OFFSET
0,2
COMMENTS
Found as a question on http://mail.python.org/mailman/listinfo/tutor poster: reavey.
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..5000
Tutor User "reavey" (reavey@nep.net) and others, How to write an algorithm for sequence, Tutor -- Discussion for learning programming with Python, 2003.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2).
FORMULA
a(2n) = 6*2^n - 5, a(2n-1) = 5*(2^n - 1). - Benoit Cloitre, Feb 16 2003
From Colin Barker, Sep 19 2012: (Start)
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3).
G.f.: (1+4*x)/((1-x)*(1-2*x^2)). (End)
MAPLE
seq(coeff(series((1+4*x)/((1-x)*(1-2*x^2)), x, n+1), x, n), n = 0..40); # G. C. Greubel, Aug 07 2019
MATHEMATICA
LinearRecurrence[{1, 2, -2}, {1, 5, 7}, 40] (* G. C. Greubel, Aug 07 2019 *)
PROG
(PARI) seq(n) = { j=a=1; p=2; print1(1" "); while(j<=n, a = a + 2^p; print1(a" "); a = a+2^(p-1); print1(a" "); p+=1; j+=2; ) }
(PARI) a(n)=if(n<0, 0, (6-n%2)*2^ceil(n/2)-5)
(Magma) I:=[1, 5, 7]; [n le 3 select I[n] else Self(n-1) +2*Self(n-2) -2*Self(n-3): n in [1..40]]; // G. C. Greubel, Aug 07 2019
(Sage)
@CachedFunction
def a(n):
if (n==0): return 1
elif (1<=n<=2): return nth_prime(n+2)
else: return a(n-1) + 2*a(n-2) - 2*a(n-3)
[a(n) for n in (0..40)] # G. C. Greubel, Aug 07 2019
(GAP) a:=[1, 5, 7];; for n in [4..30] do a[n]:=a[n-1]+2*a[n-2]-2*a[n-3]; od; a; # G. C. Greubel, Aug 07 2019
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Cino Hilliard, Feb 15 2003
STATUS
approved