A358677 - OEIS

%I #44 Jan 29 2023 05:05:16

%S 1,16,18,18,21,21,17,17,19,20,22,265549,265604,265605,265608,265681,

%T 265683,265829,265831,265831,265835,265836,265850,265850,265853,

%U 265853,265862,265873,265550,265603,265606,265607,265682,265682,265830,265830,265832,265834,265837,265849,265851,265852,265854,265861

%N Irregular triangle where row n gives the columns of A340316 whose minimum value is in row n of A340316. The lists of column indices are given in abbreviated form, using pairs (x, y) to mean the range [x..y].

%C This sequence is a spin-off from old comments of A340316 (see history there).

%C Pending availability of tighter constraints, we assume that there are no more values in row n here only after we reach a column of A340316 where the value in A340316 row n is greater than the value in A340316 row n+2.

%C Presumably, using the results from Landau as they apply to A276176, it can similarly be shown that every row here is finite. - _Peter Munn_, Dec 20 2022

%e First 2 rows are:

%e  {1..16, 18..18, 21..21} for [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,18,21];

%e  {17..17, 19..20, 22..265549, 265604..265605, 265608..265681, 265683..265829, 265831..265831, 265835..265836, 265850..265850, 265853..265853, 265862..265873}.

%e The A340316 first 2 rows being:

%e    1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18 19 20 21 22

%e   -----------------------------------------------------------------

%e    2  3  5  7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79

%e    6 10 14 15 21 22 26 33 34 35 38 39 46 51 55 57 58 62 65 69 74 77

%e      the first columns that give row 2:           ^^    ^^ ^^    ^^

%e Row 3 begins: {265550..265603, 265606..265607, 265682..265682, 265830..265830, 265832..265834, ...

%o (PARI)

%o showlist(list) = {my(slist = List()); listput(slist, list[1]); for (i=2, #list, if (list[i] != list[i-1]+1, listput(slist, list[i-1]); listput(slist, list[i]););); listput(slist, list[#list]); Vec(slist);}

%o primo(i) = factorback(primes(i));

%o ubound(nL, n) = {if (nL == 1, return(n*log(n) + n*log(log(n)))); if (nL == 2, return(n*log(n)/log(log(n)))); if (nL == 3, return(2*n*log(n)/log(log(n))^2)); if (nL == 4, return(3*n*log(n)/log(log(n))^3)); if (nL == 5, return(4*n*log(n)/log(log(n))^4));}

%o out(list1, list2, list3) = print(showlist(list1)); print(showlist(list2)); print(showlist(list3));

%o rows() = {my(nL = 3, nC = 1000000, nB=5); my(m=vector(nL, i, vector(nC))); my(vfirst = vector(nL, i, primo(i))); my(list1 = List(), list2 = List(), list3 = List()); for (nn=1, nB, my(ok=1); print("nn=", nn); for (i=1, nL, my(list = List()); my(na = vfirst[i]); my(ns = 1); if (nn==1, m[i][ns] = na; ns++); forsquarefree (k=na+1, 100*round(ubound(i,nn*nC)), if (omega(k[2]) == i, m[i][ns] = k[1]; ns++); if (ns > nC, break)); if (ns < nC, print("not enough"); out(list1, list2, list3); return;);); N = 1; for (j=1, nC, if (m[N][j] == vecmin (vector(nL, r, m[r][j])), listput(list1, j+(nn-1)*nC));); N = 2; for (j=1, nC, if (m[N][j] == vecmin (vector(nL, r, m[r][j])), listput(list2, j+(nn-1)*nC));); N = 3; for (j=1, nC, if (m[N][j] == vecmin (vector(nL, r, m[r][j])), listput(list3, j+(nn-1)*nC));); vfirst = vector(nL, i, m[i][nC]); for (i=1, nL, m[i] = vector(nC));); out(list1, list2, list3);}

%Y Cf. A276176, A340316, A346617.

%K nonn,tabf

%O 1,2

%A _Michel Marcus_, Dec 12 2022

%E Provisional rule for calculating that row n is full added by _Peter Munn_, Jan 03 2023