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A082605
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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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6
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2, 3, 5, 11, 17, 41, 65, 161, 257, 641, 1025, 2561, 4097, 10241, 16385, 40961, 65537, 163841, 262145, 655361, 1048577, 2621441, 4194305, 10485761, 16777217, 41943041, 67108865, 167772161, 268435457, 671088641, 1073741825, 2684354561
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OFFSET
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0,1
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COMMENTS
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Using this definition of a(k) we (formally) work backwards towards a(2)=5 to arrive at the formula for a(k) below.
For k >= 3, a(k) has the simple form a(k) = 2^(k-2)*(4 + 1/2*(1 + (-1)^(k+1)) + 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.
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LINKS
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FORMULA
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(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.
a(2*n) = 2^n + 1.
G.f.: (2+x-6*x^2+2*x^3-2*x^4)/((1-x)*(1-2*x)*(1+2*x)). (End)
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MAPLE
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aList := proc(len) local egf, ser, n;
egf := (exp(-2*x) + 9*exp(2*x) - 10)/4; ser := series(egf, x, len + 2);
[2, 3, 5, seq(1 + n!*coeff(ser, x, n), n = 2..len)] end:
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MATHEMATICA
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LinearRecurrence[{1, 4, -4}, {2, 3, 5, 11, 17}, 32] (* Georg Fischer, May 15 2019 *)
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PROG
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(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))
(Magma)
A082605:= func< n | n le 1 select n+2 else 2^(n-3)*(9-(-1)^n) +1 >;
(SageMath)
def A082605(n): return 1 + 2^(n-3)*(9-(-1)^n) -int(n==1)/2
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CROSSREFS
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a(0..6) and a(2*n) same as A085613(n+1).
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KEYWORD
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nonn,easy
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AUTHOR
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Johan Meyer and Ben de la Rosa (meyerjh.sci(AT)mail.uovs.ac.za), May 23 2003
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EXTENSIONS
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STATUS
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approved
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