A158923 a(1) = 2, a(n) = a(n-1) + round(log(a(n-1))) for n >= 2.

2, 3, 4, 5, 7, 9, 11, 13, 16, 19, 22, 25, 28, 31, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 99, 104, 109, 114, 119, 124, 129, 134, 139, 144, 149, 154, 159, 164, 169, 174, 179, 184, 189, 194, 199, 204, 209, 214, 219, 224, 229, 234, 239, 244, 249

Offset: 1

Author

Daniel Forgues, Mar 30 2009

Keywords

• nonn

Comments

Each interval (a(n-1), a(n)] asymptotically contains one prime power on the average.

Links

• Daniel Forgues, Table of n, a(n) for n = 1..100000

Crossrefs

Cf. A158924, "Number of prime powers - 1 in interval (A158923(n-1), A158923(n)] expressing the excess or deficit relative to the asymptotic average of 1."

Cf. A158925, "Accumulated excess or deficit of prime powers in (1, A158924(n)]" (Partial sums of A158924).

Cf. A000961, "Prime powers p^k (p prime, k >= 0)."

Cf. A025528, "Number of prime powers <= n with exponents >0."

Programs

Mathematica

NestList[# + Round@ Log[#] &, 2, 60] (* Michael De Vlieger, Nov 05 2020 *)

Python

from math import log
print(2)
a_last = n = 2
while n >= 2:
    a = a_last + int(log(a_last) + 0.5)
    print(a)
    a_last = a
    n += 1 # Ya-Ping Lu, Oct 24 2020