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a(n) = ((2*n+1)^8 - 1)/32.
3

%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