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Square array read by antidiagonals in which T(n,k) is the n-th number j with the property that the symmetric representation of sigma(j) has k parts.
+10
21
1, 2, 3, 4, 5, 9, 6, 7, 15, 21, 8, 10, 25, 27, 63, 12, 11, 35, 33, 81, 147, 16, 13, 45, 39, 99, 171, 357, 18, 14, 49, 51, 117, 189, 399, 903, 20, 17, 50, 55, 153, 207, 441, 987, 2499, 24, 19, 70, 57, 165, 243, 483, 1029, 2709, 6069, 28, 22, 77, 65, 195, 261, 513, 1113
COMMENTS
This is a permutation of the positive integers.
All odd primes are in column 2 (together with some even composite numbers) because the symmetric representation of sigma(prime(i)) is [m, m], where m = (1 + prime(i))/2, for i >= 2.
The union of all odd-indexed columns gives A071562, the positive integers that have middle divisors. The union of all even-indexed columns gives A071561, the positive integers without middle divisors. - Omar E. Pol, Oct 01 2018 Each column in the table of A357581 is a subsequence of the respective column in the table of this sequence; however, the first row in the table of A357581 is not a subsequence of the first row in the table of this sequence. - Hartmut F. W. Hoft, Oct 04 2022
EXAMPLE
Array begins:
1, 3, 9, 21, 63, 147, 357, 903, 2499, 6069, ...
2, 5, 15, 27, 81, 171, 399, 987, 2709, 6321, ...
4, 7, 25, 33, 99, 189, 441, 1029, 2793, 6325, ...
6, 10, 35, 39, 117, 207, 483, 1113, 2961, 6783, ...
8, 11, 45, 51, 153, 243, 513, 1197, 3025, 6875, ...
12, 13, 49, 55, 165, 261, 567, 1239, 3087, 6909, ...
16, 14, 50, 57, 195, 275, 609, 1265, 3249, 7011, ...
18, 17, 70, 65, 231, 279, 621, 1281, 3339, 7203, ...
20, 19, 77, 69, 255, 297, 651, 1375, 3381, 7353, ...
24, 22, 91, 75, 273, 333, 729, 1407, 3591, 7581, ...
...
MATHEMATICA
partsSRS[n_] := Length[Select[SplitBy[a341969[n], #!=0&], #[[1]]!=0&]]
widthTable[n_, {r_, c_}] := Module[{k, list=Table[{}, c], parts}, For[k=1, k<=n, k++, parts=partsSRS[k]; If[parts<=c&&Length[list[[parts]]]<r, AppendTo[list[[parts]], k]]]; Transpose[PadRight[list, {c, r}, "..."]]]
a240062[n_, r_] := Module[{arr=widthTable[n, {r, r}], vec=Table[0, PolygonalNumber[r]], i, j}, For[i=1, i<=r, i++, For[j=r-i+1, j>=1, j--, vec[[PolygonalNumber[i+j-2]+j]]=arr[[i, j]]]]; vec]
a240062T[n_, r_] := TableForm[widthTable[n, {r, r}]]
a240062[6069, 10] (* data *)
CROSSREFS
Cf. A000203, A006254, A065091, A067742, A071561, A071562, A196020, A236104, A235791, A237048, A237270, A237271, A238443, A239660, A239929, A239931- A239934, A245092, A262626, A319529, A319796, A319801, A319802.
Numbers n with the property that the symmetric representation of sigma(n) has four parts.
+10
13
21, 27, 33, 39, 51, 55, 57, 65, 69, 75, 85, 87, 93, 95, 105, 111, 115, 119, 123, 125, 129, 133, 141, 145, 155, 159, 161, 175, 177, 183, 185, 201, 203, 205, 213, 215, 217, 219, 230, 235, 237, 245, 249, 250, 253, 259, 265, 267, 287, 290, 291, 295, 301, 303, 305, 309, 310, 319, 321, 327, 329
COMMENTS
Let n = 2^k * t where k >= 0 and t is odd, and let D be the set of divisors of t less than r(n) = floor((sqrt(8*n+1) - 1)/2). The following statements are equivalent:
(1) There is exactly one d in D such that 2^(k+1) * d < e where e in D is the next odd divisor larger than d, and the largest divisor f in D satisfies 2^(k+1) * f <= r(n).
