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11, 1001, 100001, 10000001, 1000000001, 100000000001, 10000000000001, 1000000000000001, 100000000000000001, 10000000000000000001, 1000000000000000000001, 100000000000000000000001
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| These numbers are all divisible by 11. This follows from the identity
a^n-b^n = (a+b)(a^(n-1) - a^(n-2)b + ... + b^(n-1)) for odd values of n. In
this example a=10 and b=1 so a+b = 11. The sum of digits rule for divisibility
by 11 also applies.
Bisection of A000533. Also, bisection of A062397. a(n) is also A084508(n+1) written in base 2. a(n) is also A087289(n-1) written in base 2. a(n) is also the concatenation of "1", 2(n-1) digits "0" and "1". [From Omar E. Pol (info(AT)polprimos.com), Dec 13 2008]
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LINKS
| Index to sequences with linear recurrences with constant coefficients, signature (101,-100).
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FORMULA
| a(n)=100*a(n-1)-99 (with a(1)=11)[From Vincenzo Librandi, Dec 14 2010]
G.f. -11*x*(-1+10*x) / ( (100*x-1)*(x-1) ). - R. J. Mathar, Sep 01 2011
a(n) = 11*A095372(n-1). - R. J. Mathar, Sep 01 2011
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EXAMPLE
| Contribution from Omar E. Pol (info(AT)polprimos.com), Dec 14 2008: (Start)
n ....... a(n)
1 ....... 11
2 ...... 1001
3 ..... 100001
4 .... 10000001
5 ... 1000000001
(End)
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PROG
| (PARI) g(n)=forstep(x=1, n, 2, y=(10^x+1); print1(y", "))
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CROSSREFS
| Cf. A000533, A062397, A084508, A087289. [From Omar E. Pol (info(AT)polprimos.com), Dec 13 2008]
Sequence in context: A083816 A004656 A143016 * A163449 A127961 A127962
Adjacent sequences: A152574 A152575 A152576 * A152578 A152579 A152580
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KEYWORD
| nonn,easy
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AUTHOR
| Cino Hilliard (hillcino368(AT)hotmail.com), Dec 08 2008
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