

A137415


a(n) is the smallest m such that A000720(m)  A000006(m) = n.


0



1, 2, 43, 61, 83, 103, 109, 113, 151, 167, 179, 181, 197, 199, 233, 239, 251, 263, 271, 281, 283, 313, 317, 349, 353, 367, 383, 389, 401, 409, 421, 433, 443, 457, 463, 467, 487, 499, 503, 523, 547, 563, 571, 577, 593, 601, 607, 617, 619, 641, 647, 653
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OFFSET

1,2


COMMENTS

a(n) exists for all n >= 1. Proof: It's easy to show that lim inf b(m) = +oo and b(m+1)  b(m) <= 1 for all m. For every n >= 1, if b(m) = n, then there must exist some m' > m such that b(m') > n. Let m_0 be the smallest among such m', then b((m_0)1) <= n, so b(m_0) <= b((m_0)1) + 1 <= n + 1, but b(m_0) > n, so b(m_0) = n + 1. By induction every n >= 1 appears in the range of {b(m)}.
It appears that b(m) ~ m/log(m)  sqrt(m*log(m)) + o(1), so each value for {b(m)} should only appear finitely many times (e.g., b(m) = 1 only for m = 1, 10, 12, 16, 26, 27, 28, 35, 36, 40; b(m) = 0 only for some 44 m's).
It appears that {a(n)} is an increasing prime sequence, for n >= 0. (End)


LINKS



MATHEMATICA

Table[Min[Flatten[Table[If[PrimePi[m]  IntegerPart[Sqrt[Prime[m]]] == n, m, {}], {m, 1, 500}]]], {n, 1, 20}]


PROG

(PARI) a(n) = for(m=1, oo, if(primepi(m)sqrtint(prime(m))==n, return(m))) \\ Jianing Song, Feb 04 2019


CROSSREFS



KEYWORD

nonn,less


AUTHOR



EXTENSIONS



STATUS

approved



