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

1,3

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.

Links

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

Formula

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

Example

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.

Mathematica

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

Prog

(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}

Crossrefs

Cf. A061383, A059708, A010785, A165165.

Sequence in context: A275945 A061383 A059708 * A247945 A317621 A298297

Adjacent sequences: A306517 A306518 A306519 * A306521 A306522 A306523

Keyword

nonn,base

Author

R. J. Cano, Feb 21 2019

Status

approved