A368050 -id:A368050 - cp's OEIS frontend

A367634 Array read by descending antidiagonals, where row n=0 lists the natural numbers and where each new row n=1,2,... is found by taking the number n in the previous row, and "leaping" it over the next n terms to its right, keeping the other numbers fixed (see example).

1, 2, 2, 3, 1, 1, 4, 3, 3, 1, 5, 4, 2, 2, 1, 6, 5, 4, 4, 2, 1, 7, 6, 5, 5, 5, 2, 1, 8, 7, 6, 3, 3, 3, 2, 1, 9, 8, 7, 6, 6, 6, 3, 2, 1, 10, 9, 8, 7, 7, 7, 7, 3, 2, 1, 11, 10, 9, 8, 4, 4, 4, 4, 3, 2, 1, 12, 11, 10, 9, 8, 8, 8, 8, 4, 3, 2, 1, 13, 12, 11, 10, 9, 5, 5, 5, 5, 4, 3, 2, 1

Offset: 1

Wesley Ivan Hurt, Nov 25 2023

			The array begins:
  1,  2,  3,  4,  5,  6,  7,  8,  9,  10,  11,  12,  13,  ...
  2,  1,  3,  4,  5,  6,  7,  8,  9,  10,  11,  12,  13,  ...
  1,  3,  2,  4,  5,  6,  7,  8,  9,  10,  11,  12,  13,  ...
  1,  2,  4,  5,  3,  6,  7,  8,  9,  10,  11,  12,  13,  ...
  1,  2,  5,  3,  6,  7,  4,  8,  9,  10,  11,  12,  13,  ...
  1,  2,  3,  6,  7,  4,  8,  5,  9,  10,  11,  12,  13,  ...
  1,  2,  3,  7,  4,  8,  5,  9, 10,   6,  11,  12,  13,  ...
  1,  2,  3,  4,  8,  5,  9, 10,  6,  11,   7,  12,  13,  ...
  1,  2,  3,  4,  5,  9, 10,  6, 11,   7,  12,  13,   8,  ...
  1,  2,  3,  4,  5, 10,  6, 11,  7,  12,  13,   8,  14,  ...
  1,  2,  3,  4,  5,  6, 11,  7, 12,  13,   8,  14,  15,  ...
  1,  2,  3,  4,  5,  6,  7, 12, 13,   8,  14,  15,   9,  ...
  1,  2,  3,  4,  5,  6,  7, 13,  8,  14,  15,   9,  16,  ...
  ...

Cf. A000027, A368050.

A379739 Subdiagonal of the Hurt-Sada array.

Original entry on oeis.org

2, 3, 4, 3, 7, 8, 9, 6, 7, 13, 14, 9, 10, 18, 19, 12, 13, 23, 24, 25, 16, 28, 29, 30, 19, 20, 34, 35, 22, 23, 39, 40, 25, 26, 44, 45, 46, 29, 49, 50, 51, 32, 54, 55, 56, 35, 36, 60, 61, 38, 39, 65, 66, 67, 42, 70, 71, 72, 45, 75, 76, 77, 48, 49, 81, 82, 51

Offset: 1

Jeffrey Shallit, Jan 14 2025

nonn

The array is depicted in A368050.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Jeffrey Shallit, Automaton deciding whether a(n) >= n
Jeffrey Shallit, The Hurt-Sada Array and Zeckendorf Representations, arXiv:2501.08823 [math.NT], 2025. See p. 2.

