A286778 Sum of the common path length over all 2-tuples of nodes in a complete binary tree of height n.
0, 2, 22, 142, 734, 3390, 14718, 61694, 253438, 1029118, 4151294, 16683006, 66904062, 267993086, 1072791550, 4292935678, 17175543806, 68710301694, 274858508286, 1099470733310, 4397960527870, 17592005689342, 70368366690302, 281474188181502, 1125898262675454, 4503596204818430, 18014391395942398
Offset: 0
Examples
A complete binary tree of height two consists of one root node (at depth 0), two children of the root (at depth 1) and four leaf nodes (at depth 2). Notice the common path length of node u with itself, h(u,u), is simply the depth of u. The only 2-tuples to have common path length two is a leaf with itself (4 such tuples). Each child of the root with itself has common path length one (2 such tuples), as does each leaf with its sibling (4 such tuples) and each leaf with its parent (8 such tuples). All other 2-tuples have only the root as a common ancestor. Hence a(2) = 2*4 + 1*(2 + 4 + 8) + 0 = 22.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- Svante Janson, The Wiener index of simply generated random trees, Random Structures Algorithms 22 (2003), no. 4, 337-358. DOI: 10.1002/rsa.10074.
- Index entries for linear recurrences with constant coefficients, signature (9,-28,36,-16).
Crossrefs
Cf. A036799 (total path length of a binary tree of height n).
Programs
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Maple
seq( 4*2^(2*n) - (4*n+2)*2^n - 2, n=0..30); # Robert Israel, Jul 05 2017
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Mathematica
LinearRecurrence[{9,-28,36,-16},{0,2,22,142},40] (* Harvey P. Dale, Apr 30 2018 *)
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PARI
a(n) = sum(d=1, n, 2^d*(2^(n+1-d)-1)^2); \\ Michel Marcus, Jul 05 2017
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PARI
concat(0, Vec(2*x*(1 + 2*x) / ((1 - x)*(1 - 2*x)^2*(1 - 4*x)) + O(x^30))) \\ Colin Barker, Jul 05 2017
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Sage
[sum(2^d*(2^(n+1-d)-1)^2 for d in range(1,n+1)) for n in range(20)]
Formula
a(n) = Sum_{d=1..n} 2^d*(2^(n+1-d)-1)^2.
From Robert Israel, Jul 05 2017: (Start)
a(n) = 4*2^(2*n) - (4*n+2)*2^n - 2.
G.f.: 2*x*(2*x+1)/((4*x-1)*(x-1)*(2*x-1)^2).
E.g.f.: 4*exp(4*x)-(8*x+2)*exp(2*x)-2*exp(x).
(End)
a(n) = 9*a(n-1) - 28*a(n-2) + 36*a(n-3) - 16*a(n-4) for n>3. - Colin Barker, Jul 05 2017
a(n)-a(n-1) = A050488(n)*2^n . - R. J. Mathar, Aug 26 2025
Comments