A252448 -id:A252448 - cp's OEIS frontend

A249990 Start with the natural numbers, reverse the order in each pair, skip one pair, reverse the order in each triple, skip one triple, and so on.

2, 1, 6, 3, 4, 12, 5, 8, 7, 16, 9, 10, 13, 14, 22, 15, 18, 11, 24, 17, 32, 25, 26, 19, 20, 23, 30, 48, 31, 38, 21, 28, 27, 34, 33, 52, 35, 42, 29, 36, 39, 40, 49, 50, 58, 51, 54, 41, 44, 37, 60, 43, 66, 53, 84, 67, 68, 61, 62, 45, 46, 55, 56, 59, 76, 94, 77, 90

Offset: 1

Alex Ratushnyak, Nov 27 2014

nonn

A permutation of natural numbers generated by the following algorithm.

Start with the natural numbers. Reverse the order of numbers in each pair. Skip one pair. In the remainder (that is, "4, 3, 6, 5, 8, 7, 10, 9, 12, 11,...") reverse the order in each triple. Skip one triple. In the remainder (it starts with "7, 8, 5, 12, 9, 10") reverse the order in each tetrad. Skip one tetrad. And so on.

			Start with:
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...
After the first step:
2, 1, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14, 13, 16, 15, 18, ...
After the 2nd step:
2, 1, 6, 3, 4, 7, 8, 5, 12, 9, 10, 13, 14, 11, 18, 15, 16, ...
After the 3rd step:
2, 1, 6, 3, 4, 12, 5, 8, 7, 14, 13, 10, 9, 16, 15, 18, 11, ...

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Popular Computing (Calabasas, CA), Problems 195 and 196, Vol. 5 (No. 55, 1977), annotated and scanned copy of page PC55-4. See Problem 195.
Index entries for sequences that are permutations of the natural numbers

Cf. A000960, A026239, A249991.

Cf. A252448 (inverse), A252458 (fixed points).

a249990 n = a249990_list !! (n-1)
a249990_list = f 2 [1..] where
   f k xs = reverse ys ++ f (k + 1) (g zs) where
            g us = reverse vs ++ g ws where
                   (vs, ws) = splitAt k us
            (ys, zs) = splitAt k xs
-- Reinhard Zumkeller, Dec 17 2014

TOP = 100
a = list(range(TOP))
for step in range(2,TOP):
    numBlocks = (len(a)-1) // step
    if numBlocks==0:  break
    a = a[:(1+numBlocks*step)]
    for pos in range(1,len(a),step):
        a[pos:pos+step] = a[pos+step-1:pos-1:-1]
    for i in range(1, step+1):  print(str(a[i]), end=',')
    a[1:] = a[step+1:]

A135317 Let h(2n, 1) = 2n and h(2n, m) = h(2n, m-1) + 2 * d(2n, m) for m > 1, where d(2n, m) = (least multiple of m not less than h(2n, m-1)) - h(2n, m-1). Then d(2n, m) is eventually 2-periodic as a function of m, and a(n) is defined as d(2n, 2*m+1) for large m.

Original entry on oeis.org

0, 1, 2, 0, 1, 2, 3, 4, 0, 3, 4, 5, 1, 2, 3, 6, 0, 1, 4, 5, 6, 2, 5, 6, 7, 8, 0, 7, 8, 9, 3, 4, 5, 1, 2, 3, 6, 7, 10, 4, 7, 8, 0, 1, 2, 9, 10, 11, 5, 8, 9, 12, 6, 7, 10, 11, 12, 8, 9, 10, 0, 3, 4, 13, 1, 2, 5, 6, 11, 3, 4, 9, 14, 0, 1, 10, 11, 12, 2, 7, 8, 13, 5, 6, 9, 12, 13, 7, 10, 11, 14, 15, 16, 14