(2) The symmetric representation of sigma(n) consists of four parts.
The property says that the first part of the symmetric representation of n consists of the first 2^(k+1) * d - 1 legs and that the second part starts with leg e and ends with leg 2^(k+1) * f - 1 before or at the middle of the Dyck path (see A237048 and A249223) on the diagonal. Together with their symmetric pair they form the four parts. (End)
EXAMPLE
a(1) = 21 because it is the smallest number n whose symmetric representation of sigma(n) has four parts. Note that the sum of the parts is 11 + 5 + 5 + 11 = 32, equaling the sum of the divisors of 21: aigma(21) = 1 + 3 + 7 + 21 = 32.
230 = 2*5*23 is the first even number since 2^2 < 5, 2^2 * 5 < 23, and row 230 in A237048 has 20 entries with 1's in positions 1, 4, 5, and 20. Prime number 3 can be a factor for an even number in this sequence as 12246=2*3*13*157 demonstrates with the four parts 12252, 1020, 1020, and 12252 in the symmetric representation of sigma(12246) defined by 1's in positions 1, 3, 4, 12, 13, 39, 52, 156 in row 12246 of A237048; each of the four parts has maximum width 2 and the two central parts meet on the diagonal at 8492. (End)
MATHEMATICA
(* Function a237270[] and row[] are defined in A237270 *) a280107[m_, n_] := Select[Range[m, n], Length[a237270[#]]==4&]
a280107[1, 329] (* data *)
(* Implementation of the property in the Comment section *)
evenPart[n_] := Module[{f=First[FactorInteger[n]]}, If[First[f]!=2, 1, First[f]^Last[f]]]
fourPartsQ[n_] := Module[{e=evenPart[n], oddPart, r=row[n], dL}, oddPart=n/evenPart[n]; dL=Select[Divisors[oddPart], #<r&]; If[Length[dL]>1, 2*e*Last[dL]<=r && Length[Select[2*e*Most[dL]-Rest[dL], #<0&]==1, False]]
Select[Range[329], fourPartsQ] (* data *)
CROSSREFS
First differs from A264102 at a(10).
Square array read by antidiagonals of numbers whose symmetric representation of sigma consists only of parts that have width 1; column k indicates the number of parts and row n indicates the n-th number in increasing order in each of the columns.
+10
7
1, 2, 3, 4, 5, 9, 8, 7, 25, 21, 16, 10, 49, 27, 81, 32, 11, 50, 33, 625, 147, 64, 13, 98, 39, 1250, 171, 729, 128, 14, 121, 51, 2401, 207, 15625, 903, 256, 17, 169, 55, 4802, 243, 31250, 987, 3025, 512, 19, 242, 57, 14641, 261, 117649, 1029, 3249, 6875
COMMENTS
This sequence is a permutation of A174905. Numbers in the even numbered columns of the table form A241008 and those in the odd numbered columns form A241010. The first row of the table is A318843. This sequence is a subsequence of A240062 and each column in this sequence is a subsequence in the respective column of A240062.
EXAMPLE
The upper left hand 11 X 11 section of the table for a(n) <= 2*10^7:
1 2 3 4 5 6 7 8 9 10 11 ...
----------------------------------------------------------------------
1 3 9 21 81 147 729 903 3025 6875 59049
2 5 25 27 625 171 15625 987 3249 7203 9765625
4 7 49 33 1250 207 31250 1029 4761 13203 19531250
8 10 50 39 2401 243 117649 1113 6561 13527 ...
16 11 98 51 4802 261 235298 1239 7569 14013 ...
32 13 121 55 14641 275 1771561 1265 8649 14499 ...
64 14 169 57 28561 279 3543122 1281 12321 14661 ...
128 17 242 65 29282 333 4826809 1375 14161 15471 ...
256 19 289 69 57122 363 7086244 1407 15129 15633 ...
512 22 338 85 58564 369 9653618 1491 16641 15957 ...
1024 23 361 87 83521 387 19307236 1533 17689 16119 ...
...
Each column k > 1 contains odd and even numbers since, e.g., 5^(k-1) and 2 * 5^(k-1) belong to it.
Odd numbers in column 3: A001248(k), k > 1. Numbers in column 5 have the form 2^k * p^4 with p > 2 prime and 0 <= k < floor(log_2(p)).
