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%I #21 Sep 08 2022 08:45:11
%S 1,1,6,17,56,179,576,1851,5950,19125,61474,197597,635140,2041543,
%T 6562172,21092919,67799386,217928905,700493182,2251609065,7237391472,
%U 23263290299,74775653304,240352858739,772570939222,2483290023101
%N To obtain a(n+1), add the square of the n-th partial sum to the n-th partial sum of the squares, then divide this result by a(n), for all n >= 0, with a(0)=1.
%H Harvey P. Dale, <a href="/A088016/b088016.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,1,-1).
%F a(n) = 3*a(n-1) + a(n-2) - a(n-3) for n>3.
%F G.f.: (1-2*x+2*x^2-x^3) / (1-3*x-x^2+x^3).
%F G.f.: A(x) = A030186(x) * (1-x+x^2), where A030186(x) = gf of A030186.
%e G.f.: A(x) = 1 + x + 6*x^2 + 17*x^3 + 56*x^4 + 179*x^5 + 576*x^6 + ...
%e where A(x) * (1 - 3*x - x^2 + x^3) = 1 - 2*x + 2*x^2 - x^3.
%e Illustration of the initial terms: set a(0) = a(1) = 1, then
%e a(2) = ((1+1)^2 + (1^2 + 1^2))/1 = 6;
%e a(3) = ((1+1+6)^2 + (1^2 + 1^2 + 6^2))/6 = 17;
%e a(4) = ((1+1+6+17)^2 + (1^2 + 1^2 + 6^2 + 17^2))/17 = 56;
%e a(5) = ((1+1+6+17+56)^2 + (1^2 + 1^2 + 6^2 + 17^2 + 56^2))/56 = 179; ...
%p seq(coeff(series((1-2*x+2*x^2-x^3)/(1-3*x-x^2+x^3), x, n+1), x, n), n = 0 .. 40); # _G. C. Greubel_, Oct 27 2019
%t LinearRecurrence[{3,1,-1},{1,1,6,17},40] (* _Harvey P. Dale_, Nov 06 2012 *)
%o (PARI) a(n)=(sum(k=0,n-1,a(k))^2+sum(k=0,n-1,a(k)^2))/a(n-1)
%o (PARI) my(x='x+O('x^40)); Vec((1-2*x+2*x^2-x^3)/(1-3*x-x^2+x^3)) \\ _G. C. Greubel_, Oct 27 2019
%o (Magma) I:=[1,6,17]; [1] cat [n le 3 select I[n] else 3*Self(n-1) +Self(n-2) -Self(n-3): n in [1..30]]; // _G. C. Greubel_, Oct 27 2019
%o (Sage)
%o def A088016_list(prec):
%o P.<x> = PowerSeriesRing(ZZ, prec)
%o return P((1-2*x+2*x^2-x^3)/(1-3*x-x^2+x^3)).list()
%o A088016_list(40) # _G. C. Greubel_, Oct 27 2019
%o (GAP) a:=[1,6,17];; for n in [4..40] do a[n]:=3*a[n-1]+a[n-2]-a[n-3]; od; Concatenation([1], a); # _G. C. Greubel_, Oct 27 2019
%Y Cf. A030186, A087640.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Sep 18 2003
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