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Number of integer-sided triangles with sides a,b,c, a<b<c, a+b+c=n such that c - b > b - a.
2

%I #24 Jul 04 2021 03:01:22

%S 0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,1,1,0,2,1,1,2,2,1,4,2,2,4,4,2,6,4,4,6,

%T 6,4,9,6,6,9,9,6,12,9,9,12,12,9,16,12,12,16,16,12,20,16,16,20,20,16,

%U 25,20,20,25,25,20,30,25,25,30,30,25,36,30,30,36,36,30,42,36,36,42,42,36,49,42,42,49,49

%N Number of integer-sided triangles with sides a,b,c, a<b<c, a+b+c=n such that c - b > b - a.

%C Same as A025828 with zeros prepended. - _Joerg Arndt_, Nov 04 2014

%H G. C. Greubel, <a href="/A024165/b024165.txt">Table of n, a(n) for n = 1..999</a>

%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,1,1,0,1,-1,0,-1,-1,0,0,1).

%F G.f.: x^13/((1-x^3)*(1-x^4)*(1-x^6)). - _Tani Akinari_, Nov 04 2014

%F From _Robert Israel_, Nov 04 2014: (Start)

%F a(n) = a(n-3) + a(n-4) + a(n-6) - a(n-7) - a(n-9) - a(n-10) + a(n-13) for n >= 14.

%F a(6*n) = (2*n^2 - 8*n + 7 + (-1)^n)/8, n >= 1.

%F a(6*n+1) = a(6*n+4) = a(6*n+5) = (2*n^2 - 1 + (-1)^n)/8.

%F a(6*n+2) = a(6*n+3) = (2*n^2 - 4*n + 1 - (-1)^n)/8.

%F (End)

%t LinearRecurrence[{0,0,1,1,0,1,-1,0,-1,-1,0,0,1},{0,0,0,0,0,0,0,0,0,0,0,0,1},100] (* _Harvey P. Dale_, Sep 04 2017 *)

%o (Sage)

%o def A024165_list(prec):

%o P.<x> = PowerSeriesRing(QQ, prec)

%o return P( x^13/((1-x^3)*(1-x^4)*(1-x^6)) ).list()

%o a=A024165_list(100); a[1:] # _G. C. Greubel_, Jul 03 2021

%o (Magma)

%o R<x>:=PowerSeriesRing(Rationals(), 100);

%o [0,0,0,0,0,0,0,0,0,0,0,0] cat Coefficients(R!( x^13/((1-x^3)*(1-x^4)*(1-x^6)) )); // _G. C. Greubel_, Jul 03 2021

%Y Cf. A024163, A024164, A025828.

%K nonn

%O 1,19

%A _Clark Kimberling_