cp's OEIS Frontend

A026224 Numbers n such that t(n) = s(n) + 1, where s = A026136, t = A026142.

%I A026224 #17 Mar 13 2022 17:20:50
%S A026224 2,4,10,13,22,28,31,37,40,49,58,64,67,76,82,85,91,94,103,109,112,118,
%T A026224 121,130,139,145,148,157,166,172,175,184,190,193,199,202,211,220,226,
%U A026224 229,238,244,247,253,256,265,271,274,280,283
%N A026224 Numbers n such that t(n) = s(n) + 1, where s = A026136, t = A026142.
%C A026224 n is chosen to denote the numbers, as each n represents an index for sequences s and t.
%C A026224 From _Peter Munn_, Mar 08 2022: (Start)
%C A026224 2 with numbers of the form 3^i*(3k+1) + 1, i >= 1.
%C A026224 Proof:
%C A026224 n = 1 is clearly excluded by either definition, as t(1) <> s(1) + 1 and 1 is not of the form 3^i*(3k+1) + 1, i >= 1. For n >= 2 the remaining argument applies.
%C A026224 Considering the conditions s and t place on individual terms, and using basic arithmetic, it is easy to show that "s(n) > n, t(n) > n" is a necessary condition for t(n) = s(n) + 1. Taking into account the lexicographically earliest properties of s and t, it is then straightforward to show the condition is also sufficient. I omit the details.
%C A026224 Proofs in A026136 and A026142 show: s(n) > n if and only if s(n) has the form 3^i*(6k+2)+1; t(n) > n if and only if t(n) has the form (A) 3^i*4 or (B) 3^i*(6k+2), k >= 1. We consider (A) and (B) separately:
%C A026224 (A) s(n) + 1 = 3^i_1*(6k_1+2) + 2 = 3^i_2*4 = t(n)
%C A026224 Modulo 3, the left-hand side can be congruent to 1 or 2, the right-hand side to 0 or 1. Equality requires i_2 = 0, so t(n) = 4, from which we complete the solution with n = 2 and s(n) = 3.
%C A026224 or
%C A026224 (B) s(n) + 1 = 3^i_1*(6k_1+2) + 2 = 3^i_2*(6k_2+2) = t(n), k_2 >= 1
%C A026224 Modulo 3, the left-hand side can be congruent to 1 or 2, the right-hand side to 0 or 2. Equality requires i_1 >= 1, i_2 = 0.
%C A026224 So we have 3^i_1*(6k_1+2) + 2 = 6k_2+2, i_1 >= 1, k_2 >= 1. Clearly, for any i_1 >= 1 and k_1, there is a solution for k_2.
%C A026224 So for n to qualify under (B), s(n) must have the form 3^i*(6k+2) + 1, i >= 1, and therefore also the form 6j+1. If s(n) has the form 6j+1 and s(n) > n, then n = 3j+1 (see A026136) and also t(3j+1) = 6j+2 (see A026142, given j >= 1). So we need n to have the form 3^i*(3k+1) + 1, i >= 1, and for all such n there is a solution s(n) + 1 = 2*3^i*(3k+1) + 2 = t(n).
%C A026224 (End)
%Y A026224 Cf. A026136, A026142.
%K A026224 nonn,easy
%O A026224 1,1
%A A026224 _Clark Kimberling_