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A103220
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a(n) = n*(n+1)*(3*n^2+n-1)/6.
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9
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0, 1, 13, 58, 170, 395, 791, 1428, 2388, 3765, 5665, 8206, 11518, 15743, 21035, 27560, 35496, 45033, 56373, 69730, 85330, 103411, 124223, 148028, 175100, 205725, 240201, 278838, 321958, 369895, 422995, 481616, 546128, 616913, 694365, 778890
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OFFSET
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0,3
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COMMENTS
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Row sums of A103219.
From Bruno Berselli, Dec 10 2010: (Start)
a(n) = n*A002412(n) - Sum_{i=0..n-1} A002412(i). More generally: n^2*(n+1)*(2*d*n-2*d+3)/6 - (Sum_{i=0..n-1} i*(i+1)*(2*d*i-2*d+3))/6 = n * (n+1) * (3*d*n^2-d*n+4*n-2*d+2)/12; in this sequence is d=2.
The inverse binomial transform yields 0, 1, 11, 22, 12, 0, 0 (0 continued). (End)
a(n-1) is also number of ways to place 2 nonattacking semi-queens (see A099152) on an n X n board. - Vaclav Kotesovec, Dec 22 2011
Also, one-half the even-indexed terms of the partial sums of A045947. - J. M. Bergot, Apr 12 2018
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LINKS
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Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
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FORMULA
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G.f.: -x*(1+8*x+3*x^2)/(x-1)^5.
a(n) = Sum_{i=1..n} Sum_{j=1..n} max(i,j)^2. - Enrique Pérez Herrero, Jan 15 2013
a(n) = a(n-1) + (2*n-1)*n^2 with a(0)=0. - J. M. Bergot, Jun 10 2017
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MAPLE
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for(n=0, 100, print1((3*n^4+4*n^3-n)/6, ", "))
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MATHEMATICA
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CoefficientList[Series[- x (1 + 8 x + 3 x^2) / (x - 1)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 12 2013 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 13, 58, 170}, 40] (* Harvey P. Dale, Jan 23 2016 *)
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PROG
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(PARI) a(n)=n*(n+1)*(3*n^2+n-1)/6 \\ Charles R Greathouse IV, Oct 07 2015
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CROSSREFS
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Cf. A103219, A002412, A002418.
Sequence in context: A230988 A183317 A055833 * A086221 A272386 A171749
Adjacent sequences: A103217 A103218 A103219 * A103221 A103222 A103223
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KEYWORD
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easy,nonn
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AUTHOR
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Lambert Klasen (lambert.klasen(AT)gmx.de) and Gary W. Adamson, Jan 25 2005
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STATUS
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approved
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