Odd numbers in column 5: A030514(k), k > 1. Column 6: subsequence of A320511; 189 is the smallest number not in column 6. Smallest even number in column 6 is 5050.
Column 7: Numbers have the form 2^k * p^6 with p > 2 prime and 0 <= k < floor(log_2(p)).
Odd numbers in column 7: A030516(k), k > 1. Numbers in the column numbered with the n-th prime p_n have the form: 2^k * p^(p_n - 1) with p > 2 prime and 0 <= k < floor(log_2(p_n)).
MATHEMATICA
width1Table[n_, {r_, c_}] := Module[{k, list=Table[{}, c], wL, wLen, pCount, colLen}, For[k=1, k<=n, k++, wL=a341969[k]; wLen=Length[wL]; pCount=(wLen+1)/2; If[pCount<=c&&Length[list[[pCount]]]<r&&SubsetQ[{0, 1}, Union[wL]], AppendTo[list[[pCount]], k]]]; Transpose[PadRight[list, {c, r}, "..."]]]
a357581[n_, r_] := Module[{arr=width1Table[n, {r, r}], vec=Table[0, PolygonalNumber[r]], i, j}, For[i=1, i<=r, i++, For[j=r-i+1, j>=1, j--, vec[[PolygonalNumber[i+j-2]+j]]=arr[[i, j]]]]; vec]
a357581T[n_, r_] := TableForm[width1Table[n, {r, r}]]
a357581[120000, 10] (* sequence data - first 10 antidiagonals *)
a357581T[120000, 10] (* upper left hand 10x10 array *)
a357581T[20000000, 11] (* 11x11 array - very long computation time *)
CROSSREFS
Cf. A000079, A001248, A030514, A030516, A174905, A174973, A237593, A238443, A239929, A241008, A241010, A246955, A247687, A264102, A279102, A280107, A318843, A320066, A320511, A341969, A341970, A341971.
Numbers k with the property that the symmetric representation of sigma(k) has five parts.
+10
6
63, 81, 99, 117, 153, 165, 195, 231, 255, 273, 285, 325, 345, 375, 425, 435, 459, 475, 525, 561, 575, 625, 627, 665, 693, 725, 735, 775, 805, 819, 825, 875, 897, 925, 975, 1015, 1025, 1075, 1085, 1150, 1175, 1225, 1250, 1295, 1377, 1395, 1421, 1435, 1450, 1479, 1505, 1519, 1550, 1581, 1617, 1645, 1653, 1665
COMMENTS
Those numbers in this sequence with only parts of width 1 in their symmetric representation of sigma form column 5 in the table of A357581. - Hartmut F. W. Hoft, Oct 04 2022
EXAMPLE
63 is in the sequence because the 63rd row of A237593 is [32, 11, 6, 4, 2, 2, 2, 1, 2, 1, 1, 2, 1, 2, 2, 2, 4, 6, 11, 32], and the 62nd row of the same triangle is [32, 11, 5, 4, 3, 2, 1, 2, 1, 1, 1, 1, 2, 1, 2, 3, 4, 5, 11, 32], therefore between both symmetric Dyck paths there are five parts: [32, 12, 16, 12, 32]. The sums of these parts is 32 + 12 + 16 + 12 + 32 = 104, equaling the sum of the divisors of 63: 1 + 3 + 7 + 9 + 21 + 63 = 104.
(The diagram of the symmetric representation of sigma(63) = 104 is too large to include.)
MATHEMATICA
partsSRS[n_] := Length[Select[SplitBy[a341969[n], #!=0&], #[[1]]!=0&]]
a320066[n_] := Select[Range[n], partsSRS[#]==5&]
Numbers k with the property that the symmetric representation of sigma(k) has six parts.