Cf. A368050, A380079.

zeckendorf(n)=my(f1=1,f2=2,fibs=List([1]),rep=List(),i); while(f2<=n,listput(fibs,f2);[f1,f2]=[f2,f1+f2];); i=#fibs; while(i,if(fibs[i]>n, i--, listput(rep,1); n-=fibs[i]; if(i==1,break); i-=2); listput(rep,0)); Vec(rep)
a(n)=my(g=quadgen(5), Z=zeckendorf(n), state=1, new0=[1, 3, 4, 6, 3, 8, 6, 8], new1=[2, 0, 5, 7, 0, 5, 0, 2]); for(i=1, #Z, state=if(Z[i], new1, new0)[state]); if(state!=5 && state!=8, 2*n/g+1, (4-2*g)*n + (5-3*g))\1 \\ Charles R Greathouse IV, Jan 14 2025

If a(n) < n, then a(n) = floor((4-2*g)n + (5-3*g)), where g = (1+sqrt(5))/2, the golden ratio. If a(n) >= n, then a(n) = floor((2*g-2)n) + 1. There is an 8-state automaton (see the links section) that takes the Zeckendorf representation of n as input and decides whether a(n) >= 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).

Original entry on oeis.org

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, 17, 18, 31, 19, 20, 34, 21, 36, 22, 23, 39, 24, 41, 25, 26, 44, 27, 28, 47, 29, 49, 30, 31, 52, 32, 54, 33, 34, 57, 35, 36, 60, 37, 62, 38, 39, 65, 40, 41, 68, 42, 70, 43, 44, 73, 45, 75

Offset: 1

Ali Sada and David Nacin, Jan 11 2025

nonn

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
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).
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.
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.
The last appearance of k also seems to be A022342.
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

			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.

M. F. Hasler, Table of n, a(n) for n = 1..500, May 08 2025
Jeffrey Shallit, The Hurt-Sada Array and Zeckendorf Representations, arXiv:2501.08823 [math.NT], 2025. See p. 2.

Cf. A003622, A368050.

M = list()
def moves(n, L):
    if n == 0:
        return L
    return move(n, moves(n-1,L))
def move(n, L):
    pos = L.index(n)
    M.append(L[pos+1])
    return L[:pos]+L[pos+1:pos+n+1]+[n]+L[pos+n+1:]
def seq(n):
    return moves(n, list(range(1,5*n)))
seq(74)
print(M) # David Nacin via Seqfan, Jan 10 2025

def aupton(nn):
    L, out = list(range(1, 2*nn+2)), []
    for n in range(1, nn+1):
        m = L.index(n)
        out.append(L[m+1])
        L.insert(m+n+1, n)
        L.pop(m)
    return out
print(aupton(74)) # Michael S. Branicky, Jan 13 2025

From Jeffrey Shallit, Jan 13 2025: (Start)
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.
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)

A380960 Sum of n and the n-th bit of the infinite Fibonacci word.

Original entry on oeis.org

0, 2, 2, 3, 5, 5, 7, 7, 8, 10, 10, 11, 13, 13, 15, 15, 16, 18, 18, 20, 20, 21, 23, 23, 24, 26, 26, 28, 28, 29, 31, 31, 32, 34, 34, 36, 36, 37, 39, 39, 41, 41, 42, 44, 44, 45, 47, 47, 49, 49, 50, 52, 52, 54, 54, 55, 57, 57, 58, 60, 60, 62, 62, 63, 65, 65, 66

Offset: 0

Jeffrey Shallit, Feb 09 2025

nonn

Also the length of the "scrambled part" of the n-th row of the Hurt-Sada array (depicted in A368050). This is the portion of the n-th row whose i-th term differs from i. See Shallit (2025), Theorem 10.

J. Shallit, The Hurt-Sada array and Zeckendorf representations, arXiv:2501.08823 [math.NT], January 15 2025.

Cf. A003849, A368050.

a(n) = A003849(n) + n.

cp's OEIS Frontend

A367634 Array read by descending antidiagonals, where row n=0 lists the natural numbers and where each new row n=1,2,... is found by taking the number n in the previous row, and "leaping" it over the next n terms to its right, keeping the other numbers fixed (see example).

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A379739 Subdiagonal of the Hurt-Sada array.

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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).

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Python

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A380960 Sum of n and the n-th bit of the infinite Fibonacci word.

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Formula