%I #16 Sep 27 2024 13:00:53

%S 1,2,7,20,50,115,251,530,1096,2237,4529,9124,18326,36743,73591,147302,

%T 294740,589633,1179437,2359064,4718338,9436907,18874067,37748410,

%U 75497120,150994565,301989481,603979340,1207959086

%N a(n+1) = Sum_{k=1..n} (a(k) + k*(n-k)), with a(1)=1.

%C First column of matrix given by:

%C C(1,1) = 1,

%C C(n,k+1) = C(n,k) + n,

%C C(n+1,1) = Sum_{k=1..n-1} C(k, n-k+1);

%C where C(i,j) denotes cell at row i and column j

%C 1 2 3 4 5 6 ..

%C 2 4 6 8 10 ...

%C 7 10 13 16 ...

%C 20 24 28 ...

%C 50 55 ...

%C 115...

%C -------

%C Can also be seen as diagonal of the following triangle, which is obtained by shifting n-th row of the earlier mentioned matrix, by n-1 cells:

%C 1 2 3 4 5 6 ...

%C 2 4 6 8 10 ...

%C 7 10 13 16 ...

%C 20 24 28 ...

%C 50 55 ...

%C 115...

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5,-9,7,-2).

%F a(1) = 1, a(n+1) = Sum_{k=1..n} (a(k) + k*(n-k)); for n>1.

%F a(n) = 1/4 * (9*2^n - 2*n^2 - 6*n - 8); for n > 1.

%F a(n+1) = 2 * a(n) + A253145(n-1).

%F From _Stefano Spezia_, Jul 02 2020: (Start)

%F G.f.: x*(1 - 3*x + 6*x^2 - 4*x^3 + x^4)/((1 - x)^3*(1 - 2*x)).

%F a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4) for n > 5. (End)

%t a[1] = 1; a[n_] := a[n] = Sum[a[k] + k*(n - k), {k, 1, n - 1}]; Array[a, 30] (* _Amiram Eldar_, Jul 02 2020 *)

%t LinearRecurrence[{5,-9,7,-2},{1,2,7,20,50},30] (* _Harvey P. Dale_, Sep 27 2024 *)

%o (Python)

%o def a(n):

%o if n == 1: return 1

%o return sum([a(k) + k*(n-k) for k in range(1,n)])

%Y Cf. A253145.

%K nonn

%O 1,2

%A _Daniel Cieslinski_, Jul 01 2020