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%I #47 Dec 03 2023 09:10:19
%S 1,2,11,106,1457,25946,566827,14665106,438351041,14862109042,
%T 563501581931,23624177026682,1085079390005041,54185293223976266,
%U 2922842896378005707,169366580127359119906,10492171932362920604417,691986726674000405367266,48408260338825019327539531
%N a(n) = hypergeometric([n, -n], [], -1).
%C From _Peter Bala_, Mar 12 2023: (Start)
%C We conjecture that a(n+k) == a(n) (mod k) for all n and k. If true, then for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. For example, modulo 7 the sequence becomes [1, 2, 4, 1, 1, 4, 2, 1, 2, 4, 1, 1, 4, 2, ...], apparently a periodic sequence of period 7.
%C More generally, let F(x) and G(x) denote power series with integer coefficients with F(0) = G(0) = 1. Define b(n) = n! * [x^n] exp(x*G(x))*F(x)^n. Then we conjecture that b(n+k) == b(n) (mod k) for all n and k. The present sequence is the case F(x) = 1/(1 - x), G(x) = 1. Cf. A361281. (End)
%H Vincenzo Librandi, <a href="/A278070/b278070.txt">Table of n, a(n) for n = 0..370</a>
%F a(-n) = a(n).
%F a(n) = n! [x^n] exp((1-h(x))/2)*(1+h(x))/(2*h(x)) with h(x) = sqrt(1-4*x).
%F a(n) = ((2*n-1)*a(n-2) + 4*(n*(2*n-4)+1)*a(n-1))/(2*n-3) for n>=2.
%F a(n) ~ 2^(2*n-1/2) * n^n / exp(n-1/2). - _Vaclav Kotesovec_, Nov 10 2016
%F a(n) = n!*Sum_{i=0..n}(binomial(2*n-i-1,n-i)/i!). - _Vladimir Kruchinin_, Nov 23 2016
%F a(n) = n! * [x^n] exp(x)/(1 - x)^n. - _Ilya Gutkovskiy_, Sep 21 2017
%p a := n -> hypergeom([n, -n], [], -1): seq(simplify(a(n)), n=0..18);
%p # Alternatively:
%p a := proc(n) option remember; `if`(n<2, n+1,
%p ((2*n-1)*a(n-2) + 4*(n*(2*n-4)+1)*a(n-1))/(2*n-3)) end:
%p seq(a(n), n=0..18);
%t Table[HypergeometricPFQ[{n, -n}, {}, -1], {n, 0, 20}] (* _Vaclav Kotesovec_, Nov 10 2016 *)
%o (Sage)
%o def a():
%o a, b, c, d, h, e = 1, 2, 1, 8, 4, 0
%o yield a
%o while True:
%o yield b
%o e = c; c += 2
%o a, b = b, (c*a + h*b)//e
%o d += 16; h += d
%o A278070 = a()
%o [next(A278070) for _ in range(19)]
%o (Maxima)
%o a(n):=n!*sum(binomial(2*n-i-1,n-i)/i!,i,0,n); /* _Vladimir Kruchinin_, Nov 23 2016 */
%Y Cf. A278069, A278071, A361281.
%K nonn,easy
%O 0,2
%A _Peter Luschny_, Nov 10 2016
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