OFFSET
1,1
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.
FORMULA
a(n + 9!) = a(n) + 10^9 for 1 <= n <= 8! + 7!. - M. F. Hasler, Jan 08 2020, corrected Aug 11 2022
PROG
(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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
EXTENSIONS
Name clarified by Michael S. Branicky, Aug 05 2022
STATUS
approved