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A380079 Start with a list of the positive integers L in increasing order. Then, at turn n>=1, element n jumps from its current position, m, to position m+n. Then a(n) = L(m+1).

%I A380079 #57 May 09 2025 07:19:25
%S A380079 2,1,2,5,3,7,4,5,10,6,7,13,8,15,9,10,18,11,20,12,13,23,14,15,26,16,28,
%T A380079 17,18,31,19,20,34,21,36,22,23,39,24,41,25,26,44,27,28,47,29,49,30,31,
%U A380079 52,32,54,33,34,57,35,36,60,37,62,38,39,65,40,41,68,42,70,43,44,73,45,75
%N A380079 Start with a list of the positive integers L in increasing order. Then, at turn n>=1, element n jumps from its current position, m, to position m+n. Then a(n) = L(m+1).
%C A380079 It appears that a(n)=n+1 exactly when n belongs to A003622.  For all other positions, fill in the numbers 1,2,3,4, ... in increasing order. A proof should be possible using the theorem-prover Walnut, and is underway. - _Jeffrey Shallit_, Jan 12 2025
%C A380079 After infinitely many moves, the natural numbers are transformed back into their original order. This is because all numbers larger than n must jump over n, as they have a larger jump by definition. Therefore, n must return to its original place in the list. It appears that n returns to its place at A022342(n-1).
%C A380079 There seems to be an infinite number of ordered pairs of the form (k, k+1) such as (1,2), (4,5), (6,7), (9,10), (12,13), .. etc., and the first term in each pair is A003622.
%C A380079 Also, it seems that if we take the last number in the jump (i.e. L(m+n-1), we get natural numbers excluding the terms of A003622, which seems to be A022342.
%C A380079 The last appearance of k also seems to be A022342.
%C A380079 The graph has two "rays", one with slope ~ 1 on which lies about every third term, and one with slope 0.618 on which lie the other terms. - _M. F. Hasler_, May 09 2025
%H A380079 M. F. Hasler, <a href="/A380079/b380079.txt">Table of n, a(n) for n = 1..500</a>, May 08 2025
%H A380079 Jeffrey Shallit, <a href="https://arxiv.org/abs/2501.08823">The Hurt-Sada Array and Zeckendorf Representations</a>, arXiv:2501.08823 [math.NT], 2025. See p. 2.
%F A380079 From _Jeffrey Shallit_, Jan 13 2025: (Start)
%F A380079 If n belongs to A003622, then a(n) = n+1.  Otherwise, a(A022342(n)) = n-1 for n >= 1. This can be proved with the Walnut theorem prover.
%F A380079 Alternatively, let g = (1+sqrt(5))/2, and let p be the fractional part of (n+1)*g. If p < 2-g, then a(n) = n+1. If p > 2-g, then a(n) = floor((n+1)/g). (End)
%e A380079 We start with 1,2,3,4,5,6,7,8,9,10... First, 1 jumps over 2, so, a(1) = 2. In the second turn, 2 jumps over 1 and 3, so, a(2) =1. In the third turn, 3 jumps over 2,4, and 5, so a(3) = 2. And so on.
%o A380079 (Python)
%o A380079 M = list()
%o A380079 def moves(n, L):
%o A380079     if n == 0:
%o A380079         return L
%o A380079     return move(n, moves(n-1,L))
%o A380079 def move(n, L):
%o A380079     pos = L.index(n)
%o A380079     M.append(L[pos+1])
%o A380079     return L[:pos]+L[pos+1:pos+n+1]+[n]+L[pos+n+1:]
%o A380079 def seq(n):
%o A380079     return moves(n, list(range(1,5*n)))
%o A380079 seq(74)
%o A380079 print(M) # _David Nacin_ via Seqfan, Jan 10 2025
%o A380079 (Python)
%o A380079 def aupton(nn):
%o A380079     L, out = list(range(1, 2*nn+2)), []
%o A380079     for n in range(1, nn+1):
%o A380079         m = L.index(n)
%o A380079         out.append(L[m+1])
%o A380079         L.insert(m+n+1, n)
%o A380079         L.pop(m)
%o A380079     return out
%o A380079 print(aupton(74)) # _Michael S. Branicky_, Jan 13 2025
%Y A380079 Cf. A003622, A368050.
%K A380079 nonn
%O A380079 1,1
%A A380079 _Ali Sada_ and _David Nacin_, Jan 11 2025