%I #162 Jun 12 2024 17:29:54
%S 1,1,2,1,2,1,2,1,3,2,2,1,2,2,3,1,2,1,2,1,4,2,2,1,3,2,4,1,2,1,2,1,4,2,
%T 3,1,2,2,4,1,2,1,2,2,3,2,2,1,3,3,4,2,2,1,4,1,4,2,2,1,2,2,5,1,4,1,2,2,
%U 4,3,2,1,2,2,4,2,3,2,2,1,5,2,2,1,4,2,4,1,2,1
%N Number of parts in the symmetric representation of sigma(n).
%C The diagram of the symmetry of sigma has been via A196020 --> A236104 --> A235791 --> A237591 --> A237593.
%C For more information see A237270.
%C a(n) is also the number of terraces at n-th level (starting from the top) of the stepped pyramid described in A245092. - _Omar E. Pol_, Apr 20 2016
%C a(n) is also the number of subparts in the first layer of the symmetric representation of sigma(n). For the definion of "subpart" see A279387. - _Omar E. Pol_, Dec 08 2016
%C Note that the number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n. (See the second example). - _Omar E. Pol_, Dec 20 2016
%C From _Hartmut F. W. Hoft_, Dec 26 2016: (Start)
%C Using odd prime number 3, observe that the 1's in the 3^k-th row of the irregular triangle of A237048 are at index positions
%C 3^0 < 2*3^0 < 3^1 < 2*3^1 < ... < 2*3^((k-1)/2) < 3^(k/2) < ...
%C the last being 2*3^((k-1)/2) when k is odd and 3^(k/2) when k is even. Since odd and even index positions alternate, each pair (3^i, 2*3^i) specifies one part in the symmetric representation with a center part present when k is even. A straightforward count establishes that the symmetric representation of 3^k, k>=0, has k+1 parts. Since this argument is valid for any odd prime, every positive integer occurs infinitely many times in the sequence. (End)
%C a(n) = number of runs of consecutive nonzero terms in row n of A262045. - _N. J. A. Sloane_, Jan 18 2021
%C Indices of odd terms give A071562. Indices of even terms give A071561. - _Omar E. Pol_, Feb 01 2021
%C a(n) is also the number of prisms in the three-dimensional version of the symmetric representation of k*sigma(n) where k is the height of the prisms, with k >= 1. - _Omar E. Pol_, Jul 01 2021
%C With a(1) = 0; a(n) is also the number of parts in the symmetric representation of A001065(n), the sum of aliquot parts of n. - _Omar E. Pol_, Aug 04 2021
%C The parity of this sequence is also the characteristic function of numbers that have middle divisors. - _Omar E. Pol_, Sep 30 2021
%C a(n) is also the number of polycubes in the 3D-version of the ziggurat of order n described in A347186. - _Omar E. Pol_, Jun 11 2024
%H Michel Marcus, <a href="/A237271/b237271.txt">Table of n, a(n) for n = 1..5000</a>
%F a(p^k) = k + 1, where p is an odd prime and k >= 0. - _Hartmut F. W. Hoft_, Dec 26 2016
%F Theorem: a(n) <= number of odd divisors of n (cf. A001227). The differences are in A239657. - _N. J. A. Sloane_, Jan 19 2021
%F a(n) = A340846(n) - A340833(n) + 1 (Euler's formula). - _Omar E. Pol_, Feb 01 2021
%e Illustration of initial terms (n = 1..12):
%e ---------------------------------------------------------
%e n A000203 A237270 a(n) Diagram
%e ---------------------------------------------------------
%e . _ _ _ _ _ _ _ _ _ _ _ _
%e 1 1 1 1 |_| | | | | | | | | | | |
%e 2 3 3 1 |_ _|_| | | | | | | | | |
%e 3 4 2+2 2 |_ _| _|_| | | | | | | |
%e 4 7 7 1 |_ _ _| _|_| | | | | |
%e 5 6 3+3 2 |_ _ _| _| _ _|_| | | |
%e 6 12 12 1 |_ _ _ _| _| | _ _|_| |
%e 7 8 4+4 2 |_ _ _ _| |_ _|_| _ _|
%e 8 15 15 1 |_ _ _ _ _| _| |
%e 9 13 5+3+5 3 |_ _ _ _ _| | _|
%e 10 18 9+9 2 |_ _ _ _ _ _| _ _|
%e 11 12 6+6 2 |_ _ _ _ _ _| |
%e 12 28 28 1 |_ _ _ _ _ _ _|
%e ...
%e For n = 9 the sum of divisors of 9 is 1+3+9 = A000203(9) = 13. On the other hand the 9th set of symmetric regions of the diagram is formed by three regions (or parts) with 5, 3 and 5 cells, so the total number of cells is 5+3+5 = 13, equaling the sum of divisors of 9. There are three parts: [5, 3, 5], so a(9) = 3.
