OFFSET
0,2
COMMENTS
Equivalently, Sum_{i=0..n} (i+4)*A000217(i).
Sequences of the type Sum_{i=0..n} (i+k)*A000217(i):
k = 0, A001296: 0, 1, 7, 25, 65, 140, 266, 462, ...
k = 1, A000914: 0, 2, 11, 35, 85, 175, 322, 546, ...
k = 2, A050534: 0, 3, 15, 45, 105, 210, 378, 630, ... (deleting two 0)
k = 3, A215862: 0, 4, 19, 55, 125, 245, 434, 714, ...
k = 4, a(n): 0, 5, 23, 65, 145, 280, 490, 798, ...
k = 5, A239568: 0, 6, 27, 75, 165, 315, 546, 882, ...
Antidiagonal sums (without 0) give A034263: 1, 9, 39, 119, 294, ...
Diagonal: 1, 11, 45, 125, 280, 546, ... is A051740.
LINKS
Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
G.f.: x*(5 - 2*x)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = Sum_{j=0..n+2} (-1)^(n-j)*binomial(-j,-n-2)*S1(j,n), S1 Stirling cycle numbers A132393. - Peter Luschny, Apr 10 2016
EXAMPLE
a(7) = 4*0 + 5*1 + 6*3 + 7*6 + 8*10 + 9*15 + 10*21 + 11*28 = 798.
MAPLE
A241765:=n->n*(n + 1)*(n + 2)*(3*n + 17)/24; seq(A241765(n), n=0..40); # Wesley Ivan Hurt, May 09 2014
MATHEMATICA
Table[n (n + 1) (n + 2) (3 n + 17)/24, {n, 0, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 5, 23, 65, 145}, 40]
CoefficientList[Series[x (5 - 2 x)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 09 2014 *)
PROG
(Sage) [n*(n+1)*(n+2)*(3*n+17)/24 for n in (0..40)]
(Maxima) makelist(coeff(taylor(x*(5-2*x)/(1-x)^5, x, 0, n), x, n), n, 0, 40);
(Magma) /* By first comment: */ k:=4; A000217:=func<i | i*(i+1)/2>; [&+[(i+k)*A000217(i): i in [0..n]]: n in [0..40]];
(PARI) a(n)=n*(n+1)*(n+2)*(3*n+17)/24 \\ Charles R Greathouse IV, Oct 07 2015
(PARI) x='x+O('x^99); concat(0, Vec(x*(5-2*x)/(1-x)^5)) \\ Altug Alkan, Apr 10 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Apr 28 2014
STATUS
approved