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.
a(1) = 2 because 30 = 2 * 3 * 5 has third largest prime factor 2. a(2) = 2 because 42 = 2 * 3 * 7 has third largest prime factor 2. a(3) = 2 because 60 = 2 * 2 * 3 * 5 has both third and fourth largest prime factor 2. a(8) = 3 because 90 = 2 * 3 * 3 * 5 has both second and third largest prime factor 3.
b:= proc(n) option remember; local k;
if n=1 then 30
else for k from b(n-1)+1 while
nops(ifactors(k)[2])<3 do od;
k
fi
end:
a:= n-> sort(map(x-> x[1]$x[2], ifactors(b(n))[2]))[-3]:
seq(a(n), n=1..120);
b[n_] := b[n] = Module[{k}, If[n==1, 30, For[k = b[n-1]+1, PrimeNu[k] < 3, k++]; k]];
a[n_] := (Table[#[[1]], {#[[2]]}]& /@ FactorInteger[b[n]] // Flatten // Sort)[[-3]];
Array[a, 120] (* Jean-François Alcover, Nov 28 2020, after Alois P. Heinz *)
29023908435660702564000000 = 2^8 * 3^7 * 5^6 * 7^5 * 11^4 * 13^3 * 17 * 19^2 is in the sequence as there are exactly 8 distinct exponents; 1 through 8.
is(n) = #Set(factor(n)[, 2]) == 8
The number of distinct prime factors of the numbers 15, 14, 13, 12, 11, 10 are respectively 2, 2, 1, 2, 1, 2 and 2*2*1*2*1*2 = 16, hence 16 is a term.
om[n_] := om[n] = PrimeNu[n]; q[n_] := Module[{m = n, k = n - 1}, While[k > 1 && Divisible[m, om[k]], m /= om[k]; k--]; m == 1]; Select[Range[2, 10^6], q] (* Amiram Eldar, Nov 02 2021 *)
isA007304 := proc(n) if bigomega(n) = 3 and A001221(n) =3 then true; else false ; end if; end proc: # list of prime exponents pexp := proc(n) local e,pe ; e := [] ; for pe in ifactors(n)[2] do e := [op(e),op(2,pe)] ; end do: e ; end proc: isCycSchGr := proc(n) local om,nhalf ,pe; om := A001221(n) ; if om > 4 then return false; elif om = 4 then # require 2*p*q*r if type(n,'even') and type(n/2,'odd') then nhalf := n/2 ; # require nhalf =p*q*r in A007304 return isA007304(nhalf) ; else false; end if; elif om = 3 then # require p*q*r or 2*p*q^k if type(n,'even') and type(n/2,'odd') then nhalf := n/2 ; # require nhalf =p*q^k pe := pexp(nhalf) ; if nops(pe) =2 and 1 in convert(pe,set) then true; else false ; end if; elif type(n,'odd') then # require n =p*q*r if isA007304(n) then true; else false ; end if; else false; end if; elif om = 2 then # require p*q^k pe := pexp(n) ; if 1 in convert(pe,set) then true; else false; end if; else # p^k, k>=0 true ; end if; end proc: for n from 1 to 3000 do if not isCycSchGr(n) then printf("%d,",n) ; end if; end do:
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