

A020639


Lpf(n): least prime dividing n (when n > 1); a(1) = 1.


667



1, 2, 3, 2, 5, 2, 7, 2, 3, 2, 11, 2, 13, 2, 3, 2, 17, 2, 19, 2, 3, 2, 23, 2, 5, 2, 3, 2, 29, 2, 31, 2, 3, 2, 5, 2, 37, 2, 3, 2, 41, 2, 43, 2, 3, 2, 47, 2, 7, 2, 3, 2, 53, 2, 5, 2, 3, 2, 59, 2, 61, 2, 3, 2, 5, 2, 67, 2, 3, 2, 71, 2, 73, 2, 3, 2, 7, 2, 79, 2, 3, 2, 83, 2, 5, 2, 3, 2, 89, 2, 7, 2, 3, 2, 5, 2, 97
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OFFSET

1,2


COMMENTS

Also, the largest number of distinct integers such that all their pairwise differences are coprime to n.  Max Alekseyev, Mar 17 2006
The unit 1 is not a prime number (although it has been considered so in the past). 1 is the empty product of prime numbers, thus 1 has no least prime factor.  Daniel Forgues, Jul 05 2011
a(n) = least m > 0 for which n! + m and n  m are not relatively prime.  Clark Kimberling, Jul 21 2012
For n > 1, a(n) = the smallest k > 1 that divides n.  Antti Karttunen, Feb 01 2014
For n > 1, records are at prime indices.  Zak Seidov, Apr 29 2015
The initials "lpf" might be mistaken for "largest prime factor" (A009190), using "spf" for "smallest prime factor" would avoid this.  M. F. Hasler, Jul 29 2015
n = 89 is the first index > 1 for which a(n) differs from the smallest k > 1 such that (2^k + n  2)/k is an integer.  M. F. Hasler, Aug 11 2015
From Stanislav Sykora, Jul 29 2017: (Start)
For n > 1, a(n) is also the smallest k, 1 < k <= n, for which the binomial(n,k) is not divisible by n.
Proof: (A) When k and n are relatively prime then binomial(n,k) is divisible by n because k*binomial(n,k) = n*binomial(n1,k1). (B) When gcd(n,k) > 1, one of its prime factors is the smallest; let us denote it p, p <= k, and consider the binomial(n,p) = (Product_{i=0..p1}(ni))/p!. Since p is a divisor of n, it cannot be a divisor of any of the ramaining numerator factors. It follows that, denoting as e the largest e > 0 such that p^en, the numerator is divisible by p^e but not by p^(e+1). Hence, the binomial is divisible by p^(e1) but not by p^e and therefore not divisible by n. Applying (A), (B) to all considered values of k completes the proof. (End)
From Bob Selcoe, Oct 11 2017, edited by M. F. Hasler, Nov 06 2017: (Start)
a(n) = prime(j) when n == J (mod A002110(j)), n, j >= 1, where J is the set of numbers <= A002110(j) with smallest prime factor = prime(j). The number of terms in J is A005867(j1). So:
a(n) = 2 when n == 0 (mod 2);
a(n) = 3 when n == 3 (mod 6);
a(n) = 5 when n == 5 or 25 (mod 30);
a(n) = 7 when n == 7, 49, 77, 91, 119, 133, 161 or 203 (mod 210);
etc. (End)


REFERENCES

D. S. Mitrinovic et al., Handbook of Number Theory, Kluwer, Section IV.1.


LINKS

Daniel Forgues, Table of n, a(n) for n=1..100000 (terms 1..10000 from T. D. Noe)
A. E. Brouwer, Two number theoretic sums, Stichting Mathematisch Centrum. Zuivere Wiskunde, Report ZW 19/74 (1974): 3 pages. [Copy included with the permission of the author.]
OEIS Wiki, Least prime factor of n
David Singmaster, Letter to N. J. A. Sloane, Oct 3 1982.
Eric Weisstein's World of Mathematics, Least Prime Factor
Index entries for "core" sequences


FORMULA

A014673(n) = a(A032742(n)); A115561(n) = a(A054576(n)).  Reinhard Zumkeller, Mar 10 2006
A028233(n) = a(n)^A067029(n).  Reinhard Zumkeller, May 13 2006
a(n) = A027746(n,1) = A027748(n,1).  Reinhard Zumkeller, Aug 27 2011
For n > 1: a(n) = A240694(n,2).  Reinhard Zumkeller, Apr 10 2014
a(n) = A000040(A055396(n)) = n / A032742(n).  Antti Karttunen, Mar 07 2017
a(n) has average order n/(2 log n) [Brouwer]  N. J. A. Sloane, Sep 03 2017


MAPLE

A020639 := proc(n) if n = 1 then 1; else min(op(numtheory[factorset](n))) ; end if; end proc: seq(A020639(n), n=1..20) ; # R. J. Mathar, Oct 25 2010


MATHEMATICA

f[n_]:=FactorInteger[n][[1, 1]]; Join[{1}, Array[f, 120, 2]] (* Robert G. Wilson v, Apr 06 2011 *)
Join[{1}, Table[If[EvenQ[n], 2, FactorInteger[n][[1, 1]]], {n, 2, 120}]] (* Zak Seidov, Nov 17 2013 *)


PROG

(PARI) A020639(n) = { vecmin(factor(n)[, 1]) } \\ [Will yield an error for n = 1.]  R. J. Mathar, Mar 02 2012
(PARI) A020639(n)=if(n>1, if(n>n=factor(n, 0)[1, 1], n, factor(n)[1, 1]), 1) \\ Avoids complete factorization if possible. Often the smallest prime factor can be found quickly even if it is larger than primelimit. If factoring takes too long for large n, use debugging level >= 3 (\g3) to display the smallest factor as soon as it is found.  M. F. Hasler, Jul 29 2015
(Haskell)
a020639 n = spf a000040_list where
spf (p:ps)  n < p^2 = n
 mod n p == 0 = p
 otherwise = spf ps
 Reinhard Zumkeller, Jul 13 2011
(Sage)
def A020639_list(n) : return [1] + [prime_divisors(n)[0] for n in (2..n)]
A020639_list(97) # Peter Luschny, Jul 16 2012
(Scheme) (define (A020639 n) (if (< n 2) n (let loop ((k 2)) (cond ((zero? (modulo n k)) k) (else (loop (+ 1 k))))))) ;; Antti Karttunen, Feb 01 2014
(Sage) [trial_division(n) for n in (1..100)] # Giuseppe Coppoletta, May 25 2016


CROSSREFS

Cf. A090368 (bisection).
Cf. A000040, A009190, A006530, A034684, A028233, A034699, A053585.
See also A032742, A055396, A068319, A088377, A007978, A053669, A117818.
Cf. A046669 (partial sums), A072486 (partial products).
Cf. A002110, A005867.
Sequence in context: A135679 A092028 * A092067 A214606 A079879 A071889
Adjacent sequences: A020636 A020637 A020638 * A020640 A020641 A020642


KEYWORD

nonn,easy,nice,core


AUTHOR

David W. Wilson


EXTENSIONS

Deleted wrong comment from M. Lagneau in 2012, following an observation by Gionata Neri.  M. F. Hasler, Aug 11 2015
Edited by M. F. Hasler, Nov 06 2017


STATUS

approved



