P. Michael Hutchins - cp's OEIS frontend

User: P. Michael Hutchins

P. Michael Hutchins has authored 3 sequences.

A328147 a(n) = A025586(n)/4 for n>=3.

4, 1, 4, 4, 13, 2, 13, 4, 13, 4, 10, 13, 40, 4, 13, 13, 22, 5, 16, 13, 40, 6, 22, 10, 2308, 13, 22, 40, 2308, 8, 25, 13, 40, 13, 28, 22, 76, 10, 2308, 16, 49, 13, 34, 40, 2308, 12, 37, 22, 58, 13, 40, 2308, 2308, 14, 49, 22, 76, 40, 46, 2308, 2308, 16, 49, 25, 76, 17, 52, 40, 2308, 18, 2308, 28, 85, 22, 58, 76, 202, 20, 61, 2308, 2308, 21, 64, 49, 148, 22, 76, 34, 2308, 40, 70, 2308, 2308, 24, 2308, 37, 112

Offset: 3

P. Michael Hutchins, Oct 22 2019

nonn

This sequence factors out the 4 that all of the terms of A025586 for n>2 are divisible by.

			For n=3, the Collatz sequence is 3,10,5,16,8,4,2,1. The largest term is 16, so a(3) = 16/4 = 4.

Cf. A025586.

def a(n):
    if n<3: return 0
    l=[n, ]
    while True:
        if n%2==0: n/=2
        else: n = 3*n + 1
        if not n in l:
            l+=[n, ]
            if n<2: break
        else: break
    return max(l)/4

a(1)-a(2) removed from data by Michel Marcus, Nov 02 2020

A305151 a(n) = (2n+1) - A294673(n), the amount by which A294673 is less than the maximum possible for n.

Original entry on oeis.org

0, 0, 0, 3, 0, 0, 4, 10, 5, 7, 14, 0, 17, 7, 0, 25, 0, 0, 17, 0, 0, 15, 33, 11, 34, 0, 0, 19, 13, 35, 41, 56, 0, 31, 0, 11, 31, 60, 57, 27, 0, 0, 76, 27, 0, 31, 53, 0, 85, 0, 17, 37, 0, 79, 91, 74, 0, 85, 25, 0, 40, 87, 100, 119, 93, 0, 111, 0, 117, 109, 94, 83, 97, 31, 17, 51, 102, 0

Offset: 0

P. Michael Hutchins, May 26 2018

nonn

Motivation: A294673(n), which is always <= 2n+1, equals n in most cases. This sequence abstracts out that commonality to leave a cleaner signal.

Cf. A294673.

a(n) = (2*n+1) - znorder(Mod(if(n%2, 2, -2), 4*n+3)); \\ Michel Marcus, May 26 2018

A294673 Order of the "inside-out" permutation on 2n+1 letters.

Original entry on oeis.org

1, 3, 5, 4, 9, 11, 9, 5, 12, 12, 7, 23, 8, 20, 29, 6, 33, 35, 20, 39, 41, 28, 12, 36, 15, 51, 53, 36, 44, 24, 20, 7, 65, 36, 69, 60, 42, 15, 20, 52, 81, 83, 9, 60, 89, 60, 40, 95, 12, 99, 84, 66, 105, 28, 18, 37, 113, 30, 92, 119, 81, 36, 25, 8, 36, 131, 22, 135, 20, 30, 47, 60, 48, 116, 132, 100, 51, 155

Offset: 0

P. Michael Hutchins, Nov 06 2017

The "inside-out" permutation (closely related to the Mongean shuffle, see A019567) sends (t_1, t_2, ..., t_{2n+1}) to (t_{n+1}, t_{n+2}, t_{n}, t_{n+3}, t_{n-1}, ..., t_1). For n = 0, 1, 2, 3, this is (1), (2,3,1), (3,4,2,5,1), (4,5,3,6,2,7,1), whose orders are respectively 1,3,5,4.

