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A135738 Least positive integer with even digit sum in bases 2..n.

%I A135738 #17 Aug 07 2023 07:50:03
%S A135738 3,6,10,10,54,54,54,54,130,130,130,130,390,390,2000,2000,3238,3238,
%T A135738 4080,4080,7326,7326,16584,16584,17310,17310,17310,17310,17310,17310,
%U A135738 17310,17310,231000,231000,231000,231000,466352,466352,466352,466352,3020830
%N A135738 Least positive integer with even digit sum in bases 2..n.
%C A135738 The sequence is obviously increasing. It seems that a(2n+1) = a(2n) for n > 1. Is there a simple proof? Is there a simple way to construct a(n)? Notice the pattern in base N, e.g., 130 = 10000010_2 = 11211_3 = 2002_4 = 1010_5 = 334_6 = 244_7 = 202_8 = 154_9 = 109_11 = {10}{10}_12 = {10}0_13.
%H A135738 "Davar55" on mersenneforum.org, <a href="http://www.mersenneforum.org/showthread.php?p=120023">Puzzles / "Sum of digits"</a>.
%e A135738 a(2)=3 since 1=1_2, 2=10_2, so 3=11_2 is the number > 0 with even digit sum (1+1) in base 2.
%e A135738 a(3)=6 since 4=100_2, 5=12_3, so 6=20_3=110_2 is the least N > 0 with even digit sum in base 2 and in base 3.
%e A135738 a(4)=a(5)=10=1010_2=101_3=22_4=20_5 is the least N > 0 having even digit sum in bases 2 through 4 and has so also in base 5.
%o A135738 (PARI) digitsum(n,b=10,s)={n=[n];while(n=divrem(n[1],b),s+=n[2]);s}
%o A135738 A135738(Bmax,n=1)={until(!n++,for(b=2,Bmax,digitsum(n,b)%2&next(2));return(n))} /* n-th element of the sequence */
%o A135738 t=1;for(b=2,100,print(b,":",t=A135738(b,t))) /* display the list */
%Y A135738 Cf. A000120, A053735, A053737, A053824, A053827-A053836, A007953.
%K A135738 nonn,base
%O A135738 2,1
%A A135738 _M. F. Hasler_, Dec 06 2007
%E A135738 Corrected example a(3)=5 to a(3)=6 David Yablon (davar55(AT)yahoo.com), Mar 19 2010