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A368208 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. 0
3, 8, 32, 62, 138, 212, 464, 1610, 4458, 1952, 13004, 44742, 22778, 242814, 512718, 360198, 2366654, 1529030, 5532422, 13883834, 15516014, 51393768, 210568010, 271767438, 299891114, 758345724, 1204130100, 1363350560, 5171802930 (list; graph; refs; listen; history; text; internal format)
OFFSET
2,1
COMMENTS
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.
From David A. Corneth, Dec 17 2023: (Start)
An initial search for a(n) can be done over numbers one more than a prime i.e. of the form prime(m) + 1.
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).
Looking at record gaps between primes in A002386 we need "pretty large" numbers for u > 1 so one could start searching with u = 1.
For 2 <= n <= 30 we have a(n) = prime(m) + 1 for some integer m. (End)
LINKS
EXAMPLE
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.
MAPLE
f:= proc(n) local k, p;
p:= prevprime(n);
for k from 2 do
if k*p < prevprime(k*n) then return k fi;
od
end proc:
V:= Array(2..25): count:= 0:
for n from 3 while count < 24 do
v:= f(n);
if V[v] = 0 then V[v]:= n; count:= count+1 fi
od:
convert(V, list);
CROSSREFS
Sequence in context: A108492 A003470 A176304 * A180385 A148903 A148904
KEYWORD
nonn,more
AUTHOR
Robert Israel, Dec 16 2023
EXTENSIONS
a(28)..a(30) from David A. Corneth, Dec 17 2023
STATUS
approved

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Last modified June 9 09:30 EDT 2024. Contains 373239 sequences. (Running on oeis4.)