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a(0) = 0, a(1) = 1; a(2*n) = 7*a(n), a(2*n+1) = a(n) + a(n+1).
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%I #5 Mar 17 2021 07:58:03

%S 0,1,7,8,49,15,56,57,343,64,105,71,392,113,399,400,2401,407,448,169,

%T 735,176,497,463,2744,505,791,512,2793,799,2800,2801,16807,2808,2849,

%U 855,3136,617,1183,904,5145,911,1232,673,3479,960,3241,3207,19208,3249,3535,1296,5537

%N a(0) = 0, a(1) = 1; a(2*n) = 7*a(n), a(2*n+1) = a(n) + a(n+1).

%F G.f.: x * Product_{k>=0} (1 + 7*x^(2^k) + x^(2^(k+1))).

%t a[0] = 0; a[1] = 1; a[n_] := If[EvenQ[n], 7 a[n/2], a[(n - 1)/2] + a[(n + 1)/2]]; Table[a[n], {n, 0, 52}]

%t nmax = 52; CoefficientList[Series[x Product[(1 + 7 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, A342610, A342614, A342615.

%K nonn

%O 0,3

%A _Ilya Gutkovskiy_, Mar 16 2021