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%I #41 Aug 19 2022 22:57:54
%S 0,1,16,17,32,33,48,49,64,65,80,81,96,97,112,113,128,129,144,145,160,
%T 161,176,177,192,193,208,209,224,225,240,241,256,257,272,273,288,289,
%U 304,305,320,321,336,337,352,353,368,369,384,385,400,401,416,417,432,433,448,449
%N Numbers that are congruent to {0, 1} mod 16.
%C Numbers k such that k^2 - k is divisible by 16.
%H Vincenzo Librandi, <a href="/A151977/b151977.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 a(n+1) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(k)=2^(k+3) for k > 0; {b(n)} = 1,16,32,64,128,256,... - _Philippe Deléham_, Oct 17 2011
%F G.f.: x^2*(1+15*x)/((1+x)*(1-x)^2). - _Vincenzo Librandi_, Jul 11 2012
%F a(n) = (32*n - 14*(-1)^n - 46)/4. - _Vincenzo Librandi_, Jul 11 2012
%F From _David Lovler_, Aug 18 2022: (Start)
%F a(n) = A321212(n) - 2.
%F a(n) = a(n-2) + 16.
%F E.g.f.: 15 + ((16*x - 23)*exp(x) - 7*exp(-x))/2. (End)
%t CoefficientList[Series[x*(1+15*x)/((1+x)(1-x)^2),{x,0,60}],x] (* _Vincenzo Librandi_, Jul 11 2012 *)
%t LinearRecurrence[{1,1,-1},{0,1,16},80] (* _Harvey P. Dale_, Jul 24 2021 *)
%o (PARI) forstep(n=0,200,[1,15],print1(n", ")) \\ _Charles R Greathouse IV_, Oct 17 2011
%o (PARI) a(n) = (16*n - 7*(-1)^n - 23)/2 \\ _David Lovler_, Aug 18 2022
%o (Magma) [(32*n-14*(-1)^n-46)/4: n in [1..60]]; // _Vincenzo Librandi_, Jul 11 2012
%Y Cf. A321212.
%K nonn,easy
%O 1,3
%A _N. J. A. Sloane_, Aug 23 2009
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