OFFSET
1,2
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
EXAMPLE
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).
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Hal M. Switkay, Jun 06 2025
STATUS
approved