|
| |
|
|
A166942
|
|
One fifth of product plus sum of five consecutive nonnegative numbers.
|
|
4
| |
|
|
2, 27, 148, 509, 1350, 3031, 6056, 11097, 19018, 30899, 48060, 72085, 104846, 148527, 205648, 279089, 372114, 488395, 632036, 807597, 1020118, 1275143, 1578744, 1937545, 2358746, 2850147, 3420172, 4077893, 4833054, 5696095
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,1
|
|
|
COMMENTS
| a(n) = ((n*...*(n+4))+(n+...+(n+4)))/5, n >= 0.
Binomial transform of 2, 25, 96, 144, 96, 24, 0, 0, 0, 0, ....
Partial sums of A062938 where initial term 1 is replaced by 2.
|
|
|
LINKS
| Vincenzo Librandi, Table of n, a(n) for n = 0..1000
|
|
|
FORMULA
| a(n) = (n^5 + 10n^4 + 35n^3 + 50n^2 + 29n + 10)/5. [From Charles R Greathouse IV (charles.greathouse(AT)case.edu), Nov 02 2009]
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5)+24 for n > 4; a(0)=2, a(1)=27, a(2)=148, a(3)=509, a(4)=1350. [From Klaus Brockhaus, Nov 14 2009]
G.f.: (2+15*x+16*x^2-14*x^3+6*x^4-x^5)/(1-x)^6. [From Klaus Brockhaus, Nov 14 2009]
|
|
|
EXAMPLE
| a(0) = (0*1*2*3*4+0+1+2+3+4)/5 = (0+10)/5 = 2.
a(1) = (1*2*3*4*5+1+2+3+4+5)/5 = (120+15)/5 = 27.
|
|
|
MATHEMATICA
| Table[((n+4)*(n+3)*(n+2)*(n+1)*n+(n+4)+(n+3)+(n+2)+(n+1)+n)/5, {n, 0, 100}]
(Total[#]+Times@@#)/5&/@Partition[Range[0, 100], 5, 1] (* From Harvey P. Dale, Mar 5 2011 *)
|
|
|
PROG
| (MAGMA) [ (&*s + &+s)/5 where s is [n..n+4]: n in [0..29] ]; [From Klaus Brockhaus, Nov 14 2009]
|
|
|
CROSSREFS
| Cf. A001477 (nonnegative integers), A062938 (squares of the form n(n+1)(n+2)(n+3)+1), A028387 (n+(n+1)^2), A167875, A166941, A166943.
Sequence in context: A041883 A038625 A041801 * A119351 A098627 A051766
Adjacent sequences: A166939 A166940 A166941 * A166943 A166944 A166945
|
|
|
KEYWORD
| nonn,easy
|
|
|
AUTHOR
| Vladimir Orlovsky (4vladimir(AT)gmail.com), Oct 24 2009
|
|
|
EXTENSIONS
| Edited and offset corrected by Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), Nov 14 2009
|
| |
|
|
