A294090 -id:A294090 - cp's OEIS frontend

A298976 Base-6 complementary numbers: n equals the product of the 6 complement (6-d) of its base-6 digits d.

3, 10, 18, 60, 80, 108, 360, 480, 648, 2160, 2880, 3888, 12960, 17280, 23328, 77760, 103680, 139968, 466560, 622080, 839808, 2799360, 3732480, 5038848, 16796160, 22394880, 30233088, 100776960, 134369280, 181398528, 604661760, 806215680, 1088391168

Offset: 1

M. F. Hasler, Feb 09 2018

The only primitive terms of the sequence, i.e., not equal to 6 times a smaller term, are a(1) = 3, a(2) = 10 and a(5) = 80.

See A294090 for the base-10 variant, which is the main entry for this family of sequences, and A298977 for the base-7 variant.

			3 = (6-3), therefore 3 is in the sequence.
Denoting xyz[6] the base-6 expansion (i.e., x*6^2 + y*6 + z), we have:
10 = 14[6] = (6-1)*(6-4), therefore 10 is in the sequence.
18 = 30[6] = (6-3)*(6-0), therefore 18 is in the sequence.
80 = 212[6] = (6-2)*(6-1)*(6-2), therefore 80 is in the sequence.
Since the expansion of 6*x in base 6 is that of x with a 0 appended, if x is in the sequence, then 6*x = x*(6-0) is in the sequence.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,6).

Cf. A294090, A298977.

LinearRecurrence[{0, 0, 6}, {3, 10, 18, 60, 80}, 50] (* Paolo Xausa, Jul 09 2025 *)

is(n,b=6)={n==prod(i=1,#n=digits(n,b),b-n[i])}

a(n)=if(n>5,a(n%3+3)*6^(n\3-1),[3,10,18,60,80][n])

Vec(x*(3 + 10*x + 18*x^2 + 42*x^3 + 20*x^4) / (1 - 6*x^3) + O(x^60)) \\ Colin Barker, Feb 09 2018

a(n+3) = 6 a(n) for all n >= 2.

G.f.: x*(3 + 10*x + 18*x^2 + 42*x^3 + 20*x^4) / (1 - 6*x^3). - Colin Barker, Feb 09 2018

A298977 Base-7 complementary numbers: n equals the product of the 7 complement (7-d) of its base-7 digits d.

Original entry on oeis.org

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

M. F. Hasler, Feb 09 2018

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.

			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.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0, 7).

Cf. A294090, A298976.

LinearRecurrence[{0, 7}, {12, 84, 120}, 30] (* Paolo Xausa, Jul 12 2025 *)

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)
```

Vec(12*x*(1 + 7*x + 3*x^2) / (1 - 7*x^2) + O(x^60)) \\ Colin Barker, Feb 10 2018

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)

More terms from Colin Barker, Feb 10 2018

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A298976 Base-6 complementary numbers: n equals the product of the 6 complement (6-d) of its base-6 digits d.

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A298977 Base-7 complementary numbers: n equals the product of the 7 complement (7-d) of its base-7 digits d.

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