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A285525 The indices that mark the beginning of four consecutive equal terms in A285524. 2

%I #12 Feb 10 2020 05:26:19

%S 13,38,63,85,110,135,160,185,210,232,257,282,307,332,354,379,404,429,

%T 454,479,501,526,551,576,601,626,648,673,698,723,748,770,795,820,845,

%U 870,895,917,942,967,992,1017,1042,1064,1089,1114,1139,1164,1186,1211,1236,1261

%N The indices that mark the beginning of four consecutive equal terms in A285524.

%H Robert Davis, Sarah A. Nelson, T. Kyle Petersen, Bridget E. Tenner, <a href="https://arxiv.org/abs/1704.05494">The pinnacle set of a permutation</a>, arXiv:1704.05494 [math.CO], 2017.

%t b[n_] := b[n] = MaximalBy[Table[{d, d! (d+1)! 2^(n - 2d - 1) StirlingS2[n - d, d + 1]}, {d, 1, n/2}], Last][[All, 1]] // Min;

%t A285524 = Table[b[n], {n, 4, 1000}];

%t Position[If[Length[#] == 4, Join[{0}, Rest[#]], #]& /@ Split[A285524] // Flatten, 0] + 3 // Flatten (* _Jean-François Alcover_, Feb 23 2019 *)

%o (PARI) half(n) = if (n % 2, n\2, n/2 - 1);

%o a285524(n) = {v = vector(half(n), d, d!*(d + 1)!*(2^(n-2*d-1)*stirling(n-d, d+1, 2))); w = vecsort(v,,1); w[#w];}

%o lista(nn) = {v = vector(nn, n, a285524(n+3)); for (k=1, #v-4, if ((v[k] == v[k+1]) && (v[k] == v[k+2]) && (v[k] == v[k+3]), print1(k+3, ", ")););}

%Y Cf. A285524, A285526.

%K nonn

%O 1,1

%A _Michel Marcus_, Apr 20 2017

%E More terms from _Jinyuan Wang_, Feb 10 2020

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Last modified April 25 01:35 EDT 2024. Contains 371964 sequences. (Running on oeis4.)