A335927 a(n+1) = Sum_{k=1..n} (a(k) + k*(n-k)), with a(1)=1.

1, 2, 7, 20, 50, 115, 251, 530, 1096, 2237, 4529, 9124, 18326, 36743, 73591, 147302, 294740, 589633, 1179437, 2359064, 4718338, 9436907, 18874067, 37748410, 75497120, 150994565, 301989481, 603979340, 1207959086

Offset

1,2

Comments

First column of matrix given by:

C(1,1) = 1,

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

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

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

1 2 3 4 5 6 ..

2 4 6 8 10 ...

7 10 13 16 ...

20 24 28 ...

50 55 ...

115...

-------

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:

1 2 3 4 5 6 ...

2 4 6 8 10 ...

7 10 13 16 ...

20 24 28 ...

50 55 ...

115...

Links

Table of n, a(n) for n=1..29.

Index entries for linear recurrences with constant coefficients, signature (5,-9,7,-2).

Formula

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

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

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

From Stefano Spezia, Jul 02 2020: (Start)

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

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

Mathematica

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 *)
LinearRecurrence[{5, -9, 7, -2}, {1, 2, 7, 20, 50}, 30] (* Harvey P. Dale, Sep 27 2024 *)

Prog

(Python)
def a(n):
    if n == 1: return 1
    return sum([a(k) + k*(n-k) for k in range(1, n)])

Crossrefs

Cf. A253145.

Sequence in context: A360421 A123203 A309298 * A261054 A134311 A362973

Adjacent sequences: A335924 A335925 A335926 * A335928 A335929 A335930

Keyword

nonn

Author

Daniel Cieslinski, Jul 01 2020

Status

approved