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%I #15 Mar 10 2019 13:32:22
%S 1,1,2,2,3,3,4,4,4,5,5,5,6,6,6,6,7,7,7,7,7,8,8,8,8,9,9,9,9,9,9,10,10,
%T 10,10,10,10,11,11,11,11,11,11,11,12,12,12,12,12,12,12,13,13,13,13,13,
%U 13,13,13,14,14,14,14,14,14,14,15,15,15,15,15,15,15,15
%N a(n) is the maximum value of k such that A007504(k) <= prime(n).
%C Let A be A007504. The number of distinct values of k such that a(k)=r is the number of primes p in the interval A(r) <= p < A(r+1); namely: 2,2,2,3,3,4,5,4,6,6,... (see A323701). Let b(n) be the smallest r such that a(r)=n, namely: 1,3,5,7,10,13,17,22,26,... For given n, if k is the index of the smallest prime >= A(n), then b(n)=k. (The equality applies when n is a term of A013916.)
%e a(1)=1 since prime(1)=2 and 1 is max k such that A007504(k) <= 2.
%e a(5)=3 since prime(5)=11 and 3 is max k such that A007504(k) <= 11.
%e n=4 (in A013916). A(4)=17=prime(7), so b(4)=7.
%e n=7 (not in A013916). A(7)=58 < 59=prime(17), so b(7)=17.
%o (Perl) use ntheory ':all'; sub a { my $p = nth_prime($_[0]); my($s, $q) = (0, 2); while ($s <= $p) { $s += $q; $q = next_prime($q) }; prime_count($q-1)-1 }; print join(", ", map { a($_) } 1..100), "\n"; # _Daniel Suteu_, Jan 26 2019
%o (PARI) a(n) = my(k=0, p=0, s=0); while(s <= prime(n), k++; p=nextprime(p+1); s+=p); k-1; \\ _Michel Marcus_, Feb 19 2019
%Y Cf. A000040, A007504, A013916, A323701, A061568.
%K nonn
%O 1,3
%A _David James Sycamore_, Nov 12 2018
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