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A114121
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Expansion of (sqrt(1 - 4*x) + (1 - 2*x))/(2*(1 - 4*x)).
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40
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1, 2, 7, 26, 99, 382, 1486, 5812, 22819, 89846, 354522, 1401292, 5546382, 21977516, 87167164, 345994216, 1374282019, 5461770406, 21717436834, 86392108636, 343801171354, 1368640564996, 5450095992964, 21708901408216, 86492546019214
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OFFSET
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0,2
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COMMENTS
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Second binomial transform of A032443 with interpolated zeros.
a(n) is the total number of lattice points, taken over all Dyck n-paths (A000108), that (i) lie on or above ground level and (ii) lie on or directly below a peak. For example with n = 2, UUDD has 1 peak contributing 3 lattice points--(2, 0), (2, 1) and (2, 2) when the path starts at the origin--and UDUD has 2 peaks, each contributing 2 lattice points and so a(2) = 3 + 4 = 7. - David Callan, Jul 14 2006
Hankel transform is binomial(n + 2, 2). - Paul Barry, Dec 04 2007
Image of (-1)^n under the Riordan array ((1/2)(1/(1 - 4x) + 1/sqrt(1 - 4x)), c(x) - 1), c(x) the g.f. of A000108. - Paul Barry, Jun 15 2008
Also the even bisection of A116406 = 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(3) = 26 compositions are:
(6) (33) (114) (1122) (11112) (111111)
(42) (123) (1131) (11121)
(51) (132) (1221) (11211)
(213) (2112) (12111)
(222) (2121) (21111)
(231) (2211)
(312) (3111)
(321)
(411)
(End)
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LINKS
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FORMULA
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a(n) = Sum_{k=0..n} C(n, k)*2^(n-k-2)*(2^k + C(k, k/2))*(1 + (-1)^k).
G.f.: (1 - 4*x + (1 - 2*x)*sqrt(1 - 4*x))/(2*(1 - 4*x)^(3/2)).
a(n) = Sum_{k=0..n} ( Sum_{j=0..n} C(2*n, n-k-j)*(-1)^j ). (End)
a(n) = Sum_{k=0..n} C(2*n, n-k)*(1 + (-1)^k)/2. - Paul Barry, Aug 06 2009
a(n) = C(2*n-1, n-1) + (4^n + 3*0^n)/4.
Integral representation a(n) = (1/(2*pi))*(Integral_{x=0..4} x^n/sqrt(x(4 - x))) + (4^n + 0^n)/4. (End)
a(n) = Sum_{k=0..floor(n/2)} C(2*n, 2*k + (n mod 2)). - Mircea Merca, Jun 21 2011
Conjecture: n*a(n) + 2*(3 - 4*n)*a(n-1) + 8*(2*n-3)*a(n-2) = 0. - R. J. Mathar, Nov 07 2012
Conjecture verified using the differential equation (16*x^2-8*x+1)*g'(x) + (8*x-2)*g(x)-2*x=0 satisfied by the G.f. - Robert Israel, Jul 27 2020
a(n) = Sum_{i=0..n} (sum_{j=0..n} binomial(n, i+j)*binomial(n, j-i)). - Yalcin Aktar, Jan 07 2013.
0 = (1 + 2*n)*b(n) - (5 + 4*n)*b(n+1) + (4 + 2*n)*b(n+2) if n > 0 where b(n) = a(n) / 4^n. - Michael Somos, Dec 31 2013
0 = b(n+3) * (2*b(n+2) - 7*b(n+1) + 5*b(n)) + b(n+2) * (-b(n+2) + 7*b(n+1) - 7*b(n)) + b(n+1) * (-b(n+1) + 2*b(n)) if n > 0 where b(n) = a(n) / 4^n. - Michael Somos, Dec 31 2013
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EXAMPLE
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G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 99*x^4 + 382*x^5 + 1486*x^6 + 5812*x^7 + ...
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MAPLE
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seq(sum(binomial(2*n, 2*k+irem(n, 2)), k=0..floor((1/2)*n)), n=0..20)
seq(binomial(2*n-1, n)+4^(n-1)-(1/4)*0^n, n=0..20)
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MATHEMATICA
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a[ n_] := SeriesCoefficient[((1 + 1/Sqrt[1 - 4 x])/2)^2, {x, 0, n}] (* Michael Somos, Dec 31 2013 *)
ats[y_]:=Sum[(-1)^(i-1)*y[[i]], {i, Length[y]}]; Table[Length[Select[Join@@Permutations/@IntegerPartitions[n], ats[#]>=0&]], {n, 0, 15, 2}] (* Gus Wiseman, Jun 21 2021 *)
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CROSSREFS
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The case of alternating sum = 0 is A001700.
The case of alternating sum < 0 is A008549.
This is the even bisection of A116406.
The restriction to reversed partitions is A344611.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A124754 gives the alternating sum of standard compositions.
A316524 is the alternating sum of the prime indices of n.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Cf. A000041, A000070, A000302, A000346, A003242, A027306, A032443, A058622, A058696, A119899, A344607, A344610.
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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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