A050289 Zeroless pandigital numbers: numbers containing the digits 1-9 (each appearing at least once) and no 0's.

123456789, 123456798, 123456879, 123456897, 123456978, 123456987, 123457689, 123457698, 123457869, 123457896, 123457968, 123457986, 123458679, 123458697, 123458769, 123458796, 123458967, 123458976, 123459678, 123459687, 123459768, 123459786, 123459867, 123459876, 123465789

Offset: 1

Author

Eric W. Weisstein

Keywords

• nonn
• base

Comments

The first 9! = 362880 terms of this sequence are permutations of the digits 1-9 with a(9!) = 987654321. - Jeremy Gardiner, May 28 2010

First differences are given in A209280 (for the first 9! terms) or in A219664 (for at least as much initial terms). - M. F. Hasler, Mar 03 2013

A230959(a(n)) = 0. - Reinhard Zumkeller, Nov 02 2013

After the first 9! terms, 8! + 7! = 9*7! of the initial terms are repeated with a leading '1' prefixed, cf. formula. However, a(9!+8!+7!) = 1219...3 is followed by 122...9 and permutations of the last 7 digits, before 12314..9. - M. F. Hasler, Jan 08 2020, corrected Aug 11 2022 thanks to a remark from Michael S. Branicky

Links

• H. Fripertinger, Operate on "9" to display zeroless pandigitals
• James Grime and Brady Haran, Why 381,654,729 is awesome, Numberphile video (2013).
• Eric Weisstein's World of Mathematics, Pandigital Number
• Chai Wah Wu, Pandigital and penholodigital numbers, arXiv:2403.20304 [math.GM], 2024. See p. 1.

Crossrefs

Cf. A050290, A133360, A171102.

Programs

PARI

apply( {A050289(n)=if(n<=7!*81, fromdigits(Vec(numtoperm(9,n-1)))+(n-1)\9!*10^9, "not yet implemented")}, [1..25]) \\ M. F. Hasler, Jan 07 2020, corrected Aug 11 2022

Python

from itertools import count, islice, permutations, product
def c(t): return len(set(t)) == 9
def t2i(t): return int("".join(map(str, t)))
def agen():
    yield from (t2i(p) for p in permutations(range(1, 10)))
    for d in count(10):
        yield from (t2i(p) for p in product(range(1, 10), repeat=d) if c(p))
print(list(islice(agen(), 25))) # Michael S. Branicky, May 30 2022, updated Aug 05 2022

Formula

a(n + 9!) = a(n) + 10^9 for 1 <= n <= 8! + 7!. - M. F. Hasler, Jan 08 2020, corrected Aug 11 2022

Extensions

Name clarified by Michael S. Branicky, Aug 05 2022