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A237766 Least initial number of n consecutive integers that are not divisible by any of their nonzero digits. 4

%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

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Last modified August 10 12:45 EDT 2022. Contains 356039 sequences. (Running on oeis4.)