OFFSET
1,2
COMMENTS
The first two solutions of the equation a(n) = n which are not consecutive triangular numbers with odd prime indices are 1, 16. Is there a larger n? If such a number n exists, it is larger than 10^4.
Conjecture: the equation a(n) = a(n+1) has no solutions. This holds up to at least n = 10^4.
Conjecture: the constant Sum_{n >= 2} 1/a(n)! = 0.16945... is irrational.
LINKS
J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function
FORMULA
a(1) = 1.
a(p) = p*(p + 1)/2 for p prime.
a(p_1*p_2*...*p_u) = p_u*(p_u + 1)/2, where p_i's are distinct primes and p_1 < p_2 < ... < p_u.
a(P) = P, where P is a perfect number.
a(p*(p + 1)/2) = p*(p + 1)/2 for p prime.
a(n!) = 3*n + ((gpf(n!)^2 - 5*gpf(n!))/2 for n <> 4.
EXAMPLE
a(18) = 16 because:
- for r = 1: 18 does not divide (1), (1)*(2), (1)*(2)*(3), (1)*(2)*(3)*(4), (1)*(2)*(3)*(4)*(5) and divides (1)*(2)*(3)*(4)*(5)*(6), then m_min(18, 1) = 6 = A002034(18) = K(18);
- for r = 2: 18 does not divide (1*2), (1*2)*(2*3) and divides (1*2)*(2*3)*(3*4), then m_min(18, 2) = 3;
- for r = 3: 18 does not divide (1*2*3) and divides (1*2*3)*(2*3*4), then m_min(18, 3) = 2;
- for r = 4: 18 does not divide (1*2*3*4) and divides (1*2*3*4)*(2*3*4*5), then m_min(18, 4) = 2;
- for r = 5: 18 does not divide (1*2*3*4*5) and divides (1*2*3*4*5)*(2*3*4*5*6), then m_min(18, 5) = 2;
- for r = 6 = K(18): 18 divides (1*2*3*4*5*6), then m_min(18, 6) = 1, hence a(18) = 6 + 3 + 2 + 2 + 2 + 1 = 16.
PROG
(Maxima)
K(u):=(b:1, for i:1 while mod(b, u)#0 do (c:i, b:b*(i+1)), c+1);
a(n):=(s:0, for r:2 thru K(n)-1 do (z:product(j, j, 1, r), for q:1 while mod(z, n)#0 do (z:z*product(y, y, q+1, q+r), m:q+1), s:s+m), s+K(n)+1);
makelist(a(n), n, 2, 100);
CROSSREFS
KEYWORD
nonn
AUTHOR
Lechoslaw Ratajczak, May 17 2023
STATUS
approved