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A294090
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Base-10 complementary numbers: n equals the product of the 10's complement of its digits.
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3
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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FORMULA
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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
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EXAMPLE
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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.
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MATHEMATICA
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LinearRecurrence[{0, 0, 0, 10}, {5, 18, 35, 50, 180, 315}, 40] (* Harvey P. Dale, Mar 02 2024 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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