%I #15 Jan 17 2023 16:30:57
%S 0,205,12207,180150,1345210,6698715,25491585,80090332,217992420,
%T 530736345,1181964355,2447218290,4768371582,8825923015,15632700405,
%U 26652844920,43950269320,70371105957,109764982935,167250289390,249528913410,365256258675,525472668457,744102708180
%N a(n) = ((2*n+1)^8 - 1)/32.
%C a(n) and A000217(n) have the same parity.
%H Jianing Song, <a href="/A359498/b359498.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).
%F a(n) = A000217(n) * A219086(n) * A175110(n) = A219086(n) * A175110(n).
%o (PARI) a(n) = ((2*n+1)^8 - 1)/32
%o (Python)
%o def A359498(n): return ((n<<1)+1)**8-1>>5 # _Chai Wah Wu_, Jan 15 2023
%Y Cf. {((2*n+1)^2^k - 1)/2^(k+2)}: A000217 (k=1), A219086 (k=2), this sequence (k=3), A359499 (k=4).
%K nonn,easy
%O 0,2
%A _Jianing Song_, Jan 03 2023