cp's OEIS Frontend

Definition: For positive integers b (as base) and n, the positive integer (allowing initial zeros) k(n) is expomorphic relative to base b (here 9) if k(n) has exactly n decimal digits and if b^k(n) == k(n) (mod 10^n) or, equivalently, b^k(n) ends in k(n). [See Crux Mathematicorum link.]

For sequences in the OEIS, no term is allowed to begin with a digit 0 (except for the 1-digit number 0 itself). However, in the problem as defined in the Crux Mathematicorum article, leading 0 digits are allowed; under that definition a(n) = k(n) until the first k(n) which begins with digit 0. When k(n) begins with 0, then, a(n) is the next term of the sequence k(n) which doesn't begin with digit 0.

Conjecture: if k(n) is expomorphic relative to "base" b, then, the next one in the sequence, k(n+1), consists of the last n+1 digits of b^k(n).

a(n) is the backward concatenation of A133619(0) through A133619(n-1). So, a(1) = 9, a(2) = 89 and so on, with recognition of the former comments about the OEIS and terms beginning with 0. - Davis Smith, Mar 07 2019

Definition: For positive integers b (as base) and n, the positive integer (allowing initial zeros) k(n) is expomorphic relative to base b (here 5) if k(n) has exactly n decimal digits and if b^k(n) == k(n) (mod 10^n) or, equivalently, b^k(n) ends in k(n). [See Crux Mathematicorum link.]

For sequences in the OEIS, no term is allowed to begin with a digit 0 (except for the 1-digit number 0 itself). However, in the problem as defined in the Crux Mathematicorum article, leading 0 digits are allowed; under that definition a(n) = k(n) until the first k(n) which begins with digit 0. When k(n) begins with 0, then, a(n) is the next term of the sequence k(n) which doesn't begin with digit 0.

Under that definition, the term after a(4) = 3125 is not "03125" but a(5) = 203125. [Comments from Jon E. Schoenfield in A288845 and discussion with Rémy Sigrist]

Conjecture: if k(n) is expomorphic relative to "base" b, then, the next one in the sequence, k(n+1), consists of the last n+1 digits of b^k(n).

a(n) is the backward concatenation of A133615(0) through A133615(n-1). So, a(1) is 5, a(2) is 25, and so on, with recognition of the comments about the OEIS and terms beginning with 0 (for example, when n = 5, A133615(n-1) = 0, so the next nonzero digit is concatenated as well, reducing the amount subtracted from n by 1). - Davis Smith Mar 07 2019

The Crux Mathematicorum link calls these numbers "expomorphic" relative to "base" b, with here b = 8.

Under that definition, the term after a(13) = 35416895225856 is not "035416895225856" or "35416895225856" but a(14) = 7035416895225856.

Conjecture: if k(n) is "expomorphic" relative to "base" b, then the next one in the sequence, k(n+1), consists of the last n+1 digits of b^k(n).

This conjecture is true. See A133618. - David A. Corneth, Feb 10 2022