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

%I #63 Apr 22 2025 05:08:42

%S 1,2,4,8,12,16,24,32,48,64,96,128,144,192,256,288,384,512,576,768,

%T 1024,1152,1536,1728,2048,2304,3072,3456,4096,4608,6144,6912,8192,

%U 9216,12288,13824,16384,18432,20736,24576,27648,32768,36864,41472,49152,55296,65536

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

%H Dario Ch, <a href="/A317804/b317804.txt">Table of n, a(n) for n = 1..10000</a>

%F Sum_{n>=1} 1/a(n) = 24/11. - _Amiram Eldar_, Mar 29 2025

%t 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 *)

%o (Python)

%o from heapq import heappush, heappop

%o def sequence():

%o pq = [1]

%o seen = set(pq)

%o while True:

%o value = heappop(pq)

%o yield value

%o seen.remove(value)

%o for x in 2 * value, 12 * value:

%o if x not in seen:

%o heappush(pq, x)

%o seen.add(x)

%o seq = sequence()

%o finalsequence_list = [next(seq) for i in range(100)] # _Dario Ch_, Sep 01 2018

%o (Python)

%o from sympy import integer_log

%o def A317804(n):

%o def bisection(f,kmin=0,kmax=1):

%o while f(kmax) > kmax: kmax <<= 1

%o kmin = kmax >> 1

%o while kmax-kmin > 1:

%o kmid = kmax+kmin>>1

%o if f(kmid) <= kmid:

%o kmax = kmid

%o else:

%o kmin = kmid

%o return kmax

%o def f(x): return n+x-sum((x//12**i).bit_length() for i in range(integer_log(x,12)[0]+1))

%o return bisection(f,n,n) # _Chai Wah Wu_, Mar 26 2025

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

%K nonn,easy

%O 1,2

%A _Dario Ch_, Sep 01 2018