cp's OEIS Frontend

For all n, a(A226061(n+1)) = A232095(n). This works because at the positions given by each x=A226061(n+1), it holds that A219666(x) = (n+1)!-1, which has a factorial base representation (A007623) of (n,n-1,n-2,...,3,2,1) whose digit sum (A034968) is the n-th triangular number, A000217(n). This in turn is always a new record as at those points, in each significant digit position so far employed, a maximal digit value (for factorial number system) is used, and thus the preceding term, A219666(x-1) cannot have any larger digits in its factorial base representation, and so the differences between their digits (in matching positions) are all nonnegative.

This sequence gives all n at which positions the successive terms A219666(n-1) & A219666(n) in the infinite trunk of the factorial beanstalk differ only in one digit position in their factorial base representations (A007623).

Please see further comments and examples in A230410.