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A366610 Number of successive occurrences of the same first digits in A366585.

%I A366610 #25 Nov 06 2023 17:22:35
%S A366610 1,1,1,1,4,5,4,2,2,2,2,1,1,97,50,34,25,20,17,14,13,11,995,500,334,250,
%T A366610 200,167,143,125,111,9996,5000,3334,2500,2000,1667,1428,1250,1112,
%U A366610 99992,50000,33334,25000,20000,16667,14286,12500,11111,999995,500000,333334,250000,200000,166667,142857
%N A366610 Number of successive occurrences of the same first digits in A366585.
%H A366610 Chai Wah Wu, <a href="/A366610/b366610.txt">Table of n, a(n) for n = 1..2000</a>
%e A366610 a(6) = 5 because in A366585, the sixth run of same first digit of the terms lasts through five terms: 2021, 2223, 2425, 2627, 2829, each beginning with 2.
%o A366610 (PARI) lista(nn) = my(list=List(), k=1); while (k < nn, my(s="", n=digits(k)[1]); for (j=1, n, s = concat(s, Str(k+j-1));); listput(list, eval(s)); k += n;); my(w = apply(x->digits(x)[1], list)); my(list=List(), last=w[1], nb=1); for (i=2, #w, if (w[i] == last, nb++, listput(list, nb); last=w[i]; nb=1;);); Vec(list, #list-1); \\ _Michel Marcus_, Oct 15 2023
%o A366610 (Python)
%o A366610 from itertools import islice
%o A366610 def A366610_gen(): # generator of terms
%o A366610     a, c = 1, 0
%o A366610     while (c:=c+1):
%o A366610         yield c
%o A366610         c, r = divmod((w:=((t:=int(str(a:=a+int(str(a)[0]))[0]))+1)*10**(len(str(a))-1)-1)-a,t)
%o A366610         a = w-r
%o A366610 A366610_list = list(islice(A366610_gen(),30)) # _Chai Wah Wu_, Nov 04 2023
%Y A366610 Cf. A366585.
%K A366610 nonn,base
%O A366610 1,5
%A A366610 _Tamas Sandor Nagy_, Oct 14 2023
%E A366610 More terms from _Michel Marcus_, Oct 15 2023