%e From _Omar E. Pol, Dec 21 2016: (Start)
%e Illustration of the diagram of subparts (n = 1..12):
%e ---------------------------------------------------------
%e n A000203 A279391 A001227 Diagram
%e ---------------------------------------------------------
%e . _ _ _ _ _ _ _ _ _ _ _ _
%e 1 1 1 1 |_| | | | | | | | | | | |
%e 2 3 3 1 |_ _|_| | | | | | | | | |
%e 3 4 2+2 2 |_ _| _|_| | | | | | | |
%e 4 7 7 1 |_ _ _| _ _|_| | | | | |
%e 5 6 3+3 2 |_ _ _| |_| _ _|_| | | |
%e 6 12 11+1 2 |_ _ _ _| _| | _ _|_| |
%e 7 8 4+4 2 |_ _ _ _| |_ _|_| _ _ _|
%e 8 15 15 1 |_ _ _ _ _| _| _| |
%e 9 13 5+3+5 3 |_ _ _ _ _| | _| _|
%e 10 18 9+9 2 |_ _ _ _ _ _| |_ _|
%e 11 12 6+6 2 |_ _ _ _ _ _| |
%e 12 28 23+5 2 |_ _ _ _ _ _ _|
%e ...
%e For n = 6 the symmetric representation of sigma(6) has two subparts: [11, 1], so A000203(6) = 12 and A001227(6) = 2.
%e For n = 12 the symmetric representation of sigma(12) has two subparts: [23, 5], so A000203(12) = 28 and A001227(12) = 2. (End)
%e From _Hartmut F. W. Hoft_, Dec 26 2016: (Start)
%e Two examples of the general argument in the Comments section:
%e Rows 27 in A237048 and A249223 (4 parts)
%e i: 1 2 3 4 5 6 7 8 9 . . 12
%e 27: 1 1 1 0 0 1 1's in A237048 for odd divisors
%e 1 27 3 9 odd divisors represented
%e 27: 1 0 1 1 1 0 0 1 1 1 0 1 blocks forming parts in A249223
%e Rows 81 in A237048 and A249223 (5 parts)
%e i: 1 2 3 4 5 6 7 8 9 . . 12. . . 16. . . 20. . . 24
%e 81: 1 1 1 0 0 1 0 0 1 0 0 0 1's in A237048 f.o.d
%e 1 81 3 27 9 odd div. represented
%e 81: 1 0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1 blocks fp in A249223
%e (End)
%t a237271[n_] := Length[a237270[n]] (* code defined in A237270 *)
%t Map[a237271, Range[90]] (* data *)
%t (* _Hartmut F. W. Hoft_, Jun 23 2014 *)
%o (PARI) fill(vcells, hga, hgb) = {ic = 1; for (i=1, #hgb, if (hga[i] < hgb[i], for (j=hga[i], hgb[i]-1, cell = vector(4); cell[1] = i - 1; cell[2] = j; vcells[ic] = cell; ic ++;););); vcells;}
%o findfree(vcells) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
%o findxy(vcells, x, y) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[1]==x) && (vcelli[2]==y) && (vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
%o findtodo(vcells, iz) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == iz) && (vcelli[4] == 0), return (i)); ); return (0);}
%o zcount(vcells) = {nbz = 0; for (i=1, #vcells, nbz = max(nbz, vcells[i][3]);); nbz;}
%o docell(vcells, ic, iz) = {x = vcells[ic][1]; y = vcells[ic][2]; if (icdo = findxy(vcells, x-1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x+1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y-1), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y+1), vcells[icdo][3] = iz); vcells[ic][4] = 1; vcells;}
%o docells(vcells, ic, iz) = {vcells[ic][3] = iz; while (ic, vcells = docell(vcells, ic, iz); ic = findtodo(vcells, iz);); vcells;}
%o nbzb(n, hga, hgb) = {vcells = vector(sigma(n)); vcells = fill(vcells, hga, hgb); iz = 1; while (ic = findfree(vcells), vcells = docells(vcells, ic, iz); iz++;); zcount(vcells);}
%o lista(nn) = {hga = concat(heights(row237593(0), 0), 0); for (n=1, nn, hgb = heights(row237593(n), n); nbz = nbzb(n, hga, hgb); print1(nbz, ", "); hga = concat(hgb, 0););} \\ with heights() also defined in A237593; \\ _Michel Marcus_, Mar 28 2014
%Y Row lengths of A237270.
%Y Column 1 of A279387.
%Y Partial sums give A237590.
%Y Parity gives A347950.
%Y Cf. A000203, A000265, A001065, A001227, A024916, A060831, A061345, A067742, A071561, A071562, A175254, A196020, A221529, A235791, A236104, A237048, A237591, A237593, A239657, A244050, A244971, A245092, A249223, A250068, A261699, A262045, A262612, A262626, A274824, A279387, A279693, A319073, A340583, A340846, A342344, A347186.
%K nonn
%O 1,3
%A _Omar E. Pol_, Feb 25 2014