|
|
A298977
|
|
Base-7 complementary numbers: n equals the product of the 7 complement (7-d) of its base-7 digits d.
|
|
3
|
|
|
12, 84, 120, 588, 840, 4116, 5880, 28812, 41160, 201684, 288120, 1411788, 2016840, 9882516, 14117880, 69177612, 98825160, 484243284, 691776120, 3389702988, 4842432840, 23727920916, 33897029880, 166095446412, 237279209160, 1162668124884, 1660954464120
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
The only primitive terms of the sequence, i.e., not equal to 7 times a smaller term, are a(1) = 12 and a(3) = 120.
See A294090 for the base-10 variant, which is the main entry, and A298976 for the base-6 variant.
|
|
LINKS
|
|
|
FORMULA
|
a(n+2) = 7 a(n) for all n >= 1.
G.f.: 12*x*(1 + 7*x + 3*x^2) / (1 - 7*x^2).
a(n) = 12*7^(n/2) for n>1 and even.
a(n) = 120*7^((n-3)/2) for n>1 and odd.
(End)
|
|
EXAMPLE
|
Denoting xyz[7] the base-7 expansion (of n = x*7^2 + y*7 + z), we have:
12 = 15[7] = (7-1)*(7-5), therefore 12 is in the sequence.
84 = 150[7] = (7-1)*(7-5)*(7-0), therefore 84 is in the sequence.
120 = 231[7] = (7-2)*(7-3)*(7-1), therefore 120 is in the sequence.
Since the expansion of 7*x in base 7 is that of x with a 0 appended, if x is in the sequence, then 7*x = x*(7-0) is in the sequence.
|
|
PROG
|
(PARI) is(n, b=7)={n==prod(i=1, #n=digits(n, b), b-n[i])}
(PARI) a(n)=[84, 120][n%2+(n>1)]*7^(n\2-1)
(PARI) Vec(12*x*(1 + 7*x + 3*x^2) / (1 - 7*x^2) + O(x^60)) \\ Colin Barker, Feb 10 2018
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|