A322850 Number of times 2^k for k < n-1 appears as a substring within 2^n.
0, 0, 0, 0, 1, 1, 1, 3, 1, 2, 3, 3, 1, 3, 4, 3, 0, 3, 5, 5, 3, 3, 4, 4, 5, 4, 5, 6, 4, 3, 6, 8, 4, 4, 6, 3, 3, 5, 5, 6, 5, 6, 7, 5, 9, 9, 6, 8, 6, 7, 9, 9, 3, 5, 10, 5, 3, 11, 10, 8, 8, 6, 11, 7, 10, 13, 10, 10, 8, 7, 13, 16, 12, 9, 13, 8, 9, 16, 8, 9, 12, 15, 14, 7, 14, 9
Offset: 0
Examples
n = 0, a(n) = 0, 2^n = 1 - no solutions; n = 1, a(n) = 0, 2^n = 2 - no solutions; n = 2, a(n) = 0, 2^n = 4 - no solutions; n = 3, a(n) = 0, 2^n = 8 - no solutions; 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) = 4, 2^n = 16384 - solutions are 1,4,8,16; n = 15, a(n) = 3, 2^n = 32768 - solutions are 2,8,32; n = 16, a(n) = 0, 2^n = 65536 - no solutions.
Programs
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Mathematica
Array[If[# < 4, Total@ DigitCount[2^#, 10, 2^Range[0, Min[# - 1, 3]]], Total@ DigitCount[2^#, 10, {1, 2, 4, 8}]] &, 85, 0] (* Michael De Vlieger, Dec 31 2018 *) -
PARI
isp2(n) = (n==1) || (n==2) || (ispower(n,,&k) && (k==2)); a(n) = {my(d=digits(2^n), nb = 0); for (i=1, #d-1, for (j=1, #d-i+1, my(nd = vector(i, k, d[j+k-1])); if (nd[1] != 0, nb += isp2(fromdigits(nd))););); nb;} \\ Michel Marcus, Dec 30 2018
Formula
a(n) >= A322849(n), for n >= 4.
Comments