OFFSET
1,1
COMMENTS
For any n there is an infinity of numbers m for which prime(m+1) - prime(m) is greater than n.
It appears that the sequence of lengths of successive runs is equal to A053695. - Marc Bofill Janer, May 21 2019
REFERENCES
Laurențiu Panaitopol, Dinu Șerbănescu, Number theory and combinatorial problems for juniors, Ed.Gil, Zalău, (2003), ch. 1, p.7, pr. 25. (in Romanian).
FORMULA
a(2*n) = a(2*n+1) = A144309(n+1) for n>=1. - Georg Fischer, Dec 05 2022
EXAMPLE
For n = 2, prime(2) - prime(1) = 3 - 2 = 1, prime(3) - prime(2) = 5 - 3 = 2, prime(5) - prime(4) = 11 - 7 = 4, so a(2) = 4.
PROG
(MATLAB) v=primes(1000000);
for u=1:100; ss=1;
while and(v(ss+1)-v(ss)<=u, ss<length(v)-1); ss=ss+1; end;
sol(u)=ss;
end
sol
(Magma) v:=PrimesUpTo(10000000);
sol:=[];
for u in [1..60] do
for ss in [1..#v-1] do
if v[ss+1]-v[ss] gt u then
sol[u]:=ss;
break;
end if;
end for;
end for;
sol;
(PARI) a(n) = my(k=1); while(prime(k+1) - prime(k) <= n, k++); k; \\ Michel Marcus, Apr 03 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Marius A. Burtea, Apr 02 2019
STATUS
approved