Voozh

A237271

Number of parts in the symmetric representation of sigma(n).

278

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

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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

Conjecture 1: a(n) is the number of odd divisors of n except the "e" odd divisors described in A005279. Thus a(n) is the length of the n-th row of A379288. - Omar E. Pol, Dec 21 2024

The conjecture 1 was checked up n = 10000 by Amiram Eldar. - Omar E. Pol, Dec 22 2024

The conjecture 1 is true. For a proof see A379288. - Hartmut F. W. Hoft, Jan 21 2025

From Omar E. Pol, Jul 31 2025: (Start)

Conjecture 2: a(n) is the number of 2-dense sublists of divisors of n.

We call "2-dense sublists of divisors of n" to the maximal sublists of divisors of n whose terms increase by a factor of at most 2.

In a 2-dense sublist of divisors of n the terms are in increasing order and two adjacent terms are the same two adjacent terms in the list of divisors of n.

Example: for n = 10 the list of divisors of 10 is [1, 2, 5, 10]. There are two 2-dense sublists of divisors of 10, they are [1, 2] and [5, 10], so a(10) = 2.

The conjecture 2 is essentially the same as the second conjecture in the Comments of A384149. See also Peter Munn's formula in A237270.

The indices where a(n) = 1 give A174973 (2-dense numbers). See the proof there. (End)

Conjecture 3: a(n) is the number of divisors p of n such that p is greater than twice the adjacent previous divisor of n. The divisors p give the n-th row of A379288. - Omar E. Pol, Aug 02 2025

From Omar E. Pol, Oct 21 2025: (Start)

Conjecture 4: a(A000290(n)) is odd.

Conjecture 5: a(A000384(n)) is odd.

Observation : a(A002997(n)) >= 3, at least for 1 <= n <= 10000. (End)

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..5000 from Michel Marcus)

Omar E. Pol, Illustration of initial terms

FORMULA

a(n) = A001227(n) - A239657(n). - Omar E. Pol, Mar 23 2014

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

a(n) = A000005(n) - A243982(n). - Omar E. Pol, Aug 02 2025

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):

---------------------------------------------------------

n A000203 A279391 A001227 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 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 *)

a[n_] := Module[{d = Partition[Divisors[n], 2, 1]}, 1 + Count[d, _?(OddQ[#[[2]]] && #[[2]] >= 2*#[[1]] &)]]; Array[a, 100] (* Amiram Eldar, Dec 22 2024 *)

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

(Python)

from sympy import divisors

def a(n: int) -> int:

divs = list(divisors(n))

d = [divs[i:i+2] for i in range(len(divs) - 1)]

s = sum(1 for pair in d if len(pair) == 2 and pair[1] % 2 == 1 and pair[1] >= 2 * pair[0])

return s + 1

print([a(n) for n in range(1, 80)]) # Peter Luschny, Aug 05 2025

CROSSREFS

Row lengths of A237270 and of A379288.

Column 1 of A279387.

Partial sums give A237590.