This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a(7)=6 as the finite tree of 7 leaves (233, 234, 235, 238, 245, 250 and 251) and 6 internal vertices (244, 239, 232, 229, 228 and 224) rooted at 224 has the maximum depth of 5 (the path 224 -> 228 -> 232 -> 239 -> 244 -> 250/251).
Since A005187 begins 0 1 3 4 7 8 10 11 15 16 18 19 22 23 25 26 31... this sequence begins 2 5 6 9 12 13 14 17 20 21
a055938 n = a055938_list !! (n-1) a055938_list = concat $ zipWith (\u v -> [u+1..v-1]) a005187_list $ tail a005187_list -- Reinhard Zumkeller, Nov 07 2011
a[0] = 0; a[1] = 1; a[n_Integer] := a[Floor[n/2]] + n; b = {}; Do[ b = Append[b, a[n]], {n, 0, 105}]; c =Table[n, {n, 0, 200}]; Complement[c, b]
(* Second program: *)
t = Table[IntegerExponent[(2n)!, 2], {n, 0, 100}]; Complement[Range[t // Last], t] (* Jean-François Alcover, Nov 15 2016 *)
L=listcreate();for(n=1,1000,for(k=2*n-hammingweight(n)+1,2*n+1-hammingweight(n+1),listput(L,k)));Vec(L) \\ Ralf Stephan, Dec 27 2013
def a053644(n): return 0 if n==0 else 2**(len(bin(n)[2:]) - 1)
def a043545(n):
x=bin(n)[2:]
return int(max(x)) - int(min(x))
def a079559(n): return 1 if n==0 else a043545(n + 1)*a079559(n + 1 - a053644(n + 1))
print([n for n in range(1, 201) if a079559(n)==0]) # Indranil Ghosh, Jun 11 2017, after the comment by Reinhard Zumkeller
;; utilizing COMPLEMENT-macro from Antti Karttunen's IntSeq-library) (define A055938 (COMPLEMENT 1 A005187)) ;; Antti Karttunen, Aug 08 2015
a(1) = 2, as 2 is the smallest natural number such that x such that x=1+A000120(x) (as 2=1+A000120(2)=1+1). a(2) = 0, as there are no solutions for 2, because it belongs to A055938. a(11) = 14, as 14 is the smallest natural number x such that x=11+A000120(x) (as 14=11+A000120(14)=11+3).
a213723 = (* 2) . a213714 -- Reinhard Zumkeller, May 01 2015
(define (A213723 n) (A005843 (A213714 n)))
a213724 0 = 1 a213724 n = a213723 n + signum (a213723 n) -- Reinhard Zumkeller, May 01 2015
(define (A213724 n) (if (zero? n) 1 (let ((v (A213723 n))) (if (zero? v) v (+ 1 v)))))
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).
a(10)=2 because the only numbers in A055938 from which one can end to 10 by the process described in A071542/A179016 are 12 and 13 (see comment at A213717). Similarly, a(22)=3 as there are following three cases: 24 as 24-A000120(24) = 24-2 = 22, and also 28 & 29 as 28-A000120(28) = 28-3 = 25, and 29-A000120(29) = 29-4 = 25, and then 25-A000120(25) = 25-3 = 22.
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 this 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 (leaves). Starting iterating from 7 with A049820, we get 7 -> 5, 5 -> 3, 3 -> 1, and starting from 8 we get 8 -> 4, 4 -> 1, of which the former path is longer (3 steps), thus a(1) = 3+1 = 4.
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}. In this case the longest path is obtained by starting iterating from the largest of these: 79 -> 77 -> 73 -> 71 -> 69 -> 65 -> 61 -> 59 -> 57 -> 53 -> 51 -> 47 -> 45 -> 39 -> 35 -> 31 -> 29 -> 27 -> 23 -> 21 -> 17 -> 15 -> 11 -> 9, which is 23 steps long, thus a(9) = 23+1 = 24.
For 0, the nearest leaf is 2, as when we start from 2, and always subtract the binary weight, A000120, we have: 2 - A000120(2) = A011371(2) = 1, and A011371(1) = 0, thus it takes two steps to get to 0, and there are no other terms of A055938 from which it would take fewer steps), so a(0) = 2, and also a(1) = 1, because it's one step nearer to 2. a(2) = 0, because 2 is one of the terms of A055938. a(8) = 2, because 12, 13 and 14 are the three nearest leaves to 8, and A011371(12) = A011371(13) = 10, A011371(14) = 11, A011371(10) = A011371(11) = 8 (thus it takes two iterations of A011371 to reach 8 from any of those three leaves) and there are no leaves nearer. Please see also Paul Tek's illustration.
a257265 = length . us where
us n = if a079559 n == 0
then [] else () : zipWith (const $ const ())
(us $ a213723 n) (us $ a213724 n)
-- Reinhard Zumkeller, May 03 2015
;; Do a breadth-first search over the descendants, which are at each step of iteration sorted by their distance from the starting node. (define (A257265 n) (let loop ((descendants (list (cons 0 n)))) (let ((dist (caar descendants)) (node (cdar descendants))) (cond ((zero? (A079559 node)) dist) (else (loop (sort (append (list (cons (+ 1 dist) (A213724 node)) (cons (+ 1 dist) (A213723 node))) (cdr descendants)) (lambda (a b) (< (car a) (car b))))))))))
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