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A295371 a(n) = (1/(2n))*Sum_{k=0..n-1} C(n-1, k)*C(n+k, k)*C(2k, k)*(k+2)*(-3)^(n-1-k). 1
1, 3, 19, 127, 921, 6921, 53523, 422199, 3382417, 27429043, 224636259, 1854761437, 15419579761, 128941830993, 1083686483259, 9147887134119, 77520233226537, 659167237928691, 5622149927918763, 48083938099637247 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
From Mark van Hoeij, Nov 10 2022: (Start)
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(n-1)^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(n-1)^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
Heba Bou KaedBey, Mark van Hoeij, and Man Cheung Tsui, Solving Third Order Linear Difference Equations in Terms of Second Order Equations, arXiv:2402.11121 [math.AC], 2024. See p. 6.
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).
From Mark van Hoeij, Nov 10 2022: (Start)
G.f.: (hypergeom([1/2, 1/2], [1], 16*x/(1 + 3*x)^2) - 1)/4.
a(n) = (A002426(n)^2 + 3 * A002426(n-1)^2)/4. (End)
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)^(2-k) = (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[n-1, k]Binomial[n+k, k]Binomial[2k, k](k+2)(-3)^(n-1-k);
s[n_]:=a[n]=Sum[f[n, k], {k, 0, n-1}]/(2n);
Table[s[n], {n, 1, 20}]
CROSSREFS
Sequence in context: A074572 A157455 A027308 * A156069 A058860 A074568
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Nov 20 2017
EXTENSIONS
Name simplified based on the proof of Mark van Hoeij by Peter Luschny, Nov 10 2022
STATUS
approved

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Last modified March 28 17:42 EDT 2024. Contains 371254 sequences. (Running on oeis4.)