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A274766
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Multiplication of pair of contiguous repunits, i.e., (0*1, 1*11, 11*111, 111*1111, 1111*11111, ...).
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2
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0, 11, 1221, 123321, 12344321, 1234554321, 123456654321, 12345677654321, 1234567887654321, 123456789987654321, 12345679010987654321, 1234567901220987654321, 123456790123320987654321, 12345679012344320987654321, 1234567901234554320987654321
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OFFSET
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0,2
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COMMENTS
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From the second to the tenth term they look like in A259937, but it is a completely different sequence.
The inverse of sequence terms, except the first one, give a sequence of periodic terms with periods as in A002378, the sequence of oblong (or promic, or heteromecic) numbers: a(n) = n*(n+1). Digit string period L of inverse a(n) is given by L = n*(n+1).
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LINKS
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FORMULA
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O.g.f.: 11*x/((1 - x)*(1 - 10*x)*(1 - 100*x)).
E.g.f.: (1 - 11*exp(9*x) + 10*exp(99*x))*exp(x)/81.
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n>2, a(0)=0, a(1)=11, a(2)=1221.
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EXAMPLE
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a(10) = rep(10)*rep(11) = 12345679010987654321, digit string period of 1/a(10) -> L = 10*11 = 110.
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MATHEMATICA
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Table[(10^n - 1) (10^(n + 1) - 1)/81, {n, 0, 20}] (* Bruno Berselli, Jul 05 2016 *)
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PROG
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(PARI) concat(0, Vec(11*x/((1-x)*(1-10*x)*(1-100*x)) + O(x^99))) \\ Altug Alkan, Jul 05 2016
(PARI) a(n) = (1-11*10^n+10^(1+2*n))/81 \\ Colin Barker, Jul 05 2016
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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