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A180324 Vassiliev invariant of fourth order for the torus knots 5

%I #24 Sep 18 2018 15:12:55

%S 0,3,25,98,270,605,1183,2100,3468,5415,8085,11638,16250,22113,29435,

%T 38440,49368,62475,78033,96330,117670,142373,170775,203228,240100,

%U 281775,328653,381150,439698,504745,576755,656208,743600,839443

%N Vassiliev invariant of fourth order for the torus knots

%C a(n) is the Vassiliev invariant of fourth order for the torus knots. a(n) can be calculated as the number of attachments of the two arrow diagrams in the arrow diagram of the torus knot. Arrow diagram of the torus knot is 2n+1 intersecting arrows with mixing ends.

%C Antidiagonal sums of the convolution array A213847. - _Clark Kimberling_, Jul 05 2012

%C First differences of the terms produced by convolving the odd and even triangular numbers, with n>0. The sequence begins 0, 3, 28, 126, 396, 1001, 2184, 4284, 7752, 13167, 21252..starting at n=1 and has the formula (4*n^5 - 5*n^3 + 30*n)/30. - _J. M. Bergot_, Sep 09 2016

%H S. V. Allenov, <a href="http://dx.doi.org/10.1007/s10958-009-9322-5">Explicit formulas for Vassil'ev invariants of the fourth order for knots</a>, Journal of Mathematical Sciences, New York: Springer, Vol. 157, No. 3 (2009), 413-423.

%H M. Polyak, O. Viro, <a href="https://www.math.stonybrook.edu/~oleg/math/papers/1994-Polyak-Viro.pdf">Gauss diagram formulas for Vassiliev invariants</a>, Int. Math. Res. Notices, 11 (1994), 445-453.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F a(n) = (n*(n+1)*(2*n+1)^2)/6.

%F a(n) = C(2*n+2,4) + C(2*n+2,3)/2.

%F G.f.: x*(3+x)*(1+3*x)/(1-x)^5. - _Colin Barker_, Mar 17 2012

%e a(1) = 1*2*3^2/6 = 3.

%e a(2) = 2*(2+1)*(2*2+1)^2/6 = 5^2 = 25.

%p a:=n->(1/6)*n*(n+1)*(2*n+1)^2;

%p a:=n->binomial(2*n+2, 4)+binomial(2*n+2, 3)/2;

%t Table[Binomial[2n+2,4]+Binomial[2n+2,3]/2,{n,0,40}] (* _Harvey P. Dale_, Sep 18 2018 *)

%o (PARI) a(n) = n*(n+1)*(2*n+1)^2/6

%Y a(n) = (2n+1)*A000330(n).

%Y a(n) = 3*A000330(n)^2/A000217(n).

%Y a(n) = (A000330(1) + A000330(2) + … + A000330(2n-1) + A000330(2n))/2.

%K nonn,easy

%O 0,2

%A _Sergey Allenov_, Jan 18 2011

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