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Search: id:A116466
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| 1, 2, 2, 4, 2, 4, 4, 8, 10, 20, 32, 64, 112, 224, 408, 816, 1514, 3028, 5680, 11360, 21472, 42944, 81644, 163288, 311896, 623792, 1196132, 2392264, 4602236, 9204472, 17757184, 35514368, 68680170, 137360340, 266200112, 532400224, 1033703056
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Both triangles A112555 and A114700 have the property that the m-th matrix power of the triangles satisfy T^m = I + m*(T - I). So it is curious that the row squared sums of A112555 is a bisection of the unsigned row sums of A114700.
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FORMULA
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G.f.: (1+2*x)*( 2*(1+x^2)/(1-x^2) + x^2/(1-4*x^2)^(1/2) )/(2+x^2). Also, a(2*n+1) = 2*a(2*n), a(2*n) = A112556(n), where A112556 equals the row squared sums of triangle A112555.
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PROGRAM
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(PARI) {a(n)=local(x=X+X*O(X^n)); polcoeff((1+2*x)*(2*(1+x^2)/(1-x^2)+x^2/(1-4*x^2)^(1/2))/(2+x^2), n, X)} (PARI) /* a(n) as the unsigned row sums of A114700 */ {a(n)=sum(k=0, n, abs(polcoeff(polcoeff(1/(1-x*y)+ x*(1+x-2*x^2*y)/(1-x)/(1+x+x*y+x*O(x^n)+y*O(y^k))/(1-x*y), n, x), k, y)))}
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CROSSREFS
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Cf. A112556, A112555, A114700.
Sequence in context: A001316 A161831 A096865 this_sequence A116467 A079314 A060609
Adjacent sequences: A116463 A116464 A116465 this_sequence A116467 A116468 A116469
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KEYWORD
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nonn
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Feb 19 2006
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