A257679 The smallest nonzero digit present in the factorial base representation (A007623) of n, 0 if no nonzero digits present.

0, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 4, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 2, 1, 1

Offset: 0

Author

Antti Karttunen, May 04 2015

Keywords

• nonn
• base

Comments

a(0) = 0 by convention, because "0" has no nonzero digits present.

a(n) gives the row index of n in array A257503 (equally, the column index for array A257505).

Examples

			Factorial base representation (A007623) of 4 is "20", the smallest digit which is not zero is "2", thus a(4) = 2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..5040

Crossrefs

Positions of records: A001563.

Cf. A256450, A257692, A257693 (positions of 1's, 2's and 3's in this sequence).

Cf. A007623, A051683, A099563, A257680, A257684, A257687.

Cf. also A257079, A246359 and arrays A257503, A257505.

Programs

Mathematica

a[n_] := Module[{k = n, m = 2, rmin = n, r}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[0 < r < rmin, rmin = r]; m++]; rmin]; Array[a, 100, 0] (* Amiram Eldar, Jan 23 2024 *)

Python

def A(n, p=2):
    return n if n
Indranil Ghosh, Jun 19 2017

Scheme

(define (A257679 n) (let loop ((n n) (i 2) (mind 0)) (if (zero? n) mind (let ((d (modulo n i))) (loop (/ (- n d) i) (+ 1 i) (cond ((zero? mind) d) ((zero? d) mind) (else (min d mind))))))))
;; Alternative implementations based on given recurrences, using memoizing definec-macro:
(definec (A257679 n) (if (zero? (A257687 n)) (A099563 n) (min (A099563 n) (A257679 (A257687 n)))))
(definec (A257679 n) (cond ((zero? n) n) ((= 1 (A257680 n)) 1) (else (+ 1 (A257679 (A257684 n))))))

Formula

If A257687(n) = 0, then a(n) = A099563(n), otherwise a(n) = min(A099563(n), a(A257687(n))).

In other words, if n is either zero or one of the terms of A051683, then a(n) = A099563(n) [the most significant digit of its f.b.r.], otherwise take the minimum of the most significant digit and a(A257687(n)) [value computed by recursing with a smaller value obtained by discarding that most significant digit].

a(0) = 0, and for n >= 1: if A257680(n) = 1, then a(n) = 1, otherwise 1 + a(A257684(n)).

Other identities:

For all n >= 0, a(A001563(n)) = n. [n * n! gives the first position where n appears. Note also that the "digits" (placeholders) in factorial base representation may get arbitrarily large values.]

For all n >= 0, a(2n+1) = 1 [because all odd numbers end with digit 1 in factorial base].