A036557 Number of multiples of 3 in 0..2^n-1 with an even sum of base-2 digits.

1, 1, 2, 3, 6, 10, 20, 35, 70, 126, 252, 463, 926, 1730, 3460, 6555, 13110, 25126, 50252, 97223, 194446, 379050, 758100, 1486675, 2973350, 5858126, 11716252, 23166783, 46333566, 91869970, 183739940, 365088395, 730176790

Offset

0,3

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,4,-8,-3,6).

Formula

From Ralf Stephan, Aug 29 2004: (Start)

a(n) = (1/12)*(3^((n+1)/2) + 3^((n+2)/2) + 2^(n+1) + (-1)^n + 3), n > 0.

G.f.: (1 - x - 4*x^2 + 3*x^3 + 3*x^4 - x^5)/((1-x^2)*(1-2*x)*(1-3*x^2)). (End)

a(n) = 2*a(n-1) + 4*a(n-2) - 8*a(n-3) - 3*a(n-4) + 6*a(n-5). - Wesley Ivan Hurt, Apr 13 2021

Mathematica

Sum[ Sum[ Binomial[ Floor[ n/2 ], i ], {i, r, n, 6} ]*Sum[ Binomial[ Ceiling[ n/2 ], i ], {i, r, n, 6} ], {r, 0, 5} ]
Join[{1}, LinearRecurrence[{2, 4, -8, -3, 6}, {1, 2, 3, 6, 10}, 50]] (* G. C. Greubel, Dec 31 2017 *)

Prog

(PARI) x='x+O('x^30); Vec((1-x-4*x^2+3*x^3+3*x^4-x^5)/((1-x^2)*(1-2*x)*(1-3*x^2))) \\ G. C. Greubel, Dec 31 2017
(Magma) I:=[1, 2, 3, 6, 10]; [1] cat [n le 5 select I[n] else 2*Self(n-1) + 4*Self(n-2) - 8*Self(n-3) - 3*Self(n-4) + 6*Self(n-5): n in [1..30]]; // G. C. Greubel, Dec 31 2017

Crossrefs

Cf. A036555, A036556.

Sequence in context: A001405 A126930 A210736 * A173125 A047131 A231331

Adjacent sequences: A036554 A036555 A036556 * A036558 A036559 A036560

Keyword

nonn,base,easy

Author

Dan Hoey

Status

approved