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A300820
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Length of the longest sequence of consecutive primes in the prime factorization of n. a(1) = 0.
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4
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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, 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, 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
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OFFSET
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1,6
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LINKS
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FORMULA
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For n >= 0, a(A002110(n)) = n. [Primorials give the positions of the records = the first occurrence of each n.]
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EXAMPLE
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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.
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PROG
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(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));
(PARI) a(n) = {if(n == 1, return(0)); my(res = 1, f = factor(n)[, 1]~, t = 1);
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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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