%I
%S 23,37,56,56,866
%N Least initial number of n consecutive integers that are not divisible by any of their nonzero digits.
%C This sequence is complete. If a(6) were to exist, the 6 numbers would have to end in either {1,2,3,4,5,6}, {2,3,4,5,6,7}, {3,4,5,6,7,8}, {4,5,6,7,8,9}, {5,6,7,8,9,0}, {6,7,8,9,0,1}, {7,8,9,0,1,2}, {8,9,0,1,2,3}, {9,0,1,2,3,4}, or {0,1,2,3,4,5}. However, if the number has a 1 as a digit, it cannot be one of the consecutive integers. Also, if a number has a 5 as its last digit, it cannot be one of the consecutive integers. Thus, none of these sets could work.
%C If all numbers were distinct and nontrivial, a(4) would be 586 (the trivial numbers after 56 are 506 and 556).
%e 23 is the first number that is not divisible by either of its digits.
%e 37 and 38 are the first two consecutive numbers that are not divisible by any of their digits. Thus, a(2) = 37.
%e 56, 57, 58 (and 59) are the first three (and four) consecutive numbers that are not divisible by any of their digits. Thus, a(3) = a(4) = 56.
%e 866, 867, 868, 869, and 870 are the first five consecutive numbers that are not divisible by any of their digits. Thus, a(5) = 866.
%o (Python)
%o def DivDig(x):
%o ..total = 0
%o ..for i in str(x):
%o ....if i != '0':
%o ......if x/int(i) % 1 == 0:
%o ........return True
%o ..return False
%o def Nums(x):
%o ..n = 1
%o ..while n < 10**3:
%o ....count = 0
%o ....for i in range(n,n+x):
%o ......if not DivDig(i):
%o ........count += 1
%o ......else:
%o ........break
%o ....if count == x:
%o ......return n
%o ....else:
%o ......n += 1
%o x = 1
%o while x < 10:
%o ..print(Nums(x))
%o ..x += 1
%Y Cf. A038772, A005349.
%K nonn,full,fini,base
%O 1,1
%A _Derek Orr_, Feb 12 2014
