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A147656
The arithmetic mean of the n-th and (n+1)-st cubes, rounded down.
1
0, 4, 17, 45, 94, 170, 279, 427, 620, 864, 1165, 1529, 1962, 2470, 3059, 3735, 4504, 5372, 6345, 7429, 8630, 9954, 11407, 12995, 14724, 16600, 18629, 20817, 23170, 25694, 28395, 31279, 34352, 37620, 41089, 44765, 48654, 52762, 57095, 61659
OFFSET
0,2
COMMENTS
The terms of this sequence relate to intervals between cubes in the same fashion as terms of A002378 are related to intervals between squares.
FORMULA
a(n) = floor((A000578(n) + A000578(n+1))/2).
From R. J. Mathar, Nov 11 2008: (Start)
a(n) = A000578(n) + A045943(n) = n*(2n^2+3n+3)/2.
G.f.: x*(4+x+x^2)/(1-x)^4. (End)
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, May 06 2012
a(n) = A027480(n) + A006003(n). - Bruce J. Nicholson, Jun 03 2018
From A.H.M. Smeets, Sep 10 2018: (Start)
a(n) = Sum_{k=0..n-1} (n+1)^2-k for n >= 0 with empty domain of summation for n = 0.
a(n) = n*(n+1)^2 - n*(n-1)/2 for n >= 0.
Lim_{n -> inf} a(n-1)/n^3 = 1. (End)
E.g.f.: exp(x)*(8*x + 9*x^2 + 2*x^3)/2. - Stefano Spezia, Sep 12 2018
a(n) = A081435(n)-1. - R. J. Mathar, Sep 14 2018
MAPLE
seq(coeff(series(x*(x^2+x+4)/(1-x)^4, x, n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Sep 11 2018
MATHEMATICA
Table[(n^3+(n+1)^3-1)/2, {n, 0, 70}] (* Vladimir Joseph Stephan Orlovsky, May 04 2011 *)
PROG
(PARI) j=[]; for (n=0, 40, j=concat(j, n^3+floor(((n+1)^3 - n^3)/2))); j
(PARI) a(n) = n*(2*n^2+3*n+3)/2; \\ Altug Alkan, Sep 20 2018
(Magma) I:=[0, 4, 17, 45]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, May 06 2012
CROSSREFS
Cf. A000578.
Sequence in context: A162148 A166781 A376232 * A095667 A212577 A332863
KEYWORD
nonn,easy
AUTHOR
STATUS
approved