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A322849
Number of times 2^k (for k < 4) appears as a substring within 2^n.
2
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, 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, 8, 9, 8, 7, 12, 15, 10, 8, 13, 7, 8, 15, 8, 9, 12, 14, 12, 6, 13
OFFSET
0,8
COMMENTS
It appears that the only 0 in this sequence is a(16).
FORMULA
a(n) <= A322850(n), for n >= 4.
a(n) = A065712(n) + A065710(n) + A065715(n) + A065719(n). - Michel Marcus, Dec 30 2018
EXAMPLE
n = 0, a(n) = 1, 2^n = 1 - solution is 1;
n = 1, a(n) = 1, 2^n = 2 - solution is 2;
n = 2, a(n) = 1, 2^n = 4 - solution is 4;
n = 3, a(n) = 1, 2^n = 8 - solution is 8;
n = 4, a(n) = 1, 2^n = 16 - solution is 1;
n = 5, a(n) = 1, 2^n = 32 - solution is 2;
n = 6, a(n) = 1, 2^n = 64 - solution is 4;
n = 7, a(n) = 3, 2^n = 128 - solutions are 1,2,8;
n = 14, a(n) = 3, 2^n = 16384 - solutions are 1,4,8;
n = 15, a(n) = 2, 2^n = 32768 - solutions are 2,8;
n = 16, a(n) = 0, 2^n = 65536 - no solutions.
MATHEMATICA
Array[Total@ DigitCount[2^#, 10, {1, 2, 4, 8}] &, 85, 0] (* Michael De Vlieger, Dec 31 2018 *)
PROG
(Python 3.7)
import re
results = []
start_n = 0
N = 100
current_num = int(pow(2, start_n-1)) # Calculate (n-1) power. Convert to integer for better precision
for n in range(start_n, N):
if n == 0:
current_num = 1
else:
current_num += current_num
count = 0
for test_str in ["1", "2", "4", "8"]:
count += len(re.findall(test_str, str(current_num)))
results.append(count)
print(results)
(PARI) a(n) = #select(x->((x==1) || (x==2) || (x==4) || (x==8)), digits(2^n)); \\ Michel Marcus, Dec 30 2018
CROSSREFS
Cf. A065712 (1), A065710 (2), A065715 (4), A065719 (8).
Cf. A322849.
Sequence in context: A342932 A026181 A327239 * A322850 A337566 A106601
KEYWORD
base,nonn
AUTHOR
Gaitz Soponski, Dec 28 2018
STATUS
approved