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%I #13 Dec 19 2022 16:54:08
%S 11,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40,41,
%T 42,43,44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,
%U 65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,83,84,85,86,87,88,89,90,91,92,93,94,95,96,97,98,99,101
%N "False so far" sequence.
%C The last digit of a(n) is regarded as a glyph and the preceding digits as a number. So "11" reads [one "1"] and "20" [two "0"] - which are both false statements here: there is not one "1" glyph so far in the sequence when [11] is read, but two; and there are not two "0" glyph when [20] is read, but only one. The sequence is built with [a(n+1)-a(n)] minimal and a(n+1) always "false so far". This explains why the sequence doesn't begin with [10]: its statement would be true.
%C After integer [20] the sequence matches A000027 without [100] -- because [100] would be "true so far".
%C [This seems to imply that the sequence contains all numbers >= 20 except 100. - _N. J. A. Sloane_, Aug 22 2011]
%C The above is correct, and a(n) = n + 19 for n > 81. Proof: It is enough to show that all terms 101 to N are in the sequence, that N >= 101 + 100, and that the count of digits through N is at least floor(N/10) + 11 for each decimal digit. Note that, from the last two digits, the count of each digit increases by at least 20 every 100 numbers, but the required number increases by only 10. Then by a count of the digits, N = 210 suffices. It is easy to check that the remaining conditions are satisfied. - _Charles R Greathouse IV_, Dec 19 2022
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1).
%F a(n) = n + 19 for n > 81, see comments. - _Charles R Greathouse IV_, Dec 19 2022
%o (PARI) a(n)=n+if(n>81,19,n>1,18,10) \\ _Charles R Greathouse IV_, Dec 19 2022
%Y Cf. A102357.
%K base,easy,nonn
%O 1,1
%A _Eric Angelini_, Feb 26 2005
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