%I
%S 0,1,2,3,4,5,6,7,8,9,11,13,15,17,19,20,22,24,26,28,31,33,35,37,39,40,
%T 42,44,46,48,51,53,55,57,59,60,62,64,66,68,71,73,75,77,79,80,82,84,86,
%U 88,91,93,95,97,99,111,117,135,153,159,171,177,195
%N Numbers k with property that the arithmetic mean of any subset of its digits is an integer.
%C This sequence is different from A061383. Here digits in k must have all the same parity, otherwise the average of at least a pair of digits wouldn't be an integer. Note that for every 2digit term in A061383 both digits have the same parity. But not every number whose digits have all the same parity (sequence A059708) belongs here.
%F Apparently a(158+n) = A010785(35+n).
%e 17 is in this sequence because the set of digits (1,7) has an integer average: 4.
%e 159 and 195 are in this sequence because the sets of digits (1,5), (1,9), (5,9), and (1,5,9) all have integer averages, respectively: 3, 5, 7, and 5.
%o (PARI) firstTerms_vec(n)={my(v=vector(n),c,t,w:list,h);for(i=1,+oo,w=List();forsubset(i,k,listput(w,k));listpop(w,1);forvec(j=vector(i,z,[(z==1)&&(i>1),9]),h=j[1]%2;for(l=2,#j,if((j[l]%2)!=h,next(2)));for(k=1,#w,t=vecextract(j,w[k]);if(vecsum(t)%(#w[k]),next(2)));v[c++]=fromdigits(j);if(c==n,return(v))))}
%o (PARI) isok(m,{B=10})={my(w=digits(m,B));forsubset(#w,y,if(y!=Vecsmall([]),if(vecsum(vecextract(w,y))%(#y),return(0)),next));1}
%Y Cf. A061383, A059708, A010785, A165165.
%K nonn,base
%O 1,3
%A _R. J. Cano_, Feb 21 2019
