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

2, 4, 10, 13, 22, 28, 31, 37, 40, 49, 58, 64, 67, 76, 82, 85, 91, 94, 103, 109, 112, 118, 121, 130, 139, 145, 148, 157, 166, 172, 175, 184, 190, 193, 199, 202, 211, 220, 226, 229, 238, 244, 247, 253, 256, 265, 271, 274, 280, 283

Offset

1,1

Comments

n is chosen to denote the numbers, as each n represents an index for sequences s and t.

From Peter Munn, Mar 08 2022: (Start)

2 with numbers of the form 3^i*(3k+1) + 1, i >= 1.

Proof:

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.

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.

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:

(A) s(n) + 1 = 3^i_1*(6k_1+2) + 2 = 3^i_2*4 = t(n)

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.

or

(B) s(n) + 1 = 3^i_1*(6k_1+2) + 2 = 3^i_2*(6k_2+2) = t(n), k_2 >= 1

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.

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.

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).

(End)

Links

Table of n, a(n) for n=1..50.

Crossrefs

Cf. A026136, A026142.

Sequence in context: A127591 A100912 A174721 * A034233 A056718 A057283

Adjacent sequences: A026221 A026222 A026223 * A026225 A026226 A026227

Keyword

nonn,easy

Author

Clark Kimberling

Status

approved