%I #21 Mar 02 2024 15:21:33

%S 5,18,35,50,180,315,350,500,1800,3150,3500,5000,18000,31500,35000,

%T 50000,180000,315000,350000,500000,1800000,3150000,3500000,5000000,

%U 18000000,31500000,35000000,50000000,180000000,315000000,350000000,500000000,1800000000

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

%C The only primitive terms of the sequence, i.e., not equal to 10 times a smaller term, are 5, 18, 35 and 315.

%C 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}.

%C 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.

%H Colin Barker, <a href="/A294090/b294090.txt">Table of n, a(n) for n = 1..1000</a>

%H G. Villemin, <a href="http://villemin.gerard.free.fr/aNombre/TYPFORM/Compleme.htm">Nombres complémentés</a> (in French).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,10).

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

%F 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

%e 5 = (10-5), therefore 5 is in the sequence.

%e 18 = (10-1)*(10-8), therefore 18 is in the sequence.

%e 35 = (10-3)*(10-5), therefore 35 is in the sequence.

%e 315 = (10-3)*(10-1)*(10-5), therefore 315 is in the sequence.

%e If x is in the sequence, then 10*x = concat(x,0) = x*(10-0) is in the sequence.

%t LinearRecurrence[{0,0,0,10},{5,18,35,50,180,315},40] (* _Harvey P. Dale_, Mar 02 2024 *)

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

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

%o (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

%Y Cf. A298976, A298977.

%K nonn,base,easy

%O 1,1

%A _M. F. Hasler_, Feb 09 2018