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A190726
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Central coefficients of Riordan matrix A118384.
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4
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1, 6, 62, 720, 8806, 110916, 1423796, 18520788, 243289670, 3220011684, 42872967012, 573608356272, 7705343534716, 103857425975400, 1403902871946000, 19024773303675420, 258372666772083270, 3515644245559211172, 47918193512409831380
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OFFSET
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0,2
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COMMENTS
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This sequence gives the integer part of an integral approximation to log(2), thus bears strong similarity to A123178. Quality of rational approximants appears entirely sufficient to prove irrationality. - Bradley Klee, Jun 29 2018
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LINKS
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FORMULA
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a(n) = T(2*n,n), where T(n,k) = A118384(n,k).
a(n) = Sum_{k=0..n} binomial(2*n, k)*binomial(2*n, n-k)*2^k.
a(n) = Sum_{k=0..n} binomial(2*n, k)*binomial(k, n-k)*2^(n-k)*3^(2*k-n).
a(n)*log(2) - A316911(n)/A316912(n) = I_n = Integral_{t=0..1}(-1/2)^n/(1+t)*((1-t)^2*t^2/(1+t))^n*dt.
Lim_{n->oo} I_n = 0, therefore:
G.f. G(x) and derivatives G^(n)(x) = d^n/dx^n G(x) satisfy a Picard-Fuchs type differential equation, 0 = Sum_{m=0..5,n=0..3} M_{m,n} x^m G^(n)(x), with integer matrix: M = {{324,-54,0,0}, {-36,10842,-486,0}, {84,8352,14931,-243}, {0,756,19026,3024}, {0,0,672,5364}, {0,0,0,112}}.
2*(n-1)*(2*n-3)*(2*n-1)*(33*n-8)*a(n-2)+ 9*(2*n-1)*(693*n^3-1554*n^2+989*n-160)*a(n-1) -3*n*(3*n-2)*(3*n-1)*(33*n-41)*a(n)=0.
(End)
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EXAMPLE
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I_2 = Integral_{t=0..1} ((1-t)^4*t^4)/(4*(1+t)^3)*dt = 62*log(2) - 1719/40 < 10^(-3).
I_3 = Integral_{t=0..1} - ((1-t)^6*t^6)/(8*(1+t)^4)*dt = 720*log(2) - 143731/288 < 10^(-5). (End)
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MATHEMATICA
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Table[Sum[Binomial[2n, k]Binomial[2n, n-k]2^k, {k, 0, n}], {n, 0, 100}]
RecurrenceTable[{2*(n-1)*(2*n-3)*(2*n-1)*(33*n-8)*a[n-2]+ 9*(2*n-1)*(693*n^3-1554*n^2+989*n-160)*a[n-1] -3*n*(3*n-2)*(3*n-1)*(33*n-41)*a[n]==0, a[0]==1, a[1]==6}, a, {n, 0, 10}] (* Bradley Klee, Jun 29 2018 *)
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PROG
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(Maxima) makelist(sum(binomial(2*n, k)*binomial(2*n, n-k)*2^k, k, 0, n), n, 0, 12);
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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