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

Author

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

Keywords

• nonn
• easy

Comments

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 + (-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.

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).

Crossrefs

Cf. A014556, A056486.

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

Programs

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

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

Mathematica

LinearRecurrence[{1,4,-4}, {2,3,5,11,17}, 32] (* Georg Fischer, May 15 2019 *)

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))

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