A372401 Position of 210^n among 7-smooth numbers A002473.
1, 68, 547, 2119, 5817, 13008, 25412, 45078, 74409, 116147, 173379, 249532, 348375, 474018, 630922, 823885, 1058051, 1338898, 1672260, 2064302, 2521535, 3050825, 3659361, 4354687, 5144682, 6037582, 7041946, 8166692, 9421074, 10814695, 12357491, 14059744, 15932086, 17985473
Offset: 0
Keywords
Programs
-
Mathematica
Table[ Sum[Floor@ Log[7, 210^n/(2^i*3^j*5^k)] + 1, {i, 0, Log[2, 210^n]}, {j, 0, Log[3, 210^n/2^i]}, {k, 0, Log[5, 210^n/(2^i*3^j)]}], {n, 0, 12}] -
Python
import heapq from itertools import islice from sympy import primerange def A372401gen(p=7): # generator for p-smooth terms v, oldv, psmooth_primes, = 1, 0, list(primerange(1, p+1)) h = [(1, [0]*len(psmooth_primes))] idx = {psmooth_primes[i]:i for i in range(len(psmooth_primes))} loc = 0 while True: v, e = heapq.heappop(h) if v != oldv: loc += 1 if len(set(e)) == 1: yield loc oldv = v for p in psmooth_primes: vp, ep = v*p, e[:] ep[idx[p]] += 1 heapq.heappush(h, (v*p, ep)) print(list(islice(A372401gen(), 15))) # Michael S. Branicky, Jun 05 2024
-
Python
from sympy import integer_log def A372401(n): c, x = 0, 210**n for i in range(integer_log(x,7)[0]+1): for j in range(integer_log(m:=x//7**i,5)[0]+1): for k in range(integer_log(r:=m//5**j,3)[0]+1): c += (r//3**k).bit_length() return c # Chai Wah Wu, Sep 16 2024
Formula
a(n) ~ c * n^4, where c = log(210)^4/(24*log(2)*log(3)*log(5)*log(7)) = 14.282278766622... - Vaclav Kotesovec and Amiram Eldar, Sep 22 2024
Comments