A034953 -id:A034953 - cp's OEIS frontend

A324600 a(n) = (k(n)*(k(n) + 1))/2 with k = A018252 (nonprime numbers), for n >= 1.

1, 10, 21, 36, 45, 55, 78, 105, 120, 136, 171, 210, 231, 253, 300, 325, 351, 378, 406, 465, 528, 561, 595, 630, 666, 741, 780, 820, 903, 990, 1035, 1081, 1176, 1225, 1275, 1326, 1378, 1485, 1540, 1596, 1653, 1711, 1830, 1953, 2016, 2080, 2145, 2211, 2346, 2415, 2485, 2628, 2775, 2850, 2926, 3003, 3081, 3240, 3321, 3403, 3570, 3655, 3741, 3828, 3916

Offset: 1

Wolfdieter Lang, Jul 10 2019

This is a proper subsequence of the triangular numbers A000217.

The infinite sum of the reciprocals is given in A307379 (which motivated this entry).

Cf. A000217, A018252, A034953 (for prime numbers).

v:=[m:m in [1..500]| not IsPrime(m)];[v[k]*(v[k]+1)/2:k in [1..65]]; // Marius A. Burtea, Jul 12 2019

[n*(n+1)/2 : n in [1..100] | not IsPrime(n)]; // Vincenzo Librandi, Jul 12 2019

apply(x->x*(x+1)/2, select(x->!isprime(x), [1..90])) \\ Michel Marcus, Sep 18 2023

a(n) = A000217(A018252(n)), for n >= 1.

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).

Original entry on oeis.org

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

Lechoslaw Ratajczak, May 17 2023

nonn

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.

			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.

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

Cf. A002034, A006530, A034953, A048799.

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);

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.

cp's OEIS Frontend

A324600 a(n) = (k(n)*(k(n) + 1))/2 with k = A018252 (nonprime numbers), for n >= 1.

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