+10
5
147, 171, 189, 207, 243, 261, 275, 279, 297, 333, 351, 363, 369, 387, 423, 429, 465, 477, 507, 531, 549, 555, 595, 603, 605, 615, 639, 645, 657, 663, 705, 711, 715, 741, 747, 795, 801, 833, 845, 867, 873, 885, 909, 915, 927, 931, 935, 963, 969, 981, 1005, 1017, 1045, 1065, 1071, 1083, 1095, 1105, 1127
COMMENTS
Those numbers in this sequence with only parts of width 1 in their symmetric representation of sigma form column 6 in the table of A357581. - Hartmut F. W. Hoft, Oct 04 2022
EXAMPLE
147 is in the sequence because the 147th row of A237593 is [74, 25, 13, 8, 5, 4, 4, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 4, 4, 5, 8, 13, 25, 74], and the 146th row of the same triangle is [74, 25, 12, 8, 6, 4, 3, 2, 2, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 4, 6, 8, 12, 25, 74], therefore between both symmetric Dyck paths there are six parts: [74, 26, 14, 14, 26, 74]. Note that the sum of these parts is 74 + 26 + 14 + 14 + 26 + 74 = 228, equaling the sum of the divisors of 147: 1 + 3 + 7 + 21 + 49 + 147 = 228.
(The diagram of the symmetric representation of sigma(147) = 228 is too large to include.)
MATHEMATICA
partsSRS[n_] := Length[Select[SplitBy[a341969[n], #!=0&], #[[1]]!=0&]]
a320511[n_] := Select[Range[n], partsSRS[#]==6&]
CROSSREFS
Cf. A000203, A018303, A196020, A235791, A236104, A237048, A237591, A237593, A239663, A239665, A245092, A262626, A296508.
a(n) is the smallest even number k such that the symmetric representation of sigma(k) has n parts.
+10
3
2, 10, 50, 230, 1150, 5050, 22310, 106030, 510050, 2065450, 10236350
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
EXAMPLE
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.)
MATHEMATICA
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]
CROSSREFS
Cf. A000203, A018262, A005843, A196020, A235791, A236104, A237048, A237591, A237593, A239663, A239665, A240062, A245092, A262626, A296508.
Numbers n whose symmetric representation of sigma(n) consists of 3 regions with maximum width 2.
+10
2
15, 35, 45, 70, 77, 91, 110, 130, 135, 143, 154, 170, 182, 187, 190, 209, 221, 225, 238, 247, 266, 286, 299, 322, 323, 350, 374, 391, 405, 418, 437, 442, 493, 494, 506, 527, 550, 551, 572, 589, 598, 638, 646, 650, 667, 682, 703, 713, 748, 754, 782, 806, 814, 836, 850
COMMENTS
This sequence is a subsequence of A279102. The definition of the sequence excludes squares of primes, A001248, since the 3 regions of their symmetric representation of sigma have width 1 (first column in the irregular triangle of A247687). Table of numbers in this sequence arranged by the number of prime factors, counting multiplicities:
2 3 4 5 6 7 ...
------------------------------------------
15 45 135 405 1215 3645
35 70 225 1125 5625 ...
77 110 350 1750 8750 744795
91 130 550 2584 ... ...
143 154 572 2750 85455
187 170 650 3128 ...
209 182 748 3250
221 190 836 3496
247 238 850 3944
299 266 884 4216
... ... ... ...
1035 9585
... ...
The numbers in the first row of the table above are b(k) = 5*3^k, k>=1, (see A005030) so that infinitely many odd numbers occur outside of the first column. The central region of the symmetric representation of sigma(b(k)) contains 2*k-1 separate contiguous sections consisting of sequences of entire legs of width 2, k>=1 (see Lemma 2 in the link). Conjecture: The combined extent of these sections in sigma(b(k)) is 2*3^(k-1) - 1 = A048473(k-1), k>=1. Since each number n in the first column and first row has a prime factor of odd exponent a contiguous section of the symmetric representation of sigma(n) centered on the diagonal has width 2. For odd numbers n not in the first row or column in which all prime factors have even powers, such as 225 and 5625 in the second row, a contiguous section of the symmetric representation of sigma(n) centered on the diagonal has width 1 (see Lemma 1 in the link).
For each k>=3 and every prime p such that b(k-1) < 2*p < 4*b(k-2), the odd number p*b(k-1) is in the column of b(k). The two inequalities are equivalent to b(k-1) <= row(p*b(k-1)) < 2*b(k-1) ensuring that the symmetric representation of sigma(p*b(k-1)) consists of 3 regions.
45 is the only odd number in its column (see Lemma 3 in the link).
