A321578 a(n) is the maximum value of k such that A007504(k) <= prime(n).

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, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15

Offset: 1

Author

David James Sycamore, Nov 12 2018

Keywords

• nonn

Comments

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.)

Examples

			a(1)=1 since prime(1)=2 and 1 is max k such that A007504(k) <= 2.
a(5)=3 since prime(5)=11 and 3 is max k such that A007504(k) <= 11.
n=4 (in A013916). A(4)=17=prime(7), so b(4)=7.
n=7 (not in A013916). A(7)=58 < 59=prime(17), so b(7)=17.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A007504, A013916, A323701, A061568.

Programs

Mathematica

nn = 75; MapIndexed[Set[s[First[#2]], #1] &, Accumulate[Prime@ Range[nn]]]; k = 1; Table[p = Prime[n]; While[s[k] <= p, k++]; k - 1, {n, nn}] (* Michael De Vlieger, Jun 29 2025 *)

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

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