Una generación de enfoque de función de trabajo.
El ejemplo primero y luego generalizar.
2x_1+3x_2+5x_3=100
Vamos
A(z) = 1 + z^2 + z^4 + \dots + z^{2k} + \dots = \frac1{1 - z^2}
B(z) = 1 + z^3 + z^6 + \dots + z^{3k} + \dots = \frac1{1 - z^3}
C(z) = 1 + z^5 + z^{10} + \dots + z^{5k} + \dots = \frac1{1 - z^5}
F(z) = A(z)B(z)C(z) = \sum_{k=0}^{\infty} c_k z^k
c_n equals the number of ways the three terms can add up to n.
Note all the poles of F(z) are roots of unity.
We can choose an integration contour inside of these roots.
Using the residue theorem:
c_n = \frac1{2\pi i} \oint \frac{F(z)}{z^{n+1}} dz
Where the contour is around z = 0 and inside |z| = 1
n = 100 is too large for octave. Some math package capable of large floating point numbers would be required.
But n = 10 can be demonstrated.
octave code:
function r = F(z) r = (1 - z.^2).*(1-z.^3).*(1-z.^5); r = 1./r; end
[q err] = quadgk (@(z)F(z)./z.^11, (1+1i)/4, (1+1i)/4,
"WayPoints",[(-1+1i)/4,(-1-1i)/4,(1-1i)/4]) q/(2*pi*i)
ans = 4.0000e+000 + 1.4765e-011i
Manually checking for n = 10
2.5
2.2 + 3.2
2.1 + 3.1 + 5.1
5.2
There are 4 answers.
Addendum
In Maxima
taylor(1/((1-z^2)*(1-z^3)*(1-z^5)),z,0,100);
184\,z^{100}+180\,z^{99}+177\,z^{98}+173\,z^{97}+170\,z^{96}+167\,z
^{95}+163\,z^{94}+160\,z^{93}+157\,z^{92}+153\,z^{91}+151\,z^{90}+
147\,z^{89}+144\,z^{88}+141\,z^{87}+138\,z^{86}+135\,z^{85}+132\,z^{
84}+129\,z^{83}+126\,z^{82}+123\,z^{81}+121\,z^{80}+117\,z^{79}+115
\,z^{78}+112\,z^{77}+109\,z^{76}+107\,z^{75}+104\,z^{74}+101\,z^{73}
+99\,z^{72}+96\,z^{71}+94\,z^{70}+91\,z^{69}+89\,z^{68}+86\,z^{67}+
84\,z^{66}+82\,z^{65}+79\,z^{64}+77\,z^{63}+75\,z^{62}+72\,z^{61}+71
\,z^{60}+68\,z^{59}+66\,z^{58}+64\,z^{57}+62\,z^{56}+60\,z^{55}+58\,
z^{54}+56\,z^{53}+54\,z^{52}+52\,z^{51}+51\,z^{50}+48\,z^{49}+47\,z
^{48}+45\,z^{47}+43\,z^{46}+42\,z^{45}+40\,z^{44}+38\,z^{43}+37\,z^{
42}+35\,z^{41}+34\,z^{40}+32\,z^{39}+31\,z^{38}+29\,z^{37}+28\,z^{36
}+27\,z^{35}+25\,z^{34}+24\,z^{33}+23\,z^{32}+21\,z^{31}+21\,z^{30}+
19\,z^{29}+18\,z^{28}+17\,z^{27}+16\,z^{26}+15\,z^{25}+14\,z^{24}+13
\,z^{23}+12\,z^{22}+11\,z^{21}+11\,z^{20}+9\,z^{19}+9\,z^{18}+8\,z^{
17}+7\,z^{16}+7\,z^{15}+6\,z^{14}+5\,z^{13}+5\,z^{12}+4\,z^{11}+4\,z
^{10}+3\,z^9+3\,z^8+2\,z^7+2\,z^6+2\,z^5+z^4+z^3+z^2+1+\cdots
I don't know exactly how maxima does the Taylor series expansion but it is fast?
It arrives at 184 ways to get a sum of 100.
Addendum 2
Alternatively expand the denominator and divide 1:
1 - z^2 -z^3 +z^7 +z^8 -z^{10} \, \, \overline{)1 \qquad \qquad \qquad \qquad }
Then find the cofficient of c_{100} z^{100}
Generalization
\sum_{k = 1}^{r} a_k x_k = n
Let
F(z) = \prod_{k=1}^{r}\frac1{1 - z^{a_k}}
c_n = \frac1{2\pi i} \oint \frac{F(z)}{z^{n+1}} dz
Where the integration contour is inside the roots of unity.
c_n is the number of ways the terms can sum to n.
Or find the Taylor series expansion of F(z) and the c_n coeficiente.
