A047409 Numbers that are congruent to {0, 1, 4, 6} mod 8.

0, 1, 4, 6, 8, 9, 12, 14, 16, 17, 20, 22, 24, 25, 28, 30, 32, 33, 36, 38, 40, 41, 44, 46, 48, 49, 52, 54, 56, 57, 60, 62, 64, 65, 68, 70, 72, 73, 76, 78, 80, 81, 84, 86, 88, 89, 92, 94, 96, 97, 100, 102, 104, 105, 108, 110, 112, 113, 116, 118, 120, 121, 124

Offset

1,3

Comments

All squares and the products of any terms belong to the sequence. This sequence (n > 1) is closed under multiplication. - Klaus Purath, Feb 13 2023

Links

Table of n, a(n) for n=1..63.

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Formula

G.f.: x^2*(1 + 3*x + 2*x^2 + 2*x^3) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 05 2011

From Wesley Ivan Hurt, May 24 2016: (Start)

a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.

a(n) = (8*n - 9 - i^(2n) + i^(-n) + i^n)/4 where i=sqrt(-1).

a(2k) = A047452(k), a(2k-1) = A008586(k-1) for k > 0. (End)

E.g.f.: (4 + cos(x) + 4*(x - 1)*sinh(x) + (4*x - 5)*cosh(x))/2. - Ilya Gutkovskiy, May 25 2016

Sum_{n>=2} (-1)^n/a(n) = sqrt(2)*Pi/16 + (10 - sqrt(2))*log(2)/16 + sqrt(2)*log(2 + sqrt(2))/8. - Amiram Eldar, Dec 20 2021

a(n) = 2*(n-1) + floor((n+1)/4) - floor((n+2)/4). - Ridouane Oudra, Aug 19 2024

Maple

A047409:=n->(8*n-9-I^(2*n)+I^(-n)+I^n)/4: seq(A047409(n), n=1..100); # Wesley Ivan Hurt, May 24 2016

Mathematica

Select[Range[0, 110], MemberQ[{0, 1, 4, 6}, Mod[#, 8]]&] (* Harvey P. Dale, Sep 28 2011 *)

Prog

(Magma) [n : n in [0..150] | n mod 8 in [0, 1, 4, 6]]; // Wesley Ivan Hurt, May 24 2016

Crossrefs

Cf. A008586, A047452.

Sequence in context: A353355 A231573 A327204 * A161576 A327446 A100425

Adjacent sequences: A047406 A047407 A047408 * A047410 A047411 A047412

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved