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A281445 Nonnegative k for which (2*k^2 + 1)/11 is an integer. 1

%I

%S 4,7,15,18,26,29,37,40,48,51,59,62,70,73,81,84,92,95,103,106,114,117,

%T 125,128,136,139,147,150,158,161,169,172,180,183,191,194,202,205,213,

%U 216,224,227,235,238,246,249,257,260,268,271,279,282,290,293,301,304,312,315

%N Nonnegative k for which (2*k^2 + 1)/11 is an integer.

%C For prime d < 11, (2*k^2 + 1)/d can provide integers when d = 3 (A186424).

%C Corresponding values of (2*k^2 + 1)/11 are listed in A179088.

%C All k == 4 or 7 (mod 11). - _Robert Israel_, Apr 25 2017

%H Bruno Berselli, <a href="/A281445/b281445.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

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

%F E.g.f.: 4 - 5*exp(-x)/4 - 11*(1 - 2*x)*exp(x)/4.

%F a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).

%F a(n) = (22*n - 5*(-1)^n - 11)/4. Therefore: a(2*h) = 11*h - 4, a(2*h+1) = 11*h + 4.

%F If h>0,

%F h*a(n) + (6*h - 5*(-1)^h - 11)/4 = a(h*n) for odd n; otherwise:

%F h*a(n) + 4*(h - 1) = a(h*n). Some special cases:

%F h=2: 2*a(n) - 1 = a(2*n) for odd n, 2*a(n) + 4 = a(2*n) for even n;

%F h=3: 3*a(n) + 3 = a(3*n) for odd n, 3*a(n) + 8 = a(3*n) for even n;

%F h=4: 4*a(n) + 2 = a(4*n) for odd n, 4*a(n) + 12 = a(4*n) for even n;

%F h=5: 5*a(n) + 6 = a(5*n) for odd n, 5*a(n) + 16 = a(5*n) for even n, and so on.

%p seq(seq(11*i+j,j=[4,7]),i=0..50); # _Robert Israel_, Apr 25 2017

%t Select[Range[400], IntegerQ[(2*#^2 + 1)/11] &]

%o (Sage) [k for k in range(400) if ((2*k^2+1)/11).is_integer()]

%o (MAGMA) &cat [[11*n+4, 11*n+7]: n in [0..30]];

%Y Cf. A179088.

%Y Cf. A001651: nonnegative k for which (2*k^2 + 1)/3 is an integer.

%K nonn,easy

%O 1,1

%A _Bruno Berselli_, Apr 13 2017

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Last modified March 7 09:46 EST 2021. Contains 341869 sequences. (Running on oeis4.)