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A274766
Multiplication of pair of contiguous repunits, i.e., (0*1, 1*11, 11*111, 111*1111, 1111*11111, ...).
2
0, 11, 1221, 123321, 12344321, 1234554321, 123456654321, 12345677654321, 1234567887654321, 123456789987654321, 12345679010987654321, 1234567901220987654321, 123456790123320987654321, 12345679012344320987654321, 1234567901234554320987654321
OFFSET
0,2
COMMENTS
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).
FORMULA
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.
a(n) = (10^n - 1)*(10^(n+1) - 1)/81 = A002275(n)*A002275(n+1).
EXAMPLE
a(10) = rep(10)*rep(11) = 12345679010987654321, digit string period of 1/a(10) -> L = 10*11 = 110.
MATHEMATICA
Table[(10^n - 1) (10^(n + 1) - 1)/81, {n, 0, 20}] (* Bruno Berselli, Jul 05 2016 *)
PROG
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Rodolfo A. Fiorini, Jul 05 2016
EXTENSIONS
Edited and added formulae by Bruno Berselli, Jul 05 2016
Last term corrected by Colin Barker, Jul 05 2016
STATUS
approved