A285356 - OEIS

%I #18 May 07 2021 08:09:31

%S 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,

%T 31,32,36,37,38,40,41,45,46,47,48,51,55,57,58,59,66,67,70,71,78,79,80,

%U 84,92,93,94,108,109,120,136,137,155

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

%C 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.

%C Conjectures:

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

%C (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.

%C (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.

%C (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).

%H Hartmut F. W. Hoft, <a href="/A285356/a285356.pdf">Example plot and justification for the formula in conjecture 4</a>

%e 19 is in the sequence since row 19 in A237591 is 10, 4, 2, 2, 1.

%e 20 is not in the sequence since row 20 in A237591 is 11, 4, 2, 1, 2.

%t (* functions row[] and f[] are defined in A237591 *)

%t 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]

%t a285356[m_, n_] := Module[{i, sols={}}, For[i=m, i<=n, i++, If[nonincreasingQ[i], AppendTo[sols, i]]]; sols]

%t a285356[1,200] (* data *)

%o (Python)

%o import math

%o from sympy import sqrt

%o def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2))

%o def isok(n):

%o     l = [T(n, k) for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]

%o     for i in range(len(l) - 1):

%o         if l[i + 1] > l[i]: return 0

%o     return 1

%o print([n for n in range(1, 156) if isok(n)]) # _Indranil Ghosh_, Apr 20 2017

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

%K nonn

%O 1,2

%A _Hartmut F. W. Hoft_, Apr 17 2017