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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).
2
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
1,1
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
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
EXAMPLE
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.
MATHEMATICA
LinearRecurrence[{6, -13, 12, -4}, {3, 16, 53, 146}, 40] (* or *)
Array[(#+1)2^(#+1)-3#-2&, 40] (* Paolo Xausa, Oct 17 2023 *)
PROG
(Magma) [(n+1) * 2^(n+1) - 3*n - 2 : n in [1..30]]; // Vincenzo Librandi, Sep 29 2011
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
J. M. Bergot, May 17 2011
STATUS
approved