A044940 Number of runs of even length in base-9 representation of n.
0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 1
Examples
From _Antti Karttunen_, Dec 15 2017: (Start) For n = 810 = 729 + 81 = 9^3 + 9^2 thus in base 9 written as "1100", we count two runs, both of length 2, thus both even, so a(810) = 2. For n = 32805 = 5*(9^4), thus in base 9 "50000", there is one run of even length, so a(32805) = 1. For n = 65630 = 1*(9^5) + 1*(9^4) + 0*(9^3) + 0*(9^2) + 2*(9^1) + 2*(9^0) thus written as "110022" in base 9, there are three runs, all of length 2, thus all even, so a(65630) = 3. (End)
Links
- Antti Karttunen, Table of n, a(n) for n = 1..100000
Programs
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Maple
f:= proc(n) local i,j,t,Q,d; Q:= convert(n,base,9); d:= nops(Q); i:= 1: t:= 0: while i < d do for j from i+1 to d while Q[j] = Q[i] do od: if (j-i)::even then t:= t+1 fi; i:= j; od; t end proc: map(f, [$1..100]); # Robert Israel, Dec 15 2017
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Mathematica
a[n_] := Count[Length /@ Split[IntegerDigits[n, 9]], _?EvenQ]; Array[a, 120] (* Jean-François Alcover, Dec 16 2017 *)
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PARI
A044940(n) = { my(rl=1, d, prev_d = -1, s=0); while(n>0, d = (n%9); n = ((n-d)/9); if(d==prev_d, rl++, s += (1-(rl%2)); prev_d = d; rl = 1)); (s + (1-(rl%2))); }; \\ Antti Karttunen, Dec 15 2017
Extensions
More terms and secondary offset added by Antti Karttunen, Dec 15 2017