%I #31 Nov 17 2019 15:55:31
%S 1,1,1,2,1,1,1,1,3,1,1,2,1,1,1,2,1,3,1,10,1,1,1,1,5,1,1,14,1,1,1,4,1,
%T 1,1,6,1,1,1,1,1,7,1,22,3,1,1,2,7,5,1,26,1,9,1,1,1,1,1,10,1,1,3,1,1,
%U 11,1,34,1,1,1,3,1,1,5,38,1,13,1,2,3,1,1,14,1,1,1,11,1,3,1,46,1,1,1,4,1,7,33
%N Final number obtained when n is divided by its divisors starting from the smallest one in increasing order until one no longer gets an integer.
%C a(n) = 1 if n = p, n = p^3, n = p*q or n = k! for some k, or n = p*q*r where the product of two primes is more than the third, where p q and r are primes. Question: What is the longest string of ones in this sequence? Subsidiary sequence: Index of the start of the first occurrence of a string of n ones.
%C Concerning this question, see also A329549. Furthermore as a(8k+4) > 1 such a string can have at most length 7. - _David A. Corneth_, Nov 16 2019
%H Antti Karttunen, <a href="/A076933/b076933.txt">Table of n, a(n) for n = 1..16384</a>
%H Antti Karttunen, <a href="/A076933/a076933.txt">Data supplement: n, a(n) computed for n = 1..65537</a>
%e a(12) = 2: the divisors of 12 in increasing order are 1,2,3,4,6,12. and 12/1 = 12, 12/2 = 6, 6/3 = 2 that is the final integer, as the next divisor 4 > 2.
%p for i from 1 to 200 do d := sort(convert(divisors(i),list)):j := 1:g := i: while((g mod d[j])=0) do g := g/d[j]:j := j+1: if(j>nops(d)) then break:fi: od:a[i] := g:od:seq(a[k],k=1..200);
%o (PARI) A076933(n) = { my(k=n); fordiv(k,d,if(n%d,return(n),n /= d)); (n); }; \\ _Antti Karttunen_, Nov 16 2019
%Y Cf. A240694 (partial products of divisors of n), A329377 (number of iterations needed to reach the final number), A329549.
%K nonn
%O 1,4
%A _Amarnath Murthy_, Oct 18 2002
%E More terms from _Sascha Kurz_, Jan 21 2003