%I
%S 1,1,2,2,2,1,1,2,3,1,1,3,3,2,2,3,4,1,1,1,1,4,4,2,1,1,2,4,5,2,1,1,2,5,
%T 5,2,2,2,2,5,6,2,1,1,1,1,2,6,6,3,1,1,1,1,3,6,7,2,2,1,1,2,2,7,7,3,2,1,
%U 1,2,3,7,8,3,1,2,2,1,3,8,8,3,2,1,1,1,1,2,3,8
%N Triangle read by rows in which row n lists the elements of the nth row of A237591 and then the elements of the same row but in reverse order.
%C Row n is a palindromic composition of 2*n.
%C T(n,k) is also the length of the kth segment in a Dyck path on the first quadrant of the square grid, connecting the xaxis with the yaxis, from (n, 0) to (0, n), starting with a segment in vertical direction, see example.
%C Conjecture 1: the area under the nth Dyck path equals A024916(n), the sum of all divisors of all positive integers <= n.
%C If the conjecture is true then the nth Dyck path represents the boundary segments after the alternating sum of the elements of the nth row of A236104.
%C Conjecture 2: two adjacent Dyck paths never cross (checked by hand up to n = 128), hence the total area between the nth Dyck path and the (n1)st Dyck path is equal to sigma(n) = A000203(n), the sum of divisors of n.
%C The connection between A196020 and A237271 is as follows: A196020 > A236104 > A235791 > A237591 > this sequence > A239660 > A237270 > A237271.
%C PARI scripts area(n) and chkcross(n) have been written to check the 2 properties and have been run up to n=10000.  _Michel Marcus_, Mar 27 2014
%C Comments from _Franklin T. AdamsWatters_ on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)
%C The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
%C You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
%C Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n1; and the number of squares added is sigma(n).
%C Finally, look at the connected areas enclosed by n but not by n1; the size of these areas is the symmetric representation of sigma. (End)
%C Mathematica functions have been written that verified the 2 properties through n=30000.  _Hartmut F. W. Hoft_, Apr 07 2014
%C It appears that, for the nth set, the number of cells lying on the first diagonal is equal to A067742(n), the number of middle divisors of n.  _Michel Marcus_, Jun 21 2014
%C Checked _Michel Marcus_'s conjecture with two Mathematica functions up to n=100000, for more information see A240542.  _Hartmut F. W. Hoft_, Jul 17 2014
%C A003056(n) is also the number of peaks of the Dyck path related to the nth row of triangle.  _Omar E. Pol_, Nov 03 2015
%C The number of peaks of the Dyck path associated to the row A000396(n) of this triangle equals the nth Mersenne prime A000668(n), hence Mersenne primes are visible in two ways at the pyramid described in A245092.  _Omar E. Pol_, Dec 19 2016
%H Robert Price, <a href="/A237593/b237593.txt">Table of n, a(n) for n = 1..15008</a> (rows n = 1..412, flattened)
%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/polpyr01.jpg">An infinite stepped pyramid</a>
%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/polpyr02.jpg">Illustration of initial terms as an isosceles triangle (rows: 1..28)</a>
%H Omar E. Pol, <a href="http://www.polprimos.com/imagenespub/polpyr05.jpg">Perspective view of the pyramid (first 16 levels)</a>
%F Let j(n)= floor((sqrt(8n+1)1)/2) then T(n,k) = A237591(n,k), if k <= j(n); otherwise T(n,k) = A237591(n,2*j(n)+1k).  _Hartmut F. W. Hoft_, Apr 07 2014 (corrected by _Omar E. Pol_, May 31 2015)
%e Triangle begins:
%e 1 : 1, 1,
%e 2 : 2, 2,
%e 3 : 2, 1, 1, 2,
%e 4 : 3, 1, 1, 3,
%e 5 : 3, 2, 2, 3,
%e 6 : 4, 1, 1, 1, 1, 4,
%e 7 : 4, 2, 1, 1, 2, 4,
%e 8 : 5, 2, 1, 1, 2, 5,
%e 9 : 5, 2, 2, 2, 2, 5,
%e 10 : 6, 2, 1, 1, 1, 1, 2, 6,
%e 11 : 6, 3, 1, 1, 1, 1, 3, 6,
%e 12 : 7, 2, 2, 1, 1, 2, 2, 7,
%e 13 : 7, 3, 2, 1, 1, 2, 3, 7,
%e 14 : 8, 3, 1, 2, 2, 1, 3, 8,
%e 15 : 8, 3, 2, 1, 1, 1, 1, 2, 3, 8,
%e 16 : 9, 3, 2, 1, 1, 1, 1, 2, 3, 9,
%e 17 : 9, 4, 2, 1, 1, 1, 1, 2, 4, 9,
%e 18 : 10, 3, 2, 2, 1, 1, 2, 2, 3, 10,
%e 19 : 10, 4, 2, 2, 1, 1, 2, 2, 4, 10,
%e 20 : 11, 4, 2, 1, 2, 2, 1, 2, 4, 11,
%e 21 : 11, 4, 3, 1, 1, 1, 1, 1, 1, 3, 4, 11,
%e 22 : 12, 4, 2, 2, 1, 1, 1, 1, 2, 2, 4, 12,
%e 23 : 12, 5, 2, 2, 1, 1, 1, 1, 2, 2, 5, 12,
%e 24 : 13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13,
%e ...
