A082605 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.
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
Offset: 0
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (1,4,-4).
Programs
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Magma
A082605:= func< n | n le 1 select n+2 else 2^(n-3)*(9-(-1)^n) +1 >; [A082605(n): n in [0..40]]; // G. C. Greubel, Mar 23 2024
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Maple
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: aList(30); # Peter Luschny, Mar 23 2024
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Mathematica
LinearRecurrence[{1,4,-4}, {2,3,5,11,17}, 32] (* Georg Fischer, May 15 2019 *)
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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))
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SageMath
def A082605(n): return 1 + 2^(n-3)*(9-(-1)^n) -int(n==1)/2 [A082605(n) for n in range(41)] # G. C. Greubel, Mar 23 2024
Formula
(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(n) = A056486(n-1) + 1. - Ralf Stephan, Mar 19 2004
From Georg Fischer, May 15 2019: (Start)
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)
Extensions
More terms from Ralf Stephan, Mar 19 2004
Comments