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Revision History for A262522 (Bold, blue-underlined text is an addition; faded, red-underlined text is a deletion.)

Showing entries 1-10 | older changes
a(n)=0 if n is in A259934, otherwise the largest term in A045765 from which one can reach n by iterating A049820 zero or more times.
(history; published version)
#16 by N. J. A. Sloane at Sun Oct 04 13:12:21 EDT 2015
STATUS

proposed

approved

#15 by Antti Karttunen at Sun Oct 04 13:12:01 EDT 2015
STATUS

editing

proposed

#14 by Antti Karttunen at Sun Oct 04 08:22:29 EDT 2015
LINKS

Antti Karttunen, <a href="/A262522/b262522.txt">Table of n, a(n) for n = 0..17724</a>

#13 by Antti Karttunen at Sun Oct 04 07:49:46 EDT 2015
COMMENTS

If n is itself in A045765, we iterate 0 times, and then thus a(n) = n.

#12 by Antti Karttunen at Sun Oct 04 07:48:48 EDT 2015
NAME

a(n)=0 if n is in A259934, otherwise the largest term in A045765 from which one can reach n by iterating A049820. If n is itself in A045765, we iterate 0 zero or more times, and then a(n) = n.

COMMENTS

If n is itself in A045765, we iterate 0 times, and then a(n) = n.

#11 by Antti Karttunen at Sun Oct 04 07:46:18 EDT 2015
EXAMPLE

For n=1, its transitive closure (as defined by edge-relation A049820(child) = parent) is the union of {1} itself together with all its descendants: {1, 3, 4, 5, 7, 8}. We see that there are no other nodes in a subtree whose root is 1, because A049820(3) = 3 - d(3) = 1, A049820(4) = 1, A049820(5) = 3, A049820(7) = 5, A049820(8) = 4 and of these only 7 and 8 are terms of A045765. The largest term (which by necessity is always a term of A045765) is here 8, thus a(1) = 8. Note however that it is not always the largest leaf from which starts the longest path leading back to n. (In this case it is 7 instead 7, of 8, see the example in A262695).

#10 by Antti Karttunen at Sun Oct 04 07:45:09 EDT 2015
FORMULA

a(A262511(n)) = a(A262512(n)) = a(A082284(A262511(n))).

#9 by Antti Karttunen at Sun Oct 04 07:44:06 EDT 2015
FORMULA

Other identities. For all n >= 1:

For any n in A262511 but not in A259934, a(n) = a(A082284(n)).

a(A262511(n)) = a(A082284(A262511(n))).

#8 by Antti Karttunen at Sun Oct 04 07:41:26 EDT 2015
FORMULA

Other identities:

For any n in A262511 but not in A259934, a(n) = a(A082284(n)).

#7 by Antti Karttunen at Sun Oct 04 07:40:05 EDT 2015
EXAMPLE

For n=1, its transitive closure (as defined by edge-relation A049820(child) = parent) is the union of {1} itself together with all its descendants: {1, 3, 4, 5, 7, 8}. We see that there are no other nodes in a subtree whose root is 1, because A049820(3) = 3 - d(3) = 1, A049820(4) = 1, A049820(5) = 3, A049820(7) = 5, A049820(8) = 4 and of these only 7 and 8 are terms of A045765. The largest term (which by necessity is always a term of A045765) is here 8, thus a(1) = 8. Note however that it is not always the leaf from which starts the longest path leading back to n. In this case it is instead 7, see the example in A262695.