%e Illustration of initial terms as an infinite Dyck path (row n = 1..7):
%e .
%e . /\/\/\ /\ /\
%e . /\/\ /\ /\ / \ / \/\/ \
%e . /\ /\/\ / \ / \/ \ / \ / \
%e . /\/ \/ \/ \/ \/ \/ \
%e .
%e Illustration of rows 8 and 9 interpreted as Dyck paths in the first quadrant and the illustration of the symmetric representation of sigma(9) = 5 + 3 + 5 = 13, see below:
%e .
%e y y
%e . .
%e . ._ _ _ _ _ _ _ _ _ _ 5
%e ._ _ _ _ _ .  _ _ _ _ _
%e .  . _ _ _ _ 3
%e . _ .  _ 
%e . _ _ . _ _ __ _ 5
%e .  .   
%e . Area = 56  . Area = 69   
%e .  .   
%e .  .   
%e . . . . . . . .  . x . . . . . . . . .  . x _
%e .
%e . Fig. 1 Fig. 2 Fig. 3
%e .
%e Figure 1. For n = 8 the 8th row of triangle is [5, 2, 1, 1, 2, 5] and the area under the symmetric Dyck path is equal to A024916(8) = 56.
%e Figure 2. For n = 9 the 9th row of triangle is [5, 2, 2, 2, 2, 5] and the area under the symmetric Dyck path is equal to A024916(9) = 69.
%e Figure 3. The symmetric representation of sigma(9): between both symmetric Dyck paths there are three regions (or parts) of sizes [5, 3, 5].
%e The sum of divisors of 9 is 1 + 3 + 9 = A000203(9) = 13. On the other hand the difference between the areas under the Dyck paths equals the sum of the parts of the symmetric representation of sigma(9) = 69  56 = 5 + 3 + 5 = 13, equaling the sum of divisors of 9.
%e .
%e Illustration of initial terms as Dyck paths in the first quadrant:
%e (row n = 1..28)
%e ._ _ _ _ _ _ _ _ _ _ _ _ _ _ _
%e _ _ _ _ _ _ _ _ _ _ _ _ _ _ 
%e _ _ _ _ _ _ _ _ _ _ _ _ _ _ 
%e _ _ _ _ _ _ _ _ _ _ _ _ _  
%e _ _ _ _ _ _ _ _ _ _ _ _ _  
%e _ _ _ _ _ _ _ _ _ _ _ _   _ _ _
%e _ _ _ _ _ _ _ _ _ _ _ _  _ _ _ 
%e _ _ _ _ _ _ _ _ _ _ _   _ _  _
%e _ _ _ _ _ _ _ _ _ _ _  _ _ _ _ _
%e _ _ _ _ _ _ _ _ _ _   _ _ _
%e _ _ _ _ _ _ _ _ _ _  _ _ _ _ _ _ _
%e _ _ _ _ _ _ _ _ _  _ _ _ _  _ _ 
%e _ _ _ _ _ _ _ _ _  _ _ _ __ _  
%e _ _ _ _ _ _ _ _  _ _ _ __    _ _ _ _ _
%e _ _ _ _ _ _ _ _    _ _  __ _ _ _ _ 
%e _ _ _ _ _ _ _  _ _ _ _   _ _ _ _ _   
%e _ _ _ _ _ _ _ _ _ _ _ _   _ _ _ _ _     
%e _ _ _ _ _ _  _ _ _  __ _ _ _       
%e _ _ _ _ _ _ _ _ _  _ _ _ _         
%e _ _ _ _ _  _ _  _ _ _ _           
%e _ _ _ _ _ _  __ _ _             
%e _ _ _ _ _ __ _ _ _               
%e _ _ _ _ _  _ _ _                 
%e _ _ _ _ __ _                   
%e _ _ _ _ _                     
%e _ _ _ _                       
%e _ __                         
%e _                           
%e ____________________________
%e .
%e n:1 2 3 4 5 6 7 8..10..12..14..16..18..20..22..24..26..28
%e .
%e It appears that the total area (also the total number of cells) in the first n set of symmetric regions of the diagram is equal to A024916(n), the sum of all divisors of all positive integers <= n.
