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A237766 Least initial number of n consecutive integers that are not divisible by any of their nonzero digits. 4
23, 37, 56, 56, 866 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

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.

If all numbers were distinct and nontrivial, a(4) would be 586 (the trivial numbers after 56 are 506 and 556).

LINKS

Table of n, a(n) for n=1..5.

EXAMPLE

23 is the first number that is not divisible by either of its digits.

37 and 38 are the first two consecutive numbers that are not divisible by any of their digits. Thus, a(2) = 37.

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.

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.

PROG

(Python)

def DivDig(x):

..total = 0

..for i in str(x):

....if i != '0':

......if x/int(i) % 1 == 0:

........return True

..return False

def Nums(x):

..n = 1

..while n < 10**3:

....count = 0

....for i in range(n, n+x):

......if not DivDig(i):

........count += 1

......else:

........break

....if count == x:

......return n

....else:

......n += 1

x = 1

while x < 10:

..print(Nums(x))

..x += 1

CROSSREFS

Cf. A038772, A005349.

Sequence in context: A124888 A141521 A080906 * A215163 A089685 A186883

Adjacent sequences:  A237763 A237764 A237765 * A237767 A237768 A237769

KEYWORD

nonn,full,fini,base

AUTHOR

Derek Orr, Feb 12 2014

STATUS

approved

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Last modified June 24 20:32 EDT 2022. Contains 354830 sequences. (Running on oeis4.)