cp's OEIS Frontend

A328570 Index of the least significant zero digit in the primorial base expansion of n, when the rightmost digit is in the position 1.

%I A328570 #30 Oct 20 2019 21:46:43
%S A328570 1,2,1,3,1,3,1,2,1,4,1,4,1,2,1,4,1,4,1,2,1,4,1,4,1,2,1,4,1,4,1,2,1,3,
%T A328570 1,3,1,2,1,5,1,5,1,2,1,5,1,5,1,2,1,5,1,5,1,2,1,5,1,5,1,2,1,3,1,3,1,2,
%U A328570 1,5,1,5,1,2,1,5,1,5,1,2,1,5,1,5,1,2,1,5,1,5,1,2,1,3,1,3,1,2,1,5,1,5,1,2,1,5
%N A328570 Index of the least significant zero digit in the primorial base expansion of n, when the rightmost digit is in the position 1.
%C A328570 Index of the least prime not dividing A276086(n), where A276086 converts the primorial base expansion of n into its prime product form.
%C A328570 Starting from x = n, repeatedly divide x by prime(1) (discarding the remainder), and set x to the integer quotient floor(x/prime(1)), then divide x with prime(2) (again discarding the remainder, and setting x to the integer quotient), etc., stopping as soon one of the primes is a divisor of the previous integer quotient (leaving zero remainder). a(n) is then the index of that prime, equal to 1 + the number of iterations done.
%H A328570 Antti Karttunen, <a href="/A328570/b328570.txt">Table of n, a(n) for n = 0..32768</a>
%H A328570 <a href="/index/Pri#primorialbase">Index entries for sequences related to primorial base</a>
%F A328570 a(n) = A000720(A326810(n)) = A257993(A276086(n)) = A055396(A276087(n)).
%F A328570 For all n >= 0, A002110(a(n)) = A328580(n), a(A276086(n)) = A328578(n).
%F A328570 For all odd n, A000040(a(n)) = A326810(n) > A276088(n).
%F A328570 For all n >= 0, A276086(n)/A002110(a(n)-1) = A328475(n) and A276086(n)-A002110(a(n)-1) = A328476(n).
%e A328570 For n = 2, we divide it with A000040(1) = 2, and it leaves zero remainder, so we have finished on the first round (needing no actual iterations), and thus a(2) = 1. Note that 2 in primorial base (A049345) is written as "10", and indeed the first zero from the right occurs at the position 1.
%e A328570 For n = 5, we first divide 5 with prime(1) = 2, and discarding the remainder, we are left with floor(5/2) = 2. Then we divide that 2 with prime(2) = 3, leaving floor(2/3) = 0 and remainder 2. And finally we divide 0 with prime(3) = 5, and that doesn't leave any remainder, thus we are finished on the third round, and a(5) = 3. Note that 5 in primorial base is written as "21", and allowing here a leading zero, written as "021", we see that it is in this case the least significant zero occurring at position 3 from the right.
%e A328570 For n = 43, we first divide it with prime(1) = 2, leaving a remainder 1 and integer quotient 21. Then we divide 21 with prime(2) = 3, which doesn't leave any remainder, thus we are finished on the second round, and a(43) = 2. Note that 43 is written as "1201" in primorial base, with the least significant zero occurring in the position 2.
%o A328570 (PARI) A328570(n) = { my(i=1, p=2); while(n && (n%p), i++; n = n\p; p = nextprime(1+p)); (i); };
%Y A328570 Cf. A000720, A002110, A049345, A053669, A055396, A143293, A257993, A276086, A276087, A276088, A326810, A328475, A328476, A328569, A328578, A328580.
%K A328570 nonn,base
%O A328570 0,2
%A A328570 _Antti Karttunen_, Oct 20 2019