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Non-monotonic "True so far" sequence: In the first n terms, the digit (a(n) mod 10) occurs floor(a(n)/10) times; a(n) is the smallest such number.
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%I #14 Nov 21 2019 04:39:59

%S 10,12,13,14,15,16,17,18,19,20,23,24,25,26,27,28,29,30,34,35,36,37,38,

%T 39,40,45,46,47,48,49,50,56,57,58,59,60,67,68,69,70,78,79,80,89,90,

%U 102,103,104,105,106,107,108,109,112,113,114,115,116,117,118,119,123,124

%N Non-monotonic "True so far" sequence: In the first n terms, the digit (a(n) mod 10) occurs floor(a(n)/10) times; a(n) is the smallest such number.

%C Sequence has 5191475 terms. The numbers of occurrences of digits 0-9 are 3589309, 4812817, 4977431, 4564762, 3741602, 3738734, 3599425, 3599878, 3598956, 3589537.

%C This sequence first differs from the original "True so far" sequence A102357 at a(351) = 920 because this is the first term that is less than the previous term, 1002.

%C The sequence is injective (no term appears twice) as consequence of the definition, while this is imposed through monotonicity in A102357. - _M. F. Hasler_, Nov 18 2019

%H M. F. Hasler, <a href="/A102850/b102850.txt">Table of n, a(n) for n = 1..10000</a>, Nov 18 2019

%e a(10) = 20 because up to this point there are two 0 digits in the sequence, including the 0 in 20.

%e a(5191476) doesn't exist. 35893100 would yield a total of 3589311 0's, while 35893110 or 35893120 would yield 3589310 0's. Similar reasons prevent other terms ending with other digits.

%o (PARI) c=Vec(0,10); for(n=1,351, a=vecmin(c)*10+10; while(a\10<=c[a%10+1] || a\10 != c[a%10+1]+#select(d->d==a%10,digits(a)), a++); [c[d+1]++|d<-digits(a)]; print1(a",")) \\ _M. F. Hasler_, Nov 18 2019

%Y Cf. A102357.

%K base,easy,fini,nonn,less

%O 1,1

%A _David Wasserman_, Feb 28 2005