Offset: 0

Alex Abercrombie, Feb 15 2008

nonn

Sequence yielding an ordering of N*N derived from a family of recurrences.
For any integer k define h(k,1)=k and for m>1 define h(k,m)=h(k,m-1)+2*((-h(k,m-1)) mod m) where "r mod s" denotes least nonnegative residue of r modulo s [informally, h(k,m) is got by "reflecting" h(k,m-1) in the least multiple of m that is >=h(k,m-1)]. Then for fixed k>=0 there are integers c(k), b(k), m(k) such that for all m>m(k) we have h(k,2*m+1)-h(k,2*m)=2*c(k) and h(k,2*m+2)-h(k,2*m+1)=2*b(k). [In the terms of the function d defined in the Name, c(k) = d(k, 2*m+1) and b(k) = d(k, 2*m+2) for all m>m(k).]
For all n we have c(2*n+1)=c(2*n) and b(2*n+1)=1+b(2*n). Moreover b(2*n) is even for all n. The function n->(c(2*n),b(2*n)/2) is a bijection from the nonnegative integers N to N*N [as well as n->(b(n),c(n))]. It is "monotone" in the sense that n<=n' whenever c(2*n)<=c(2*n') and b(2*n)<=b(2*n').
This sequence is a(n) = c(2*n).
The results on which the definition is based are not yet proved, but they are plausible and overwhelmingly supported by numerical evidence.
For each fixed m, k->h(k,m) is a bijection Z->Z (this is easy!). However for k<0 the sequence h(k,m) does not have the pseudo-periodic property we have used in defining c(k) and b(k).
m(k) appears to be O(sqrt k).
From Andrey Zabolotskiy, Dec 28 2024: (Start)
See my Github link for the proof that the sequence is indeed well-defined.
That fact is equivalent to the quantity d(k,m) + d(k,m+1) eventually becoming constant. That constant value can be first reached when m is odd (case B) or even (case C).
On the plane (b(k), c(k)), the points from case B (resp. case C) fall in the region which is approximately the octant b(k) > c(k) (resp. c(k) > b(k)).
On the plane (x=n, y=a(n)), the graph of this sequence fills in a certain region in the plane. It's bounded from below by the line y=0 and from above, it seems, by the curve y = sqrt(Pi*x). That region, it seems, is further divided into two parts: the one below the curve y = sqrt((4/Pi)*x) contains points from case B, the one above it contains points from case C. The latter part looks more dense on the graph. (End)

			h(18,m) for m>=1 goes 18,18,18,22,28,32,38,42,48...so we can take m(18) = 2, then 2*c(18)=28-22=38-32=48-42=...=6, so a(9) = c(18) = 3 and similarly b(18) = 2.

Andrey Zabolotskiy, Table of n, a(n) for n = 0..20000
Jon Maiga, Computer-generated formulas for A135317, Sequence Machine.
Andrey Zabolotskiy, Abercrombie's sequence, Github (formalized Lean proof that this sequence is well-defined).

Cf. A319571, A319572, A319573, A252448.

a[n_] := Module[{h = 2 n, b, c = 0, m = 1},
   While[Ceiling[h/(m+1)] != Floor[h/m],
    m++; b = Mod[-h, m]; h += 2 b;
    m++; c = Mod[-h, m]; h += 2 c];
   c];
Table[a[n], {n, 0, 93}] (* Andrey Zabolotskiy, Apr 29 2023, Dec 25 2024 *)

def a(n):
    h, m, c = 2*n, 1, 0
    while (h+m)//(m+1) != h//m:
        m += 1; b = (-h) % m; h += 2*b
        m += 1; c = (-h) % m; h += 2*c
    return c
print([a(n) for n in range(30)]) # Andrey Zabolotskiy, Dec 25 2024

It appears that a(n) = A319573(A252448(2*n+1)) [discovered by Sequence Machine], moreover, c(n) = A319573(n') and b(n) = A319572(n') with n' = A252448(n), where b and c are described above. - Andrey Zabolotskiy, Apr 28 2023

Edited by Andrey Zabolotskiy, Apr 29 2023

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A252458 Fixed points of permutations A249990 and A252448.

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A249990 Start with the natural numbers, reverse the order in each pair, skip one pair, reverse the order in each triple, skip one triple, and so on.

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A135317 Let h(2n, 1) = 2n and h(2n, m) = h(2n, m-1) + 2 * d(2n, m) for m > 1, where d(2n, m) = (least multiple of m not less than h(2n, m-1)) - h(2n, m-1). Then d(2n, m) is eventually 2-periodic as a function of m, and a(n) is defined as d(2n, 2*m+1) for large m.

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cp's OEIS Frontend

A252458 Fixed points of permutations A249990 and A252448.

Views

Author

Keywords

Comments

Crossrefs

Programs

Haskell

A249990 Start with the natural numbers, reverse the order in each pair, skip one pair, reverse the order in each triple, skip one triple, and so on.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Python

A135317 Let h(2*n, 1) = 2*n and h(2*n, m) = h(2*n, m-1) + 2 * d(2*n, m) for m > 1, where d(2*n, m) = (least multiple of m not less than h(2*n, m-1)) - h(2*n, m-1). Then d(2*n, m) is eventually 2-periodic as a function of m, and a(n) is defined as d(2*n, 2*m+1) for large m.

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A135317 Let h(2n, 1) = 2n and h(2n, m) = h(2n, m-1) + 2 * d(2n, m) for m > 1, where d(2n, m) = (least multiple of m not less than h(2n, m-1)) - h(2n, m-1). Then d(2n, m) is eventually 2-periodic as a function of m, and a(n) is defined as d(2n, 2*m+1) for large m.