cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

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.

Original entry on oeis.org

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

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Author

Hartmut F. W. Hoft, Aug 07 2014

Keywords

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.
		

Crossrefs

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

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

A377654 Numbers m^2 for which the center part (containing the diagonal) of its symmetric representation of sigma, SRS(m^2), has width 1 and area m.

Original entry on oeis.org

1, 9, 25, 49, 81, 121, 169, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1369, 1521, 1681, 1849, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 4225, 4489, 4761, 5041, 5329, 6241, 6561, 6889, 7225, 7569, 7921, 8649, 9025, 9409, 10201, 10609, 11449, 11881, 12321, 12769, 13225, 14161, 14641, 15129, 15625
Offset: 1

Views

Author

Hartmut F. W. Hoft, Nov 03 2024

Keywords

Comments

Since for numbers m^2 in the sequence the width at the diagonal of SRS(m^2) is 1, the area m of its center part is odd so that this sequence is a proper subsequence of A016754 and since SRS(m^2) has an odd number of parts it is a proper subsequence of A319529. The smallest odd square not in this sequence is 225 = 15^2. SRS(225) is {113, 177, 113}, its center part has maximum width 2, its width at the diagonal is 1.
The k+1 parts of SRS(p^(2k)), p an odd prime and k >= 0, through the diagonal including the center part have areas (p^(2k-i) + p^i)/2 for 0 <= i <= k. They form a strictly decreasing sequence. Since p^(2k) has 2k+1 divisors and SRS(p^(2k)) has 2k+1 parts, all of width 1 (A357581), the even powers of odd primes form a proper subsequence of A244579. For the subsequence of squares of odd primes p, SRS(p^2) consists of the 3 parts { (p^2 + 1)/2, p, (p^2 + 1)/2 } see A001248, A247687 and A357581.
The areas of the parts of SRS(m^2) need not be in descending order through the diagonal as a(112) = 275^2 = 75625 with SRS(75625) = (37813, 7565, 3443, 1525, 715, 738, 275, 738, 715, 1525, 3443, 7565, 37813) demonstrates.
An equivalent description of the sequence is: The center part of SRS(m^2) has width 1, m is odd, and A249223(m^2, m-1) = 0.
Conjectures (true for all a(n) <= 10^8):
(1) The central part of SRS(a(n)) is the minimum of all parts of SRS(a(n)), 1 <= n.
(2) The terms in this sequence are the squares of the terms in A244579.

Examples

			The center part of SRS(a(3)) = SRS(25) has area 5, all 3 parts have width 1, and 25 with 3 divisors also belongs to A244579.
The center part of SRS(a(7)) = SRS(169) has area 13, all 3 parts have width 1, and 169 with 3 divisors also belongs to A244579.
The center part of SRS(a(10)) = SRS(441) has area 21 and width 1, but the maximum width of SRS(441) is 2. Number 441 has 9 divisors and SRS(441) has 7 parts while 21 has 4 divisors and SRS(21) has 4 parts so that 21 is in A244579 while 441 is not.
		

Crossrefs

Programs

  • Mathematica
    (* t237591 and partsSRS compute rows in A237270 and A237591, respectively *)
    (* t249223 and widthPattern are also defined in A376829 *)
    row[n_] := Floor[(Sqrt[8 n+1]-1)/2]
    t237591[n_] := Map[Ceiling[(n+1)/#-(#+1)/2]-Ceiling[(n+1)/(#+1)-(#+2)/2]&, Range[row[n]]]
    partsSRS[n_] := Module[{widths=t249223[n], legs=t237591[n], parts, srs}, parts=widths legs; srs=Map[Apply[Plus, #]&, Select[SplitBy[Join[parts, Reverse[parts]], #!=0&], First[#]!=0&]]; srs[[Ceiling[Length[srs]/2]]]-=Last[widths]; srs]
    t249223[n_] := FoldList[#1+(-1)^(#2+1)KroneckerDelta[Mod[n-#2 (#2+1)/2, #2]]&, 1, Range[2, row[n]]]
    widthPattern[n_] := Map[First, Split[Join[t249223[n], Reverse[t249223[n]]]]]
    centerQ[n_] := Module[{pS=partsSRS[n]}, Sqrt[n]==pS[[(Length[pS]+1)/2]]]/;OddQ[n]
    widthQ[n_] := Module[{wP=SplitBy[widthPattern[n], #!=0&]}, wP[[(Length[wP]+1)/2]]]=={1}/;OddQ[n]
    a377654[m_, n_] := Select[Map[#^2&, Range[m, n, 2]], centerQ[#]&&widthQ[#]&]/;OddQ[m]
    a377654[1, 125]

