A079360 Sequence of sums of alternating increasing powers of 2.

1, 5, 7, 15, 19, 35, 43, 75, 91, 155, 187, 315, 379, 635, 763, 1275, 1531, 2555, 3067, 5115, 6139, 10235, 12283, 20475, 24571, 40955, 49147, 81915, 98299, 163835, 196603, 327675, 393211, 655355, 786427, 1310715, 1572859, 2621435, 3145723

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

Cf. A079361, A079362, A048488 (bisection).

Sequence in context: A309292 A050851 A058918 * A314365 A314366 A068580

Adjacent sequences: A079357 A079358 A079359 * A079361 A079362 A079363

Keyword

easy,nonn

Author

Cino Hilliard, Feb 15 2003

Status

approved