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A101499
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A Chebyshev transform of the Catalan numbers.
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2
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1, 1, 1, 3, 9, 25, 73, 223, 697, 2217, 7161, 23427, 77457, 258417, 868881, 2941311, 10016241, 34289041, 117935473, 407344771, 1412307481, 4913508489, 17148100569, 60018592735, 210619695913, 740910077497, 2612194773481
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,4
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COMMENTS
| A Chebyshev transform of A000108. Under the Chebyshev transform, we map a g.f. g(x) to (1/(1+x^2))g(x/(1+x^2)). Also equivalent to a Catalan transform followed by the Chebyshev transform to 1/(1-x), where the Catalan transform maps h(x)->h(xc(x)), c(x) the g.f. of A000108.
a(n) is the number of peakless Motzkin paths of length n in which the (1,0)-steps at level >=1 come in 2 colors. Example: a(4)=9 because, denoting u=(1,1), h=(1,0), and d=(1,-1), we have 1 path of shape hhhh, 2 paths of shape huhd, 2 paths of shape uhdh, and 2^2=4 paths of shape uhhd. [Emeric Deutsch, May 3 2011]
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FORMULA
| G.f.: (sqrt(1+x^2)-sqrt(1-4x+x^2))/(2x*sqrt(1+x^2)); a(n)=sum{k=0..floor(n/2), binomial(n-k, k)C(n-2k)}; a(n)=sum{k=0..floor(n/2), sum{i=0..n-2k, sum{j=0..n-2k, ((2i+1)/(n-2k+i+1))(-1)^(i-j)C(2n-4k, n-2k-i)C(i, j)}}}.
Given g.f. A(x) then B(x)=x*A(x) satisfies 0=f(x, B(x)) where f(x, y)= x-(1+x^2)*(y-y^2) . - Michael Somos Sep 18 2006
Given g.f. A(x) then B(x)=x*A(x) satisfies 0=f(B(x), B(x^2), B(x^4)) where f(u, v, w)= w -v^2*w^2 -(1-v)*w*(v+w) +(u-u^2)^2*(v^2+w^2-v-w) . - Michael Somos Sep 18 2006
Given g.f. A(x) then B(x)=x*A(x) satisfies 0=f(B(x), B(x^2)) where f(u, v)= (v-v^2) -(u-u^2)^2*(1+2*(v-v^2)) . - Michael Somos Sep 18 2006
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PROG
| (PARI) {a(n)=local(A); if(n<0, 0, n++; A=serreverse(x-x^2+x*O(x^n)); polcoeff( subst(A, x, x/(1+x^2)), n))} /* Michael Somos Sep 18 2006 */
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CROSSREFS
| Sequence in context: A101197 A101168 A079857 * A004665 A196431 A183111
Adjacent sequences: A101496 A101497 A101498 * A101500 A101501 A101502
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KEYWORD
| easy,nonn
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AUTHOR
| Paul Barry (pbarry(AT)wit.ie), Dec 04 2004
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