%e It appears that the total area (also the total number of cells) in the nth set of symmetric regions of the diagram is equal to sigma(n) = A000203(n) (checked by hand up n = 128).
%e From _Omar E. Pol_, Aug 18 2015: (Start)
%e The above diagram is also the top view of the stepped pyramid described in A245092 and it is also the top view of the staircase described in A244580, in both cases the figure represents the first 28 levels of the structure. Note that the diagram contains (and arises from) a hidden pattern which is shown below.
%e .
%e Illustration of initial terms as an isosceles triangle:
%e Row _ _
%e 1 _11_
%e 2 _2 __ 2_
%e 3 _2 11 2_
%e 4 _3 _11_ 3_
%e 5 _3 2 __ 2 3_
%e 6 _4 _1111_ 4_
%e 7 _4 2 11 2 4_
%e 8 _5 _2 _11_ 2_ 5_
%e 9 _5 2 2 __ 2 2 5_
%e 10 _6 _2 1111 2_ 6_
%e 11 _6 3 _1111_ 3 6_
%e 12 _7 _2 2 11 2 2_ 7_
%e 13 _7 3 2 _11_ 2 3 7_
%e 14 _8 _3 _12 __ 21_ 3_ 8_
%e 15 _8 3 2 1111 2 3 8_
%e 16 9 3 2 1111 2 3 9
%e ...
%e This diagram is the simpler representation of the sequence.
%e The number of horizontal line segments in the nth level in each side of the diagram equals A001227(n), the number of odd divisors of n.
%e The number of horizontal line segments in the left side of the diagram plus the number of the horizontal line segment in the right side equals A054844(n).
%e The total number of vertical line segments in the nth level of the diagram equals A131507(n).
%e Note that this symmetric pattern also emerges from the front view of the stepped pyramid described in A245092, which is related to sigma A000203, the sumofdivisors function, and other related sequences. The diagram represents the first 16 levels of the pyramid. (End)
%t row[n_]:=Floor[(Sqrt[8n+1]1)/2]
%t s[n_,k_]:=Ceiling[(n+1)/k(k+1)/2]Ceiling[(n+1)/(k+1)(k+2)/2]
%t f[n_,k_]:=If[k<=row[n],s[n,k],s[n,2 row[n]+1k]]
%t TableForm[Table[f[n,k],{n,1,50},{k,1,2 row[n]}]] (* _Hartmut F. W. Hoft_, Apr 08 2014 *)
%o (PARI) row(n) = {my(orow = row237591(n)); vector(2*#orow, i, if (i <= #orow, orow[i], orow[2*#orowi+1]));}
%o area(n) = {my(rown = row(n)); surf = 0; h = n; odd = 1; for (i=1, #row, if (odd, surf += h*rown[i], h = rown[i];); odd = !odd;); surf;}
%o heights(v, n) = {vh = vector(n); ivh = 1; h = n; odd = 1; for (i=1, #v, if (odd, for (j=1, v[i], vh[ivh] = h; ivh++), h = v[i];); odd = !odd;); vh;}
%o isabove(hb, ha) = {for (i=1, #hb, if (hb[i] < ha[i], return (0));); return (1);}
%o chkcross(nn) = {hga = concat(heights(row(1), 1), 0); for (n=2, nn, hgb = heights(row(n), n); if (! isabove(hgb, hga), print("pb cross at n=", n)); hga = concat(hgb, 0););} \\ _Michel Marcus_, Mar 27 2014
%o (Python)
%o from sympy import sqrt
%o import math
%o def row(n): return int(math.floor((sqrt(8*n + 1)  1)/2))
%o def s(n, k): return int(math.ceil((n + 1)/k  (k + 1)/2))  int(math.ceil((n + 1)/(k + 1)  (k + 2)/2))
%o def T(n, k): return s(n, k) if k<=row(n) else s(n, 2*row(n) + 1  k)
%o for n in range(1, 11): print [T(n, k) for k in range(1, 2*row(n) + 1)] # _Indranil Ghosh_, Apr 21 2017
%Y Row n has length 2*A003056(n).
%Y Row sums give A005843, n >= 1.
%Y Column k starts in row A008805(k1).
%Y Column 1 = right border = A008619, n >= 1.
%Y Bisections are in A259176, A259177.
%Y Cf. A000203, A000217, A001227, A024916, A054844, A067742, A131507, A196020, A235791, A236104, A237048, A237270, A237271, A237590, A237591, A239660, A239931A239934, A244050, A244580, A245092, A249351, A261350, A261699, A262611, A262612, A262626, A279387, A280850, A280851, A286000, A286001, A296508, A335616.
%K nonn,tabf,look
%O 1,3
%A _Omar E. Pol_, Feb 22 2014
