|
| |
|
|
A291013
|
|
p-INVERT of (1,1,1,1,1,...), where p(S) = (1 - S^2)^3.
|
|
2
|
|
|
|
0, 3, 6, 15, 36, 85, 198, 456, 1040, 2352, 5280, 11776, 26112, 57600, 126464, 276480, 602112, 1306624, 2826240, 6094848, 13107200, 28114944, 60162048, 128450560, 273678336, 581959680, 1235222528, 2617245696, 5536481280, 11693719552, 24662507520, 51942260736
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
Suppose s = (c(0), c(1), c(2),...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
See A291000 for a guide to related sequences.
|
|
|
LINKS
|
|
|
|
FORMULA
|
G.f.: x*(3 - 12*x + 15*x^2 - 6*x^3 + x^4)/(1 - 2*x)^3.
a(n) = 6*a(n-1) - 12*a(n-2) + 8*a(n-3) for n >= 7.
a(n) = 2^(n-6) * (60 + 17*n + n^2) for n>2. - Colin Barker, Aug 23 2017
E.g.f.: -(15/16) + (9/16)*x - x^2/16 + (1/16)*(15 +9*x +x^2)*exp(2*x). - G. C. Greubel, Jun 05 2023
|
|
|
MATHEMATICA
|
z = 60; s = x/(1-x); p = (1 - s^2)^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1] (* A000012 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A291013 *)
LinearRecurrence[{6, -12, 8}, {0, 3, 6, 15, 36, 85}, 41] (* G. C. Greubel, Jun 05 2023 *)
|
|
|
PROG
|
(PARI) concat(0, Vec(x*(3 -12*x +15*x^2 -6*x^3 +x^4)/(1-2*x)^3 + O(x^50))) \\ Colin Barker, Aug 23 2017
(Magma) [0, 3, 6] cat [2^(n-6)*(60 +17*n +n^2): n in [3..40]]; // G. C. Greubel, Jun 05 2023
(SageMath) [(2^(n-2)*(60 +17*n +n^2) -15*int(n==0) + 9*int(n==1))//16 for n in range(41)] # G. C. Greubel, Jun 05 2023
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
