A322849 Number of times 2^k (for k < 4) appears as a substring within 2^n.
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
Examples
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.
Programs
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Mathematica
Array[Total@ DigitCount[2^#, 10, {1, 2, 4, 8}] &, 85, 0] (* Michael De Vlieger, Dec 31 2018 *) -
PARI
a(n) = #select(x->((x==1) || (x==2) || (x==4) || (x==8)), digits(2^n)); \\ Michel Marcus, Dec 30 2018
Comments