A291125 "Doubly A289484" numbers: numbers with prime factorization p1^e1 * p2^e2 * ... * pk^ek such that there exist i < j < k with p1^e1 * p2^e2 * ... pi^ei > p(i+1) and p1^e1 * p2^e2 * ... pj^ej > p(j+1).

60, 84, 120, 132, 168, 180, 210, 240, 252, 264, 280, 300, 312, 315, 330, 336, 360, 390, 396, 408, 420, 440, 456, 468, 480, 495, 504, 510, 520, 528, 540, 552, 560, 570, 585, 588, 600, 612, 616, 624, 630, 660, 672, 680, 684, 690, 693, 720, 728, 756, 760, 765, 770, 780

Offset: 1

Author

Richard Locke Peterson, Aug 17 2017

Keywords

• nonn

Comments

These form a subsemigroup and a subsequence of the sequence A289484.

Density: Only 4.3% of the integers between 1 and 400 are doubly A289484.divisible by at least 3 primes. If a term in the sequence is squarefree, it must be divisible by at least 4 primes. If a number n is in the sequence, then every multiple is also in it. Using Wolfram Alpha, about 48% of the integers between 10^40+1 to 10^40+62 were found to be doubly A289484.

Examples

			60=2^2*3*5 is a term because 2^2 > 3 and 2^2*3 > 5.
315=3^2*5*7 is a term because 3^2 > 5 and 3^2*5 > 7.
		

Crossrefs

Cf. A289484.

Programs

Maple

isA291125 := proc(n)
    local pset,p,pprodidx,pprod,nu,falls ;
    pset := sort(convert(numtheory[factorset](n),list)) ;
    pprod := 1;
    falls := 0 ;
    for pprodidx from 1 to nops(pset)-1 do
        p := pset[pprodidx] ;
        nu := padic[ordp](n,p) ;
        pprod := pprod*p^nu ;
        if pprod > pset[pprodidx+1] then
            falls := falls+1 ;
            if falls >= 2 then
                return true;
            end if;
        end if;
    end do:
    return false ;
end proc:
for n from 1 to 3000 do
    if isA291125(n) then
        printf("%d,",n) ;
    end if;
end do: # R. J. Mathar, Oct 20 2017

PARI

is(n,f=factor(n))=my(p=1,t,s); for(i=1,#f~, t=f[i,1]^f[i,2]; if(p>t,s++); p*=t); s>1 \\ Charles R Greathouse IV, Jun 10 2020

Extensions

New name from Charles R Greathouse IV, Jun 11 2020