

A001276


Smallest k such that the product of q/(q1) over the primes from prime(n) to prime(n+k1) is greater than 2.
(Formerly M2650 N1057)


6



2, 3, 7, 15, 27, 41, 62, 85, 115, 150, 186, 229, 274, 323, 380, 443, 509, 577, 653, 733, 818, 912, 1010, 1114, 1222, 1331, 1448, 1572, 1704, 1845, 1994, 2138, 2289, 2445, 2609, 2774, 2948, 3127, 3311, 3502, 3699, 3900, 4112, 4324, 4546, 4775, 5016, 5255, 5493
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OFFSET

1,1


COMMENTS

A perfect (or abundant) number with prime(n) as its lowest prime factor must be divisible by at least a(n) distinct primes.
In fact, a(n) is the least possible number of distinct prime factors for a (prime(n))rough abundant number: (prime(n))^(e_n) * ... * (prime(n+a(n)1))^(e_(n+a(n)1)) is abundant for sufficiently large e_n, ..., e_(n+a(n)1).  Jianing Song, Apr 13 2021


REFERENCES

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).


LINKS



FORMULA

a(n) = li(prime(n)^2) + O(n^2/exp((log n)^(4/7  e))) for any e > 0.


EXAMPLE

Every odd abundant number has at least 3 distinct prime factors, and 945 = 3^3 * 5 * 7 has exactly 3, so a(2) = 3.  Jianing Song, Apr 13 2021


MATHEMATICA

a[n_] := Module[{p = Prime[n], r = 1, k = 0}, While[r <= 2, r *= p/(p  1); p = NextPrime[p]; k++]; k]; Array[a, 50] (* Amiram Eldar, Jul 12 2019 *)


PROG

(PARI) a(n)=my(pr=1., k=0); forprime(p=prime(n), default(primelimit), pr*=p/(p1); k++; if(pr>2, return(k))) \\ Charles R Greathouse IV, May 09 2011


CROSSREFS

Cf. A108227 (least number of prime factors for a (prime(n))rough abundant number, counted with multiplicity).


KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



