A362070 - OEIS

%I #22 Aug 28 2023 08:21:21

%S 1,3,6,9,15,6,28,10,16,15,66,9,91,28,15,16,153,16,190,15,28,66,276,10,

%T 33,91,29,28,435,15,496,24,66,153,28,16,703,190,91,15,861,28,946,66,

%U 18,276,1128,16,54,33,153,91,1431,29,66,28,190

%N Let m_min(n, k) be the smallest m such that n divides Product_{t=1..m} RisingFactorial(t, k). a(n) = Sum_{r=1..K(n)} m_min(n, r), where K(n) is the Kempner number A002034(n).

%C 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.

%C Conjecture: the equation a(n) = a(n+1) has no solutions. This holds up to at least n = 10^4.

%C Conjecture: the constant Sum_{n >= 2} 1/a(n)! = 0.16945... is irrational.

%H J. Sondow and E. W. Weisstein, <a href="http://mathworld.wolfram.com/SmarandacheFunction.html">MathWorld: Smarandache Function</a>

%F a(1) = 1.

%F a(p) = p*(p + 1)/2 for p prime.

%F 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.

%F a(P) = P, where P is a perfect number.

%F a(p*(p + 1)/2) = p*(p + 1)/2 for p prime.

%F a(n!) = 3*n + ((gpf(n!)^2 - 5*gpf(n!))/2 for n <> 4.

%e a(18) = 16 because:

%e - 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);

%e - 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;

%e - for r = 3: 18 does not divide (1*2*3) and divides (1*2*3)*(2*3*4), then m_min(18, 3) = 2;

%e - 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;

%e - 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;

%e - 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.

%o (Maxima)

%o K(u):=(b:1, for i:1 while mod(b,u)#0 do (c:i, b:b*(i+1)), c+1);

%o 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);

%o makelist(a(n),n,2,100);

%Y Cf. A002034, A006530, A034953, A048799.

%K nonn

%O 1,2

%A _Lechoslaw Ratajczak_, May 17 2023