%I #28 Apr 23 2018 08:34:42
%S 1,48,161856,39002646528,674708032182398976,
%T 839431510934341028210638848,75178263784150214825106859877233852416,
%U 484905075185415831301477770434885768003422223597568,225327830550164300895512117291590826401931052058453494726924435456,7544971365077550026405694467600069733983243666195122776655161969325034606646263808
%N Determinant of the (n+1) X (n+1) Hankel-type matrix with (i,j)-entry equal to A005259(i+j) for all i,j = 0,...,n.
%C Conjecture: a(n)/24^n is always a positive integer. Similarly, if b(n) denotes the (n+1) X (n+1) Hankel-type determinant with (i,j)-entry equal to A005258(i+j) for all i,j = 0,...,n, then b(n)/10^n is always a positive integer; also, if p is a prime with floor(p/10) odd and p is not congruent to 31 or 39 modulo 40, then p divides b((p-1)/2).
%C Conjecture: if A(x) = 1 + 48*x + 161856*x^2 + ... denotes the o.g.f. then A(x/3)^(1/8) has integer coefficients (checked up to x^30). - _Peter Bala_, Apr 22 2018
%e a(0) = 1 since A005259(0+0) = 1.
%e A(x/3)^(1/8) = 1 + 2*x + 2234*x^2 + 180536476*x^3 + 1041213553880806*x^4 + 431806318205326490858140*x^5 + 12890648790962619413782473229673892*x^6 + 27715196341006992690056202634389754569453086008*x^7 + 4292939920556011562306504817069205738464230629574745210785030*x^8 + 47915532217380103151430239883031701095737468980424637791531495548671526291244*x^9 + .... - _Peter Bala_, Apr 22 2018
%t A[n_]:=Sum[Binomial[n,k]^2*Binomial[n+k,k]^2,{k,0,n}]; a[n_]:=Det[Table[A[i+j],{i,0,n},{j,0,n}]]; Table[a[n],{n,0,10}]
%Y Cf. A005258, A005259, A225776.
%K nonn,easy
%O 0,2
%A _Zhi-Wei Sun_, Aug 14 2013