A298976 -id:A298976

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A294090

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

+10
3

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

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.

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).

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

EXAMPLE

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.

MATHEMATICA

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

PROG

(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

CROSSREFS

Cf. A298976, A298977.

KEYWORD

nonn,base,easy

AUTHOR

M. F. Hasler, Feb 09 2018

STATUS

approved

A298977

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

+10
3

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

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.

LINKS

Colin Barker, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (0, 7).

FORMULA

a(n+2) = 7 a(n) for all n >= 1.

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)

EXAMPLE

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.

PROG

(PARI) 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)

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

CROSSREFS

Cf. A294090, A298976.