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Using Euler's 6-term sequence A014556, we define the partial recurrence relation a(0)=2, a(1)=3, a(2)=5; a(k) = 2*a(k-1) - 1 - (-2)^(k-2), 3 <= k <= 5.
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%I #28 Aug 24 2024 01:53:38

%S 2,3,5,11,17,41,65,161,257,641,1025,2561,4097,10241,16385,40961,65537,

%T 163841,262145,655361,1048577,2621441,4194305,10485761,16777217,

%U 41943041,67108865,167772161,268435457,671088641,1073741825,2684354561

%N Using Euler's 6-term sequence A014556, we define the partial recurrence relation a(0)=2, a(1)=3, a(2)=5; a(k) = 2*a(k-1) - 1 - (-2)^(k-2), 3 <= k <= 5.

%C Using this definition of a(k) we (formally) work backwards towards a(2)=5 to arrive at the formula for a(k) below.

%C For k >= 3, a(k) has the simple form a(k) = 2^(k-2)*(4 + (1 + (-1)^(k+1))/2) + 1; and it follows by induction that a(k) is congruent to 17 (mod 24) for all k >= 4. Direct calculations show that for k >= 3, the discriminants of the polynomials x^2 + x + a(k), D(k) = 1 - 4*a(k), satisfy the functional equation -D(k) = a(k+2) + 2.

%H G. C. Greubel, <a href="/A082605/b082605.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,4,-4).

%F (a(k))_(k>=0) = 2^(k-2)*(4 + Sum_{r=2..k-1} (-1)^r) + 1, the empty sums corresponding to k=0, 1, 2 of course taken to be zero.

%F a(n) = A056486(n-1) + 1. - _Ralf Stephan_, Mar 19 2004

%F From _Georg Fischer_, May 15 2019: (Start)

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

%F G.f.: (2+x-6*x^2+2*x^3-2*x^4)/((1-x)*(1-2*x)*(1+2*x)). (End)

%p aList := proc(len) local egf, ser, n;

%p egf := (exp(-2*x) + 9*exp(2*x) - 10)/4; ser := series(egf, x, len + 2);

%p [2, 3, 5, seq(1 + n!*coeff(ser,x, n), n = 2..len)] end:

%p aList(30); # _Peter Luschny_, Mar 23 2024

%t LinearRecurrence[{1,4,-4}, {2,3,5,11,17}, 32] (* _Georg Fischer_, May 15 2019 *)

%o (PARI) a(n)=if(n<2,if(n<1,2,3),if(n%2==0,4^(n/2)+1,5/2*4^((n-1)/2)+1))

%o (Magma)

%o A082605:= func< n | n le 1 select n+2 else 2^(n-3)*(9-(-1)^n) +1 >;

%o [A082605(n): n in [0..40]]; // _G. C. Greubel_, Mar 23 2024

%o (SageMath)

%o def A082605(n): return 1 + 2^(n-3)*(9-(-1)^n) -int(n==1)/2

%o [A082605(n) for n in range(41)] # _G. C. Greubel_, Mar 23 2024

%Y Cf. A014556, A056486.

%Y a(0..6) and a(2*n) same as A085613(n+1).

%K nonn,easy

%O 0,1

%A Johan Meyer and Ben de la Rosa (meyerjh.sci(AT)mail.uovs.ac.za), May 23 2003

%E More terms from _Ralf Stephan_, Mar 19 2004