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A237271
Number of parts in the symmetric representation of sigma(n).
225
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, 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, 4, 3, 2, 1, 2, 2, 4, 2, 3, 2, 2, 1, 5, 2, 2, 1, 4, 2, 4, 1, 2, 1
OFFSET
1,3
COMMENTS
The diagram of the symmetry of sigma has been via A196020 --> A236104 --> A235791 --> A237591 --> A237593.
For more information see A237270.
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
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
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
From Hartmut F. W. Hoft, Dec 26 2016: (Start)
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
3^0 < 2*3^0 < 3^1 < 2*3^1 < ... < 2*3^((k-1)/2) < 3^(k/2) < ...
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)
a(n) = number of runs of consecutive nonzero terms in row n of A262045. - N. J. A. Sloane, Jan 18 2021
Indices of odd terms give A071562. Indices of even terms give A071561. - Omar E. Pol, Feb 01 2021
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
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
The parity of this sequence is also the characteristic function of numbers that have middle divisors. - Omar E. Pol, Sep 30 2021
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
LINKS
FORMULA
a(p^k) = k + 1, where p is an odd prime and k >= 0. - Hartmut F. W. Hoft, Dec 26 2016
Theorem: a(n) <= number of odd divisors of n (cf. A001227). The differences are in A239657. - N. J. A. Sloane, Jan 19 2021
a(n) = A340846(n) - A340833(n) + 1 (Euler's formula). - Omar E. Pol, Feb 01 2021
EXAMPLE
Illustration of initial terms (n = 1..12):
---------------------------------------------------------
n A000203 A237270 a(n) Diagram
---------------------------------------------------------
. _ _ _ _ _ _ _ _ _ _ _ _
1 1 1 1 |_| | | | | | | | | | | |
2 3 3 1 |_ _|_| | | | | | | | | |
3 4 2+2 2 |_ _| _|_| | | | | | | |
4 7 7 1 |_ _ _| _|_| | | | | |
5 6 3+3 2 |_ _ _| _| _ _|_| | | |
6 12 12 1 |_ _ _ _| _| | _ _|_| |
7 8 4+4 2 |_ _ _ _| |_ _|_| _ _|
8 15 15 1 |_ _ _ _ _| _| |
9 13 5+3+5 3 |_ _ _ _ _| | _|
10 18 9+9 2 |_ _ _ _ _ _| _ _|
11 12 6+6 2 |_ _ _ _ _ _| |
12 28 28 1 |_ _ _ _ _ _ _|
...
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.
From _Omar E. Pol, Dec 21 2016: (Start)
Illustration of the diagram of subparts (n = 1..12):
---------------------------------------------------------
---------------------------------------------------------
. _ _ _ _ _ _ _ _ _ _ _ _
1 1 1 1 |_| | | | | | | | | | | |
2 3 3 1 |_ _|_| | | | | | | | | |
3 4 2+2 2 |_ _| _|_| | | | | | | |
4 7 7 1 |_ _ _| _ _|_| | | | | |
5 6 3+3 2 |_ _ _| |_| _ _|_| | | |
6 12 11+1 2 |_ _ _ _| _| | _ _|_| |
7 8 4+4 2 |_ _ _ _| |_ _|_| _ _ _|
8 15 15 1 |_ _ _ _ _| _| _| |
9 13 5+3+5 3 |_ _ _ _ _| | _| _|
10 18 9+9 2 |_ _ _ _ _ _| |_ _|
11 12 6+6 2 |_ _ _ _ _ _| |
12 28 23+5 2 |_ _ _ _ _ _ _|
...
For n = 6 the symmetric representation of sigma(6) has two subparts: [11, 1], so A000203(6) = 12 and A001227(6) = 2.
For n = 12 the symmetric representation of sigma(12) has two subparts: [23, 5], so A000203(12) = 28 and A001227(12) = 2. (End)
From Hartmut F. W. Hoft, Dec 26 2016: (Start)
Two examples of the general argument in the Comments section:
Rows 27 in A237048 and A249223 (4 parts)
i: 1 2 3 4 5 6 7 8 9 . . 12
27: 1 1 1 0 0 1 1's in A237048 for odd divisors
1 27 3 9 odd divisors represented
27: 1 0 1 1 1 0 0 1 1 1 0 1 blocks forming parts in A249223
Rows 81 in A237048 and A249223 (5 parts)
i: 1 2 3 4 5 6 7 8 9 . . 12. . . 16. . . 20. . . 24
81: 1 1 1 0 0 1 0 0 1 0 0 0 1's in A237048 f.o.d
1 81 3 27 9 odd div. represented
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
(End)
MATHEMATICA
a237271[n_] := Length[a237270[n]] (* code defined in A237270 *)
Map[a237271, Range[90]] (* data *)
(* Hartmut F. W. Hoft, Jun 23 2014 *)
PROG
(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; }
findfree(vcells) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == 0) && (vcelli[4] == 0), return (i)); ); return (0); }
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); }
findtodo(vcells, iz) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == iz) && (vcelli[4] == 0), return (i)); ); return (0); }
zcount(vcells) = {nbz = 0; for (i=1, #vcells, nbz = max(nbz, vcells[i][3]); ); nbz; }
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; }
docells(vcells, ic, iz) = {vcells[ic][3] = iz; while (ic, vcells = docell(vcells, ic, iz); ic = findtodo(vcells, iz); ); vcells; }
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); }
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
KEYWORD
nonn
AUTHOR
Omar E. Pol, Feb 25 2014
STATUS
approved