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A099429
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A Jacobsthal-Lucas convolution.
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3
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0, 0, 2, 3, 12, 25, 66, 147, 344, 765, 1710, 3751, 8196, 17745, 38234, 81915, 174768, 371365, 786438, 1660239, 3495260, 7340025, 15379122, 32156323, 67108872, 139810125, 290805086, 603979767, 1252698804, 2594876065, 5368709130, 11095332171, 22906492256
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OFFSET
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0,3
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COMMENTS
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Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 02 2014
If we concatenate the lexicographically ordered bit strings of length n, then a(n) is the number of times 11 appears as a substring, if overlapping substrings are not considered as being separate. - John M. Campbell, Jan 18 2019
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LINKS
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FORMULA
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G.f.: x^2*(2-x)/(1-x-2*x^2)^2. [Typo corrected by Colin Barker, Jun 16 2012]
a(n) = Sum_{k=0..n} J(n-k)*(2^(k-1) -(-1)^k +0^k/2).
a(n) = Sum_{k=0..n+1} J(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2.
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EXAMPLE
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G.f. = 2*x^2 + 3*x^3 + 12*x^4 + 25*x^5 + 66*x^6 + 147*x^7 + 344*x^8 + ...
If we concatenate the lexicographically ordered bit strings of length 4, we obtain the expression 0000000100100011010001010110011110001001101010111100110111101111, and we see that the substring 11 appears a total of a(4) = 12 times, with overlapping substrings not being considered as being separate. - John M. Campbell, Jan 18 2019
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MATHEMATICA
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CoefficientList[Series[x^2*(2-x)/(1-x-2x^2)^2, {x, 0, 32}], x] (* Michael De Vlieger, Jan 18 2019 *)
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PROG
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(PARI) {a(n) = if( n>=0, polcoeff( x^2*(2-x)/((1+x)*(1-2*x))^2 + x*O(x^n), n), polcoeff( x*(1-2*x)/((1+x)*(2-x))^2 + x*O(x^-n), -n) )}; /* Michael Somos, Jun 02 2014 */
(Magma) m:=40; R<x>:=PowerSeriesRing(Integers(), m); [0, 0] cat Coefficients(R!( x^2*(2-x)/(1-x-2*x^2)^2 )); // G. C. Greubel, Feb 25 2019
(Sage) (x^2*(2-x)/(1-x-2*x^2)^2).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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