A090196 Odd integers with two divisors a, b such that a < b <= 2a.

Original entry on oeis.org

15, 35, 45, 63, 75, 77, 91, 99, 105, 117, 135, 143, 153, 165, 175, 187, 189, 195, 209, 221, 225, 231, 245, 247, 255, 273, 285, 297, 299, 315, 323, 325, 345, 351, 357, 375, 385, 391, 399, 405, 425, 429, 435, 437, 441, 455, 459, 465, 475, 483, 493, 495, 513, 525, 527, 539, 551, 555
Offset: 1

Views

Author

Steven Finch, Jan 22 2004

Keywords

Comments

Clearly all even integers have two such divisors a, b. Consider the set S of all integers satisfying this property. Maier & Tenenbaum proved Erdős' conjecture that S has asymptotic density 1.
A244579 and the present sequences are complements in the sequence of odd numbers. - Hartmut F. W. Hoft, Dec 10 2016
From Omar E. Pol, Jan 10 2017: (Start)
Odd numbers k with the property that the number of parts in the symmetric representation of sigma(k) is not equal to the number of divisors of k.
Odd numbers that are not in A244579.
All terms are composites. (End)
The subsequence of semiprimes is A082663. - Bernard Schott, Apr 17 2022

References

  • R. R. Hall and G. Tenenbaum, Divisors, Cambridge Univ. Press, 1988, pp. 95-99.

Crossrefs

Programs

  • Mathematica
    Select[Range[1, 999, 2], (Divisors[#] /. {_, a_, _, b_, _} /; a < b <= 2a -> True) === True&] (* Jean-François Alcover, Nov 05 2016 *)
  • PARI
    is(n)=my(d=divisors(n));for(i=2,#d\2+1,if(d[i]<2*d[i-1], return(n%2))); 0 \\ Charles R Greathouse IV, Jun 20 2013

Formula

a(n) ~ 2n. - Charles R Greathouse IV, Jun 20 2013

Extensions

Corrected by Charles R Greathouse IV, Jul 23 2012
Corrected by Jean-François Alcover and Charles R Greathouse IV, Jun 20 2013

A355875 For n >= 1, k >= 1; a(n) = Sum_{k=1..n} f(n,k). f(n,k) = n*k/(n + k) if n*k divides (n+k), 0 otherwise.

Original entry on oeis.org

0, 1, 0, 2, 0, 5, 0, 4, 0, 5, 0, 13, 0, 7, 6, 8, 0, 15, 0, 14, 0, 11, 0, 26, 0, 13, 0, 26, 0, 42, 0, 16, 0, 17, 10, 39, 0, 19, 0, 43, 0, 41, 0, 22, 38, 23, 0, 52, 0, 25, 0, 26, 0, 45, 0, 59, 0, 29, 0, 111, 0, 31, 14, 32, 0, 85, 0, 34, 0, 76, 0, 86, 0, 37, 30, 38
Offset: 1

Views

Author

Ctibor O. Zizka, Jul 19 2022

Keywords

Comments

a(n) = n/2 for n in A125499.
a(n) = 0 for n in A244579.

Examples

			n = 6; a(6) = Sum_{k=1..6} f(6,k) = 0 + 0 + 2 + 0 + 0 + 3 = 5.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[If[Divisible[n*k, n + k], n*k/(n + k), 0], {k, 1, n}]; Array[a, 100] (* Amiram Eldar, Jul 21 2022 *)
Showing 1-4 of 4 results.