A054383 Number of (zeroless) pandigital fractions for 1/n.

0, 12, 2, 4, 12, 3, 7, 46, 3, 0, 0, 4, 3, 8, 2, 3, 27, 1, 2, 0, 0, 1, 3, 2, 0, 9, 4, 1, 2, 0, 0, 1, 0, 0, 5, 0, 1, 2, 0, 0, 0, 0, 1, 5, 0, 1, 0, 0, 0, 0, 0, 1, 4, 0, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• base

Comments

a(n) is also the number of k such that k and n*k, taken together, are zeroless pandigital. - Nathaniel Johnston, Jun 25-26 2011

There are 179540 nonzero terms in the sequence. The largest n for which a(n) > 0 is 98765432 representing the pandigital fraction 1/98765432. The largest a(n) is a(8) = 46. - Chai Wah Wu, May 23 2015

Examples

			a(3)=2 since there are 2 such pandigital fractions for 1/3: 5823/17469 and 5832/17496.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Pandigital Fraction.

Crossrefs

Cf. A064160, A115927, A115929, A115930, A115931, A115932.

Programs

Python

from itertools import permutations
l = {}
for d in permutations('123456789', 9):
    for i in range(8):
        s1, s2 = int(''.join(d[:i+1])), int(''.join(d[i+1:]))
        q, r = divmod(s1,s2)
        if not r:
            if q in l:
                l[q] += 1
            else:
                l[q] = 1
A054383_list = [0]*max(l)
for d in l:
    A054383_list[d-1] = l[d] # Chai Wah Wu, May 23 2015

Extensions

More terms from Vit Planocka (planocka(AT)mistral.cz), Sep 21 2003