cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A380438 Integers k that are the product of 3 distinct primes, the smallest of which is larger than the 5th root of k: k = p*q*r, where p, q, r are primes and k^(1/5) < p < q < r.

Original entry on oeis.org

30, 105, 165, 195, 231, 385, 455, 595, 665, 715, 805, 935, 1001, 1015, 1045, 1085, 1105, 1235, 1265, 1295, 1309, 1435, 1463, 1495, 1505, 1547, 1595, 1615, 1645, 1705, 1729, 1771, 1855, 1885, 1955, 2015, 2035, 2065, 2093, 2135, 2185, 2233, 2255, 2261, 2345, 2365, 2387, 2405, 2431, 2465, 2485
Offset: 1

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Author

Matthew Goers, Jan 24 2025

Keywords

Comments

This subsequence of the sphenics (A007304) is similar to A362910 or A138109 for semiprimes. Ishmukhametov and Sharifullina defined semiprimes n = p*q where each prime is greater than n^(1/4) as strongly semiprime. This sequence defines sphenic numbers with an analogous 'strength' as a product of 3 distinct primes k = p*q*r where each prime is greater than k^(1/5), or, alternately, k < p^5.
The only even term is 30 = 2*3*5.
As there are many equivalent ways of expressing Ishmukhametov and Sharifullina's "strongly semiprime" criterion, it is not obvious how it should most appropriately be extended to measure an equivalent "strength" of numbers with more prime factors. Here we follow a comparison of the least prime factor, p, to the factored number, k; but we could instead compare the greatest prime factor, r, to k; or p to r; or measure the variance/standard deviation of the prime factors (more precisely, after twice taking the logarithm of each factor as is done in A379271). Furthermore, it looks clear that the comparison used here (p against k^(1/5)) could be shown to give a substantially lower density asymptotically within the sphenics than Ishmukhametov and Sharifullina's equivalent for semiprimes. - Peter Munn, Feb 18 2025 and May 13 2025

Examples

			231 = 3*7*11 and 231^(1/5) < 3, so 231 is in the sequence.
255 = 3*5*17 but 255^(1/5) > 3, so 255 is not in the sequence.
		

Crossrefs

Subsequence of A253567, A290965, A379271, and A007304.
A046301 is a subsequence (product of 3 successive primes).
Cf. A115957, A138109, A251728, A362910 (strong semiprimes), A380995.

Programs

  • Mathematica
    q[k_] := Module[{f = FactorInteger[k]}, f[[;; , 2]] == {1, 1, 1} && f[[1, 1]]^5 > k]; Select[Range[2500], q] (* Amiram Eldar, Feb 14 2025 *)
  • PARI
    isok(k) = my(f=factor(k)); (bigomega(f)==3) && (omega(f)==3) && (k < vecmin(f[,1])^5); \\ Michel Marcus, Jan 27 2025
    
  • PARI
    list(lim)=my(v=List()); forprime(p=2,sqrtnint(lim\=1,3), forprime(q=p+1,min(sqrtint(lim\p),p^2), forprime(r=q+2,min(lim\(p*q),p^4\q), listput(v,p*q*r)))); Set(v) \\ Charles R Greathouse IV, May 20 2025
    
  • Python
    from math import isqrt
    from sympy import primepi, primerange, integer_nthroot
    def A380438(n):
        def bisection(f,kmin=0,kmax=1):
            while f(kmax) > kmax: kmax <<= 1
            kmin = kmax >> 1
            while kmax-kmin > 1:
                kmid = kmax+kmin>>1
                if f(kmid) <= kmid:
                    kmax = kmid
                else:
                    kmin = kmid
            return kmax
        def f(x): return n+x-sum(max(0,primepi(min(x//(p*q),p**4//q))-b) for a,p in enumerate(primerange(integer_nthroot(x,3)[0]+1),1) for b,q in enumerate(primerange(p+1,isqrt(x//p)+1),a+1))
        return bisection(f,n,n) # Chai Wah Wu, Mar 28 2025