

A085354


3*4^n(n+4)*2^(n1).


2



1, 7, 36, 164, 704, 2928, 11968, 48448, 195072, 783104, 3138560, 12567552, 50298880, 201256960, 805158912, 3220914176, 12884246528, 51538231296, 206155546624, 824627691520, 3298522300416, 13194113318912, 52776503607296
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OFFSET

0,2


COMMENTS

Binomial transform of A060188.
The depth i nodes of a perfect binary tree are numbered 2^i through 2^(i+1)  1, so that the root has number 1, depth 1 nodes have numbers 2 and 3, depth 2 nodes have numbers 4, 5, 6 and 7 and so on. We sum all the numbers in the path connecting a leaf node to the root. For a height n tree, a(n) is the sum of these sums for all leaves nodes. So for instance a height 1 tree has paths 1, 2 and 1, 3 connecting the root to the leaves, and (1+2) + (1+3) = a(1) = 7. This interpretation suggests a recursive formula for computing a(n) by completing the paths covered in a(n1) and adding the leaves.  Jean M. Morales, Oct 24 2013


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index to sequences with linear recurrences with constant coefficients, signature (8,20,16).


FORMULA

a(n) = sum{m = 2^n..2^(n+1)} A005187(m). a(n) = 2^n*(2^(n+1)1) + sum_{k = 0..(n1)} a(k) .  Philippe Deléham, Feb 19 2004
G.f.: (1x)/((14*x)*(12*x)^2).  Bruno Berselli, Sep 05 2011
a(n) = 2*a(n1) + 3*2^(2n1)  2^(n1), a(0) = 1  Jean M. Morales, Oct 24 2013


MATHEMATICA

Table[3 * 4^n  (n + 4) * 2^(n  1), {n, 0, 19}] (* Alonso del Arte, Oct 23 2013 *)


PROG

(MAGMA) [3*4^n(n+4)*2^(n1): n in [0..30]]; // Vincenzo Librandi, Sep 05 2011


CROSSREFS

Sequence in context: A080420 A181292 A026018 * A051198 A003516 A095931
Adjacent sequences: A085351 A085352 A085353 * A085355 A085356 A085357


KEYWORD

nonn,easy


AUTHOR

Paul Barry, Jun 24 2003


STATUS

approved



