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A343839
Semi-one numbers: Positive integers k such that exactly half of the integers 1..k have a 1 in their decimal expansion.
2
2, 16, 24, 160, 270, 272, 1456, 3398, 3418, 3420, 3422, 13120, 44686, 118096, 674934, 1062880
OFFSET
1,1
COMMENTS
There are only these 16 semi-one numbers. That there can only be finitely many is fairly easy to show: consider how many 100-billion-digit numbers have no 1s in them. Eventually the proportion of 1-less numbers drops below 50% and stays there. 5217031 numbers up to 9999999 have a 1 in them so proportion of 1-ful numbers can't drop below 50% for numbers with more digits. Hence the search program can stop at 10 million.
LINKS
Gary Gordon and Glen Whitney, The Playground, Math Horizons vol XXV issue 4, April 2018, Problem C21, semi-one numbers (and other semi-n).
EXAMPLE
16 is semi-1 because 1,10,11,12,13,14,15,16 have a 1 in them, there are 8 such numbers, and 8 is half of 16. 2 is semi-1 because 1 has a 1 in it and 2 does not.
MATHEMATICA
s = {}; c = 0; Do[If[DigitCount[n, 10, 1] > 0, c++]; If[n == 2*c, AppendTo[s, n]], {n, 1, 1062880}]; s (* Amiram Eldar, May 01 2021 *)
With[{nn=11*10^5}, Select[Partition[Riffle[Range[nn], Accumulate[Table[If[DigitCount[n, 10, 1]>0, 1, 0], {n, nn}]]], 2], #[[1]]==2#[[2]]&]][[;; , 1]] (* Harvey P. Dale, Jun 23 2023 *)
PROG
(Perl)
for (1..10000000) {
if (/1/) {
$s++;
}
if ($_==2*$s) {
print $_."\n";
}
}
(PARI) lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, #va, va[n] = va[n-1] + (#select(x->(x==1), digits(n)) > 0); ); for (n=1, nn, if (va[n] == n/2 , print1(n, ", ")); ); } \\ Michel Marcus, May 02 2021
CROSSREFS
Sequence in context: A337426 A295822 A212765 * A019317 A355714 A344989
KEYWORD
nonn,base,fini,full
AUTHOR
Adam Atkinson, May 01 2021
STATUS
approved