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.

A347980 a(n) is the smallest odd number k whose symmetric representation of sigma(k) has maximum width n.

Original entry on oeis.org

1, 15, 315, 2145, 3465, 17325, 45045, 51975, 225225, 405405, 315315, 765765, 1576575, 2297295
Offset: 1

Views

Author

Hartmut F. W. Hoft, Sep 22 2021

Keywords

Comments

The sequence is not increasing with the maximum width of the symmetric representation just like A347979.
Observation: a(2)..a(14) ending in 5. - Omar E. Pol, Sep 23 2021

Examples

			The pattern of maximum widths of the parts in the symmetric representation of sigma for the first four terms in the sequence is:
   a(n) parts  successive widths
     1:   1          1
    15:   3        1 2 1
   315:   3        1 3 1
  2145:   7    1 2 3 4 3 2 1
		

Crossrefs

Programs

  • Mathematica
    a262045[n_] := Module[{a=Accumulate[Map[If[Mod[n - # (#+1)/2, #]==0, (-1)^(#+1), 0] &, Range[Floor[(Sqrt[8n+1]-1)/2]]]]}, Join[a, Reverse[a]]]
    a347980[n_, mw_] := Module[{list=Table[0, mw], i, v}, For[i=1, i<=n, i+=2, v=Max[a262045[i]]; If [list[[v]]==0, list[[v]]=i]]; list]
    a347980[2500000,14] (* long evaluation time *)

A348171 Square array read by upward antidiagonals in which T(w,p) is the smallest number k whose symmetric representation of sigma(k) consists of p parts with maximum width w occurring in at least one of its p parts.

Original entry on oeis.org

1, 6, 3, 60, 78, 9, 120, 7620, 15, 21, 360, 28920, 315, 75, 81, 840, 261720, 1326, 495, 63, 147, 3360, 1422120, 3465, 22542, 525, 189, 729, 2520, 22622880, 17325, 44574, 5005, 1275, 357, 903, 5040, 12728520, 45045, 199578, 6435, 16575, 1287, 1197, 3025, 10080, 50858640, 51975, 7734558, 34034, 131835, 2145, 3861, 2499, 6875
Offset: 1

Views

Author

Hartmut F. W. Hoft, Oct 04 2021

Keywords

Comments

The first row of the table below is A318843 and the first column is A250070.
T(1,k+1) <= 3^k, for all k>=0, since for k=2j the (j+1)-st part in the symmetric representation of sigma(3^k) extends across the diagonal, and for k=2j+1 the (j+1)-st part is completed before the diagonal.
The data computed so far for a partially filled table of 15 rows and 15 columns, show that all rows, all columns (except column 4 for n <= 6 *10^7), and the diagonal are nonmonotonic.

Examples

			The 10x10 section of the table with dashes indicating values greater than 6*10^7; rows w denote the maximum width and columns p the number of parts in the symmetric representation of sigma(T(w,p)).
w\p | 1     2        3      4       5       6       7       8        9   ...
----------------------------------------------------------------------------
  1 | 1     3        9      21      81      147     729     903      3025
  2 | 6     78       15     75      63      189     357     1197     2499
  3 | 60    7620     315    495     525     1275    1287    3861     3591
  4 | 120   28920    1326   22542   5005    16575   2145    29325    11583
  5 | 360   261720   3465   44574   6435    131835  76125   24225    82593
  6 | 840   1422120  17325  199578  34034   83655   196707  468027   62985
  7 | 3360  22622880 45045  7734558 153153  442442  314925  1108965  471975
  8 | 2520  12728520 51975     -    205275  2067065 1429275 2359875  557175
  9 | 5040  50858640 225225    -    646646  2863718 2395197 5353725  2785875
  10| 10080    -     405405    -    1990989 2124694 6500375 36535499 7753875
   ...
The symmetric representation of sigma for T(2,3) = 15 consists of the three parts (8, 8, 8) of maximum widths (1, 2, 1), and that of T(3,3) = 315 consists of the three parts (158, 308, 158) of maximum widths (1, 3, 1).
		

Crossrefs

Programs

  • Mathematica
    (* function a341969 is defined in A341969 *)
    a348171[n_,  {w_, p_}] := Module[{list=Table[0, {i, w}, {j, p}], k, s, c, u}, For[k=1, k<=n, k++, s=Map[Max, Select[SplitBy[a341969[k], # != 0 &], #[[1]] != 0 &]]; c = Length[s]; u = Max[s]; If[u<=w && c<=p, If[list[[u, c]] == 0, list[[u, c]] = k ]]]; list]
    table=a348171[60000000, {15, 15}] (* 15x15 table; very long computation time *)
    p[n_] := n-row[n-1](row[n-1]+1)/2
    w[n_] := row[n-1]-p[n]+2
    Map[table[[w[#], p[#]]]&, Range[55]] (* sequence data *)

Formula

a((w+p-2)(w+p-1)/2 + p) = T(w,p), for all w, p >= 1.
T(w(n), p(n)) = a(n), for all n >= 1, where p(n) = n - r(n-1) * (r(n-1) + 1)/2, w(n) = r(n-1) - p(n) + 2, and r(n) = floor((sqrt(8*n+1) - 1)/2).

A320521 a(n) is the smallest even number k such that the symmetric representation of sigma(k) has n parts.

