearing in their natural order.
%C A361534 The scatterplot resembles a "fine-toothed comb" similar to that of A361133.
%H A361534 Michael De Vlieger, <a href="/A361534/b361534.txt">Table of n, a(n) for n = 1..10000</a>
%H A361534 Michael De Vlieger, <a href="/A361534/a361534.png">Log log scatterplot of a(n)</a>, n = 1..2^20, showing primes in red.
%e A361534 a(1,2,3) = 1,2,3 and every prime divisor in R divides only one term (S = 1). Since 3 is the greatest such prime, a(4) = 6, the least novel multiple of 3.
%e A361534 a(2,3,4) = 2,3,6 and S = 2 since both 2 and 3 divide two terms in U and no prime divides more than that. Since 3 is the greatest such prime a(5) must be 9.
%e A361534 a(3,4,5) = 3,6,9 and S = 1 because 2|6 but no other term in {3,6,9}, therefore a(6) = 4, the least novel multiple of 2.
%t A361534 nn = 120; c[_] = False; q[_] = 1;
%t A361534 Array[Set[{a[#], c[#]}, {#, True}] &, 3];
%t A361534 Set[{h, i, j}, {a[1], a[2], a[3]}]; u = 4;
%t A361534 w = Array[FactorInteger[#][[All, 1]] &, 3];
%t A361534 Do[g = Reverse@ Union@ Flatten[w];
%t A361534  p = First[
%t A361534     Join @@ Map[ReverseSort,
%t A361534       SplitBy[SortBy[
%t A361534         Table[{p, Count[{h, i, j}, _?(Divisible[#, p] &)]}, {p, g}],
%t A361534         Last], Last]]][[1]] ;
%t A361534  (k = q[#]; While[c[k #], k++]; k *= #; While[c[# q[#]], q[#]++]) &[p];
%t A361534  Set[{a[n], c[k], w, h, i, j}, {k, True,
%t A361534    Append[w[[2 ;; -1]], FactorInteger[k][[All, 1]]], i, j, k}];
%t A361534  If[k == u, While[c[u], u++]], {n, 4, nn}];
%t A361534 Array[a, nn] (* _Michael De Vlieger_, Mar 15 2023 *)
%Y A361534 Cf. A007947, A361133.
%K A361534 nonn
%O A361534 1,2
%A A361534 _David James Sycamore_, Mar 15 2023