We remark that the choice of 1026 + 1 in the proof is suitable merely because
Sp(1026 + 1) mod 7 = 3. In fact, one may generate similar Smith numbers of the
form 4PN ×(10e+1)×10k provided that Sp(10e +1) mod 7 = 3, e.g., with e = 23
or 24, and with the exponent k for the power of ten adjusted accordingly.