login
Numbers that are congruent to {1, 5} mod 20.
2

%I #24 Sep 06 2022 10:29:30

%S 1,5,21,25,41,45,61,65,81,85,101,105,121,125,141,145,161,165,181,185,

%T 201,205,221,225,241,245,261,265,281,285,301,305,321,325,341,345,361,

%U 365,381,385,401,405,421,425,441,445,461,465,481,485,501,505,521,525

%N Numbers that are congruent to {1, 5} mod 20.

%H Colin Barker, <a href="/A228141/b228141.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F a(n) = -3*(4+(-1)^n) + 10*n.

%F a(n) = a(n-1) + a(n-2) - a(n-3).

%F G.f.: x*(15*x^2+4*x+1) / ((x-1)^2*(x+1)).

%F E.g.f.: 15 + (10*x - 12)*exp(x) - 3*exp(-x). - _David Lovler_, Sep 05 2022

%t Flatten[Table[20n + {1, 5}, {n, 0, 24}]] (* _Alonso del Arte_, Aug 12 2013 *)

%o (PARI) Vec(x*(15*x^2+4*x+1)/((x-1)^2*(x+1)) + O(x^100))

%Y Cf. A216815, A216816, A228142.

%K nonn,easy

%O 1,2

%A _Colin Barker_, Aug 12 2013