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A306520
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Numbers k with property that the arithmetic mean of any subset of its digits is an integer.
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1
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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
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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FORMULA
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Apparently a(158+n) = A010785(35+n).
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EXAMPLE
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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.
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MATHEMATICA
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Select[Range[0, 200], AllTrue[Mean/@Subsets[IntegerDigits[#], {2, IntegerLength[ #]}], IntegerQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 09 2020 *)
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PROG
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(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}
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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