The energy allocated at the first step of synthesis is equal

          E1 = E_2H2+E_2H2 = 2224.749  KeV + 2224.749  KeV = 4449.749  KeV

At the second step there is one reaction of synthesis

      ₂¹H+ ₁⁰n = ₃¹H + 6257.426  KeV.

The energy allocated at the second step of synthesis is equal

         E2 = E_2H3 =6257.426  KeV.

At the third step there is one reaction of synthesis

      ₃¹H+₂¹H=₅²He + 16699.28  KeV.

The energy allocated at the third step of synthesis is equal

           E3 = E_2H3 =16699.28  KeV.

The sums of energies allocated in the previous steps can be defined under the formulas:

    E1sum= E1=4449.749  KeV — for the first step.

    E2sum= E1+E2=4449.749  KeV +6257.426  KeV=10706.924  KeV — for the second step.

    E3sum= E1+E2+E3=4449.749  KeV +6257.426  KeV+16699.28  KeV=27406.204 KeV -for the third step.

Let’s make the table of allocation of energy at the synthesis of a kernel of helium ₅²He

                                                                                  (15) Table № T-3.2.

                                              Allocation of energy at the synthesis of a kernel of helium ₅²He                   

According to table № T-3.2 we shall construct the schedule № G-3.1 describing the allocation of energy E in steps and total allocation of energy E_sum.                                                                                           

The schedule of allocation of energy in steps E
And total allocation of energy Esum, at synthesis of a kernel of helium ₅²He.
(16) Schedule № G-3.1.

  After construction of the schedule we move to the analysis of the received data and the schedule.
In this analysis a full chain of synthesis is made, considering all the nuclide and calculated allocation of energy for all kernels.            
     — A choice and the power analysis of chains of synthesis, kernels of atoms
  And so, the program of the analysis is certain, we move to the following stage.
Let’s choose chains of synthesis from protons and neutrons up to the heaviest kernels limited by the periodic table with Z=110, A=272 and Z=111, A=272.
 The chain was chosen by the rate of its maximal length and by the presence of data in table № A-1, for elements of this chain.
The chosen chain of synthesis (₁¹H — a kernel with Z=111, A=272), for two kernels, is represented on schemes № S-3.2, № S-3.3, № S-3.4.
   Chains of synthesis (₁¹H — a kernel with Z=111, A=272): for three kernels — on schemes № S-3.7, № S-3.8; for four kernels — № S-3.9; for ten kernels — № S-3.10.
    Chains of synthesis (₁¹H — a kernel with Z=110, A=272) for two kernels are represented on schemes № S-3.5, № S-3.6.
Let’s disassemble a chain of synthesis for two kernels or nuclides

Variant — (₁¹H — a kernel with Z=111, A=272) on schemes № S-3.2, № S-3.3, № S-3.4.

(17) Scheme № S-3.2.

(18) Table № T-3.3, a line 16.

(19) Scheme № S-3.3.

(20) Table № T-3.3, a line 15.