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A274743
Repunits with odd indices multiplied by 99, i.e., 99*(1, 111, 11111, 1111111, ...).
3
99, 10989, 1099989, 109999989, 10999999989, 1099999999989, 109999999999989, 10999999999999989, 1099999999999999989, 109999999999999999989, 10999999999999999999989, 1099999999999999999999989, 109999999999999999999999989, 10999999999999999999999999989
OFFSET
1,1
COMMENTS
It is apparent that the reciprocals of the terms in the sequence give an increasing sequence of periodic terms similar to A095372, but with the initial term equal to "01". The leading zero is important (see links). Furthermore, the reciprocals of the terms give a sequence of even growing periods, starting from 2, with delta = 4 (i.e., 2, 6, 10, 14, 18, ...).
Adding "11" to each term gives the binary representation of the n-th iteration of "Rule 14" elementary cellular automaton starting with a single ON (black cell) as in A266299.
FORMULA
a(n) = 101*a(n-1) - 100*a(n-2) for n>2, with a(0)= 99 and a(1)= 10989.
a(n) = 99*A100706(n-1).
G.f.: 99*x*(1 + 10*x)/((1 - x)*(1 - 100*x)). - Ilya Gutkovskiy, Jul 04 2016
a(n) = 11*(10^(2*n-1)-1). - Wesley Ivan Hurt, Jul 04 2016
E.g.f.: 11*(9 - 10*exp(x) + exp(100*x))/10. - Stefano Spezia, Aug 05 2024
EXAMPLE
a(2) = 101*10989 - 100*99 = 1099989.
MAPLE
A274743:=n->11*(10^(2*n-1)-1): seq(A274743(n), n=1..20); # Wesley Ivan Hurt, Jul 04 2016
MATHEMATICA
Array[99(10^(2 # - 1) - 1)/9 &, 15] (* Michael De Vlieger, Jul 04 2016 *)
99*Table[FromDigits[PadRight[{}, 2n+1, 1]], {n, 0, 15}] (* Harvey P. Dale, Jul 22 2019 *)
PROG
(Magma) [11*(10^(2*n-1)-1) : n in [1..20]]; // Wesley Ivan Hurt, Jul 04 2016
(PARI) Vec(99*x*(1+10*x)/((1-x)*(1-100*x)) + O(x^99)) \\ Altug Alkan, Jul 05 2016
(PARI) a002275(n) = (10^n-1)/9
a(n) = 99*a002275(2*n-1) \\ Felix Fröhlich, Jul 05 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Rodolfo A. Fiorini, Jul 04 2016
STATUS
approved