The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A016029

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A016029

a(1) = a(2) = 1, a(2n + 1) = 2*a(2n) and a(2n) = 2*a(2n - 1) + (-1)^n.
(history; published version)

#32 by Joerg Arndt at Sat Dec 30 23:42:48 EST 2023

STATUS

editing

approved

#31 by Paolo P. Lava at Sat Dec 30 12:04:12 EST 2023

FORMULA

a(n) = (3/5)*2^(n-1) + (1/5 + (1/10)*i)*i^(n-1) + (1/5-(1/10)*i)*(-i)^(n-1), with n > 0 and i=sqrt(-1). - Paolo P. Lava, Jun 10 2008

STATUS

approved

editing

#30 by Michael De Vlieger at Sun Jul 10 09:43:58 EDT 2022

STATUS

reviewed

approved

#29 by Michel Marcus at Sun Jul 10 02:50:34 EDT 2022

STATUS

proposed

reviewed

#28 by G. C. Greubel at Sun Jul 10 00:04:48 EDT 2022

STATUS

editing

proposed

#27 by G. C. Greubel at Sun Jul 10 00:03:52 EDT 2022

FORMULA

From G. C. Greubel, Jul 08 2022: (Start)

a(n) = round( 3*2^(n-1)/5 ).

E.g.f.: (1/10)*(3*exp(2*x) + 4*sin(x) + 2*cos(x) - 5). - _G. C. Greubel_, Jul 08 2022(End)

PROG

(Magma) [n le Round(3 select Floor((n+1)/2) else *2*Self^(n-1) - Self(n-2) +2*Self(n-3/5): n in [1..41]]; // G. C. Greubel, Jul 08 2022

STATUS

proposed

editing

Discussion

Sun Jul 10

00:04

G. C. Greubel: Made suggested changes; added the suggested formula.

#26 by G. C. Greubel at Fri Jul 08 15:17:34 EDT 2022

STATUS

editing

proposed

Discussion

Fri Jul 08

19:18

Kevin Ryde: I don't speak magma, but does it run recurrence like it seems?  Would it be easier or faster by round(3*2^(n-1)/5) ?

#25 by G. C. Greubel at Fri Jul 08 15:17:15 EDT 2022

FORMULA

From Ralf Stephan, Jan 12 2005: (Start)

G.f.: x*(1-x+x^2)/((1-2*x)*(1+x^2)). - _Ralf Stephan_, Jan 12 2005(End)

STATUS

proposed

editing

#24 by G. C. Greubel at Fri Jul 08 03:03:51 EDT 2022

STATUS

editing

proposed

Discussion

Fri Jul 08

03:12

Michel Marcus: in history #4 we see (1/10) {3*2^n + 3*(-1)^[n/2] - (-1)^[(n+1)/2] }. G.f.: x(1-x+x^2)/[(1-2x)(1+x^2)]. - R. Stephan, Jan 12 2005; so ??

#23 by G. C. Greubel at Fri Jul 08 03:03:01 EDT 2022

LINKS

G. C. Greubel, <a href="/A016029/b016029.txt">Table of n, a(n) for n = 1..1000</a>

FORMULA

a(n) = (1/10)*(3*2^n + 3*(-1)^floor(n/2) - (-1)^floor((n+1)/2)). G.f.: x(1-x+x^2)/((1-2x)*(1+x^2)). - _Ralf Stephan_, Jan 12 2005

G.f.: x*(1-x+x^2)/((1-2*x)*(1+x^2)). - Ralf Stephan, Jan 12 2005

a(n) = (3/5)*2^(n-1) + (1/5 + (1/10)*i)*i^n + (3/5)*2^n -1) + (1/5-(1/10)*i)*(-i)^(n, -1), with n >= 0 and i=sqrt(-1). - Paolo P. Lava, Jun 10 2008

E.g.f.: (1/10)*(3*exp(2*x) + 4*sin(x) + 2*cos(x) - 5). - G. C. Greubel, Jul 08 2022

MATHEMATICA

LinearRecurrence[{2, -1, 2}, {1, 1, 2}, 31] (* Ray Chandler, Sep 23 2015 *)

PROG

(Magma) [n le 3 select Floor((n+1)/2) else 2*Self(n-1) - Self(n-2) +2*Self(n-3): n in [1..41]]; // G. C. Greubel, Jul 08 2022

(SageMath) [(1/10)*(3*2^n + 2*i^n*(((n+1)%2) - 2*i*(n%2))) for n in (1..40)] # G. C. Greubel, Jul 08 2022

STATUS

approved

editing