Since the factors of n = p*q satisfy 2 < p < q < 2*p the first column in the table above is a subsequence of A082663 and of A087718 (see Lemma 4 in the link). Each of the two outer regions consists of a single leg of width 1 and length (1 + p*q)/2. The center region of size p+q consists of two subparts (see A196020 & A280851) of width 1 of sizes 2*p-q and 2*q-p, respectively (see Lemma 5 in the link). The table below arranges the first column in the table above according to the length 2*p-q of their single contiguous extent of width 2 in the center region: 1 3 5 7 9 11 13 15 ...
------------------------------------------------------
15 35 187 247 143 391 2257 323
91 77 493 589 221 1363 3139 437
703 209 943 2479 551 2911 6649 713
1891 299 1537 3397 851 3901 ... 1247
2701 527 4183 8509 1643 6313 1457
... ... ... ... ... ... ....
Sequences with larger differences 2*p - q are not in OEIS.
EXAMPLE
a(6) = 91 = 7*13 is in the sequence and in the 2-column of the first table since 1 < 2 < 7 < 13 = row(91) representing the 4 odd divisors 1 - 91 - 7 - 13 (see A237048) results in the following pattern for the widths of the legs (see A249223): 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2 for 3 regions with width not exceeding 2. It also is in the 1-column of the second table since it has a single area of width 2 which is 1 unit long. a(29) = 405 = 5*3^4 is in the sequence and in the 5-column of the first table since 1 < 2 < 3 < 5 < 6 < 9 < 10 < 15 < 18 < 27 = row(405) representing the 10 odd divisors 1 - 405 - 3 - 5 - 135 - 9 - 81 - 15 - 45 - 27 results in the following pattern for the widths of the legs: 1, 0, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2 for 3 regions with width not exceeding 2, and 7 = 2*4 - 1 sections of width 2 in the central region.
a(35) = 506 = 2*11*23 is in the sequence since positions 1 < 4 < 11 < 23 < row(506) = 31 representing the 4 odd divisors 1 - 253 - 11 - 23 results in the following pattern for the widths of the legs: 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2 for 3 regions with width not exceeding 2, with the two outer regions consisting of 3 legs of width 1, and a single area of width 2 in the central region.
MATHEMATICA
(* Functions path and a237270 are defined in A237270 *) maxDiagonalLength[n_] := Max[Map[#[[1]]-#[[2]]&, Transpose[{Drop[Drop[path[n], 1], -1], path[n-1]}]]]
a338486[m_, n_] := Module[{r, list={}, k}, For[k=m, k<=n, k++, r=a237270[k]; If[Length[r]== 3 && maxDiagonalLength[k]==2, AppendTo[list, k]]]; list]
a338486[1, 850]
CROSSREFS
Cf. A001248, A005030, A048473, A082663, A087718, A129521, A196020, A226755, A235791, A237048, A237270, A237271, A237591, A237593, A247687, A249223, A279102, A280107, A280851.
Numbers k with the property that the symmetric representation of sigma(k) has seven parts.
+10
1
357, 399, 441, 483, 513, 567, 609, 621, 651, 729, 759, 777, 783, 837, 861, 891, 957, 999, 1023, 1053, 1089, 1107, 1131, 1161, 1209, 1221, 1269, 1287, 1323, 1353, 1419, 1431, 1443, 1521, 1551, 1595, 1599, 1677, 1705, 1749, 1815, 1833, 1887, 1947, 1989, 2013, 2035, 2067, 2091, 2145, 2193, 2223, 2255
EXAMPLE
357 is in the sequence because the 357th row of A237593 is [179, 60, 31, 18, 12, 9, 7, 6, 4, 4, 3, 3, 2, 3, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 3, 2, 3, 3, 4, 4, 6, 7, 9, 12, 18, 31, 60, 179], and the 356th row of the same triangle is [179, 60, 30, 18, 13, 9, 6, 6, 4, 4, 3, 3, 3, 2, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 3, 3, 3, 4, 4, 6, 6, 9, 13, 18, 30, 60, 179], therefore between both symmetric Dyck paths there are seven parts: [179, 61, 29, 38, 29, 61, 179]. Note that the sum of these parts is 179 + 61 + 29 + 38 + 29 + 61 + 179 = 576, equaling the sum of the divisors of 357: 1 + 3 + 7 + 17 + 21 + 51 + 119 + 357 = 576.
(The diagram of the symmetric representation of sigma(357) = 576 is too large to include.)
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