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A327977
Breadth-first reading of the subtree rooted at 7 of the tree where each parent node is the arithmetic derivative (A003415) of all its children.
7
7, 10, 21, 25, 18, 38, 46, 65, 77, 217, 361, 129, 205, 493, 529, 98, 426, 718, 170, 254, 462, 982, 1501, 2077, 2257, 2105, 2933, 6953, 11513, 14393, 16469, 17813, 19769, 21653, 24053, 25769, 27413, 29993, 34553, 35369, 41273, 42233, 42869, 44969, 45113, 45173, 11917, 27757, 38881, 45937, 62317, 76897, 84781, 102637, 111457, 114481, 117217, 118477, 120781, 127117, 128881, 501, 1141
OFFSET
1,1
COMMENTS
Permutation of A328117.
The branching degree of vertex v is given by A099302(v).
Leaves form a subsequence of A098700.
For any number k at level n (where 7 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. As an example of possible beginning of such a sequence they give: 1 ← 7 ← 10 ← 25 ← 46 ← 129 ← 170 ← 501 ← 414 ← 2045.
LINKS
Victor Ufnarovski and Bo Ahlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003.
EXAMPLE
The 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 7 onward:
(0)
|
(1)
|
7
|
10______________________________
| |
21________ 25
| | |
18___ 38_____ 46_________________________________
| | | | | | | |
65 77 217 361____ 129____ 205 493_____ 529
| | | | | | |
98 426 718 170 254 462 982
| | | | | | |
[3] [21] [15] [9] [9] [28] [17]
On the last level illustrated above, the numbers in brackets [ ] tell how many children the node has. E.g, there are three for 98: 1501, 2077, 2257, as A003415(1501) = A003415(2077) = A003415(2257) = 98, and nine for 170: 501, 1141, 2041, 2869, 4309, 5461, 6649, 6901, 7081.
PROG
(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
A327977list(e) = { my(lista=List([7]), f); for(i=1, e, f = lista[i]; for(k=1, 1+A002620(f), if(A003415(k)==f, listput(lista, k)))); Vec(lista); };
(PARI)
\\ With precomputed large A328117, use this:
v328117 = readvec("a328117.txt");
A327977list(e) = { my(lista=List([7]), f, i); for(n=1, e, f = lista[n]; print("n=", n, " #lista=", #lista, " A002620(", f, ")=", A002620(f)); my(u=1+A002620(f)); if(u>=v328117[#v328117], print("Not enough precomputed terms of A328117 as search upper limit ", u, " > ", v328117[#v328117], " (the last item in v328117). Number of expansions so far=", n); return(1/0)); i=1; while(v328117[i]<u, if(A003415(v328117[i])==f, listput(lista, v328117[i])); i++)); Vec(lista); };
v327977 = A327977list(114);
A327977(n) = v327977[n];
for(n=1, #v327977, write("b327977.txt", n, " ", A327977(n)));
(Sage) # uses[A003415]
def A327977():
'''Breadth-first reading of irregular subtree rooted at 7, defined by the edge-relation A003415(child) = parent. Starts giving terms from 7 onward, after a(0) = 0 and a(1) = 1.'''
yield 7
for x in A327977():
for k in [1 .. 1+floor((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(52, A327977())
CROSSREFS
Cf. A327975 for the subtree starting from 5, and also A263267 for another similar tree.
Sequence in context: A249942 A300021 A363292 * A097634 A120312 A074377
KEYWORD
nonn,tabf
AUTHOR
Antti Karttunen, Oct 02 2019
STATUS
approved