A370455 a(n) = greatest m such that 2^m divides prime(n+1)*prime(n+2) - prime(n)*prime(n+3).

0, 1, 2, 3, 2, 3, 3, 1, 4, 1, 3, 3, 3, 1, 2, 4, 1, 4, 1, 1, 2, 2, 1, 1, 3, 2, 3, 3, 3, 2, 1, 2, 3, 2, 2, 1, 2, 1, 2, 2, 3, 1, 2, 4, 1, 3, 1, 3, 7, 1, 2, 2, 2, 3, 2, 4, 1, 3, 1, 2, 1, 3, 3, 3, 1, 2, 2, 1, 5, 2, 2, 1, 1, 2, 2, 1, 5, 1, 1, 3, 3, 2, 1, 2, 2, 1

Offset: 1

Author

Clark Kimberling, Feb 26 2024

Keywords

• nonn

Examples

			prime(4)*prime(5) - prime(3)*prime(6) = 7*11 - 5*13 = 12, which is divisible by 2^2 but not 2^3, so a(3) = 2.
		

Crossrefs

Cf. A007814, A117301.

Programs

Mathematica

p[n_] := Prime[n];
u = Table[p[n + 1] p[n + 2] - p[n] p[n + 3], {n, 1, 2000}];  (* A117302 *)
s[n_] := Last[Select[Range[15], IntegerQ[u[[n]]/2^#] &]];
Table[s[n], {n, 1, 200}]

PARI

a(n) = valuation(prime(n+1)*prime(n+2) - prime(n)*prime(n+3), 2); \\ Michel Marcus, Mar 01 2024

Python

from sympy import prime
def A370455(n): return (~(m:=prime(n+1)*prime(n+2)-prime(n)*prime(n+3)) & m-1).bit_length() # Chai Wah Wu, Mar 02 2024