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A015446
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Generalized Fibonacci numbers: a(n) = a(n-1) + 10*a(n-2).
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16
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1, 1, 11, 21, 131, 341, 1651, 5061, 21571, 72181, 287891, 1009701, 3888611, 13985621, 52871731, 192727941, 721445251, 2648724661, 9863177171, 36350423781, 134982195491, 498486433301, 1848308388211, 6833172721221
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OFFSET
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0,3
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COMMENTS
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The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=2, 11*a(n-2) equals the number of 11-colored compositions of n with all parts >=2, such that no adjacent parts have the same color. - Milan Janjic, Nov 26 2011
For a(n) = [(1+(4m+1)^1/2)^n)-(1-(4m+1)^1/2))^n)]/[(2^n)(4m+1)^1/2), a(n)/a(n-1) appears to converge to (1+sqrt(4m+1))/2. Here with m = 10, the numbers in the sequence are congruent with those of the Fibonacci sequence modulo m-1 = 9. For example, F(8) = 21 (Fibonacci) corresponds to a(8) = 5061 (here) because 2+1 and 5+0+1+6 are congruent. - Maleval Francis, Nov 12 2013
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LINKS
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FORMULA
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a(n) = (((1+sqrt(41))/2)^(n+1) - ((1-sqrt(41))/2)^(n+1))/sqrt(41).
a(n) = Sum_{k=0..n} binomial((n+k)/2, k)*(1+(-1)^(n-k))*10^((n-k)/2)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*10^k. (End)
a(n) is the entry (M^n)_1,1 where the matrix M = [1,2;5,0]. - Simone Severini, Jun 22 2006
a(n) = (sum{1<=k<=n+1, k odd}C(n+1,k)*41^((k-1)/2))/2^n. - Vladimir Shevelev, Feb 05 2014
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MATHEMATICA
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CoefficientList[Series[1/(1-x-10*x^2), {x, 0, 50}], x] (* G. C. Greubel, Apr 30 2017 *)
LinearRecurrence[{1, 10}, {1, 1}, 30] (* Harvey P. Dale, Dec 12 2018 *)
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PROG
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(Sage) [lucas_number1(n, 1, -10) for n in range(1, 25)] # Zerinvary Lajos, Apr 22 2009
(Magma) [ n eq 1 select 1 else n eq 2 select 1 else Self(n-1)+10*Self(n-2): n in [1..30] ]; // Vincenzo Librandi, Aug 23 2011
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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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STATUS
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approved
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