A306520 Numbers k with property that the arithmetic mean of any subset of its digits is an integer.

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, 42, 44, 46, 48, 51, 53, 55, 57, 59, 60, 62, 64, 66, 68, 71, 73, 75, 77, 79, 80, 82, 84, 86, 88, 91, 93, 95, 97, 99, 111, 117, 135, 153, 159, 171, 177, 195

Offset: 1

Author

R. J. Cano, Feb 21 2019

Keywords

• nonn
• base

Comments

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 2-digit term in A061383 both digits have the same parity. But not every number whose digits have all the same parity (sequence A059708) belongs here.

Examples

			17 is in this sequence because the set of digits (1,7) has an integer average: 4.
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.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..177 (all terms up to 1 million)

Crossrefs

Cf. A061383, A059708, A010785, A165165.

Programs

Mathematica

Select[Range[0,200],AllTrue[Mean/@Subsets[IntegerDigits[#],{2, IntegerLength[ #]}],IntegerQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 09 2020 *)

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))))}

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}

Formula

Apparently a(158+n) = A010785(35+n).