A Generalized Smarandache Function
Created | Updated Dec 19, 2008
The Smarandache function is defined to be the smallest positive integer m such that n divides m! and the pseudo-Smarandache function is defined to be the smallest positive integer m such that n divides m(m+1)/2 (the sum of the positive integers up to and including m). It would be better to define the pseudo-Smarandache function to be the smallest positive integer m such that n divides m(m-1)/2 (the binomial coefficient) so that the relationship with the Smarandache function could be maintained. For typographical convenience, let c(m,k) denote m "choose" k (a binomial coefficient). Then c(m,k)=m!/((m-k)!k!). Let q be a prime. Let f(m,q) denote the larger power of q that divides m! and let g(m,k,q) denote the largest power of q that divides c(m,k). It's easy to prove that (m-1)/(q-1)≥f(m,q)>(m-1)/(q-1)-1-(q/(q-1))[logqm] where [] denotes the greatest integer function. Then (2q-1)/(q-1)+(q/(q-1)){[logq(m-k)]+[logqk]}>g(m,k,q)>(2-q)/(q-1)-(q/(q-1))[logqm]. Since the binomial coefficients are non-negative integers, this can be improved to (2q-1)/(q-1)+(q/(q-1)){[logq(m-k)]+[logqk]}>g(m,k,q)≥0. As expected, g(m,k,q) increases logarithmically with m and k. Let h(n) denote the first binomial coefficient that n divides. Properties of Pascal's triangle can be used to easily deduce certain h(n) values (as well as certain re-defined pseudo-Smarandache values.) (Recall that the next row in Pascal's triangle can be generated from the current row by adding adjacent elements of the current row.) Let {1} be row 0, {1,1} be row 1, {1,2,1} be row 2, et cetera. Then if q is prime, every element of row q other than c(q,0) and c(q,q) is divisible by q (and h(q)=(q,1)). Then every element in the equilateral triangle bounded by c(q,1), c(q,q-1), c(2(q-1),q-1) is divisible by q and no element in the equilateral triangles bounded by c(q,0), c(2(q-1),0), c(2(q-1),q-2) and c(q,q), c(2(q-1),q), c(2(q-1),2(q-1)) is divisible by q. If n=q1q2 where q1 and q2 are primes, 2(q1-1)>q2>q1, then h(n)=c(q2,q2-q1+1). The condition 2(q1-1)>q2>q1 assures intersection of the appropriate equilateral triangles associated with q1 and q2. If this condition is not met, h(n), n=q1q2, can be found by considering the intersection of equilateral triangles associated with multiples of q1 and q2. For example, the equilateral triangle bounded by c(5,1), c(5,4), c(8,4) intersects the equilateral triangle bounded by c(6,1), c(6,2), c(7,2) and every element of the first triangle is divisible by 5 and every element of the second triangle is divisible by 3. In general, h(n) has many interesting properties.
