A231717 After a(0)=0, a(n) = A231713(A219666(n),A219666(n-1)).

0, 1, 2, 2, 3, 3, 3, 3, 1, 6, 3, 3, 3, 2, 1, 6, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 2, 3, 10, 3, 3, 3, 2, 1, 6, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 1, 3, 5, 5, 3, 10, 2, 3, 1, 3, 5, 3, 1, 3, 6, 2, 1, 2, 4, 2, 4, 5, 3, 3, 9, 3, 1, 3, 6, 2, 1, 2, 4, 2, 4, 5, 3, 2, 4, 3, 10

Offset: 0

Author

Antti Karttunen, Nov 12 2013

Keywords

• nonn

Comments

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.

Links

• Antti Karttunen, Table of n, a(n) for n = 0..3149

Crossrefs

A231718 gives the positions of ones.

Cf. also A230410, A231719, A232095.

Programs

Scheme

(define (A231717 n) (if (zero? n) n (A231713bi (A219666 n) (A219666 (- n 1)))))

Formula

a(0)=0, and for n>=1, a(n) = A231713(A219666(n),A219666(n-1)).