A383942 a(n) = (8*10^(2n) - 10^(n+1) + 2) / 9.

78, 8778, 887778, 88877778, 8888777778, 888887777778, 88888877777778, 8888888777777778, 888888887777777778, 88888888877777777778, 8888888888777777777778, 888888888887777777777778, 88888888888877777777777778, 8888888888888777777777777778

Offset: 1

Author

David Radcliffe, Aug 18 2025

Keywords

• nonn
• base
• easy
• changed

Comments

This is one of four infinite families of triangular numbers consisting of two different digits. The other three families are A319170, A037156 (n>1), and A309597 (n>2).

Links

• Paolo Xausa, Table of n, a(n) for n = 1..495
• Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Crossrefs

Cf. A000217, A037156, A062691, A073551, A256340, A309597, A319170.

Subsequence of: A119216, A119230, A119236.

Programs

Mathematica

A383942[n_] := (8*10^(2*n) - 10^(n+1) + 2)/9; Array[A383942, 15] (* or *)
LinearRecurrence[{111, -1110, 1000}, {78, 8778, 887778}, 15] (* Paolo Xausa, Aug 27 2025 *)

Python

def A383942(n): return (8*10**(2*n)-10**(n+1)+2)//9

Formula

a(n) = A000217(A073551(n+1)).

G.f.: 6*x*(13 + 20*x)/((1 - x)*(1 - 10*x)*(1 - 100*x)). - Stefano Spezia, Aug 19 2025