A298977 -id:A298977 - cp's OEIS frontend

A294090 Base-10 complementary numbers: n equals the product of the 10's complement of its digits.

5, 18, 35, 50, 180, 315, 350, 500, 1800, 3150, 3500, 5000, 18000, 31500, 35000, 50000, 180000, 315000, 350000, 500000, 1800000, 3150000, 3500000, 5000000, 18000000, 31500000, 35000000, 50000000, 180000000, 315000000, 350000000, 500000000, 1800000000

Offset: 1

M. F. Hasler, Feb 09 2018

The only primitive terms of the sequence, i.e., not equal to 10 times a smaller term, are 5, 18, 35 and 315.
For base 2, 3, 4 and 5, the corresponding sequences are less interesting: b = 2 yields powers of 2, A000079; b = 3 yields 4 times powers of 3, A003946 \ {1}; b = 4 yields {2, 6}*{4^k, k>=0} = A122756 = 2*A084221; b = 5 yields 8*{5^k, k>=0} = A128625 \ {1}.
See A298976 for base-6 complementary numbers. Base 7 yields {12, 120}*{7^k, k>=0}, cf. A298977. The linked web page (in French) gives also examples for base-100 complementary numbers, e.g., 198 = (100 - 1)*(100 - 98), 1680 = (100 - 16)*(100 - 80), ..., and for base-1000 complementary numbers.

			5 = (10-5), therefore 5 is in the sequence.
18 = (10-1)*(10-8), therefore 18 is in the sequence.
35 = (10-3)*(10-5), therefore 35 is in the sequence.
315 = (10-3)*(10-1)*(10-5), therefore 315 is in the sequence.
If x is in the sequence, then 10*x = concat(x,0) = x*(10-0) is in the sequence.

Colin Barker, Table of n, a(n) for n = 1..1000
G. Villemin, Nombres complémentés (in French).
Index entries for linear recurrences with constant coefficients, signature (0,0,0,10).

Cf. A298976, A298977.

LinearRecurrence[{0,0,0,10},{5,18,35,50,180,315},40] (* Harvey P. Dale, Mar 02 2024 *)

is(n,b=10)={n==prod(i=1,#n=digits(n,b),b-n[i])}

a(n)=if(n>6,a((n-3)%4+3)*10^((n-3)\4),[5,18,35,50,180,315][n])

Vec(x*(5 + 18*x + 35*x^2 + 50*x^3 + 130*x^4 + 135*x^5) / (1 - 10*x^4) + O(x^60)) \\ Colin Barker, Feb 09 2018

a(n+4) = 10 a(n) for all n >= 3.

G.f.: x*(5 + 18*x + 35*x^2 + 50*x^3 + 130*x^4 + 135*x^5) / (1 - 10*x^4). - Colin Barker, Feb 09 2018

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

Original entry on oeis.org

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

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A294090 Base-10 complementary numbers: n equals the product of the 10's complement of its digits.

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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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