A298977 Base-7 complementary numbers: n equals the product of the 7 complement (7-d) of its base-7 digits d.
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
Offset: 1
Examples
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.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (0, 7).
Programs
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Mathematica
LinearRecurrence[{0, 7}, {12, 84, 120}, 30] (* Paolo Xausa, Jul 12 2025 *)
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PARI
is(n,b=7)={n==prod(i=1,#n=digits(n,b),b-n[i])}
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PARI
a(n)=[84,120][n%2+(n>1)]*7^(n\2-1)
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PARI
Vec(12*x*(1 + 7*x + 3*x^2) / (1 - 7*x^2) + O(x^60)) \\ Colin Barker, Feb 10 2018
Formula
a(n+2) = 7 a(n) for all n >= 2.
From Colin Barker, Feb 10 2018: (Start)
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)
Extensions
More terms from Colin Barker, Feb 10 2018
Comments