A372400 Position of 30^n among 5-smooth numbers A051037.

1, 18, 83, 228, 486, 888, 1466, 2255, 3283, 4583, 6189, 8134, 10445, 13158, 16305, 19916, 24027, 28667, 33870, 39665, 46086, 53166, 60937, 69429, 78675, 88709, 99561, 111263, 123849, 137347, 151793, 167219, 183658, 201139, 219695, 239359, 260165, 282141, 305320

Offset: 0

Author

Michael De Vlieger, Jun 03 2024

Keywords

• nonn

Comments

Also position of 30^(n+1) in A143207.

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A002110, A051037, A143207, A202821, A372401, A372402.

Programs

Mathematica

Table[Sum[Floor@ Log[5, 30^n/(2^i*3^j)] + 1, {i, 0, Log[2, 30^n]}, {j, 0, Log[3, 30^n/2^i]}], {n, 0, 38}]

PARI

a(n)=my(t=30^n,u=5*t); sum(a=0,logint(t,5), u\=5; sum(b=0,logint(u,3), logint(u\3^b,2)+1)) \\ Charles R Greathouse IV, Sep 18 2024

Python

# uses imports/function in A372401
print(list(islice(A372401gen(p=5), 40))) # Michael S. Branicky, Jun 05 2024

Python

from sympy import integer_log
def A372400(n):
    c, x = 0, 30**n
    for i in range(integer_log(x,5)[0]+1):
        for j in range(integer_log(y:=x//5**i,3)[0]+1):
            c += (y//3**j).bit_length()
    return c # Chai Wah Wu, Sep 16 2024

Formula

a(n) = k*n^3 + (3k/2)*n^2 + O(n) where k = (log 30)^3/(6 log 2 log 3 log 5) = 5.35057081984.... - Charles R Greathouse IV, Sep 19 2024