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%I #12 Apr 24 2024 05:40:54
%S 1,6,22,76,262,916,3260,11800,43334,161028,604052,2283048,8681116,
%T 33171144,127260088,489870896,1891057222,7317881444,28378110628,
%U 110251755656,429040567732,1672032067544,6524678847688,25490986350416,99696437839132,390298689482216
%N Expansion of (1 + 2*x) / ((1 - 2*x)*sqrt(1 - 4*x)).
%C Conjecture: For p Pythagorean prime (A002144), a(p) - 6 == 0 (mod p).
%C Conjecture: For p prime of the form 4*k + 3 (A002145), a(p) + 2 == 0 (mod p).
%F a(n) = 5*A000984(n) - 4* A029759(n) = binomial(2*n,n) + 4*Sum_{k=0..n-1} 2^(n-k-1)*binomial(2*k,k).
%F a(n) = 2*a(n-1) + A028283(n) = 2*a(n-1) + 2*binomial(2n-2, n-1)*(3*n-1)/n for n >= 1.
%F a(n) = 2*A082590(n-1) + A082590(n) for n >= 1.
%F a(n) = 2*A188622(n) - A126966(n).
%F D-finite with recurrence n*a(n) +2*(-2*n-1)*a(n-1) +4*(-n+6)*a(n-2) +8*(2*n-5)*a(n-3)=0. - _R. J. Mathar_, Apr 24 2024
%p a := n -> binomial(2*n,n) + 4*add(2^(n-k-1)*binomial(2*k,k), k = 0 .. n-1):
%p seq(a(n), n = 0 .. 25);
%p # Second program:
%p a:= proc(n) option remember; `if`(n=0,1,2*a(n-1)+2*binomial(2*n-2, n-1)*(3*n-1)/n) end: seq(a(n), n = 0 .. 25);
%p # Recurrence:
%p a := proc(n) option remember; if n < 2 then return [1, 6][n + 1] fi;
%p ((-18*(n - 2)^2 - 42*n + 66)*a(n - 1) + 4*(3*n - 1)*(2*n - 3)*a(n - 2)) / (n*(4 - 3*n)) end: seq(a(n), n = 0..25); # _Peter Luschny_, Apr 23 2024
%Y Cf. A002144, A002145, A000984, A028283, A029759, A082590, A126966, A188622.
%K nonn
%O 0,2
%A _Mélika Tebni_, Apr 23 2024
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