cp's OEIS Frontend

A366585 Write down the positive integers. To obtain the terms of the sequence, concatenate groups of these so the first digit of each term is the number of consecutive numbers that are concatenated.

%I A366585 #22 Nov 06 2023 07:18:27
%S A366585 1,23,4567,89101112131415,16,17,18,19,2021,2223,2425,2627,2829,303132,
%T A366585 333435,363738,394041,42434445,46474849,5051525354,5556575859,
%U A366585 606162636465,666768697071,72737475767778,79808182838485,8687888990919293,949596979899100101102,103,104,105,106,107
%N A366585 Write down the positive integers. To obtain the terms of the sequence, concatenate groups of these so the first digit of each term is the number of consecutive numbers that are concatenated.
%C A366585 Terms beginning with 1 will appear in runs of long linear succession. These are interspersed with clusters of values that tend to increase at other, uneven rates in the sequence.
%o A366585 (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;); Vec(list); \\ _Michel Marcus_, Oct 14 2023
%o A366585 (Python)
%o A366585 from itertools import islice
%o A366585 def A366585_gen(): # generator of terms
%o A366585     a = 1
%o A366585     while True:
%o A366585         c = a+int(str(a)[0])
%o A366585         yield int(''.join(str(i) for i in range(a,c)))
%o A366585         a = c
%o A366585 A366585_list = list(islice(A366585_gen(),20)) # _Chai Wah Wu_, Nov 04 2023
%Y A366585 Cf. A366610, A053067.
%K A366585 nonn,base
%O A366585 1,2
%A A366585 _Tamas Sandor Nagy_, Oct 14 2023