A233656 - OEIS

%I #21 Mar 24 2022 04:21:47

%S 1,3,7,17,45,127,371,1101,3289,9851,29535,88585,265733,797175,2391499,

%T 7174469,21523377,64570099,193710263,581130753,1743392221,5230176623,

%U 15690529827,47071589437,141214768265,423644304747,1270932914191,3812798742521,11438396227509,34315188682471

%N a(n) = 3*a(n-1) - 2*(n-1), with a(0) = 1.

%C Inverse binomial transform of this sequence is A090129.

%C Conjecture: Last digit of a(n) has repeating pattern of 20 digits as follows: {1, 3, 7, 7, 5, 7, 1, 1, 9, 1, 5, 5, 3, 5, 9, 9, 7, 9, 3, 3}, with an equal frequency of the five odd digits.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (5,-7,3).

%F a(n) = 3*a(n-1) - 2*(n-1), with a(0) = 1.

%F (a(n+1) - a(n))/2 = A007051(n).

%F From _Colin Barker_, Dec 17 2013: (Start)

%F a(n) = (1 + 3^n + 2*n)/2.

%F a(n) = 5*a(n-1) - 7*a(n-2) + 3*a(n-3).

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

%F E.g.f.: exp(x)*(1 + exp(2*x) + 2*x)/2. - _Stefano Spezia_, Mar 20 2022

%e a(1) = 3*1 - 2*0 = 3;

%e a(2) = 3*3 - 2*1 = 7.

%t LinearRecurrence[{5,-7,3},{1,3,7},30] (* _Harvey P. Dale_, Feb 13 2015 *)

%o (PARI) Vec((x^2+2*x-1)/((x-1)^2*(3*x-1)) + O(x^100)) \\ _Colin Barker_, Dec 17 2013

%Y Cf. A007051, A090129.

%K nonn,easy

%O 0,2

%A _Richard R. Forberg_, Dec 14 2013