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A272263
a(n) = numerator of A000032(n) - 1/2^n.
2
1, 1, 11, 31, 111, 351, 1151, 3711, 12031, 38911, 125951, 407551, 1318911, 4268031, 13811711, 44695551, 144637951, 468058111, 1514668031, 4901568511, 15861809151, 51329892351, 166107021311, 537533612031, 1739495309311, 5629125066751, 18216231370751
OFFSET
0,3
COMMENTS
A000032(n), Lucas numbers, and 1/2^n are autosequences of the second kind.
Then a(n)/2^n is also an autosequence of the second kind.
Their corresponding autosequences of the first kind are A000045(n) and n/2^n, the Oresme numbers.
Difference table of A000032(n) - 1/2^n:
1, 1/2, 11/4, 31/8, 111/16, 351/32, 1151/64, ...
9/4, 9/8, 49/16, 129/32, 449/64, 1409/128, ...
31/16, 31/32, 191/64, 511/128, 1791/256, ...
129/64, 129/128, 769/256, ...
511/256, 511/256, ...
2049/1024, ... .
The first upper diagonal is A140323(n)/A004171(n). The main diagonal is the double, i.e. A140323(n)/A000302(n). The inverse binomial transform is the signed sequence.
Quintisections from a(2):
11, 31, 111, 351, 1151,
3711, 12031, 38911, 125951, 407551,
1318911, 4268031, 13811711, 44695551, 144637951,
etc.
FORMULA
a(n) = a(n-1) + 10*A085449(n), for n>0, a(0)=1.
a(n) = A087131(n) - 1.
From Colin Barker, Apr 24 2016: (Start)
a(n) = (-1+(1-sqrt(5))^n+(1+sqrt(5))^n).
a(n) = 3*a(n-1)+2*a(n-2)-4*a(n-3) for n>2.
G.f.: (1-2*x+6*x^2) / ((1-x)*(1-2*x-4*x^2)).
(End)
EXAMPLE
Numerators of a(0) =2-1=1, a(1)=1-1/2=1/2, a(2)=3-1/4=11/4, a(3)=4-1/8=31/8, ... .
MATHEMATICA
CoefficientList[Series[(1 - 2*x + 6*x^2)/((1 - x)*(1 - 2*x - 4*x^2)), {x, 0, 30}], x] (* Robert Price, Apr 24 2016 *)
Table[Numerator[LucasL@ n - 1/2^n], {n, 0, 26}] (* Michael De Vlieger, Apr 24 2016 *)
PROG
(PARI) Vec((1-2*x+6*x^2)/((1-x)*(1-2*x-4*x^2)) + O(x^50)) \\ Colin Barker, Apr 24 2016
KEYWORD
nonn,frac,easy
AUTHOR
Paul Curtz, Apr 24 2016
STATUS
approved