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%I #23 Jul 16 2020 11:51:14
%S 1,10,268,7576,652726,913180816
%N The least k such that i*k + 1 is a product of i different primes for i = 1..n.
%C Since for k = 5000000000420503488, i*k+1 is a product of i different primes for i = 1..7, a(7) <= 5000000000420503488.
%C a(n) is the least parameter k such that equation tau(x^k) = x has at least A005425(n) solutions in positive integers.
%H Vladimir Letsko, <a href="https://www.facebook.com/notes/%D0%B7%D0%B0%D0%B4%D0%B0%D1%87%D0%B8-%D0%B8-math-problems-/%D0%BA%D0%BE%D0%BD%D0%BA%D1%83%D1%80%D1%81-2019-%D0%BA%D0%BE%D1%80%D0%BE%D1%82%D0%BA%D0%B8%D0%B5-%D0%B7%D0%B0%D0%B4%D0%B0%D1%87%D0%B8-%D1%81-%D1%80%D0%B5%D1%88%D0%B5%D0%BD%D0%B8%D1%8F%D0%BC%D0%B8/1398390813670090/">Problem K7</a> (in Russian).
%e 268 is in the sequence because 268 + 1 is prime, 2*268 + 1 is a product of 2 different primes, 3*268 + 1 is a product of 3 different primes, and 268 is the least number with such properties.
%p nn := 1; for kk to 6 do
%p n := nn; do n := nextprime(n); f := true;
%p for k from 2 to kk do a := k*(n-1)+1; if not IsSquareFree(a) or nops(ifactors(a)[2]) <> k then f := false; break end if end do;
%p if f then nn := n-1; print(nn); break end if end do end do
%o (PARI) isok(k,n) = {for (i=1, n, if ((omega(i*k+1) != i) || (bigomega(i*k+1) != i), return (0));); return (1);}
%o a(n) = {my(k=1); while(!isok(k,n), k++); k;} \\ _Michel Marcus_, Jul 15 2020
%Y Cf. A000005, A005425, A336028.
%K nonn,more
%O 1,2
%A _Vladimir Letsko_, Jul 05 2020
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