%I #19 May 06 2021 11:04:47
%S 0,0,0,0,1,0,0,0,0,0,0,0,2,0,0,0,3,0,4,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,
%T 0,0,0,0,0,0,0,0,0,0,0,0,0,0,6,0,0,0,7,0,0,0,0,0,0,0,8,0,0,0,9,0,10,0,
%U 0,0,11,0,12,0,0,0,13,0,0,0,0,0,0,0,14,0,0,0,15,0,16,0,0,0,17,0,18,0,0,0,19,0,0,0,0,0,0,0,20,0,0,0,21,0,22,0,0,0,23,0,0
%N Left inverse for injection A255411: a(0) = 0, after which, if n = A255411(k) for some k, then a(n) = k, otherwise a(n) = 0.
%H Antti Karttunen, <a href="/A257685/b257685.txt">Table of n, a(n) for n = 0..10080</a>
%F a(0) = 0, after which, if n = A255411(k) for some k, then a(n) = k, otherwise a(n) = 0.
%F a(n) = (1-A257680(n)) * A257684(n).
%F Other identities:
%F For all n >= 0, a(A255411(n)) = n. [This sequence works as a left inverse of A255411.]
%t Position[Select[Range[0, 120], ! MemberQ[IntegerDigits[#, MixedRadix[Reverse@ Range@ 12]], 1] &], #] - 1 & /@ Range[0, 120] /. {} -> 0 // Flatten (* _Michael De Vlieger_, May 30 2016, Version 10.2 *)
%o (Scheme) (define (A257685 n) (* (- 1 (A257680 n)) (A257684 n)))
%o (Python)
%o from sympy import factorial as f
%o def a007623(n, p=2):
%o return n if n<p else a007623(n//p, p+1)*10 + n%p
%o def a257684(n):
%o x=str(a007623(n))[:-1]
%o y="".join(str(int(i) - 1) if int(i)>0 else '0' for i in x)[::-1]
%o return 0 if n==1 else sum([int(y[i])*f(i + 1) for i in range(len(y))])
%o def a257680(n): return 1 if '1' in str(a007623(n)) else 0
%o def a(n): return (1 - a257680(n))*a257684(n)
%o print([a(n) for n in range(101)]) # _Indranil Ghosh_, Jun 21 2017
%Y Cf. A255411, A257680, A257681, A257684.
%K nonn
%O 0,13
%A _Antti Karttunen_, May 04 2015