A241010 Numbers n with the property that the number of parts in the symmetric representation of sigma(n) is odd, and that all parts have width 1.

1, 2, 4, 8, 9, 16, 25, 32, 49, 50, 64, 81, 98, 121, 128, 169, 242, 256, 289, 338, 361, 484, 512, 529, 578, 625, 676, 722, 729, 841, 961, 1024, 1058, 1156, 1250, 1369, 1444, 1681, 1682, 1849, 1922, 2048, 2116, 2209, 2312, 2401, 2738, 2809, 2888, 3025, 3249, 3362, 3364, 3481, 3698, 3721, 3844

Offset: 1

Author

Hartmut F. W. Hoft, Aug 07 2014

Keywords

• nonn

Comments

The first eight entries in A071562 but not in this sequence are 6, 12, 15, 18, 20, 24, 28 & 30.

The first eight entries in A238443 but not in this sequence are 6, 12, 18, 20, 24, 28, 30 & 36.

The union of A241008 and of this sequence equals A174905 (for a proof see link in A174905).

Let n = 2^m * product(p_i^e_i, i=1,...,k) = 2^m * q with m >= 0, k >= 0, 2 < p_1, ...< p_k primes and e_i >= 1, for all 1 <= i <= k. For each number n in this sequence all e_i are even, and for any two odd divisors f < g of n, 2^(m+1) * f < g. The sum of the areas of the regions r(n, z) equals sigma(n). For a proof of the characterization and the formula see the theorem in the link below.

Numbers 3025 = 5^2 * 11^2 and 510050 = 2^1 * 5^2 * 101^2 are the smallest odd and even numbers, respectively, in the sequence with two distinct odd prime factors.

Among the 706 numbers in the sequence less than 1000000 (see link to the table) there are 143 that have two different odd prime factors, but none with three. All numbers with three different odd prime factors are larger than 500000000. Number 4450891225 = 5^2 * 11^2 * 1213^2 is in the sequence, but may not be the smallest one with three different odd prime factors. Note that 1213 is the first prime that extends the list of divisors of 3025 while preserving the property for numbers in this sequence.

The subsequence of numbers n = 2^(k-1) * p^2 satisfying the constraints above is A247687.

n = 3^(2*h) = 9^h = A001019(h), h>=0, is the smallest number for which the symmetric representation of sigma(n) has 2*h+1 regions of width one, for example for h = 1, 2, 3 and 5, but not for h = 4 in which case 3025 = 5^2 * 11^2 < 3^8 = 6561 is the smallest (see A318843). [Comment corrected by Hartmut F. W. Hoft, Sep 04 2018]

Computations using this characterization are more efficient than those of Dyck paths for the symmetric representations of sigma(n), e.g., the Mathematica code below.

Examples

			This irregular triangle presents in each column those elements of the sequence that have the same factor of a power of 2.
  row/col      2^0    2^1   2^2   2^3    2^4    2^5  ...
   2^k:          1      2     4     8     16     32  ...
   3^2:          9
   5^2:         25     50
   7^2:         49     98
   3^4:         81
  11^2:        121    242   484
  13^2:        169    338   676
  17^2:        289    578  1156  2312
  19^2:        361    722  1444  2888
  23^2:        529   1058  2116  4232
   5^4:        625   1250
   3^6:        729
  29^2:        841   1682  3364  6728
  31^2:        961   1922  3844  7688
  37^2:       1369   2738  5476 10952 21904
  41^2:       1681   3362  6724 13448 26896
  43^2:       1849   3698  7396 14792 29584
  47^2:       2209   4418  8836 17672 35344
   7^4:       2401   4802
  53^2:       2809   5618 11236 22472 44944
  5^2*11^2:   3025
  3^2*19^2:   3249
  59^2:       3481   6962 13924 27848 55696
  61^2:       3721   7442 14884 29768 59536
  67^2:       4489   8978 17956 35912 71824 143648
  3^2*23^2:   4761
  71^2:       5041
  ...
  5^2*101^2:225025 510050
  ...
Number 3025 = 5^2 * 11^2 is in the sequence since its divisors are 1, 5, 11, 25, 55, 121, 275, 605 and 3025. Number 6050 = 2^1 * 5^2 * 11^2 is not in the sequence since 2^2 * 5 > 11 while 5 < 11.
Number 510050 = 2^1 * 5^2 * 101^2 is in the sequence since its 9 odd divisors 1, 5, 25, 101, 505, 2525, 10201, 51005 and 225025 are separated by factors larger than 2^2. The areas of its 9 regions are 382539, 76515, 15339, 3939, 1515, 3939, 15339, 76515 and 382539. However, 2^2 * 5^2 * 101^2 is not in the sequence.
The first row is A000079.
The rows, except the first, are indexed by products of even powers of the odd primes satisfying the property, sorted in increasing order.
The first column is a subsequence of A244579.
A row labeled p^(2*h), h>=1 and p>=3 with p = A000040(n), has A098388(n) entries.
Starting with the second column, dividing the entries of a column by 2 creates a proper subsequence of the prior column.
See A259417 for references to other sequences of even powers of odd primes that are subsequences of column 1.
The first entry greater than 16 in column labeled 2^4 is 21904 since 37 is the first prime larger than 2^5. The rightmost entry in the row labeled 19^2 is 2888 in the column labeled 2^3 since 2^4 < 19 < 2^5.
		

Links

• Hartmut F. W. Hoft, Table of n, a(n) for n = 1..563 (values less than 1000000)
• Hartmut F. W. Hoft, Proof of characterization theorem
• Hartmut F. W. Hoft, Table of n, a(n) for n = 1...706(values less than 1000000)
• Hartmut F. W. Hoft, Illustration of the symmetric representation of sigma(a(n)), for n=1..15
• Hartmut F. W. Hoft, Proof of property for n and formula for regions of sigma(n)

Crossrefs

Cf. A000203, A174905, A236104, A237270 (symmetric representation of sigma(n)), A237271, A237593, A238443, A241008, A071562, A246955, A247687, A250068, A250070, A250071.

Cf. A318843

Programs

Mathematica

(* path[n] and a237270[n] are defined in A237270 *)
atmostOneDiagonalsQ[n_] := SubsetQ[{0, 1}, Union[Flatten[Drop[Drop[path[n], 1], -1] - path[n-1], 1]]]
Select[Range[1000], atmostOneDiagonalsQ[#] && OddQ[Length[a237270[#]]]&] (* data *)
(* more efficient code based on numeric characterization *)
divisorPairsQ[m_, q_] := Module[{d = Divisors[q]}, Select[2^(m + 1)*Most[d] - Rest[d], # >= 0 &] == {}]
a241010AltQ[n_] := Module[{m, q, p, e}, m=IntegerExponent[n, 2]; q=n/2^m; {p, e} = Transpose[FactorInteger[q]]; q==1||(Select[e, EvenQ]==e && divisorPairsQ[m, q])]
a241010Alt[m_,n_] := Select[Range[m, n], a241010AltQ]
a241010Alt[1,4000] (* data *)

Formula

Formula for the z-th region in the symmetric representation of n = 2^m * q in this sequence, 1 <= z <= sigma_0(q) and q odd: r(n, z) = 1/2 * (2^(m+1) - 1) * (d_z + d_(2*x+2-z)) where 1 = d_1 < ... < d_(2*x+1) = q are the odd divisors of n.

Extensions

More terms and further edited by Hartmut F. W. Hoft, Jun 26 2015 and Jul 02 2015 and corrected Oct 11 2015