A336535 - OEIS

%I #25 Oct 21 2022 21:26:17

%S 1,6,28,91,231,496,946,1653,2701,4186,6216,8911,12403,16836,22366,

%T 29161,37401,47278,58996,72771,88831,107416,128778,153181,180901,

%U 212226,247456,286903,330891,379756,433846,493521,559153,631126,709836,795691,889111,990528,1100386,1219141,1347261,1485226,1633528,1792671

%N a(n) = (m(n)^2 + 3)*(m(n)^2 + 7)/32, where m(n) = 2*n - 1.

%C For m(n) = 3,5,11, and 181, the perfect numbers (A000396), 6, 28, 496, and 33550336 are produced, respectively. 3,5,11, and 181 are the numbers m(n) such that (m(n)^2+7) is a power of 2. cf A038198.

%D David M. Burton, Elementary Number Theory, McGraw-Hill (2011), 25.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F From _Stefano Spezia_, Jul 25 2020: (Start)

%F O.g.f.: x*(1 + x + 8*x^2 + x^3 + x^4)/(1 - x)^5.

%F a(n) = (1 - n + n^2)*(2 - n + n^2)/2.

%F a(n) = A002061(n)*A014206(n-1)/2.

%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5. (End)

%e m(2) = 2*2-1 = 3 and (3^2+3)*(3^2+7)/32 = 6, so 6 is in the sequence.

%t Table[((n^2+3)*(n^2+7))/32, {n,1,100,2}]

%o (PARI) a(n)=(1-n+n^2)*(2-n+n^2)/2 \\ _Charles R Greathouse IV_, Oct 21 2022

%Y Cf. A000396, A038198.

%Y Cf. A002061, A014206.

%K nonn,easy

%O 1,2

%A _Jeff Brown_, Jul 24 2020