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A291125
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"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).
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0
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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MAPLE
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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;
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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