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A132658 a(6n+k) = 3a(6n+k-1)-3a(6n+k-2)+2a(6n+k-3), k = 0, 1, 3, 4, 5; a(6n+2) = 3a(6n+1)-3a(6n). a(0) = a(1) = 0, a(2) = 1.

%I A132658 #13 Aug 13 2025 23:44:48
%S A132658 0,0,1,3,6,11,21,42,63,105,210,441,903,1806,2709,4515,9030,18963,
%T A132658 38829,77658,116487,194145,388290,815409,1669647,3339294,5008941,
%U A132658 8348235,16696470,35062587,71794821,143589642,215384463,358974105,717948210
%N A132658 a(6n+k) = 3a(6n+k-1)-3a(6n+k-2)+2a(6n+k-3), k = 0, 1, 3, 4, 5; a(6n+2) = 3a(6n+1)-3a(6n). a(0) = a(1) = 0, a(2) = 1.
%C A132658 The third differences are 0, 0, 1, 3, 6, -11, 21, 42, 63, 105, 210, -441, ..., equal to the original sequence if each 6th term is negated.
%p A132658 A132658 := proc(n)
%p A132658     option remember;
%p A132658     if n <=1 then
%p A132658         0;
%p A132658     elif n = 2 then
%p A132658         1;
%p A132658     elif modp(n,6) = 2 then
%p A132658         3*procname(n-1)-3*procname(n-2);
%p A132658     else
%p A132658         3*procname(n-1)-3*procname(n-2)+2*procname(n-3) ;
%p A132658     end if;
%p A132658 end proc:
%p A132658 seq(A132658(n),n=0..80) ; # _R. J. Mathar_, Aug 07 2017
%t A132658 a[n_] := a[n] = Which[n <= 1, 0, n == 2, 1, Mod[n, 6] == 2, 3a[n-1] - 3a[n-2], True, 3a[n-1] - 3a[n-2] + 2a[n-3]];
%t A132658 Table[a[n], {n, 0, 36}] (* _Jean-François Alcover_, Oct 27 2023, after _R. J. Mathar_ *)
%Y A132658 Cf. A024495.
%K A132658 nonn
%O A132658 0,4
%A A132658 _Paul Curtz_, Nov 15 2007
%E A132658 Edited by _R. J. Mathar_, Jul 07 2008