|
|
A161704
|
|
a(n) = (3*n^5 - 35*n^4 + 145*n^3 - 235*n^2 + 152*n + 30)/30.
|
|
20
|
|
|
1, 2, 3, 6, 9, 18, 59, 190, 513, 1186, 2435, 4566, 7977, 13170, 20763, 31502, 46273, 66114, 92227, 125990, 168969, 222930, 289851, 371934, 471617, 591586, 734787, 904438, 1104041, 1337394, 1608603, 1922094, 2282625, 2695298, 3165571, 3699270
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
{a(k): 0 <= k < 6} = divisors of 18:
|
|
LINKS
|
|
|
FORMULA
|
a(n) = C(n,0) + C(n,1) + 2*C(n,3) - 4*C(n,4) + 12*C(n,5).
G.f.: ( 1-4*x+6*x^2-2*x^3-7*x^4+18*x^5 ) / (x-1)^6. - R. J. Mathar, Jul 12 2016
|
|
EXAMPLE
|
Differences of divisors of 18 to compute the coefficients of their interpolating polynomial, see formula:
1 2 3 6 9 18
1 1 3 3 9
0 2 0 6
2 -2 6
-4 8
12
|
|
MAPLE
|
|
|
MATHEMATICA
|
CoefficientList[Series[(1 - 4*x + 6*x^2 - 2*x^3 - 7*x^4 + 18*x^5)/(x - 1)^6, {x, 0, 50}], x] (* G. C. Greubel, Jul 16 2017 *)
|
|
PROG
|
(Magma) [(3*n^5 - 35*n^4 + 145*n^3 - 235*n^2 + 152*n + 30)/30: n in [0..50]]; // Vincenzo Librandi, Dec 27 2010
(PARI) a(n)=n*(3*n^4-35*n^3+145*n^2-235*n+152)/30+1
|
|
CROSSREFS
|
Cf. A000124, A000125, A000127, A002522, A005408, A006261, A016813, A018251, A058331, A080856, A086514, A161701, A161702, A161703, A161706, A161707, A161708, A161710, A161711, A161712, A161713, A161715, A161856.
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|