1,2

Commonly named after Florentin Smarandache, although studied 60 years earlier by Aubrey Kempner.

Kempner gave an algorithm to compute a(n) from the prime factorization of n. Partial solutions were given earlier by Lucas in 1883 and Neuberg in 1887. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Dec 23 2004

a(n) = degree of lowest degree monic polynomial over Z that vanishes identically on the integers mod n [Newman]

Smallest k such that n divides product of k consecutive integers starting with n+1. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Oct 26 2002

If m and n are any integers with n > 1, then |e - m/n| > 1/(a(n)+1)!. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Sep 04 2006

C. Dumitrescu, A brief history of the "Smarandache function". Bull. Pure Appl. Sci. Sec. E, Math., 12 (1993), no. 1-2, 77-82.

P. Erdos and I. Kastanas, Problem/Solution 6674:The smallest factorial that is a multiple of n, Amer. Math. Monthly 101 (1994) 179.

A. J. Kempner, Miscellanea, Amer. Math. Monthly, 25 (1918), 201-210. See Section II, "Concerning the smallest integer m! divisible by a given integer n".

E. Lucas, Question Nr. 288, Mathesis 3 (1883), 232.

R. Muller, Unsolved problems related to Smarandache Function, Number Theory Publishing Company, Phoenix, AZ, ISBN 1-879585-37-5, 1993.

J. Neuberg, Solutions des questions proposees, Question Nr. 288, Mathesis 7 (1887), 68-69.

D. J. Newman, A Problem Seminar. Problem 17, Spinger-Verlag 1982.

S. M. Ruiz, A Congruence with Smarandache's Function, Smarandache Notions Journal, Vol. 10, No. 1-2-3, 1999, 130-132.

F. Smarandache, A Function in the Number Theory, Analele Univ. Timisoara, Fascicle 1, Vol. XVIII, 1980, pp. 79-88; Smarandache Function J. 1 (1990), no. 1, 3-17.

J. Sondow, A geometric proof that e is irrational and a new measure of its irrationality, Amer. Math. Monthly 113 (2006) 637-641.

T. D. Noe, Table of n, a(n) for n=1..1000

C. Ashbacher, An Introduction to the Smarandache Function, Erhus Univ. Press, Vail, 62 pages, 1995.

C. Dumitrescu and V. Seleacu, The Smarandache Function, Erhus Univ. Press, Vail, 137 pages, 1996.

A. Ivic (2004), On a problem of Erdos involving the largest prime factor of n

M. L. Perez et al., eds., Smarandache Notions Journal

J. Perry, Calculating the Smarandache Numbers [Broken link]

Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.

Index entries for sequences related to factorial numbers.

Clearly a(n) >= P(n), the largest prime factor of n (= A006530). a(n) = P(n) for almost all n (Erdos and Kastanas 1994, Ivic 2004). The exceptions are A057109. a(n) = P(n) if and only if a(n) is prime because if a(n) > P(n) and a(n) were prime, then since n divides a(n)!, n would also divide (a(n)-1)!, contradicting minimality of a(n). - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Jan 10 2005

It appears that if p is prime then a(p^k)=k*p for 0<=k<=p. Hence it appears also that if n = 2^m*p(1)^e(1)*...*p(r)^e(r) and if there exists b, 1<=b<= r, such that Max( 2*m+2, p(i)*e(i), 1<=i<=r ) = p(b)*e(b) with e(b)<=p(b) then a(n)=e(b)*p(b). E.g.: a(2145986896455317997802121296896)=a(2^10*3^3*7^9*11^9*13^8) = 13*8 = 104, since 8*13 = Max (2*10+2, 3*3, 7*9, 11*9, 13*8) and 8<=13. - Benoit Cloitre (benoit7848c(AT)orange.fr), Sep 01 2002

It appears that a(2^m-1) = largest prime factor of 2^m - 1 (A005420).

a(n!) = n for all n > 0, and a(p) = p if p is prime. - Jonathan Sondow (jsondow(AT)alumni.princeton.edu), Dec 23 2004

a(8) = 4 because 8 divides 4! and 8 does not divide k! for k < 4.

a:=proc(n) local b: b:=proc(m) if type(m!/n, integer) then m else fi end: [seq(b(m), m=1..100)][1]: end: seq(a(n), n=1..84); (Emeric Deutsch (deutsch(AT)duke.poly.edu), Aug 01 2005)

Do[m = 1; While[ !IntegerQ[m!/n], m++ ]; Print[m], {n, 1, 85}] (* or for larger n's *)

Smarandache[1] := 1; Smarandache[n_] := Max[Smarandache @@@ FactorInteger[n]]; Smarandache[p_, 1] := p; Smarandache[p_, alpha_] := Smarandache[p, alpha] = Module[{a, k, r, i, nu, k0 = alpha(p - 1)}, i = nu = Floor[Log[p, 1 + k0]]; a[1] = 1; a[n_] := (p^n - 1)/(p - 1); k[nu] = Quotient[alpha, a[nu]]; r[nu] = alpha - k[nu]a[nu]; While[r[i] > 0, k[i - 1] = Quotient[r[i], a[i - 1]]; r[i - 1] = r[i] - k[i - 1]a[i - 1]; i-- ]; k0 + Plus @@ k /@ Range[i, nu]]; Table[ Smarandache[n], {n, 85}] (from Eric Weisstein, based on a formula of Kempner's, May 17 2004)

(PARI) a(n)=if(n<0, 0, s=1; while(s!%n>0, s++); s)

Cf. A007672, A064759, A000142, A094371, A094372, A046022, A094404.

Cf. also A006530, A057109, A001113, A122378, A122379, A122416, A122417.

Sequence in context: A025492 A077004 A064760 this_sequence A088491 A140271 A141295

Adjacent sequences: A002031 A002032 A002033 this_sequence A002035 A002036 A002037

nonn,nice,easy

njas

Error in 45th term corrected by David W. Wilson (davidwwilson(AT)comcast.net) May 15 1997