login
A262522
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.
8
0, 8, 0, 7, 8, 7, 0, 7, 8, 79, 20, 79, 0, 13, 20, 79, 24, 79, 0, 19, 20, 79, 0, 79, 24, 25, 40, 79, 28, 79, 0, 79, 40, 33, 0, 79, 36, 37, 140, 79, 40, 43, 0, 43, 50, 79, 0, 79, 140, 49, 50, 79, 52, 79, 0, 55, 56, 79, 0, 79, 140, 79, 0, 63, 64, 79, 66, 67, 68, 79, 0, 79, 140, 79, 74, 75, 123, 79, 0, 79, 88, 123, 98, 123, 140, 85, 98, 123, 88, 103, 0, 123, 98, 103, 0, 123
OFFSET
0,2
COMMENTS
If n is itself in A045765, we iterate 0 times, and thus a(n) = n.
LINKS
FORMULA
If A262693(n) = 1 [when n is in A259934],
then a(n) = 0,
otherwise, if A060990(n) = 0 [when n is one of the leaves, A045765],
then a(n) = n,
otherwise:
a(n) = Max_{k = A082284(n) .. A262686(n)} [A049820(k) = n] * a(k).
(In the last clause [ ] stands for Iverson bracket, giving as its result 1 only when A049820(k) = n, and 0 otherwise).
Other identities. For all n >= 1:
a(A262511(n)) = a(A262512(n)) = a(A082284(A262511(n))).
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 of 8, see the example in A262695).
For n=9, its transitive closure is {9, 11, 13, 15, 16, 17, 19, 21, 23, 24, 27, 29, 31, 33, 35, 36, 37, 39, 41, 43, 45, 47, 51, 53, 55, 57, 59, 61, 63, 64, 65, 67, 69, 71, 73, 75, 77, 79}. The largest term is 79, thus a(9) = 79.
PROG
(Scheme, with memoization-macro definec)
(definec (A262522 n) (cond ((= 1 (A262693 n)) 0) ((= 0 (A060990 n)) n) (else (let loop ((m 0) (k (A262686 n))) (cond ((<= k n) m) ((= n (A049820 k)) (loop (max m (A262522 k)) (- k 1))) (else (loop m (- k 1))))))))
KEYWORD
nonn
AUTHOR
Antti Karttunen, Oct 04 2015
STATUS
approved