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a(0) = 0, a(1) = 1; a(2*n) = 5*a(n), a(2*n+1) = a(n) + a(n+1).
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%I #5 Mar 17 2021 07:57:57

%S 0,1,5,6,25,11,30,31,125,36,55,41,150,61,155,156,625,161,180,91,275,

%T 96,205,191,750,211,305,216,775,311,780,781,3125,786,805,341,900,271,

%U 455,366,1375,371,480,301,1025,396,955,941,3750,961,1055,516,1525,521,1080,991

%N a(0) = 0, a(1) = 1; a(2*n) = 5*a(n), a(2*n+1) = a(n) + a(n+1).

%F G.f.: x * Product_{k>=0} (1 + 5*x^(2^k) + x^(2^(k+1))).

%t a[0] = 0; a[1] = 1; a[n_] := If[EvenQ[n], 5 a[n/2], a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 55}]

%t nmax = 55; CoefficientList[Series[x Product[(1 + 5 x^(2^k) + x^(2^(k + 1))), {k, 0, Floor[Log[2, nmax]] + 1}], {x, 0, nmax}], x]

%Y Cf. A002487, A178569, A178590, A237711, A244643, A342611, A342614, A342615.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Mar 16 2021