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A272360
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Numbers n such that the sum of the digits of the numbers from 1 to n divides the sum of the numbers from 1 to n.
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0
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1, 2, 3, 4, 5, 6, 7, 8, 9, 80, 119, 180, 398, 399, 998, 999, 1055, 1809, 2715, 4063, 4529, 11374, 18180, 19199, 27269, 54519, 110549, 113695, 168399, 294999, 454511, 624591, 636349, 639999, 1090719, 1818180, 2350079, 9639999, 17576999, 17914111, 54545436, 61484399, 81818169, 91728090, 466359999, 909091519, 909113679, 909156319, 911363679, 915636319, 999999998, 999999999
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OFFSET
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1,2
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COMMENTS
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The ratios for the numbers in the list are: 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 7, 10, 19, 19, 37, 37, 40, 67, 97, 136, 151, 325, 505, 526, 737, 1363, 2497, 2584, 3789, 6313, 9469, 12507, 12727, 12784, 20451, 33670, 43214, 154481, 280413, 284734, 814111, 905187
The sequence is infinite, since it contains all the numbers of the form 10^(3^k)-1 and 10^(3^k)-2. For these numbers the ratio is (10^(3^k)-1)/(9*3^k), which is integer because in general x^(3^k)-1 can be factored into (x-1)(x^2+x+1)(x^6+x^3+1)(x^18+x^9+1)...(x^(2*3^(k-1))+x^(3^(k-1))+1) and since here x=10, x-1=9 and each of the following k factors is divisible by 3 because its sum of digits is 3, thus 10^(3^k)-1 is divisible by 9*3^k. - Giovanni Resta, Apr 27 2016
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LINKS
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FORMULA
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EXAMPLE
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The sum of the digits of the numbers from 1 to 80 is 648; the sum of the numbers from 1 to 80 is 80*81/2 = 3240 and 3240 / 648 = 5
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MAPLE
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P:=proc(q) local b, k; global a, n; a:=0; for n from 1 to q do
b:=n; for k from 1 to ilog10(n)+1 do a:=a+(b mod 10); b:=trunc(b/10); od;
if type(n*(n+1)/(2*a), integer) then print(n); fi; od; end: P(10^9);
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MATHEMATICA
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With[{nn=10^9}, Position[Thread[{Accumulate[Range[nn]], Accumulate[ Table[ Total[ IntegerDigits[n]], {n, nn}]]}], _?(Divisible[#[[1]], #[[2]]]&), 1, Heads->False]]//Flatten (* Harvey P. Dale, Sep 30 2017 *)
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PROG
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(PARI) list(lim)=my(v=List(), s, t); for(n=1, lim, s+=n; t+=sumdigits(n); if(s%t==0, listput(v, n))); Vec(v) \\ Charles R Greathouse IV, Apr 29 2016
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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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EXTENSIONS
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STATUS
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approved
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