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a(5n) = a(n), a(5n+1) = a(5n+2) = a(5n+3) = a(5n+4) = a(n) + 1, a(0) = 0.
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%I #18 Sep 08 2016 09:31:25

%S 0,1,1,1,1,1,2,2,2,2,1,2,2,2,2,1,2,2,2,2,1,2,2,2,2,1,2,2,2,2,2,3,3,3,

%T 3,2,3,3,3,3,2,3,3,3,3,2,3,3,3,3,1,2,2,2,2,2,3,3,3,3,2,3,3,3,3,2,3,3,

%U 3,3,2,3,3,3,3,1,2,2,2,2,2,3,3,3,3,2,3,3,3,3,2,3,3,3,3,2,3,3,3,3,1,2,2,2,2,2,3,3,3,3,2,3,3,3,3,2,3,3,3,3,2

%N a(5n) = a(n), a(5n+1) = a(5n+2) = a(5n+3) = a(5n+4) = a(n) + 1, a(0) = 0.

%C Number of nonzero digits in the base 5 representation of n.

%C Fixed point of the mapping 0 -> 01111, 1 -> 12222, 2 -> 23333, ...

%C Self-similar or fractal sequence (underlining every fifth term, reproduce the original sequence).

%H Robert Israel, <a href="/A276134/b276134.txt">Table of n, a(n) for n = 0..10000</a>

%H Michael Gilleland, <a href="/selfsimilar.html">Some Self-Similar Integer Sequences</a>

%H Ilya Gutkovskiy, <a href="/A276134/a276134.jpg">Illustration (mapping 0 -> 01111, 1 -> 12222, 2 -> 23333, ...)</a>

%F a(5^k) = 1.

%F a(5^k-1) = k.

%F a(5^k-m) = k, k>0, m = 2,3,4.

%F a(5^k+m) = 2, k>0, m = 1,2,3,4.

%F a(5^k-a(5^k)) = k.

%F a(5^k+(-1)^k) = (k + (-1)^k*(k - 1) + 3)/2.

%F a(5^k+(-1)^k-1) = A093178(k).

%F a(5^k+(-1)^k+1) = A000034(k+1), k>0.

%F G.f. g(x) satisfies g(x) = (1+x+x^2+x^3+x^4)*g(x^5) + (x+x^2+x^3+x^4)/(1-x^5). - _Robert Israel_, Sep 07 2016

%e The evolution starting with 0 is: 0 -> 01111 -> 0111112222122221222212222 -> ...

%e ...

%e a(0) = 0;

%e a(1) = a(5*0+1) = a(0) + 1 = 1;

%e a(2) = a(5*0+2) = a(0) + 1 = 1;

%e a(3) = a(5*0+3) = a(0) + 1 = 1;

%e a(4) = a(5*0+4) = a(0) + 1 = 1;

%e a(5) = a(5*1+0) = a(1) = 1;

%e a(6) = a(5*1+1) = a(1) + 1 = 2, etc.

%e ...

%e Also a(10) = 1, because 10 (base 10) = 20 (base 5) and 20 has 1 nonzero digit.

%p f:= n -> nops(subs(0=NULL,convert(n,base,5))):

%p map(f, [$0..100]); # _Robert Israel_, Sep 07 2016

%t Join[{0}, Table[IntegerLength[n, 5] - DigitCount[n, 5, 0], {n, 120}]]

%Y Cf. A000034, A000120, A093178, A160384, A160385.

%K nonn,base,look

%O 0,7

%A _Ilya Gutkovskiy_, Aug 21 2016