A218548 Positions of records in A213726.
0, 2, 10, 22, 47, 208, 224, 974, 1934, 1972, 3765, 3893, 8058, 15738, 16122, 65080, 129592, 130104
Offset: 1
Examples
See explanation at A218549.
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See explanation at A218549.
A218548(7)=224, A213726(224)=7 thus a(7)=7 as there are the following seven leaves of beanstalk (i.e. terms of A055938): 233, 234, 235, 238, 245, 250 and 251 that will pass through 224, when an iterative process of x-A000120(x) (see A071542) is applied to them several times.
a(10) = 3 because we include 10 itself ("1010" in binary) and the two numbers n for which it is true that n - A000120(n) = 10, i.e., 12 and 13 ("1100" and "1101" in binary). Furthermore, there do not exist any such numbers for 12 or 13, as both are members of A055938 (see also the comment at A213717).
Similarly, a(22) = 5 as there are the following five cases: 22 itself, 24 as 24-A000120(24) = 24-2 = 22 (note that 24 is in A055938), 25 as 25-A000120(25) = 25-3 = 22, and the two terminal nodes (leaves) branching from 25, that is, 28 & 29 (as 28-A000120(28) = 28-3 = 25, and 29-A000120(29) = 29-4=25).
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. Thus a(1) = 2.
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}, of which only thirteen members: {13, 19, 24, 33, 36, 37, 43, 55, 63, 64, 67, 75, 79} are leaves (in A045765), thus a(9) = 13.
The first four tendrils of the beanstalk sprout at 2, 5, 6 and 9, (the first four nonzero terms of A213730) which are all leaves (i.e., in A055938), thus the first four terms of this sequence are all 0's. The next term A213730(5)=10, which is not leaf, but branches to two leaf-branches (12 and 13, as with both we have: 12-A000120(12)=10 and 13-A000120(13)=10, and both 12 and 13 are found from A055938, so the tendril at 10 is a binary tree of one internal vertex (and two leaves), i.e., \/, thus a(5)=1.
From 11 sprouts the following finite side-tree of "factorial beanstalk":
18 19
\ /
14 15
\ /
11
Its leaves are the numbers 14, 18 and 19 (which all occur in A219658), whose factorial base representations (see A007623) are '210', '300' and '301' respectively. The corresponding parent nodes are obtained by subtracting the sum of factorial base digits, thus we get 18-3 = 15 and also 19-4 = 15, thus 15 ('211' in factorial base) is the parent of 18 and 19. For 14 and 15 we get 14-3 = 15-4 = 11, thus 11 is the parent of both 14 and 15, and the common ancestor of all numbers 11, 14, 15, 18 and 19.
For numbers not occurring in A219666 this sequence gives the number of leaves in such subtrees. Thus a(11)=3, a(14)=1 (counting just the leaf 14 itself), a(15)=2 and a(18) = a(19) = 1.
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