Original entry on oeis.org

2, 10, 50, 230, 1150, 5050, 22310, 106030, 510050, 2065450, 10236350
Offset: 1

Views

Author

Omar E. Pol, Oct 14 2018

Keywords

Comments

It appears that a(n) = 2 * q where q is odd and that the symmetric representation of sigma(a(n)/2) has the same number of parts as that for a(n). Number a(12) > 15000000. - Hartmut F. W. Hoft, Sep 22 2021

Examples

			a(1) = 2 because the second row of A237593 is [2, 2], and the first row of the same triangle is [1, 1], therefore between both symmetric Dyck paths there is only one part: [3], equaling the sum of the divisors of 2: 1 + 2 = 3. See below:
.
.     _ _ 3
.    |_  |
.      |_|
.
.
a(2) = 10 because the 10th row of A237593 is [6, 2, 1, 1, 1, 1, 2, 6], and the 9th row of the same triangle is [5, 2, 2, 2, 2, 5], therefore between both symmetric Dyck paths there are two parts: [9, 9]. Also there are no even numbers k < 10 whose symmetric representation of sigma(k) has two parts. Note that the sum of these parts is 9 + 9 = 18, equaling the sum of the divisors of 10: 1 + 2 + 5 + 10 = 18. See below:
.
.     _ _ _ _ _ _ 9
.    |_ _ _ _ _  |
.              | |_
.              |_ _|_
.                  | |_ _ 9
.                  |_ _  |
.                      | |
.                      | |
.                      | |
.                      | |
.                      |_|
.
a(3) = 50 because the 50th row of A237593 is [26, 9, 4, 3, 3, 1, 2, 1, 1, 1, 1, 2, 1, 3, 3, 4, 9, 26], and the 49th row of the same triangle is [25, 9, 5, 3, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 3, 5, 9, 25], therefore between both symmetric Dyck paths there are three parts: [39, 15, 39]. Also there are no even numbers k < 50 whose symmetric representation of sigma(k) has three parts. Note that the sum of these parts is 39 + 15 + 39 = 93, equaling the sum of the divisors of 50: 1 + 2 + 5 + 10 + 25 + 50 = 93. (The diagram of the symmetric representation of sigma(50) = 93 is too large to include.)
		

Crossrefs

Row 1 of A320537.
Cf. A237270 (the parts), A237271 (number of parts), A174973 (one part), A239929 (two parts), A279102 (three parts), A280107 (four parts), A320066 (five parts), A320511 (six parts).

Programs

  • Mathematica
    (* support functions are defined in A341969, A341970 & A341971 *)
    a320521[n_, len_] := Module[{list=Table[0, len], i, v}, For[i=2, i<=n, i+=2, v=Count[a341969[i], 0]+1;If[list[[v]]==0, list[[v]]=i]]; list]
    a320521[15000000,11] (* Hartmut F. W. Hoft, Sep 22 2021 *)

Extensions

a(6)-a(11) from Hartmut F. W. Hoft, Sep 22 2021

A348142 Square array read by upward antidiagonals in which T(w,p) is the smallest number k whose symmetric representation of sigma(k) consists of p parts with maximum width w occurring in everyone of its p parts.

Original entry on oeis.org

1, 6, 3, 60, 78, 9, 120, 7620, 1014, 21, 360, 28920, 967740, 12246, 81, 840, 261720, 6969720, 116136420, 171366, 147, 3360, 1422120
Offset: 1

Views

Author

Hartmut F. W. Hoft, Oct 04 2021

Keywords

Comments

It appears that the first row is A318843 and that the first column is A250070.
Columns 1 and 2 both are identical with those of the table in A348171 and row 1 is identical with that of A348171.
In the remainder of the 7th antidiagonal a(24..26) > 120*10^6, a(27) = 1922622, and a(28) = 903.

Examples

			The 10x8 section of the table T(w,p) with dashes indicating values greater than 120*10^6; rows w denote the common maximum width in all parts and columns p the number of parts in the symmetric representation of sigma(T(w,p)).
w\p | 1     2         3        4         5        6        7        8  ...
--------------------------------------------------------------------------
  1 | 1     3         9        21        81       147     729      903
  2 | 6     78        1014     12246     171366   1922622 28960854  -
  3 | 60    7620      967740   116136420   -         -       -
  4 | 120   28920     6969720    -
  5 | 360   261720      -
  6 | 840   1422120     -
  7 | 3360  22622880    -
  8 | 2520  12728520    -
  9 | 5040  50858640    -
  10| 10080    -
   ...
The symmetric representation of sigma for T(2,2) = 78 consists of the two parts (84, 84) of maximum widths (2, 2), and that of T(2,3) = 1014 consists of the three parts (1020, 156, 1020) of maximum widths (2, 2, 2).
		

Crossrefs

Programs

  • Mathematica
    (* function a341969 is defined in A341969 *)
    a348142[n_, {w_, p_}] := Module[{list=Table[0, {i, w}, {j, p}], k, s, c, u}, Monitor[For[k=1, k<=n, k++, s=Map[Max, Select[SplitBy[a341969[k], #!=0&], #[[1]]!=0&]]; c=Length[s]; u=Union[s]; If[Length[u]==1&&u[[1]]<=w&&c<=p, If[list[[u[[1]], c]]==0, list[[u[[1]], c]]=k]]], list]; list]
    table=a348142[120000000, {10, 10}] (* 10x10 table; very long computation time *)
    p[n_] := n-row[n-1](row[n-1]+1)/2
    w[n_] := row[n-1]-p[n]+2
    Map[table[[w[#], p[#]]]&,  Range[23]] (* sequence data *)

Formula

a((w+p-2)(w+p-1)/2 + p) = T(w,p), for all w, p >= 1.
T(w(n),p(n)) = a(n), for all n >= 1, where p(n) = n - r(n-1) * (r(n-1) + 1)/2, w(n) = r(n-1) - p(n) + 2, and r(n) = floor((sqrt(8*n+1) - 1)/2).
Showing 1-4 of 4 results.