

A295371


a(n) = (1/(2n))*Sum_{k=0..n1} C(n1, k)*C(n+k, k)*C(2k, k)*(k+2)*(3)^(n1k).


1



1, 3, 19, 127, 921, 6921, 53523, 422199, 3382417, 27429043, 224636259, 1854761437, 15419579761, 128941830993, 1083686483259, 9147887134119, 77520233226537, 659167237928691, 5622149927918763, 48083938099637247
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OFFSET

1,2


COMMENTS

The fact that a(n) is an integer follows from the formula for the generating function. It is not difficult to show that f(x) := (hypergeom([1/2, 1/2], [1], 16*x)  1)/4 is an element of Z[[x]]. Substituting x > x/(1 + 3*x)^2 then shows that the given g.f. is in Z[[x]] as well. The fact that this formula is indeed the g.f. follows from the recurrence.
One can also show that a(n) is odd, as follows. Reducing f(x) in Z[[x]] modulo 2 gives: x + x^2 + x^4 + x^8 + x^16 + ... Again substitute x > x/(1 + 3*x)^2, which modulo 2 is: x + x^3 + x^5 + x^7 + ... Then use the fact that (a+b)^(2^i) is congruent to a^(2^i) + b^(2^i) modulo 2 to see that f(x/(1 + 3*x)^2) is congruent to x + x^2 + x^3 + x^4 + ... modulo 2, so every a(n) is congruent to 1 modulo 2.
The formula a(n) = (A002426(n)^2 + 3*A002426(n1)^2)/4 gives a second proof that a(n) is an odd integer. The numbers A002426(n) are odd, and so their squares are congruent to 1 modulo 8. Hence A002426(n)^2 + 3*A002426(n1)^2 is congruent to 1 + 3 * 1 modulo 8. Since a(n) is that number divided by 4, it follows that a(n) is an odd integer. (End)


LINKS



FORMULA

Via the Zeilberger algorithm we find that the sequence the following recurrence: (2n + 1)*(n + 3)^2*a(n + 3) = (2n + 1)*(7n^2 + 38n + 52)*a(n + 2) + 3*(2n + 5)*(7n^2 + 4n + 1)*a(n + 1)  27*(2n + 5)*n^2*a(n).
G.f.: (hypergeom([1/2, 1/2], [1], 16*x/(1 + 3*x)^2)  1)/4.
G.f.: EllipticK((4*sqrt(x))/(3*x + 1))/(2*Pi)  (1/4).  Peter Luschny, Nov 10 2022


EXAMPLE

a(3) = 19 since (1/6)*Sum_{k=0,1,2} binomial(2,k)*binomial(3+k,k)*binomial(2k,k)*(k+2)*(3)^(2k) = (2*(3)^2 + 2*4*2*3*(3) + 10*6*4)/6 = 19.


MAPLE

ogf := EllipticK((4*sqrt(x))/(3*x + 1))/(2*Pi)  (1/4); ser := series(ogf, x, 22): seq(coeff(ser, x, n), n = 1..20); # Peter Luschny, Nov 10 2022


MATHEMATICA

f[n_, k_]:=f[n, k]=Binomial[n1, k]Binomial[n+k, k]Binomial[2k, k](k+2)(3)^(n1k);
s[n_]:=a[n]=Sum[f[n, k], {k, 0, n1}]/(2n);
Table[s[n], {n, 1, 20}]


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



