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%I #16 Feb 28 2018 03:14:21
%S 2,3,4,5,6,7,8,9,33,4,5,6,7,8,9,44,5,6,7,8,9,55,6,7,8,9,66,7,8,9,77,8,
%T 9,88,9,99,444,5,6,7,8,9,55,6,7,8,9,66,7,8,9,77,8,9,88,9,99,555,6,7,8,
%U 9,66,7,8,9,77,8,9,88,9,99,666,7,8
%N Take the sequence of nonnegative integers whose decimal digits are not in strictly increasing order. Partition the sequence into subsequences whose elements are consecutive integers. Then a(n) is the number of elements in the n-th partition.
%C Only defined as an integer for a(1) through a(255), as a(256) references the infinite partition (123456790, 123456791, ..., 999999999, 1000000000, 1000000001, ...). No integer greater than 123456789 has a strictly increasing sequence of digits (itself being the only case for 9 digits, and by the pigeonhole principle, a >9-digit number must have a digit repeated and is thus not strictly increasing).
%H Gunnar Lee Johnson, <a href="/A295638/b295638.txt">Table of n, a(n) for n = 1..255</a>
%e For a(1)=2 through a(8)=9, these correspond to the consecutive subsequences (10, 11), (20, 21, 22), ..., (80, 81, 82, ..., 88). The jumps at e.g. a(9)=33 or a(37)=444 correspond to (90, 91, ..., 122) and (790, 791, ..., 1233), where 89 and 123, and 789 and 1234, are the values partitioning the subsequences.
%o (Python) def a(n):
%o (x,i,count,switch) = (0,0,1,True)
%o while True:
%o if switch == (list(sorted(set(str(i)))) == list(str(i))):
%o count += 1
%o else:
%o if not switch: x += 1
%o if x == n: return count
%o (count, switch) = (1, not switch)
%o i += 1
%o (PARI) is(n) = my(d=digits(n)); d != vecsort(d,,8);
%o lista(nn) = {my(w = select(n->is(n), vector(nn, k, k))); my(dw = vector(#w-1, k, w[k+1] - w[k])); my(k = 1); for (n=1, #dw, if (dw[n] == 1, k++, print1(k, ", "); k = 1););} \\ _Michel Marcus_, Jan 08 2018
%Y The nonnegative integers minus A009993 is the sequence that is partitioned.
%K nonn,fini,less,base
%O 1,1
%A _Gunnar Lee Johnson_, Nov 24 2017