|
%I M1814 N0718 #90 Mar 04 2023 04:24:48
%S 1,1,2,7,56,2212,2595782,3374959180831,5695183504489239067484387,
%T 16217557574922386301420531277071365103168734284282
%N a(0) = a(1) = 1; for n > 0, a(n+1) = a(n)*(a(0) + ... + a(n-1)) + a(n)*(a(n) + 1)/2.
%C Number of planted trees in which every node has degree <= 3 and of height n; or products of height n when multiplication is commutative but non-associative.
%C Also called planted 3-trees or planted unary-binary trees.
%C The next term (which was incorrectly given) is in fact too large to include. See the b-file.
%C Comment from _Marc LeBrun_: Maximum possible number of distinct new values after applying a commuting operation N times to a single initial value.
%C Divide the natural numbers in sets of consecutive numbers, starting with {1}, each set with number of elements equal to the sum of elements of the preceding set. The number of elements in the n-th (n>0) set gives a(n). The sets begin {1}, {2}, {3,4}, {5,6,7,8,9,10,11}, ... - _Floor van Lamoen_, Jan 16 2002
%C Consider the free algebraic system with one binary commutative (x+y) operator and one generator A. The number of elements of height n is a(n) where the height of A is zero and the height of (x+y) is one more than the maximum height of x and y. - _Michael Somos_, Mar 06 2012
%C Sergey Zimnitskiy, May 08 2013, provided an illustration for A006894 and A002658 in terms of packing circles inside circles. The following description of the figure was supplied by _Allan Wilks_. Label a blank page "1" and draw a black circle labeled "2". Subsequent circles are labeled "3", "4", ... . In the black circle put two red circles (numbered "3" and "4"); two because the label of the black circle is "2". Then in each of the red circles put blue circles in number equal to the labels of the red circles. So these get labeled "5", ..., "11". Then in each of the blue circles, starting with circle "5", place a set of green (say) circles, equal in number to the label of the enclosing blue circle. When all of the green circles have been drawn, they will be labeled "12", ..., "67". If you take the maximum circle label at each colored level, you get 1,2,4,11,67,2279,..., which is A006894, which itself is the partial sums of A002658. The picture is a visualization of _Floor van Lamoen_'s comment above.
%C See A067338 for a variant of the integer partitioning construction, starting with {1,2}, {3,4,5}, ... - _M. F. Hasler_, Jan 21 2015
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H David Wasserman, <a href="/A002658/b002658.txt">Table of n, a(n) for n = 0..13</a>
%H Mayfawny Bergmann, <a href="http://scholar.rose-hulman.edu/rhumj/vol15/iss2/1/">Efficiency of Lossless Compression of a Binary Tree via its Minimal Directed Acyclic Graph Representation</a>. Rose-Hulman Undergraduate Mathematics Journal: Vol. 15 : Iss. 2, Article 1. (2014).
%H I. M. H. Etherington, <a href="http://www.jstor.org/stable/3605743">Non-associate powers and a functional equation</a>, Math. Gaz. 21 (1937), 36-39; addendum 21 (1937), 153.
%H I. M. H. Etherington, <a href="/A000108/a000108_13.pdf">On non-associative combinations</a>, Proc. Royal Soc. Edinburgh, 59 (Part 2, 1938-39), 153-162. [Annotated scanned copy]
%H I. M. H. Etherington, <a href="http://dx.doi.org/10.1017/S0950184300002639">Some problems of non-associative combinations (I)</a>, Edinburgh Math. Notes, 32 (1940), pp. i-vi. Part II is by A. Erdelyi and I. M. H. Etherington, and is on pages vii-xiv of the same issue.
%H A. Erdelyi and I. M. H. Etherington, <a href="http://dx.doi.org/10.1017/S0950184300002640">Some problems of non-associative combinations (II)</a>, Edinburgh Math. Notes, 32 (1940), pp. vii-xiv.
%H F. Harary, et al., <a href="http://cobweb.cs.uga.edu/~rwr/publications/binary.pdf">Counting free binary trees admitting a given height</a>, J. Combin. Inform. System Sci. 17 (1992), no. 1-2, 175--181. MR1216977 (94c:05039)
%H Harary, Frank; Palmer, Edgar M.; Robinson, Robert W., <a href="/A005588/a005588.pdf">Counting free binary trees admitting a given height</a>, J. Combin. Inform. System Sci. 17 (1992), no. 1-2, 175-181. (Annotated scanned copy)
%H Z. A. Melzak, <a href="http://dx.doi.org/10.4153/CMB-1968-012-1">A note on homogeneous dendrites</a>, Canad. Math. Bull., 11 (1968), 85-93.
%H Sergey Zimnitskiy, <a href="/A006894/a006894.jpg">Illustration of initial terms of A006894 and A002658</a>
%H <a href="/index/Ro#rooted">Index entries for sequences related to rooted trees</a>
%H <a href="/index/Tra#trees">Index entries for sequences related to trees</a>
%H <a href="/index/Cor#core">Index entries for "core" sequences</a>
%F a(n + 1) = a(n) * (a(n) / a(n-1) + (a(n) + a(n-1)) / 2) [equation (5) on page 87 of Melzak 1968 with a() instead of his f()].
%F a(n) ~ 2 * c^(2^n), where c = 1.2460208329836625089431529441999359284665241772983812581752523573774108242448... . - _Vaclav Kotesovec_, May 21 2015
%p s := proc(n) local i,j,ans; ans := [ 1 ]; for i to n do ans := [ op(ans),ans[ i ]*(add(j,j=ans)-ans[ i ])+ans[ i ]*(ans[ i ]+1)/2 ] od; RETURN(ans); end; t1 := s(10); A002658 := n->t1[n];
%t Clear[a, b]; a[0] = a[1] = 1; b[0] = b[1] = 1; b[n_] := b[n] = b[n-1] + a[n-1]; a[n_] := a[n] = (a[n-1]+1)*a[n-1]/2 + a[n-1]*b[n-1]; Table[a[n], {n, 0, 9}] (* _Jean-François Alcover_, Jan 31 2013, after Frank Harary *)
%t RecurrenceTable[{a[n] == a[n-1]*(a[n-1]/a[n-2]+(a[n-1]+a[n-2])/2), a[0]==1, a[1]==1},a,{n,0,10}] (* _Vaclav Kotesovec_, May 21 2015 *)
%o (PARI) {a(n) = local(a1, a2); if( n<2, n>=0, a2 = a(n-1); a1 = a(n-2); a2 * (a2 / a1 + (a1 + a2) / 2))} /* _Michael Somos_, Mar 06 2012 */
%o (PARI) print1(s=a=1);for(i=1,9,print1(","a*=(1-a)/2+s);s+=a) \\ _M. F. Hasler_, Jan 21 2015
%o (Haskell)
%o a002658 n = a002658_list !! n
%o a002658_list = 1 : 1 : f [1,1] where
%o f (x:xs) = y : f (y:x:xs') where y = x * sum xs + x * (x + 1) `div` 2
%o -- _Reinhard Zumkeller_, Apr 10 2012
%o (Python)
%o from itertools import islice
%o def agen():
%o yield 1
%o an = s = 1
%o while True:
%o yield an
%o an1 = an*s + an*(an+1)//2
%o an, s = an1, s+an
%o print(list(islice(agen(), 10))) # _Michael S. Branicky_, Nov 14 2022
%Y Cf. A006894, A005588. First differences of A072638.
%K nonn,easy,core,nice
%O 0,3
%A _N. J. A. Sloane_
%E Corrected by _David Wasserman_, Nov 20 2006
|