This is the odd bisection of A238371 and also the odd bisection of A003558 (see Joseph L. Wetherell's comment below).

			For n=2: Iterating the "inside-out" permutation of a string of length 2n+1=5:
12345
34251
25413
41532
53124
12345
...
which has order a(2) = 5.

Robert Israel, Table of n, a(n) for n = 0..10000
N. J. A. Sloane, Table of n, a(n) for n = 0..32683 (computed using Robert Israel's Maple program)

Cf. A003558, A019567, A163778, A238371, A294434.

f:=func;
[f(n): n in [0..100]]; // Joseph L. Wetherell, Nov 12 2017

[Order(Integers(4*n+3)!-2*(-1)^n): n in [0..100]]; // Joseph L. Wetherell, Nov 15 2017

f:= proc(n)
  ilcm(op(map(nops,convert(map(op, [[n+1],seq([n+1+i,n+1-i],i=1..n)]),disjcyc))))
end proc:
map(f, [$0..100]); # Robert Israel, Nov 09 2017

a[n_] := MultiplicativeOrder[-2(-1)^n, 4n+3];
a /@ Range[0, 100] (* Jean-François Alcover, Apr 07 2020 *)

Follow(s,f)={my(t=f(s),k=1); while(t>s, k++; t=f(t)); if(s==t, k, 0)}
CyclePoly(n,x)={my(p=0); for(i=1, 2*n+1, my(l=Follow(i, j->n+1+(-1)^j*ceil((j-1)/2) )); if(l,p+=x^l)); p}
a(n)={my(p=CyclePoly(n,x), m=1); for(i=1,poldegree(p),if(polcoeff(p,i), m=lcm(m,i))); m} \\ Andrew Howroyd, Nov 08 2017

a(n)=znorder(Mod(if(n%2,2,-2),4*n+3)) \\ See Wetherell formula; Charles R Greathouse IV, Nov 15 2017

The permutation sends i (1 <= i <= 2n+1) to p(i) = n + 1 + f(i), where f(i) = (-1)^i*ceiling((i-1)/2).
a(n) = minimal k>0 such that p^k() = p^0().
a((A163778(n)-1)/2) = A163778(n). - Andrew Howroyd, Nov 11 2017.
From Joseph L. Wetherell, Nov 14 2017: (Start)
a(n) is equal to the order of multiplication-by-2 acting on the set of nonzero elements in (Z/(4n+3)Z), modulo the action of +-1. To be precise, identify i=1,2,...,2*n+1 with the odd representatives J=1,3,...,4*n+1 of this set, via the map J = 2*i-1. It is not hard to show that the induced permutation on the set of J values is given on integer representatives by J -> (4*n+3+J)/2 if i=(J+1)/2 is even and J -> (4*n+3-J)/2 if i=(J+1)/2 is odd. It follows that this induces the permutation J -> +-J/2 (mod 4*n+3), from which we immediately see that the order is as stated.
Note that the order of 2 acting on (Z/(4n+3)Z)/{+-1} is the same as the order of either 2 or -2 acting on (Z/(4n+3)Z), depending on which of these is a quadratic residue modulo 4n+3. Thus an equivalent (and often easier) way to compute a(n) is as the order of -2*(-1)^n acting on (Z/(4n+3)Z).
Among other things, the lower and upper bounds log_2(n) + 2 < a(n) <= 2*n+1 follow immediately.
(End)
It appears that the upper bound a(n) = 2n+1 occurs iff 2n+1 belongs to A163778 or equivalently iff n belongs to A294434. This almost (but not quite) follows from the above comments by Andrew Howroyd and Joseph L. Wetherell. - N. J. A. Sloane, Nov 16 2017

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User: P. Michael Hutchins

A328147 a(n) = A025586(n)/4 for n>=3.

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A305151 a(n) = (2n+1) - A294673(n), the amount by which A294673 is less than the maximum possible for n.

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A294673 Order of the "inside-out" permutation on 2n+1 letters.

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