|
| |
|
|
A294070
|
|
a(n) = (1/4)*(n^2 - 2*n)^2 + (9/4)*(n^2 - 2*n) + 6.
|
|
1
|
|
|
|
4, 6, 15, 40, 96, 204, 391, 690, 1140, 1786, 2679, 3876, 5440, 7440, 9951, 13054, 16836, 21390, 26815, 33216, 40704, 49396, 59415, 70890, 83956, 98754, 115431, 134140, 155040, 178296, 204079, 232566, 263940, 298390, 336111, 377304, 422176, 470940, 523815
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,1
|
|
|
LINKS
|
SESC NSU Correspondence School, First assignments for 2018/2019 (in Russian), Kvant, 2018, No. 7, p. 42, Mathematics section, 6th grade, exercise no. 2. "Calculate and show in a reduced fraction form the following sum: 1/(2*3) + 2/(3*5) + 3/(5*8) + 4/(8*12) + 5/(12*17)."
|
|
|
FORMULA
|
a(n) = (n^2 - 3*n + 6)*(n^2 - n + 4)/4.
G.f.: x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.
a(n) = (24 - 18*n + 13*n^2 - 4*n^3 + n^4)/4. (End)
E.g.f.: (1/4)*exp(x)*(16 + 8*x + 14*x^2 + 6*x^3 + x^4). - Stefano Spezia, Nov 30 2018
|
|
|
EXAMPLE
|
2*2, 2*3, 3*5, 5*8, 8*12, 12*17, 17*23, 23*30, 30*38, ...
|
|
|
MAPLE
|
b:=n->(n^2-3*n+6)/2: seq(b(n)*b(n+1), n=1..40); # Muniru A Asiru, Aug 16 2018
|
|
|
MATHEMATICA
|
Times@@@Partition[Array[(#^2 -3# +6)/2 &, 40], 2, 1] (* Michael De Vlieger, Sep 24 2018 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {4, 6, 15, 40, 96}, 40] (* G. C. Greubel, Feb 10 2019 *)
|
|
|
PROG
|
(GAP) List([1..40], n->(n^2-3*n+6)*(n^2-n+4)/4); # Muniru A Asiru, Aug 16 2018
(PARI) Vec(x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1-x)^5 + O(x^40)) \\ Colin Barker, Nov 26 2018
(Sage) [(n^2-3*n+6)*(n^2-n+4)/4 for n in (1..40)] # G. C. Greubel, Feb 10 2019
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
