A384669 Positive integers setting a new record for the sum of the square roots of the prime exponents.

1, 2, 4, 6, 12, 24, 30, 60, 120, 180, 210, 360, 420, 840, 1260, 1680, 2520, 3360, 4620, 6720, 7560, 9240, 13860, 18480, 27720, 36960, 55440, 73920, 83160, 110880, 120120, 180180, 221760, 240240, 360360, 480480, 720720, 960960, 1081080, 1441440, 2042040, 2882880, 4084080, 5765760, 6126120

Offset: 1

Author

Hal M. Switkay, Jun 06 2025

Keywords

• nonn

Comments

This sequence is the special case x = 1/2 of a class of sequences A_x indexed by real numbers x, defined as follows. Let k be a natural number with prime factorization k = Product_{i=1..r} (p_i)^(e_i), where p_i are distinct primes, and e_i are natural numbers. Define f_x(k) = Sum_{i=1..r} (e_i)^x. Then the sequence A_x consists of natural numbers k where f_x(k) sets a new record.

We may define g_k(x) = f_x(k), so that g_k consists of real functions indexed by natural numbers. Each g_k is a sum of exponential functions, whose bases are natural numbers.

We provide facts, conjectures, and questions about the class of sequences A_x.

Facts:

For all real x, A_x is an infinite, increasing sequence of numbers of least prime signature (A025487) starting with 1 and 2.

For all x <= 0, A_x coincides with the primorials (A002110). For all x >= 1, A_x coincides with the powers of 2 (A000079). Thus the interesting sequences A_x have 0 <= x <= 1.

For all x > 0, A_x(n+1) <= 2*A_x(n).

1, 2, 4, 6, 12, 24, and no other natural numbers, are terms of every A_x with 0 < x < 1.

Conjectures:

The sequences A_x appear to be distinct in the following sense. Given 0 <= x < y <= 1, not only are A_x and A_y distinct, their ranges appear to have finite intersection.

Define A_0+, respectively A_1- as the limit of A_x as x approaches 0 from the right, respectively as x approaches 1 from the left. Then A_0+ appears to coincide with A168264, and A_1- appears to coincide with A029744 without the term 3.

If 0 < x < 1 and d is a natural number, then d divides some term of A_x.

If 0 < x < 1 and d is a natural number, then d divides infinitely many terms of A_x.

If 0 < x < 1 and d is a natural number, then d divides all sufficiently large terms of A_x.

Define S_n to be the set of x for which A_x(n) achieves a minimum value as a function of x (well-defined because the values are integers). Then for fixed n >= 10, S_n appears to be a subset of the interval log(2)/log(3) <= x <= log(3)/log(4), although in general S_n itself is not an interval.

Questions:

Does every number of least prime signature appear in at least one A_x for some x? 216 is not a term in any A_x I have examined.

Are there any x for which A_x contains infinitely many highly composite numbers (A002182), respectively infinitely many deeply composite numbers (A095848)?

Does the sequence of sets S_n have any limit points x in [0,1]? If x is such a limit point, A_x would presumably grow more slowly than other A_x. Examining A_x for x rational with denominator 720, A_x appears to contain a maximum number of terms (80) less than 2 * 10! when x = 507/720, 511/720, 515/720, and 517/720, all between 0.70 and 0.72.

Answer to the first question above: No, 216 is not a term of A_x for any x. If it were, we would have g_216(x) > g_210(x) and g_216(x) > g_192(x), i.e., 2*3^x > 4 and 2*3^x > 6^x+1. The first inequality holds only if x > x0 = log(2)/log(3). But 2*3^x0 < 6^x0+1 and it is easily verified that the function 2*3^x - (6^x+1) is decreasing for x > x0, so the second equality cannot hold when x > x0. - Pontus von Brömssen, Jun 13 2025

Additional facts from Hal M. Switkay, Jun 29 2025: (Start)

If k1 and k2 have the same prime signature, then the functions g_k1(x) and g_k2(x) are identical. Hence g_k(x) is identical to g_A046523(k)(x), where A046523(k) is the smallest number with the same prime signature as k. This is why the terms of A_x are all numbers of least prime signature.

For all k, lim_{x->-oo} g_k(x) = A056169(k) = the number of unitary prime divisors of k.

For all k, g_k(0) = omega(k) = A001221(k) = the number of distinct prime divisors of k.

For all k, g_k(1) = bigomega(k) = A001222(k) = the number of prime divisors of k counted with multiplicity.

For all k > 1, lim_{x->oo} [log(g_k(x))/x] = log(A051903(k)), where A051903(k) = the maximum exponent in the prime factorization of k. When k has least prime signature, A051903(k) = A007814(k), the exponent of the largest power of 2 dividing k. (End)

Using the notation of A385722, this sequence is B_1. - Hal M. Switkay, Jul 27 2025

Examples

			f_(1/2)(24) = sqrt(3) + sqrt(1), because 24 = (2^3)(3^1). This is a record value for f_(1/2), so 24 is in the sequence. f_(1/2)(30) = sqrt(1) + sqrt(1) + sqrt(1) (because 30 = (2^1)(3^1)(5^1)), which is larger still, putting 30 in the sequence. However, f_(1/2)(32) = sqrt(5) (because 32 = 2^5), smaller than the previous value, so 32 is not in the sequence.
g_216(x) = 3^x + 3^x, because 216 = (2^3)(3^3).
		

Crossrefs

Cf. A000079, A001221, A001222, A002110, A002182, A007814, A025487, A029744, A046523, A051903, A056169, A095848, A168264, A385722.

Programs

Mathematica

s[n_] := Total[Sqrt[FactorInteger[n][[;; , 2]]]]; s[1] = 0; With[{lps = Cases[Import[ "https://oeis.org/A025487/b025487.txt", "Table"], {, }][[;; , 2]]}, sm = -1; seq = {}; Do[s1 = s[lps[[i]]]; If[s1 > sm, sm = s1; AppendTo[seq, lps[[i]]]], {i, 1, Length[lps]}]; seq] (* Amiram Eldar, Jun 08 2025 *)

PARI

s(n) = my(f=factor(n)); sum(k=1, #f~, sqrt(f[k,2]));
lista(nn) = my(r=-oo, list=List()); for (n=1, nn, my(ss=s(n)); if (ss > r, r = ss; listput(list, n));); Vec(list); \\ Michel Marcus, Jun 15 2025