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A246923
Expansion of g.f.: 1 / AGM(1-9*x, sqrt((1-x)*(1-81*x))).
7
1, 25, 1089, 60025, 3690241, 241025625, 16359689025, 1140463805625, 81081830657025, 5852177325225625, 427465780890020929, 31528177440967935225, 2344153069158724611841, 175473167541934734763225, 13211212029033949825064769, 999630716942846408773325625
OFFSET
0,2
COMMENTS
In general, the g.f. of the squares of coefficients in g.f. 1/sqrt((1-p*x)*(1-q*x)) is given by
1/AGM(1-p*q*x, sqrt((1-p^2*x)*(1-q^2*x))) = Sum_{n>=0} x^n*( Sum_{k=0..n} p^(n-k)*((q-p)/4)^k*C(n,k)*C(2*k,k) )^2,
and consists of integer coefficients when 4|(q-p).
Here AGM(x,y) = AGM((x+y)/2, sqrt(x*y)) is the arithmetic-geometric mean.
LINKS
FORMULA
a(n) = A084771(n)^2 = ( Sum_{k=0..n} 2^k*C(n,k)*C(2*k,k) )^2.
G.f.: 1 / AGM((1-x)*(1+9*x), (1+x)*(1-9*x)) = Sum_{n>=0} a(n)*x^(2*n).
a(n) ~ 3^(4*n+2) / (8*Pi*n). - Vaclav Kotesovec, Sep 27 2019
a(n) = 9^n * ( LegendreP(n, 5/3) )^2. - G. C. Greubel, May 30 2023
EXAMPLE
G.f.: A(x) = 1 + 25*x + 1089*x^2 + 60025*x^3 + 3690241*x^4 + 241025625*x^5 +...
where the square-root of each term yields A084771:
[1, 5, 33, 245, 1921, 15525, 127905, 1067925, ...],
the g.f. of which is 1/sqrt((1-x)*(1-9*x)).
MATHEMATICA
a[n_] := Sum[2^k * Binomial[n, k] * Binomial[2k, k], {k, 0, n}]^2; Array[a, 17, 0] (* Amiram Eldar, Dec 11 2018 *)
Table[9^n*LegendreP[n, 5/3]^2, {n, 0, 40}] (* G. C. Greubel, May 30 2023 *)
PROG
(PARI) {a(n, p=1, q=9)=polcoeff( 1 / agm(1-p*q*x, sqrt((1-p^2*x)*(1-q^2*x) +x*O(x^n))), n) }
for(n=0, 20, print1(a(n), ", "))
(PARI) {a(n, p=1, q=9)=polcoeff( 1 / sqrt((1-p*x)*(1-q*x) +x*O(x^n)), n)^2 }
for(n=0, 20, print1(a(n), ", "))
(PARI) {a(n, p=1, q=9)=sum(k=0, n, p^(n-k)*((q-p)/4)^k*binomial(n, k)*binomial(2*k, k))^2 }
for(n=0, 20, print1(a(n), ", "))
(Magma) [9^n*Evaluate(LegendrePolynomial(n), 5/3)^2 : n in [0..40]]; // G. C. Greubel, May 30 2023
(SageMath) [9^n*gen_legendre_P(n, 0, 5/3)^2 for n in range(41)] # G. C. Greubel, May 30 2023
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 07 2014
STATUS
approved