OFFSET
1,2
COMMENTS
The sequences A_x were defined in A384669; please see that sequence for more details.
Because of the continuity of the functions g_k(x) defined in A384669, if k is a term of A_y, then k is a term of A_x with x rational sufficiently close to y; so it suffices to study A_x for x rational, 0 < x < 1.
Let x = p/q, with p and q natural numbers, p < q. Then B_n is the ordered union of the ranges of A_x, where q has Euler totient <= n (that is, A000010(q) <= n). B_1 is just the sequence A_(1/2), that is, A384669. The present sequence B_2 is the ordered union of the ranges of A_x, where x = 1/6, 1/4, 1/3, 1/2, 2/3, 3/4, 5/6.
8 is the first term in B_2 (this sequence) that is not in B_1 = A384669.
15120 appears to be the first term in B_4 that is not in B_2 (this sequence).
More discussion about B_n and B_oo appears in a linked pdf.
LINKS
Pontus von Brömssen, Table of n, a(n) for n = 1..2000 (terms 1..186 from Hal M. Switkay)
Hal M. Switkay, More information about B_n and B_oo
EXAMPLE
8 is a term of A_(2/3) and therefore of this sequence as well as A_x for x > 2/3, even though 8 is not a term of A384669, because 3^(2/3) (corresponding to 8) > 2 = 1^(2/3) + 1^(2/3) (corresponding to 6). Thus 8 qualifies to be a term in B_2, this sequence.
PROG
(PARI) s(n, q) = my(f=factor(n)); sum(k=1, #f~, f[k, 2]^q);
listaq(nn, q) = my(r=-oo, list=List()); for (n=1, nn, my(ss=s(n, q)); if (ss > r, r = ss; listput(list, n)); ); Vec(list);
putlist(list, elems) = for (i=1, #elems, listput(list, elems[i])); list;
lista(nn) = my(list=List(), vq=[1/6, 1/4, 1/3, 1/2, 2/3, 3/4, 5/6]); for (i=1, #vq, list = putlist(list, listaq(nn, vq[i]))); Set(Vec(list)); \\ Michel Marcus, Jul 08 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Hal M. Switkay, Jul 07 2025
STATUS
approved
