A362070 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).

1, 3, 6, 9, 15, 6, 28, 10, 16, 15, 66, 9, 91, 28, 15, 16, 153, 16, 190, 15, 28, 66, 276, 10, 33, 91, 29, 28, 435, 15, 496, 24, 66, 153, 28, 16, 703, 190, 91, 15, 861, 28, 946, 66, 18, 276, 1128, 16, 54, 33, 153, 91, 1431, 29, 66, 28, 190

Offset: 1

Author

Lechoslaw Ratajczak, May 17 2023

Keywords

• nonn

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.

Examples

			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.
		

Links

• J. Sondow and E. W. Weisstein, MathWorld: Smarandache Function

Crossrefs

Cf. A002034, A006530, A034953, A048799.

Programs

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);

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.