One knows that prior to life appearing on planet Earth, there were inert molecules called pre-biotic, made of carbon (C), hydrogen (H), nitrogen (N), and oxygen (O).   These molecules combine to give birth to families of molecules which are necessary for life: amino-acids, proteins, nucleotides, etc. Among those, the four lighter ones are the nucleotides. To calculate the transition from the chemically inert state to the living state with the tree structure, let us begin by setting pre-biotic molecules as B, H, D, M, C, W.   Then, since they are the lightest ones, the 4 nucleotides are set as BM, BC, BW, HM.   These settings enable us to determine the value of pre-biotic molecules that shape the tree structure’s matrix. Then, we calculate the other molecules and carry on with by increasing the initial tree structure.   Calculating the matrix   As it is impossible, initially, to know the place of each nulceaotide, it is followed by a question mark. What we get is the following system of equals:   B+M= C5H5N5 ? B+C= C5H6N2O2 ? B+W= C4H5N3O? H+M= C5H5N5O?   One adds on both sides of the equal sign. The number of atoms is written on the right. 3B + H + 2M + C + W = 19C + 21H + 15N +4O   The carbon atoms of the B molecule are thus marked BC, hydrogen atoms, BH, etc.   Which results in 3BC + HC + 2MC + CC + WC = 19 3BH + HH + MH + CH + WH = 21 3 BN + HN + 2MN + CN + WN = 15 3 BO + HO + 2MO + CO + WO = 4   To begin with, one studies the various distributions of oxygen atoms as there are less of them B+M=BM   Which gives us the following diagram:
 Calculating the transition from the chemically inert to the living state
 Which gives 4 possibilities: (The double arrow between C and W – and therefore between Bc and Bw – means that their value can be inverted.)
 It is already obvious that there is not but one solution but several. The aim is not to calculate them all but to single out one that will match the mathematical criteria formerly defined.   Choosing the second possibility
 Will give us :   B+M = C5H5N5 H+M= C5H6N2O2   i.e.
 Let us carry on with nitrogen atoms.   We already have the following:
 The distribution will necessarily follow the pattern
 There is one solution for this distribution
 Let us carry on with carbon atoms.   We already have the following:
 The distribution will necessarily follow the pattern
 With Y ≥ 1 and consequently X  ≤ 4   At least 1 carbon atom must indeed be set in BM, as there is 1 less in BC or BW.   Which gives the following sequences:
 Bc
 Hc
 Let us pick the 4th
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 1 B= C3H4N3 H=C3H5O2 D=? M=C2HN2 C=C2HN2O W=CHO 2 B=C3H4N3 H=C3H5O2 D=? M=C2HN2 C=CHO W=C2HN2O
 B= C3H4N3 H=C3H5O2 D=C3H5N3O2 M=C2HN2 C=CHO W=C2HN2O
 B’ B’ B’ H’ H’ H’ D’ D’ D’ M’ C’ W’ M’ C’ W’ M’ C’ W’
 B = 3 B H = 7 H D = 9 D M = 10 M C = 12 C W = 12 W