A190730 Let b(n,0) = n and b(n,k) = 2*b(n,k-1) + 1 for k > 0. Then a(n) = b(n,1) + b(n,2) + ... + b(n,n).

3, 16, 53, 146, 367, 876, 2025, 4582, 10211, 22496, 49117, 106458, 229335, 491476, 1048529, 2228174, 4718539, 9961416, 20971461, 44040130, 92274623, 192937916, 402653113, 838860726, 1744830387, 3623878576, 7516192685, 15569256362, 32212254631, 66571992996

Offset: 1

Author

J. M. Bergot, May 17 2011

Keywords

• nonn
• easy

Comments

It turns out that b(n,k) = A087322(n,k) = (n + 1)*2^k - 1 for 1 <= k <= n (without the 0th column). - Petros Hadjicostas, Feb 15 2021

Examples

			One way to view it is to begin with n = 5, then 5 + 6 = 11 --> 11 + 12 = 23 --> 23 + 24 = 47 --> 47 + 48 = 95 --> 95 + 96 = 191. There are n steps, in this case 5, that give the sum 11 + 23 + 47 + 95 + 191 = 367. This is the same as (2*5+1) + (4*5+3) + (8*5+7) + (16*5+15) + (32*5+31). The formula gives (5+1)*2^(5+1) - 3*5 - 2 = 6*64 - 17 = 367.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (6,-13,12,-4).

Crossrefs

Cf. A016789, A036289, A087322.

Programs

Magma

[(n+1) * 2^(n+1) - 3*n - 2 : n in [1..30]]; // Vincenzo Librandi, Sep 29 2011

Mathematica

LinearRecurrence[{6,-13,12,-4},{3,16,53,146},40] (* or *)
Array[(#+1)2^(#+1)-3#-2&,40] (* Paolo Xausa, Oct 17 2023 *)

Formula

a(n) = (n+1) * 2^(n+1) - 3*n - 2 = A036289(n+1) - A016789(n).

G.f.: -x*(-3 + 2*x + 4*x^2) / ( (2*x-1)^2*(x-1)^2 ). - R. J. Mathar, May 29 2011

E.g.f.: exp(x)*(2*exp(x)*(1 + 2*x) - 2 - 3*x). - Stefano Spezia, Oct 16 2023