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%I #21 Sep 08 2022 08:45:14
%S 1,2,3,4,6,8,10,12,14,16,19,22,25,29,33,37,41,45,49,54,59,64,70,76,82,
%T 88,94,100,107,114,121,129,137,145,153,161,169,178,187,196,206,216,
%U 226,236,246,256,267,278,289,301,313,325,337,349,361,374,387,400,414,428
%N a(n+1) = a(n) + number of squares so far; a(1) = 1.
%C Conjecture: a(n) = m^2 iff m mod 3 > 0.
%C a(n) is a square iff n is congruent to {1, 4} mod 9. - _Vladeta Jovovic_, Aug 30 2004
%H Reinhard Zumkeller, <a href="/A097602/b097602.txt">Table of n, a(n) for n = 1..10000</a>
%H <a href="/index/Rec#order_11">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,0,0,0,0,0,1,-2,1).
%F a(9*n+1) = (3*n+1)^2; a(9*n+4) = (3*n+2)^2. - _Vladeta Jovovic_, Aug 30 2004
%F G.f.: x*(1+x^4-x^9+x^10)/((1+x+x^2)*(1+x^3+x^6)*(1-x)^3). - _Vladeta Jovovic_, Aug 30 2004
%F a(n+1) = a(n) + Sum_{k=1..n} A010052(a(k)). - _Reinhard Zumkeller_, Nov 15 2011
%e a(2) = a(1) + #{1} = 1 + 1 = 2;
%e a(3) = a(2) + #{1} = 2 + 1 = 3;
%e a(4) = a(3) + #{1} = 3 + 1 = 4;
%e a(5) = a(4) + #{1,4} = 4 + 2 = 6;
%e a(6) = a(5) + #{1,4} = 6 + 2 = 8;
%e a(7) = a(6) + #{1,4} = 8 + 2 = 10;
%e a(8) = a(7) + #{1,4} = 10 + 2 = 12;
%e a(9) = a(8) + #{1,4} = 12 + 2 = 14;
%e a(10) = a(9) + #{1,4} = 14 + 2 = 16;
%e a(11) = a(10) + #{1,4,16} = 16 + 3 = 19;
%e a(12) = a(11) + #{1,4,16} = 19 + 3 = 22.
%t LinearRecurrence[{2,-1,0,0,0,0,0,0,1,-2,1}, {1,2,3,4,6,8,10,12,14,16,19}, 70] (* _G. C. Greubel_, Jan 14 2019 *)
%o (Haskell)
%o a097602 n = a097602_list !! (n-1)
%o a097602_list = 1 : f 1 1 where
%o f x c = y : f y (c + a010052 y) where y = x + c
%o -- _Reinhard Zumkeller_, Nov 15 2011
%o (PARI) my(x='x+O('x^70)); Vec(x*(1+x^4-x^9+x^10)/((1+x+x^2)*(1+x^3+x^6)*(1-x)^3)) \\ _G. C. Greubel_, Jan 14 2019
%o (Magma) m:=70; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!( x*(1+x^4-x^9+x^10)/((1+x+x^2)*(1+x^3+x^6)*(1-x)^3) )); // _G. C. Greubel_, Jan 14 2019
%o (Sage) a=(x*(1+x^4-x^9+x^10)/((1+x+x^2)*(1+x^3+x^6)*(1-x)^3)).series(x, 70).coefficients(x, sparse=False); a[1:] # _G. C. Greubel_, Jan 14 2019
%o (GAP) a:=[1,2,3,4,6,8,10,12,14,16,19];; for n in [12..70] do a[n]:= 2*a[n-1]-a[n-2]+a[n-9]-2*a[n-10]+a[n-11]; od; a; # _G. C. Greubel_, Jan 14 2019
%Y Cf. A101135, A096777, A131073.
%K nonn
%O 1,2
%A _Reinhard Zumkeller_, Aug 30 2004
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