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A006846
Hammersley's polynomial p_n(1).
(Formerly M1807)
13
1, 1, 2, 7, 41, 376, 5033, 92821, 2257166, 69981919, 2694447797, 126128146156, 7054258103921, 464584757637001, 35586641825705882, 3136942184333040727, 315295985573234822561, 35843594275585750890976, 4575961401477587844760793, 651880406652100451820206941
OFFSET
0,3
COMMENTS
Equals column 0 of triangle A104027. Also equals column 0 of triangle A104030 (offset 1). Both A104027 and A104030 involve the trinomial coefficients. - Paul D. Hanna, Mar 06 2005
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
J. M. Hammersley, An undergraduate exercise in manipulation, Math. Scientist, 14 (1989), 1-23.
J. M. Hammersley, An undergraduate exercise in manipulation, Math. Scientist, 14 (1989), 1-23. (Annotated scanned copy)
FORMULA
a(n) = Sum_{k>=0} (-1)^(n+k)*A065547(n, k) = Sum_{k>=0} A085707(n, k). - Philippe Deléham, Feb 26 2004
E.g.f.: cosh(sqrt(3)*x/2)/cos(x/2) = Sum_{n>=0} a(n)*x^(2n)/(2n)!. - Paul D. Hanna, Feb 27 2005
a(n) = (-1)^n*A104027(n, 0). a(n+1) = (-1)^(n+1)*A104030(n, 0). - Paul D. Hanna, Mar 06 2005
G.f.: 1/(1-x/(1-x/(1-3x/(1-4x/(1-7x/(1-.../(1-ceiling((n+1)^2/4)*x/(1-... (continued fraction). - Paul Barry, Feb 24 2010
a(n) ~ 4*cosh(sqrt(3)*Pi/2) * (2*n)! / Pi^(2*n+1). - Vaclav Kotesovec, Jun 07 2021
MAPLE
A006846 := proc(n)
option remember ;
if n =0 then
return 1;
else
add(binomial(2*n, 2*m)*procname(m)/(-4)^(n-m), m=0..n-1) ;
(3/4)^n-% ;
end if
end proc:
seq(A006846(n), n=0..20) ; # R. J. Mathar, Jan 10 2018
MATHEMATICA
h[n_, x_] := Sum[c[k] x^k, {k, 0, n}]; eq[n_] := SolveAlways[h[n, x*(x-1)] == EulerE[2*n, x], x]; a[n_] := Sum[(-1)^(n+k)*c[k], {k, 0, n}] /. eq[n] // First; Table[a[n], {n, 0, 15}] (* Jean-François Alcover, Oct 02 2013, after Philippe Deléham *)
PROG
(PARI) {a(n)=local(X=x+x*O(x^(2*n))); round((2*n)!*polcoeff(cosh(sqrt(3)*X/2)/cos(X/2), 2*n))} \\ Paul D. Hanna
(Julia)
function A006846list(len::Int) # Algorithm of L. Seidel (1877)
R = Array{BigInt}(len)
A = fill(BigInt(0), len+1); A[1] = 1
for n in 1:len
for k in n:-1:2 A[k] += A[k+1] end
for k in 2:1:n A[k] += A[k-1] end
R[n] = A[n]
end
return R
end
println(A006846list(20)) # Peter Luschny, Jan 02 2018
CROSSREFS
KEYWORD
nonn
STATUS
approved