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A178485
(A178475(n)-6)/9.
5
1371, 1372, 1381, 1383, 1392, 1393, 1471, 1472, 1491, 1494, 1502, 1504, 1581, 1583, 1591, 1594, 1613, 1614, 1692, 1693, 1702, 1704, 1713, 1714, 2371, 2372, 2381, 2383, 2392, 2393, 2571, 2572, 2601, 2605, 2612, 2615, 2681, 2683, 2701, 2705, 2723, 2725
OFFSET
1,1
COMMENTS
There are 5!=120 terms in this finite sequence. Its origin is the fact that numbers whose decimal expansion is a permutation of 12345 are all of the form 9k+6.
LINKS
Nathaniel Johnston, Table of n, a(n) for n = 1..120 (full sequence)
FORMULA
a(n) + a(5!+1-n) = 7406.
a(n) == 1, 2, 3, 4 or 5 (mod 10).
a(n+6)-a(n) is an element of { 100, 110, 111, 200, 220, 222, 679 }.
a(n+6)-a(n) = 679 iff (n-1)%24 > 17, where % denotes the remainder upon division.
a(n+6)-a(n) = 200, 220 or 222 iff (n-1)%30 > 23, i.e. n==25,...,30 (mod 30).
PROG
(PARI) v=vector(5, i, 10^(i-1))~; vecsort(vector(5!, i, numtoperm(5, i)*v))
is_A178475(x)= { vecsort(Vec(Str(x)))==Vec("12345") }
forstep( m=12345, 54321, 9, is_A178475(m) & print1(m", "))
KEYWORD
fini,full,nonn,base,easy
AUTHOR
M. F. Hasler, May 28 2010
STATUS
approved