A385722 a(n) = the n-th term of the sequence B_2 defined as the ordered union of the ranges of A_x, where the sequence A_x is defined in A384669, and x is a proper positive fraction with denominator 2, 3, 4, or 6.

1, 2, 4, 6, 8, 12, 16, 24, 30, 32, 36, 48, 60, 64, 72, 96, 120, 128, 144, 180, 192, 210, 240, 256, 288, 360, 384, 420, 480, 512, 576, 720, 768, 840, 960, 1024, 1152, 1260, 1440, 1536, 1680, 1920, 2048, 2304, 2310, 2520, 2880, 3072, 3360, 3840, 4096, 4320, 4608, 4620, 5040

Offset: 1

Author

Hal M. Switkay, Jul 07 2025

Keywords

• nonn

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.

Examples

			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.
		

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

Crossrefs

Cf. A000010, A000079, A002110, A002182, A025487, A095848, A384669.

Programs

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