%I #49 Mar 26 2020 07:06:48
%S 5,6,9,14,33,49,62,94,177,817,961,445,913,1633,2173,2209,1146,886,
%T 1822,4414,19193,25829,32393,41033,47429,57929,64133,88229,101753,
%U 111173,116729,129413,138233,148553,160229,173093,183929,188453,208613,216773,232229,235913,244229,249929,257573,262793,272633,278153,282533,288329,294473,304613,316229,320933,322853,323429,327653,328313,1155,2649
%N Breadth-first reading of the subtree rooted at 5 of the tree where each parent node is the arithmetic derivative (A003415) of all its children.
%C Permutation of A328115.
%C The branching degree of vertex v is given by A099302(v).
%C Leaves form a subsequence of A098700.
%C Most terms of A189760 (apart from 0, 1, 2, 414, ...) seem to be located in this tree, in positions where they have no smaller siblings.
%C For any number k at level n (where 5 is at level 2), we have A256750(k) = A327966(k) = n.
%C Question: Does this subtree contain infinitely long paths? How many? Cf. conjecture number 8 in Ufnarovski and Ahlander paper, and a similar tree starting from 7, A327977.
%H Victor Ufnarovski and Bo Ahlander, <a href="http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Ufnarovski/ufnarovski.html">How to Differentiate a Number</a>, J. Integer Seqs., Vol. 6, 2003.
%e Because we have A003415(5) = 1, A003415(6) = 5, A003415(9) = 6, A003415(14) = 9, A003415(33) = A003415(49) = 14, A003415(62) = 33, etc, this subtree is laid out as below. The terms of this sequence are obtained by scanning each successive level of the tree from left to right, from the node 5 onward:
%e (0)
%e |
%e (1)
%e |
%e 5
%e |
%e 6
%e |
%e 9
%e |
%e 14________________
%e | |
%e 33 49
%e | |
%e 62________ 94_____________________________
%e | | | | | | | |
%e | | | | | | | |
%e 177 817 961 445 913 1633 2173 2209
%e | | | |
%e | | | |
%e 1146 886 1822 4414
%e | | | |
%e | | | |
%e (19193, (1155, (19921, ..., 829921) (22045, ..., 4870849)
%e 25829, 2649, [49 children for 4414]
%e ..., ..., [27 children for 1822]
%e 328313) 196249)
%e [19 children for 886]
%e [38 children
%e for 1146]
%e The row lengths thus start as: 1, 1, 1, 1, 2, 2, 8, 4, 133 (= 38+19+27+49), ...
%o (PARI)
%o A002620(n) = ((n^2)>>2);
%o A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
%o A327975list(e) = { my(lista=List([5]), f); for(n=1, e, f = lista[n]; for(k=1,1+A002620(f),if(A003415(k)==f, listput(lista,k)))); Vec(lista); };
%o v328975 = A327975list(21);
%o A327975(n) = v328975[n];
%o (Sage) # uses[A003415]
%o def A327975():
%o '''Breadth-first reading of irregular subtree rooted at 5, defined by the edge-relation A003415(child) = parent.'''
%o yield 5
%o for x in A327975():
%o for k in [1 .. 1+(x*x)//2]:
%o if A003415(k) == x: yield k
%o def take(n, g):
%o '''Returns a list composed of the next n elements returned by generator g.'''
%o z = []
%o if 0 == n: return z
%o for x in g:
%o z.append(x)
%o if n > 1: n = n-1
%o else: return(z)
%o take(60, A327975())
%Y Cf. A003415, A098699, A098700, A099302, A099303, A099307, A099308, A189760, A256750, A327966, A327968, A328115.
%Y Cf. A327977 for the subtree starting from 7, and also A263267 for another similar tree.
%K nonn,tabf
%O 1,1
%A _Antti Karttunen_, Oct 02 2019