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%I #48 May 31 2022 12:54:17
%S 2,6,30,210,43890,510510,510510,3967173210,134748093480,530514844860,
%T 4201942828713630,1706257740074998110,125050509312845636520,
%U 511284700554162118403820,2695009287439086535873235280,206794067314254446263154097180,86753583273488685998907289046220
%N a(n) is the least oblong number that is divisible by the first n primes.
%F From _Michael S. Branicky_, May 25 2022: (Start)
%F a(n) <= (m-1)*m, where m = A002110(n).
%F a(n) = m*(m+1), where m = A344005(A002110(n)).
%F (End)
%F a(n) = A118478(n)*(A118478(n)+1). - _Chai Wah Wu_, May 31 2022
%e 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.
%e The first oblong number that is a multiple of primorial(5) = 2310 is 19*2310 = 43890, so a(5) = 43890.
%o (Python)
%o from sympy import integer_nthroot, primorial
%o def oblong(n): r = integer_nthroot(n, 2)[0]; return r*(r+1) == n
%o def a(n):
%o m = psharp = primorial(n)
%o while not oblong(m): m += psharp
%o return m
%o print([a(n) for n in range(1, 11)]) # _Michael S. Branicky_, May 25 2022
%o (Python) # faster alternative using Python 3.8+ A344005(n) by _Chai Wah Wu_
%o from sympy import primorial
%o def a(n): return (m := A344005(primorial(n)))*(m+1)
%o print([a(n) for n in range(1, 18)]) # _Michael S. Branicky_, May 26 2022
%o (PARI) a002110(n) = prod(i=1,n, prime(i)) \\ after _Washington Bomfim_ in A002110
%o a344005(n) = for(m=1, oo, if((m*(m+1))%n==0, return(m)))
%o a(n) = my(m=a344005(a002110(n))); m*(m+1) \\ _Felix Fröhlich_, May 31 2022
%Y Cf. A000040, A002378, A002110, A118478, A344005.
%K nonn
%O 1,1
%A _Ali Sada_, May 25 2022
%E a(9)-a(17) from _Michael S. Branicky_, May 26 2022
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