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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
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
Sequence in context: A141521 A080906 A358421 * A215163 A089685 A186883
KEYWORD
nonn,full,fini,base
AUTHOR
Derek Orr, Feb 12 2014
STATUS
approved

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Last modified June 20 23:53 EDT 2024. Contains 373535 sequences. (Running on oeis4.)