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%I #32 Jan 18 2022 06:55:05
%S 1,1,2,3,7,11,26,42,99,163,382,638,1486,2510,5812,9908,22819,39203,
%T 89846,155382,354522,616666,1401292,2449868,5546382,9740686,21977516,
%U 38754732,87167164,154276028,345994216,614429672,1374282019,2448023843
%N Expansion of ((1 + x - 2x^2) + (1+x)*sqrt(1-4x^2))/(2(1-4x^2)).
%C Interleaving of A114121 and A032443. Row sums of A116405. Binomial transform is A116409.
%C Appears to be the number of n-digit binary numbers not having more zeros than ones; equivalently, the number of unrestricted Dyck paths of length n not going below the axis. - _Ralf Stephan_, Mar 25 2008
%C From _Gus Wiseman_, Jun 20 2021: (Start)
%C Also the number compositions of n with alternating sum >= 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. The a(0) = 1 through a(5) = 11 compositions are:
%C () (1) (2) (3) (4) (5)
%C (11) (21) (22) (32)
%C (111) (31) (41)
%C (112) (113)
%C (121) (122)
%C (211) (212)
%C (1111) (221)
%C (311)
%C (1121)
%C (2111)
%C (11111)
%C (End)
%C From _J. Stauduhar_, Jan 14 2022: (Start)
%C Also, for n >= 2, first differences of partial row sums of Pascal's triangle. The first ceiling(n/2)+1 elements of rows n=0 to n=4 in Pascal's triangle are:
%C 1
%C 1 1
%C 1 2
%C 1 3 3
%C 1 4 6
%C ...
%C The cumulative sums of these partial rows form the sequence 1,3,6,13,24,..., and its first differences are a(2),a(3),a(4),... in this sequence.
%C (End)
%F a(n) = A114121(n/2)*(1+(-1)^n)/2 + A032443((n-1)/2)*(1-(-1)^n)/2.
%F a(n) = Sum_{k=0..floor(n/2)} binomial(n-1,k). - _Paul Barry_, Oct 06 2007
%F Conjecture: n*(n-3)*a(n) +2*(-n^2+4*n-2)*a(n-1) -4*(n-2)^2*a(n-2) +8*(n-2)*(n-3)*a(n-3)=0. - _R. J. Mathar_, Nov 28 2014
%F a(n) ~ 2^(n-2) * (1 + (3+(-1)^n)/sqrt(2*Pi*n)). - _Vaclav Kotesovec_, May 30 2016
%F a(n) = 2^(n-1) - A294175(n). - _Gus Wiseman_, Jun 27 2021
%t CoefficientList[Series[((1+x-2x^2)+(1+x)Sqrt[1-4x^2])/(2(1-4x^2)),{x,0,40}],x] (* _Harvey P. Dale_, Aug 16 2012 *)
%t ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],ats[#]>=0&]],{n,0,15}] (* _Gus Wiseman_, Jun 20 2021 *)
%Y The alternating sum = 0 case is A001700 or A088218.
%Y The alternating sum > 0 case appears to be A027306.
%Y The bisections are A032443 (odd) and A114121 (even).
%Y The alternating sum <= 0 version is A058622.
%Y The alternating sum < 0 version is A294175.
%Y The restriction to reversed partitions is A344607.
%Y A103919 counts partitions by sum and alternating sum (reverse: A344612).
%Y A124754 gives the alternating sum of standard compositions.
%Y A344610 counts partitions by sum and positive reverse-alternating sum.
%Y A344616 lists the alternating sums of partitions by Heinz number.
%Y Cf. A000041, A000070, A000097, A003242, A006330, A028260, A058696, A119899, A239830, A344605, A344611, A344650, A344739.
%K easy,nonn
%O 0,3
%A _Paul Barry_, Feb 13 2006
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