A295638 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.
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, 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, 9, 66, 7, 8, 9, 77, 8, 9, 88, 9, 99, 666, 7, 8
Offset: 1
Examples
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.
Links
- Gunnar Lee Johnson, Table of n, a(n) for n = 1..255
Crossrefs
The nonnegative integers minus A009993 is the sequence that is partitioned.
Programs
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PARI
is(n) = my(d=digits(n)); d != vecsort(d,,8); 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 -
Python
def a(n): (x,i,count,switch) = (0,0,1,True) while True: if switch == (list(sorted(set(str(i)))) == list(str(i))): count += 1 else: if not switch: x += 1 if x == n: return count (count, switch) = (1, not switch) i += 1
Comments