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%I #9 Mar 07 2020 01:21:19
%S 0,0,1,5,26,133,708,3861,21604,123266,715221,4206956,25032840,
%T 150413348,911379384,5562367173,34164355848,211015212580,
%U 1309815397995,8166460799805,51120054233490,321156223592865,2024257417812240,12797201858645100
%N a(n) = A047752(12n+5).
%C Also A047754(4n+1). The proof is elementary and involves reducing both expressions via the various linear expansions with A001764, A047753, A047751 and A047749 which are found in these sequences. - _R. J. Mathar_, Oct 18 2008
%K nonn
%O 0,4
%A _N. J. A. Sloane_
%E Edited by _N. J. A. Sloane_, Jan 17 2009 at the suggestion of _R. J. Mathar_
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