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A050289
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Zeroless pandigital numbers: numbers containing the digits 1-9 (each appearing at least once) and no 0's.
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48
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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
(list;
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refs;
listen;
history;
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internal format)
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OFFSET
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1,1
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COMMENTS
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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
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
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LINKS
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FORMULA
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a(n + 9!) = a(n) + 10^9 for 1 <= n <= 8! + 7!. - M. F. Hasler, Jan 08 2020, corrected Aug 11 2022
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PROG
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(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))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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