A322849 - OEIS

%I #29 Feb 11 2019 19:47:30

%S 1,1,1,1,1,1,1,3,1,2,3,3,1,3,3,2,0,3,5,5,3,3,4,4,3,3,4,6,4,3,6,7,4,4,

%T 6,3,3,5,5,6,4,5,7,5,8,8,5,7,6,7,9,9,3,5,10,5,3,11,10,7,8,6,10,7,8,11,

%U 8,9,8,7,12,15,10,8,13,7,8,15,8,9,12,14,12,6,13

%N Number of times 2^k (for k < 4) appears as a substring within 2^n.

%C It appears that the only 0 in this sequence is a(16).

%F a(n) <= A322850(n), for n >= 4.

%F a(n) = A065712(n) + A065710(n) + A065715(n) + A065719(n). - _Michel Marcus_, Dec 30 2018

%e n =  0, a(n) = 1, 2^n =     1 - solution is 1;

%e n =  1, a(n) = 1, 2^n =     2 - solution is 2;

%e n =  2, a(n) = 1, 2^n =     4 - solution is 4;

%e n =  3, a(n) = 1, 2^n =     8 - solution is 8;

%e n =  4, a(n) = 1, 2^n =    16 - solution is 1;

%e n =  5, a(n) = 1, 2^n =    32 - solution is 2;

%e n =  6, a(n) = 1, 2^n =    64 - solution is 4;

%e n =  7, a(n) = 3, 2^n =   128 - solutions are 1,2,8;

%e n = 14, a(n) = 3, 2^n = 16384 - solutions are 1,4,8;

%e n = 15, a(n) = 2, 2^n = 32768 - solutions are 2,8;

%e n = 16, a(n) = 0, 2^n = 65536 - no solutions.

%t Array[Total@ DigitCount[2^#, 10, {1, 2, 4, 8}] &, 85, 0] (* _Michael De Vlieger_, Dec 31 2018 *)

%o (Python 3.7)

%o import re

%o results = []

%o start_n = 0

%o N = 100

%o current_num = int(pow(2, start_n-1))    # Calculate (n-1) power. Convert to integer for better precision

%o for n in range(start_n, N):

%o     if n == 0:

%o         current_num = 1

%o     else:

%o         current_num += current_num

%o     count = 0

%o     for test_str in ["1", "2", "4", "8"]:

%o         count += len(re.findall(test_str,  str(current_num)))

%o     results.append(count)

%o print(results)

%o (PARI) a(n) = #select(x->((x==1) || (x==2) || (x==4) || (x==8)), digits(2^n)); \\ _Michel Marcus_, Dec 30 2018

%Y Cf. A065712 (1), A065710 (2), A065715 (4), A065719 (8).

%Y Cf. A322849.

%K base,nonn

%O 0,8

%A _Gaitz Soponski_, Dec 28 2018