

A050289


Zeroless pandigital numbers: numbers containing the digits 19 (each appearing at least once) and no 0's.


48



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
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OFFSET

1,1


COMMENTS

The first 9! = 362880 terms of this sequence are permutations of the digits 19 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
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



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, n1)))+(n1)\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))


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



