OFFSET
0,2
COMMENTS
LINKS
FORMULA
a(0) = 1, and for n >= 1, if A257511(1+a(n-1)) > 0, then a(n) = a(n-1) + 1, otherwise a(n-1) + 2. [In particular, if the previous term is 2k, then the next term is 2k+1, because all odd numbers are members.]
Other identities:
From Antti Karttunen, May 26 2015: (Start)
Alternative recurrence for the same sequence:
Set k = A258198(n), d = n - A258199(n) and f = A000142(k+1) = (k+1)! If d < f then b(n) = f+d, otherwise b(n) = ((2+floor((d-f)/A258199(n))) * f) + b((d-f) mod A258199(n)). For offset=1 sequence, define a(n) = b(n-1).
(End)
MATHEMATICA
Select[Range@ 101, MemberQ[IntegerDigits[#, MixedRadix[Reverse@ Range@ 12]], 1] &] (* Michael De Vlieger, May 30 2016, Version 10.2 *)
r = MixedRadix[Reverse@ Range[2, 12]]; Select[Range@ 101, Min[IntegerDigits[#, r] /. 0 -> Nothing] == 1 &] (* Michael De Vlieger, Aug 14 2016, Version 10.2 *)
PROG
(Scheme, with Antti Karttunen's IntSeq-library)
;; Alternatively, as a naive recurrence:
(definec (A256450 n) (if (zero? n) 1 (let ((prev (A256450 (- n 1)))) (cond ((even? prev) (+ 1 prev)) ((> (A257511 (+ 1 prev)) 0) (+ 1 prev)) (else (+ 2 prev))))))
;; Faster recurrence May 26 2015:
(definec (A256450 n) (let* ((k (A258198 n)) (d (- n (A258199 n))) (f (A000142 (+ 1 k)))) (cond ((< d f) (+ f d)) (else (+ (* f (+ 2 (floor->exact (/ (- d f) (A258199 n))))) (A256450 (modulo (- d f) (A258199 n))))))))
(Python)
def A(n, p=2): return n if n<p else A(n//p, p+1)*10 + n%p
print([n for n in range(1, 151) if str(A(n)).count("1")>=1]) # Indranil Ghosh, Jun 19 2017
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Antti Karttunen, Apr 27 2015
EXTENSIONS
Starting offset changed from 1 to 0 by Antti Karttunen, May 30 2016
STATUS
approved