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a(n) = s(1)*s(n) + s(2)*s(n-1) + ... + s(k)*s(n+1-k), where k = floor((n+1)/2), s = (natural numbers >= 3).
17

%I #18 Aug 23 2024 19:08:24

%S 9,12,31,38,70,82,130,148,215,240,329,362,476,518,660,712,885,948,

%T 1155,1230,1474,1562,1846,1948,2275,2392,2765,2898,3320,3470,3944,

%U 4112,4641,4828,5415,5622,6270,6498,7210,7460,8239,8512,9361,9658,10580,10902,11900,12248,13325

%N a(n) = s(1)*s(n) + s(2)*s(n-1) + ... + s(k)*s(n+1-k), where k = floor((n+1)/2), s = (natural numbers >= 3).

%H G. C. Greubel, <a href="/A024312/b024312.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1).

%F From _G. C. Greubel_, Jan 17 2022: (Start)

%F a(n) = ( (75 + 182*n + 63*n^2 + 4*n^3) - 3*(25 + 10*n + n^2)*(-1)^n )/48.

%F G.f.: x*(9 + 3*x - 8*x^2 - 2*x^3 + 2*x^4)/((1-x)^4 * (1+x)^3).

%F a(n) = (-60 - 18*n + (14 + 3*n)*f(n) + 3*(4+n)*f(n)^2 - 2*f(n)^3)/6, where f(n) = floor((n+5)/2). (End)

%t Table[Sum[j*(n-j+5), {j, 3, Floor[(n+5)/2]}], {n, 50}] (* _G. C. Greubel_, Jan 17 2022 *)

%t LinearRecurrence[{1,3,-3,-3,3,1,-1},{9,12,31,38,70,82,130},60] (* _Harvey P. Dale_, Aug 23 2024 *)

%o (Sage) [( (75 +182*n +63*n^2 +4*n^3) - 3*(25 +10*n +n^2)*(-1)^n )/48 for n in (1..50)] # _G. C. Greubel_, Jan 17 2022

%o (Magma) [(&+[j*(n+5-j): j in [3..Floor((n+5)/2)]]) : n in [1..50]]; // _G. C. Greubel_, Jan 17 2022

%o (PARI) a(n)=((75+182*n+63*n^2+4*n^3)-3*(25+10*n+n^2)*(-1)^n)/48 \\ _Charles R Greathouse IV_, Oct 21 2022

%Y Cf. A024313, A024314, A024315, A024316, A024317, A024318, A024319, A024320, A024321, A024322, A024323, A024324, A024325, A024326, A024327.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_