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Hermite's problem: number of positive integral solutions to x + y + z = n subject to x <= y + z, y <= z + x and z <= x + y.
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%I #33 May 09 2022 23:08:51

%S 0,0,0,1,3,3,7,6,12,10,18,15,25,21,33,28,42,36,52,45,63,55,75,66,88,

%T 78,102,91,117,105,133,120,150,136,168,153,187,171,207,190,228,210,

%U 250,231,273,253,297,276,322,300,348,325,375,351,403,378,432

%N Hermite's problem: number of positive integral solutions to x + y + z = n subject to x <= y + z, y <= z + x and z <= x + y.

%D G. Pólya and G. Szegő, Problems and Theorems in Analysis I, Springer-Verlag, Part I, Chap. 1, Problem 31.

%H William A. Tedeschi, <a href="/A069981/b069981.txt">Table of n, a(n) for n = 0..10000</a>

%H G. E. Andrews, P. Paule and A. Riese, <a href="http://www.risc.uni-linz.ac.at/research/combinat/risc/publications/#ppaule">MacMahon's partition analysis III. The Omega package</a>, p. 16.

%H G. E. Andrews, P. Paule and A. Riese, <a href="http://dx.doi.org/10.1006/eujc.2001.0527">MacMahon's Partition Analysis: The Omega Package</a>, Europ. J. Combin., 22 (2001), 887-904.

%F G.f.: x^3*(1 + 2*x - 2*x^2)/(1 - x)/(1 - x^2)^2.

%F a(n) = (n+8)*(n-2)/8 for n even, (n^2-1)/8 for n odd.

%F a(n) = (2*n^2 + 6*n - 17 + 3*(2*n - 5)*(-1)^n)/16 for n>0. - _Luce ETIENNE_, Jun 29 2015

%F E.g.f.: (16 + (x^2 + x - 16)*cosh(x) + (x^2 + 7*x - 1)*sinh(x))/8. - _Stefano Spezia_, May 09 2022

%t f[n_]:=If[EvenQ[n],((n+8)(n-2))/8,(n^2-1)/8];Join[{0},Array[f,60]] (* _Harvey P. Dale_, Jul 26 2011 *)

%o (PARI) for(n=0, 60, print1(if(n==0, 0, (2*n^2 + 6*n - 17 + 3*(2*n - 5)*(-1)^n)/16), ", ")) \\ _G. C. Greubel_, Jun 10 2018

%o (Magma) [0] cat [(2*n^2 + 6*n - 17 + 3*(2*n - 5)*(-1)^n)/16: n in [1..60]]; // _G. C. Greubel_, Jun 10 2018

%Y Cf. A005044.

%K nonn,easy,nice

%O 0,5

%A _N. J. A. Sloane_, May 06 2002