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%I #33 Jan 01 2023 12:35:51
%S 1,1,2,0,2,2,1,3,4,-2,2,0,2,2,4,0,4,4,-1,3,2,2,4,6,1,7,8,-6,2,-4,4,0,
%T 4,-2,2,0,2,2,4,0,4,4,2,6,8,-4,4,0,4,4,8,0,8,8,-5,3,-2,4,2,6,-1,5,4,0,
%U 4,4,2,6,8,2,10,12,-5,7,2,6,8,14,1,15,16,-14,2,-12,8,-4,4,-6,-2,-8,8,0,8,-4,4,0,4,4,8,-6,2
%N Antidiagonal sums of a Legendre-binomial triangle for p = 3.
%C Diagonal sums of A117947.
%H Antti Karttunen, <a href="/A117963/b117963.txt">Table of n, a(n) for n = 0..16384</a> (terms 0..5000 from Alois P. Heinz)
%F a(n) = a(3n+2)/a(2).
%F a(n) = Sum_{k=0..floor(n/2)} L(C(n-k,k)/3) where L(j/p) is the Legendre symbol of j and p.
%F From _Paul D. Hanna_, Jul 11 2006: (Start)
%F G.f. satisfies: A(x) = A(x^3)*(1 - 4*x^3 - x^6)/(1 - x - x^2).
%F a(n) == Fibonacci(n+1) (mod 3);
%F a(n) == a(n-1) + a(n-2) (mod 3). (End)
%e The triangle mentioned in the name starts:
%e {1},
%e {1, 1},
%e {1,-1, 1},
%e {1, 0, 0, 1},
%e {1, 1, 0, 1, 1},
%e {1,-1, 1, 1,-1, 1},
%e {1, 0, 0,-1, 0, 0, 1}.
%t a[n_] := Sum[JacobiSymbol[Binomial[n - k, k], 3], {k, 0, n/2}];
%t a /@ Range[0, 100] (* _Jean-François Alcover_, Oct 14 2019 *)
%o (PARI) {a(n)=local(A=1+x+x*O(x^n)); for(i=1,#binary(n), A=subst(A,x,x^3+x*O(x^n)) *(1-4*x^3-x^6)/(1-x-x^2+x*O(x^n))); polcoeff(A,n,x)} \\ _Paul D. Hanna_, Jul 11 2006
%o (PARI) A117963list(upto_n) = { my(A=1+x+x*O(x^upto_n)); for(i=1, #binary(upto_n), print1(i,", "); A=subst(A, x, x^3+x*O(x^upto_n)) *(1-4*x^3-x^6)/(1-x-x^2+x*O(x^upto_n))); print(); vector(upto_n,n,polcoeff(A,n-1)); }; \\ _Antti Karttunen_, Jan 01 2023, after _Paul D. Hanna_'s program above.
%o (PARI) A117963(n) = sum(k=0,n\2,kronecker(binomial(n-k,k),3)); \\ _Antti Karttunen_, Jan 01 2023
%Y Cf. A117947, A117964 [= a(n) mod 2].
%K easy,sign
%O 0,3
%A _Paul Barry_, Apr 05 2006
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