Stieltjes continued fractions related to the paperfolding sequence and Rudin-Shapiro sequence

Abstract

We investigate two Stieltjes continued fractions given by the paperfolding sequence and the Rudin-Shapiro sequence. By explicitly describing certain subsequences of the convergents P n (x)/Q n (x) modulo 4, we give the formal power series expansions (modulo 4) of these two continued fractions and prove that they are congruent modulo 4 to algebraic series in Z [[x]]. Therefore, the coefficient sequences of the formal power series expansions are 2-automatic. Write Q n (x) = i≥0 a n,i x i . Then (Q n (x)) n≥0 defines a two-dimensional coefficient sequence (a n,i ) n,i≥0 . We prove that the coefficient sequences (a n,i mod 4) n≥0 introduced by both (Q n (x)) n≥0 and (P n (x)) n≥0 are 2-automatic for all i ≥ 0. Moreover, the pictures of these two dimensional coefficient sequences modulo 4 present a kind of self-similar phenomenon.

On the algebraicity of Thue-Morse and period-doubling continued fractions

“….17) for s = A(i, 0 2k ). We find that (A(i, 0 24 ), E 24 ) = (A(i, 0 16 ), E 16 ), so that we only have to verify that identities (2.12) through (2.17) hold for 2 ≤ k ≤ 12, which turns out to be true.…”

“…The authors [10] proved that the Stieltjes continued fractions defined by the Thue-Morse sequence and the perioddoubling sequence in Z [[x]] are congruent, modulo 4, to algebraic series in Z [[x]]. In 2020, Wu [16] obtained similar results concerning the Stieltjes continued fractions defined by the paperfolding sequence and the Golay-Shaprio-Rudin sequence.…”

“…15 [13, 17] 1 [3,4] 6 [9,4] 11 [8,13] 16 [18, 4] 2 [5,6] 7 [10,4] 12 [14,4] 17 [19,12] 3 [1,7] 8 [11,6] 13 [15,16] 18 [16,6] 4 [4,4] 9 [6,12] 14 [12,16] 19 [17,8] and output function n → τ (n): For all k ≥ 1, a 2 2k−1 = a + 1. Therefore we only have to check that identity (3.7) holds for k = 3, 5, 7, which turns out to be true.…”

“…) = (a 18 + a 16 + a 2 + 1)x 24 + (a 17 + a16 + a + 1)x 22 + (a16 + a 15 + a 14 + a 13 + a 11 + a 10 + a 9 + a 7 + a 6 + a 5 + a 3 + a 2 + a + 1)x 20 + (a 15 + a 14 + a 13 + a 12 + a 11 + a 10 + a 9 + a 8 + a 7 + a 6 + a 5 + a 4 + a 3 + a 2 + a + 1)x 18 + (a 13 + a 10 + a 9 + a 8 + a 6 + a 5 + a 4 + a)x 16 + (a 11 + a 10 + a 9 + a 8 + a 5 + a 4 + a 3 + a 2 )x 14 + (a 9 + a 8 + a 6 + a 4 + a 3 )x 12 , p 3 (x) = (a 14 + a 12 + a 10 + a 8 + a 6 + a 4 + a 2 + 1)x 14 + (a 13 + a 12 + a 9 + a 8 + a 5 + a 4 + a + 1)x 13 + (a 10 + a 8 + a 2 + 1)x 10 + (a 9 + a 8 + a + 1)x 9 , p 6 (x) = (a 12 + a 8 + a 4 + 1)x 12 + (a 10 + a 8 + a 2 + 1)x 10 + (a 8 + 1)x 8 + (a 6 + a 4 + a 2 + 1)x 6 , p 9 (x) = (a 10 + a 8 + a 2 + 1)x 10 + (a 9 + a 8 + a + 1)x 9 + (a 8 + 1)x 8 + (a 7 + a 6…”