cp's OEIS Frontend

A300820 Length of the longest sequence of consecutive primes in the prime factorization of n. a(1) = 0.

%I A300820 #12 Mar 22 2018 17:56:43
%S A300820 0,1,1,1,1,2,1,1,1,1,1,2,1,1,2,1,1,2,1,1,1,1,1,2,1,1,1,1,1,3,1,1,1,1,
%T A300820 2,2,1,1,1,1,1,2,1,1,2,1,1,2,1,1,1,1,1,2,1,1,1,1,1,3,1,1,1,1,1,2,1,1,
%U A300820 1,2,1,2,1,1,2,1,2,2,1,1,1,1,1,2,1,1,1,1,1,3,1,1,1,1,1,2,1,1,1,1,1,2,1,1,3
%N A300820 Length of the longest sequence of consecutive primes in the prime factorization of n. a(1) = 0.
%H A300820 Antti Karttunen, <a href="/A300820/b300820.txt">Table of n, a(n) for n = 1..65537</a>
%H A300820 <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>
%F A300820 For n > 1, a(n) = A038374(A087207(n)).
%F A300820 For n >= 0, a(A002110(n)) = n. [Primorials give the positions of the records = the first occurrence of each n.]
%e A300820 For n = 350 = 2 * 5^2 * 7 = prime(1) * prime(3)^2 * prime(4), the longest stretch of consecutive primes is from prime(3) to prime(4), with length 2, thus a(350) = 2.
%o A300820 (PARI) A300820(n) = if(omega(n)<=1, omega(n), my(pis=apply(p->primepi(p),factor(n)[,1]),el=1,m=1); for(i=2,#pis,if(pis[i] == (1+pis[i-1]),el++; m = max(m,el), el=1)); (m));
%o A300820 (PARI) a(n) = {if(n == 1, return(0)); my(res = 1, f = factor(n)[, 1]~, t = 1);
%o A300820 for(i = 1, #f - 1, if(f[i+1]==nextprime(f[i]+1), t++, res = max(res, t);  t = 1)); max(res, t)} \\ _David A. Corneth_, Mar 21 2018
%Y A300820 Cf. A038374, A087207, A104210 (positions of terms > 1), A296210.
%Y A300820 Cf. also A167447.
%K A300820 nonn
%O A300820 1,6
%A A300820 _Antti Karttunen_, Mar 21 2018