A285356 Numbers n such that the entries in the n-th row of the irregular triangle A237591 are in nonincreasing order.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 23, 24, 26, 28, 29, 30, 31, 32, 36, 37, 38, 40, 41, 45, 46, 47, 48, 51, 55, 57, 58, 59, 66, 67, 70, 71, 78, 79, 80, 84, 92, 93, 94, 108, 109, 120, 136, 137, 155

Offset

1,2

Comments

For the numbers n in the sequence the lengths of the steps in the (first half of the) associated Dyck path of A237593 are nonincreasing.

Conjectures:

(1) The sequence consists of the 59 numbers listed above; tested through 5000000.

(2) The values f(n,k) in the n-th row of triangle A237591 are either 1 or 2 for all k with ceiling((sqrt(4*n+1)-1)/2) <= k <= floor((sqrt(8*n+1)-1)/2) = r(n), the length of the n-th row, though the lower bound need not be minimal; tested through 2500000.

(3) For every n > 155 there is an inversion 1 = f(n,k-1) < f(n,k) = 2 where k >= ceiling((sqrt(4*n+1)-1)/2, except the inversions for n = 174 at k = 12 and for n = 231 at k = 14; tested through 2500000.

(4) For all n > 231 = A066370(2), the position of the rightmost inversion in the n-th row is given by the formula r(n) - r( Binomial( r(n) + 2, 2) - 1 - n); tested through 2500000. Expressed in terms of A-numbers the formula is: A003056(n) - A003056(A000217(A003056(n) + 1) - 1 - n).

Links

Table of n, a(n) for n=1..59.

Hartmut F. W. Hoft, Example plot and justification for the formula in conjecture 4

Example

19 is in the sequence since row 19 in A237591 is 10, 4, 2, 2, 1.
20 is not in the sequence since row 20 in A237591 is 11, 4, 2, 1, 2.

Mathematica

(* functions row[] and f[] are defined in A237591 *)
nonincreasingQ[n_] := Module[{i=2, b=row[n], good=True}, While[good && i<=b, good=good && (f[n, i]<=f[n, i-1]); i++]; good]
a285356[m_, n_] := Module[{i, sols={}}, For[i=m, i<=n, i++, If[nonincreasingQ[i], AppendTo[sols, i]]]; sols]
a285356[1, 200] (* data *)

Prog

(Python)
import math
from sympy import sqrt
def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2))
def isok(n):
    l = [T(n, k) for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]
    for i in range(len(l) - 1):
        if l[i + 1] > l[i]: return 0
    return 1
print([n for n in range(1, 156) if isok(n)]) # Indranil Ghosh, Apr 20 2017

Crossrefs

Cf. A000217, A003056, A064322, A066370, A237591, A237593, A285426 (complement).

Sequence in context: A236866 A122526 A120401 * A127034 A348960 A367613

Adjacent sequences: A285353 A285354 A285355 * A285357 A285358 A285359

Keyword

nonn

Author

Hartmut F. W. Hoft, Apr 17 2017

Status

approved