A317804 Numbers of form 2^i*12^j, with i, j >= 0.

1, 2, 4, 8, 12, 16, 24, 32, 48, 64, 96, 128, 144, 192, 256, 288, 384, 512, 576, 768, 1024, 1152, 1536, 1728, 2048, 2304, 3072, 3456, 4096, 4608, 6144, 6912, 8192, 9216, 12288, 13824, 16384, 18432, 20736, 24576, 27648, 32768, 36864, 41472, 49152, 55296, 65536

Offset

1,2

Links

Dario Ch, Table of n, a(n) for n = 1..10000

Formula

Sum_{n>=1} 1/a(n) = 24/11. - Amiram Eldar, Mar 29 2025

Mathematica

With[{max = 10^5}, Flatten[Table[2^i*12^j, {i, 0, Log2[max]}, {j, 0, Log[12, max/2^i]}]] // Sort] (* Amiram Eldar, Mar 29 2025 *)

Prog

(Python)
from heapq import heappush, heappop
def sequence():
    pq = [1]
    seen = set(pq)
    while True:
        value = heappop(pq)
        yield value
        seen.remove(value)
        for x in 2 * value, 12 * value:
            if x not in seen:
                heappush(pq, x)
                seen.add(x)
seq = sequence()
finalsequence_list = [next(seq) for i in range(100)]  # Dario Ch, Sep 01 2018
(Python)
from sympy import integer_log
def A317804(n):
    def bisection(f, kmin=0, kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum((x//12**i).bit_length() for i in range(integer_log(x, 12)[0]+1))
    return bisection(f, n, n) # Chai Wah Wu, Mar 26 2025

Crossrefs

Cf. A025612, A003596, A107326, A003597, A107364, A025616, A108201, A108238.

Sequence in context: A181818 A363063 A336496 * A328524 A322447 A170892

Adjacent sequences: A317801 A317802 A317803 * A317805 A317806 A317807

Keyword

nonn,easy

Author

Dario Ch, Sep 01 2018

Status

approved