A292531 - OEIS

%I #15 Oct 01 2017 00:48:57

%S 0,0,8,0,24,32,48,0,80,96,120,128,168,192,224,0,288,320,360,384,440,

%T 480,528,512,624,672,728,768,840,896,960,0,1088,1152,1224,1280,1368,

%U 1440,1520,1536,1680,1760,1848,1920,2024,2112,2208,2048,2400,2496,2600,2688

%N a(n) = 0 if n is a power of 2. Otherwise, product of 2 numbers nearest n that have more 2's in their prime factorization than n.

%C 1) For all odd n, a(n) = n^2 - 1.

%C 2) All numbers in sequence are divisible by 8.

%C 3) a(n) is not divisible by 16 if and only if n = 8k+3 or n = 8k+5.

%C Proposition 2) is true. Proof: 0 mod 2 = 0, so the conjecture trivially holds when n is a power of 2. For n not a power of 2, a(n) has by definition the repeated prime factor 2^2 and so is divisible by 8 when a(n) > 4. - _Felix Fröhlich_, Sep 19 2017

%F a(n) = p^2 * ceiling(n/p) * floor(n/p), where p = A171977(n). - _Giovanni Resta_, Sep 19 2017

%e a(40) = 1536 because 40 has three 2's in its prime factorization, and the closest integers to 40 that have at least four 2's are 32 and 48, and 32 times 48 = 1536.

%t a[n_] := Block[{p = 2 2^IntegerExponent[n, 2]}, Floor[n/p] Ceiling[n/p] p^2]; Array[a, 60] (* _Giovanni Resta_, Sep 19 2017 *)

%Y Cf. A000290.

%K nonn,easy

%O 1,3

%A _J. Lowell_, Sep 18 2017

%E More terms from _Giovanni Resta_, Sep 19 2017