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A307153 Sequence gives pair of terms giving the numbers of previous even digits and previous odd digits; a(0)=0.

%I A307153 #25 Aug 01 2019 00:28:49
%S A307153 0,1,1,1,3,1,5,1,7,1,9,1,11,1,14,2,15,3,18,4,19,5,22,7,23,8,24,11,26,
%T A307153 13,28,15,30,16,32,18,34,20,35,22,37,24,39,26,41,29,43,31,46,33,48,35,
%U A307153 50,36,52,38,54,40,55,42,57,44,59,46,61,49,63,51,66,53,68
%N A307153 Sequence gives pair of terms giving the numbers of previous even digits and previous odd digits; a(0)=0.
%C A307153 Up to n = 10^5, any integer generally appears 0, 1 or 2 times. Only 248, 428 and 806 appear 3 times and 1 appears 8 times.
%C A307153 Are there any numbers that appear 4, 5 or more times?
%C A307153 From _Giovanni Resta_, Apr 01 2019: (Start)
%C A307153 4 times: 15711971, 22606282, 22826268, ...
%C A307153 5 times: 42862042, 44464482, 82802082, ...
%C A307153 6 times: 224026426, 224028040, 224042062, ...
%C A307153 7 times: 242620882, 244220442, 260088080, ...
%C A307153 Therefore, the first terms that appear n times, with n >= 0, are 6, 0, 3, 248, 15711971, 42862042, 224026426, 242620882, 1, ... (End)
%H A307153 Paolo P. Lava, <a href="/A307153/b307153.txt">Table of n, a(n) for n = 0..10000</a>
%F A307153 a(2n+1) = total number of even digits from a(0) to a(2n).
%F A307153 a(2n+2) = total number of odd digits from a(0) to a(2n+1).
%e A307153 a(1) = 1 because there is only one even digit before a(1): a(0) = 0.
%e A307153 a(2) = 1 because there is only one odd digit before a(2): a(1) = 1. Etc.
%p A307153 P:=proc(q) local a,b,d,d1,k,n,p,p1; a:=[0]: p:=1; d:=0;
%p A307153 for n from 2 to q do a:=[op(a),p]: b:=[op(convert(p,base,10))]:
%p A307153 p1:=0: d1:=0: for k from 1 to nops(b) do if b[k] mod 2=0
%p A307153 then p1:=p1+1: else d1:=d1+1: fi; od; d:=d+d1: p:=p+p1:
%p A307153 a:=[op(a),d]: b:=[op(convert(d,base,10))]: p1:=0: d1:=0:
%p A307153 for k from 1 to nops(b) do if b[k] mod 2=0 then p1:=p1+1:
%p A307153 else d1:=d1+1: fi; od; d:=d+d1: p:=p+p1: od; op(a); end: P(35);
%o A307153 (PARI) nb = [0,0]; for (n=1, 71, print1 (v=nb[1+n%2]", "); apply(d -> nb[1+d%2]++, if (v, digits(v), [0]))) \\ _Rémy Sigrist_, May 04 2019
%Y A307153 Cf. A196563, A196564.
%K A307153 nonn,base,easy
%O A307153 0,5
%A A307153 _Paolo P. Lava_, Mar 27 2019