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 A085354 a(n) = 3*4^n - (n+4)*2^(n-1). 2
 1, 7, 36, 164, 704, 2928, 11968, 48448, 195072, 783104, 3138560, 12567552, 50298880, 201256960, 805158912, 3220914176, 12884246528, 51538231296, 206155546624, 824627691520, 3298522300416, 13194113318912, 52776503607296 (list; graph; refs; listen; history; text; internal format)
 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(n-1) and adding the leaves. - Jean M. Morales, Oct 24 2013 LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..1000 Index entries for 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..(n-1)} a(k). - Philippe Deléham, Feb 19 2004 G.f.: (1-x)/((1-4*x)*(1-2*x)^2). - Bruno Berselli, Sep 05 2011 a(n) = 2*a(n-1) + 3*2^(2n-1) - 2^(n-1), 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 *) LinearRecurrence[{8, -20, 16}, {1, 7, 36}, 30] (* Harvey P. Dale, Apr 08 2019 *) PROG (Magma) [3*4^n-(n+4)*2^(n-1): n in [0..30]]; // Vincenzo Librandi, Sep 05 2011 CROSSREFS Cf. A005187, A060188. Sequence in context: A243037 A181292 A026018 * A051198 A003516 A095931 Adjacent sequences: A085351 A085352 A085353 * A085355 A085356 A085357 KEYWORD nonn,easy AUTHOR Paul Barry, Jun 24 2003 STATUS approved

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Last modified May 28 19:55 EDT 2024. Contains 372919 sequences. (Running on oeis4.)