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
Keywords
Links
- Index entries for linear recurrences with constant coefficients, signature (5,-9,7,-2).
Crossrefs
Cf. A253145.
Programs
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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 *) -
Python
def a(n): if n == 1: return 1 return sum([a(k) + k*(n-k) for k in range(1,n)])
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)
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