A354411 a(n) is the least oblong number that is divisible by the first n primes.

2, 6, 30, 210, 43890, 510510, 510510, 3967173210, 134748093480, 530514844860, 4201942828713630, 1706257740074998110, 125050509312845636520, 511284700554162118403820, 2695009287439086535873235280, 206794067314254446263154097180, 86753583273488685998907289046220

Offset: 1

Author

Ali Sada, May 25 2022

Keywords

• nonn

Examples

			2, 3, and 5 are the first three primes. The first oblong number that is a multiple of all three first primes is 30. So, a(3) = 30.
The first oblong number that is a multiple of primorial(5) = 2310 is 19*2310 = 43890, so a(5) = 43890.
		

Crossrefs

Cf. A000040, A002378, A002110, A118478, A344005.

Programs

PARI

a002110(n) = prod(i=1,n, prime(i)) \\ after Washington Bomfim in A002110
a344005(n) = for(m=1, oo, if((m*(m+1))%n==0, return(m)))
a(n) = my(m=a344005(a002110(n))); m*(m+1) \\ Felix Fröhlich, May 31 2022

Python

from sympy import integer_nthroot, primorial
def oblong(n): r = integer_nthroot(n, 2)[0]; return r*(r+1) == n
def a(n):
    m = psharp = primorial(n)
    while not oblong(m): m += psharp
    return m
print([a(n) for n in range(1, 11)]) # Michael S. Branicky, May 25 2022

Python

# faster alternative using Python 3.8+ A344005(n) by Chai Wah Wu
from sympy import primorial
def a(n): return (m := A344005(primorial(n)))*(m+1)
print([a(n) for n in range(1, 18)]) # Michael S. Branicky, May 26 2022

Formula

From Michael S. Branicky, May 25 2022: (Start)

a(n) <= (m-1)*m, where m = A002110(n).

a(n) = m*(m+1), where m = A344005(A002110(n)).

(End)

a(n) = A118478(n)*(A118478(n)+1). - Chai Wah Wu, May 31 2022

Extensions

a(9)-a(17) from Michael S. Branicky, May 26 2022