|
%I #28 Dec 17 2023 10:28:11
%S 3,8,32,62,138,212,464,1610,4458,1952,13004,44742,22778,242814,512718,
%T 360198,2366654,1529030,5532422,13883834,15516014,51393768,210568010,
%U 271767438,299891114,758345724,1204130100,1363350560,5171802930
%N a(n) is the least k such that, if p is the greatest prime less than k, there is a prime between n*p and n*k, but for 1 < j < n there is no prime between j*p and j*k.
%C a(n) is the least k such that A049711(n*k) < n*A049711(k) but A049711(j*k) >= A049711(j*k) for 1 < j < n.
%C From _David A. Corneth_, Dec 17 2023: (Start)
%C An initial search for a(n) can be done over numbers one more than a prime i.e. of the form prime(m) + 1.
%C If a(n) is of the form prime(m) + u where there is no prime p between (exclusive) prime(m) and prime(m) + u then there is no prime between (n-1)*prime(m) and (n-1)*(prime(m) + u).
%C Looking at record gaps between primes in A002386 we need "pretty large" numbers for u > 1 so one could start searching with u = 1.
%C For 2 <= n <= 30 we have a(n) = prime(m) + 1 for some integer m. (End)
%e a(4) = 32 because 31 is the greatest prime less than 32, and there are no primes between 2*31 = 62 and 2*32 = 64 and no primes between 3*31 = 93 and 3*32 = 96, but there is a prime between 4*31 = 124 and 4*32 = 128, namely 127.
%p f:= proc(n) local k,p;
%p p:= prevprime(n);
%p for k from 2 do
%p if k*p < prevprime(k*n) then return k fi;
%p od
%p end proc:
%p V:= Array(2..25): count:= 0:
%p for n from 3 while count < 24 do
%p v:= f(n);
%p if V[v] = 0 then V[v]:= n; count:= count+1 fi
%p od:
%p convert(V,list);
%Y Cf. A002386, A008864, A049711, A367035, A151799.
%K nonn,more
%O 2,1
%A _Robert Israel_, Dec 16 2023
%E a(28)..a(30) from _David A. Corneth_, Dec 17 2023
| 