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

Author

M. F. Hasler, Feb 09 2018

Keywords

• nonn
• base
• easy

Comments

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.

Examples

			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.
		

Links

• 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).

Crossrefs

Cf. A298976, A298977.

Programs

Mathematica

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

PARI

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

PARI

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

PARI

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

Formula

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