A327975 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.

5, 6, 9, 14, 33, 49, 62, 94, 177, 817, 961, 445, 913, 1633, 2173, 2209, 1146, 886, 1822, 4414, 19193, 25829, 32393, 41033, 47429, 57929, 64133, 88229, 101753, 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

Offset: 1

Author

Antti Karttunen, Oct 02 2019

Keywords

• nonn
• tabf

Comments

Permutation of A328115.

The branching degree of vertex v is given by A099302(v).

Leaves form a subsequence of A098700.

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.

For any number k at level n (where 5 is at level 2), we have A256750(k) = A327966(k) = n.

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.

Examples

			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:
   (0)
    |
   (1)
    |
    5
    |
    6
    |
    9
    |
    14________________
    |                 |
    33               49
    |                 |
    62________       94_____________________________
    |    |    |       |       |      |      |       |
    |    |    |       |       |      |      |       |
   177  817  961     445     913   1633   2173    2209
              |       |       |                     |
              |       |       |                     |
            1146     886    1822                  4414
              |       |       |                     |
              |       |       |                     |
            (19193,  (1155,  (19921, ..., 829921)  (22045, ..., 4870849)
             25829,   2649,                        [49 children for 4414]
               ...,  ...,    [27 children for 1822]
            328313)  196249)
                     [19 children for 886]
        [38 children
         for 1146]
The row lengths thus start as: 1, 1, 1, 1, 2, 2, 8, 4, 133 (= 38+19+27+49), ...
		

Links

• Victor Ufnarovski and Bo Ahlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003.

Crossrefs

Cf. A003415, A098699, A098700, A099302, A099303, A099307, A099308, A189760, A256750, A327966, A327968, A328115.

Cf. A327977 for the subtree starting from 7, and also A263267 for another similar tree.

Programs

PARI

A002620(n) = ((n^2)>>2);
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
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); };
v328975 = A327975list(21);
A327975(n) = v328975[n];

Sage

# uses[A003415]
def A327975():
  '''Breadth-first reading of irregular subtree rooted at 5, defined by the edge-relation A003415(child) = parent.'''
  yield 5
  for x in A327975():
    for k in [1 .. 1+(x*x)//2]:
      if A003415(k) == x: yield k
def take(n, g):
  '''Returns a list composed of the next n elements returned by generator g.'''
  z = []
  if 0 == n: return z
  for x in g:
    z.append(x)
    if n > 1: n = n-1
    else: return(z)
take(60, A327975())