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A285333
a(n) = A048675(A285332(n)).
7
0, 1, 2, 2, 3, 4, 4, 3, 6, 4, 9, 6, 5, 8, 8, 4, 15, 8, 12, 5, 14, 10, 27, 8, 10, 6, 25, 12, 7, 16, 16, 5, 210, 16, 45, 10, 35, 16, 18, 5, 105, 16, 28, 11, 462, 28, 81, 10, 21, 12, 20, 7, 154, 26, 125, 16, 30, 8, 49, 24, 11, 32, 32, 6, 10659, 212, 420, 17, 910, 46, 75, 10, 78, 36, 175, 20, 33, 20, 24, 6, 3094, 106, 315, 18, 385, 32, 56, 17, 780045
OFFSET
0,3
COMMENTS
Following A285332, also this sequence can be represented in a form of a binary tree:
0
|
...................1...................
2 2
3......../ \........4 4......../ \........3
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
6 4 9 6 5 8 8 4
15 8 12 5 14 10 27 8 10 6 25 12 7 16 16 5
etc.
LINKS
FORMULA
a(n) = A048675(A285332(n)).
For all n >= 1, a(2n) = A285332(n).
a(2^n) = A109162(1+n). [The left edge of the tree.]
a(A000225(n)) = n. [The right edge of tree.]
PROG
(PARI)
A019565(n) = {my(j, v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
A048675(n) = my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; \\ Michel Marcus, Oct 10 2016
A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
A065642(n) = { my(r=A007947(n)); if(1==n, n, n = n+r; while(A007947(n) <> r, n = n+r); n); };
A285332(n) = { if(n<=1, n+1, if(!(n%2), A019565(A285332(n/2)), A065642(A285332((n-1)/2)))); };
A285333(n) = if(!n, n, if(!(n%2), A285332(n/2), A048675(A285332(n))));
(Scheme) (define (A285333 n) (A048675 (A285332 n)))
CROSSREFS
Cf. A001477, A048675, A109162, A285325, A285330, A285332 (even bisection).
Sequence in context: A085430 A246794 A334030 * A086416 A168148 A147968
KEYWORD
nonn,tabf
AUTHOR
Antti Karttunen, Apr 19 2017
STATUS
approved