A272896 Difference between the number of odd and even digits in the decimal expansion of 2^n.
1, -1, -1, -1, 0, 0, -2, -1, -1, 1, -2, -4, -2, 0, -1, -1, 1, 2, -4, -4, -1, 1, -1, -5, 2, 2, -4, 1, -3, 1, 0, -4, -2, 2, 3, 3, 1, 4, -2, 2, 5, 3, -1, -5, -2, -2, -2, 1, -1, 3, -4, 0, 2, 2, -1, -1, 5, 2, 2, -4, -3, 1, -5, -1, 0, 0, -6, 3, 5, 5, 2, -10, -8, 2, -3, 7, 9, 0, 0
Offset: 0
Examples
2^10 = 1024, 2^11 = 2048, 2^12 = 4096, 2^13 = 8192. So a(10) = 1 - 3 = -2, a(11) = 0 - 4 = -4, a(12) = 1 - 3 = -2, a(13) = 2 - 2 = 0.
Links
- Indranil Ghosh, Table of n, a(n) for n = 0..10000
Programs
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Mathematica
Table[Count[#, ?OddQ] - Count[#, ?EvenQ] &@ IntegerDigits[2^n], {n, 0, 100}] (* Michael De Vlieger, May 09 2016 *)
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PARI
a(n) = #select(x -> x%2, digits(2^n)) - #select(x -> !(x%2), digits(2^n)); for(n=0, 78, print1(a(n),", ")) \\ Indranil Ghosh, Mar 13 2017
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Python
def A272896(n): x=y=0 for i in str(2**n): if int(i)%2: x+=1 else: y+=1 return x - y # Indranil Ghosh, Mar 13 2017
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Ruby
def a(n) str = (2 ** n).to_s str.size - str.split('').map(&:to_i).select{|i| i % 2 == 0}.size * 2 end (0..n).each{|i| p a(i)}
Comments