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A071238
a(n) = n*(n+1)*(2*n^2+1)/6.
7
0, 1, 9, 38, 110, 255, 511, 924, 1548, 2445, 3685, 5346, 7514, 10283, 13755, 18040, 23256, 29529, 36993, 45790, 56070, 67991, 81719, 97428, 115300, 135525, 158301, 183834, 212338, 244035, 279155, 317936, 360624, 407473, 458745, 514710, 575646, 641839
OFFSET
0,3
COMMENTS
Binomial transform of [1, 8, 21, 22, 8, 0, 0, 0, ...]. - Gary W. Adamson, Dec 28 2007
For n > 0, a(n) is the n-th antidiagonal sum of the convolution arrays A213752 and A213836). - Clark Kimberling, Jun 20 2012
The first differences are given in A277229, as a convolution of the odd-indexed triangular numbers A000217(2*n+1) and the squares A000290(n), n >= 0. - J. M. Bergot, Sep 14 2016
REFERENCES
T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
FORMULA
G.f.: x*(1+x)*(1+3*x)/(1-x)^5. - Colin Barker, Mar 22 2012
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 4, a(0)=0, a(1)=1, a(2)=9, a(3)=38, a(4)=110. - Yosu Yurramendi, Sep 03 2013
E.g.f.: (1/6)*x*(6 + 21*x + 14*x^2 + 2*x^3)*exp(x). - G. C. Greubel, Sep 17 2016
a(n) = n*A000292(n) + (n-1)*A000292(n-1). - Bruno Berselli, Sep 22 2016
a(n) = A002417(n-1) + A002417(n). - Yasser Arath Chavez Reyes, Feb 15 2024
MAPLE
A071238:=n->n*(n+1)*(2*n^2+1)/6: seq(A071238(n), n=0..60); # Wesley Ivan Hurt, Sep 24 2016
MATHEMATICA
Table[n (n + 1) (2 n^2 + 1)/6, {n, 0, 37}] (* or *)
CoefficientList[Series[x (1 + x) (1 + 3 x)/(1 - x)^5, {x, 0, 37}], x] (* Michael De Vlieger, Sep 14 2016 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 9, 38, 110}, 40] (* Harvey P. Dale, Oct 02 2021 *)
PROG
(Magma) [n*(n+1)*(2*n^2+1)/6: n in [0..40]]; // Vincenzo Librandi, Jun 14 2011
(PARI) a(n)=n*(n+1)*(2*n^2+1)/6; \\ Joerg Arndt, Sep 04 2013
CROSSREFS
Cf. A000292, A002417, A071270, A277229 (first differences).
Sequence in context: A117085 A120780 A071229 * A213583 A343521 A050854
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Jun 12 2002
STATUS
approved