|
|
A167875
|
|
One third of product plus sum of three consecutive nonnegative integers; a(n)=(n+1)(n^2+2n+3)/3.
|
|
10
|
|
|
1, 4, 11, 24, 45, 76, 119, 176, 249, 340, 451, 584, 741, 924, 1135, 1376, 1649, 1956, 2299, 2680, 3101, 3564, 4071, 4624, 5225, 5876, 6579, 7336, 8149, 9020, 9951, 10944, 12001, 13124, 14315, 15576, 16909, 18316, 19799, 21360, 23001, 24724, 26531
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
a(n) = ((n*(n+1)*(n+2))+(n+(n+1)+(n+2)))/3, n >= 0.
Inverse binomial transform of A080930.
a(n) = A002061(n+1)+a(n-1) for n > 0.
a(n) = A005894(n)-a(n-1) for n > 0.
|
|
LINKS
|
|
|
FORMULA
|
a(n) = (n^3+3*n^2+5*n+3)/3.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3)+2 for n > 3; a(0)=1, a(1)=4, a(2)=11, a(3)=24.
G.f.: (1+x^2)/(1-x)^4.
a(n)-4*a(n-1)+6*a(n-2)-4*a(n-3)+a(n-4)=0 for n>3. - Bruno Berselli, May 26 2010
|
|
EXAMPLE
|
a(0) = (0*1*2+0+1+2)/3 = (0+3)/3 = 1.
a(1) = (1*2*3+1+2+3)/3 = (6+6)/3 = 4.
a(6)-4*a(5)+6*a(4)-4*a(3)+a(2) = 119-4*76+6*45-4*24+11 = 0. - Bruno Berselli, May 26 2010
|
|
MATHEMATICA
|
(Times@@#+Total[#])/3&/@Partition[Range[0, 65], 3, 1] (* Harvey P. Dale, Mar 14 2011 *)
|
|
PROG
|
(Magma) [ (&*s + &+s)/3 where s is [n..n+2]: n in [0..42] ];
|
|
CROSSREFS
|
Cf. A001477 (nonnegative integers),
A167876 (1, 3, 4, 2, 0, 0, 0, 0, ...),
A014820 ((1/3)*(n^2+2*n+3)*(n+1)^2),
A054602 (Sum_{d|3} phi(d)*n^(3/d)),
A005894 (centered tetrahedral numbers),
A004277 (1 and the positive even numbers),
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|