%I #38 Jan 24 2024 01:49:46

%S 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,

%T 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,

%U 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

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

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

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

%H Antti Karttunen, <a href="/A257679/b257679.txt">Table of n, a(n) for n = 0..5040</a>

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

%F 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].

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

%F Other identities:

%F 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.]

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

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

%t 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 *)

%o (Scheme)

%o (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))))))))

%o ;; Alternative implementations based on given recurrences, using memoizing definec-macro:

%o (definec (A257679 n) (if (zero? (A257687 n)) (A099563 n) (min (A099563 n) (A257679 (A257687 n)))))

%o (definec (A257679 n) (cond ((zero? n) n) ((= 1 (A257680 n)) 1) (else (+ 1 (A257679 (A257684 n))))))

%o (Python)

%o def A(n, p=2):

%o return n if n<p else A(n//p, p+1)*10 + n%p

%o def a(n):

%o return 0 if n==0 else min(int(i) for i in str(A(n)) if i !='0')

%o print([a(n) for n in range(201)]) # _Indranil Ghosh_, Jun 19 2017

%Y Positions of records: A001563.

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

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

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

%K nonn,base

%O 0,5

%A _Antti Karttunen_, May 04 2015