Displaying 1-10 of 23 results found.
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.
+0
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 number of parts in the symmetric representation of sigma(n) is even, and that all parts have width 1.
+0
21
3, 5, 7, 10, 11, 13, 14, 17, 19, 21, 22, 23, 26, 27, 29, 31, 33, 34, 37, 38, 39, 41, 43, 44, 46, 47, 51, 52, 53, 55, 57, 58, 59, 61, 62, 65, 67, 68, 69, 71, 73, 74, 76, 79, 82, 83, 85, 86, 87, 89, 92, 93, 94, 95, 97
COMMENTS
The first eight entries in A071561 but not in this sequence are 75, 78, 102, 105, 114, 138, 174 & 175. The first eight entries in A239929 but not in this sequence are 21, 27, 33, 39, 51, 55, 57 & 65. Let n = 2^m * Product_{i=1..k} p_i^e_i = 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 k > 0, at least one e_i is odd, and for any two odd divisors f < g of n, 2^(m+1) * f < g. Let the odd divisors of n be 1 = d_1 < ... < d_2x = q where 2x = sigma_0(q). The z-th region of the symmetric spectrum of n has area a_z = 1/2 * (2^(m+1) - 1) *(d_z + d_(2x+1-z)), for 1 <= z <= 2x. Therefore, the sum of the area of the regions equals sigma(n). For a proof see Theorem 6 in the link of A071561. - Hartmut F. W. Hoft, Sep 09 2015, Sep 04 2018
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[100], atmostOneDiagonalsQ[#] && EvenQ[Length[a237270[#]]]&] (* data *)
CROSSREFS
Cf. A000203, A071561, A071562, A174905, A236104, A237270, A237271, A237593, A238443, A241010, A246955.
Numbers j for which the symmetric representation of sigma(j) has two parts, each of width one.
+0
21
3, 5, 7, 10, 11, 13, 14, 17, 19, 22, 23, 26, 29, 31, 34, 37, 38, 41, 43, 44, 46, 47, 52, 53, 58, 59, 61, 62, 67, 68, 71, 73, 74, 76, 79, 82, 83, 86, 89, 92, 94, 97, 101, 103, 106, 107, 109, 113, 116, 118, 122, 124, 127, 131, 134, 136, 137, 139, 142, 146, 148, 149, 151, 152, 157, 158, 163, 164, 166, 167, 172, 173, 178, 179, 181, 184, 188, 191, 193, 194, 197, 199
COMMENTS
The sequence is the intersection of A239929 (sigma(j) has two parts) and of A241008 (sigma(j) has an even number of parts of width one). The numbers in the sequence are precisely those defined by the formula for the triangle, see the link. The symmetric representation of sigma(j) has two parts, each part having width one, precisely when j = 2^(k - 1) * p where 2^k <= row(j) < p, p is prime and row(j) = floor((sqrt(8*j + 1) - 1)/2). Therefore, the sequence can be written naturally as a triangle as shown in the Example section.
The symmetric representation of sigma(j) = 2*j - 2 consists of two regions of width 1 that meet on the diagonal precisely when j = 2^(2^m - 1)*(2^(2^m) + 1) where 2^(2^m) + 1 is a Fermat prime (see A019434). This subsequence of numbers j is 3, 10, 136, 32896, 2147516416, ...[?]... ( A191363). The k-th column of the triangle starts in the row whose initial entry is the first prime larger than 2^(k+1) (that sequence of primes is A014210, except for 2). Observation: at least the first 82 terms coincide with the numbers j with no middle divisors whose largest divisor <= sqrt(j) is a power of 2, or in other words, coincide with the intersection of A071561 and A365406. - Omar E. Pol, Oct 11 2023
FORMULA
Formula for the triangle of numbers associated with the sequence:
P(n, k) = 2^k * prime(n) where n >= 2, 0 <= k <= floor(log_2(prime(n)) - 1).
EXAMPLE
We show portions of the first eight columns, 0 <= k <= 7, of the triangle.
0 1 2 3 4 5 6 7
3
5 10
7 14
11 22 44
13 26 52
17 34 68 136
19 38 76 152
23 46 92 184
29 58 116 232
31 62 124 248
37 74 148 296 592
41 82 164 328 656
43 86 172 344 688
47 94 188 376 752
53 106 212 424 848
59 118 236 472 944
61 122 244 488 976
67 134 268 536 1072 2144
71 142 284 568 1136 2272
. . . . . .
. . . . . .
127 254 508 1016 2032 4064
131 262 524 1048 2096 4192 8384
137 274 548 1096 2192 4384 8768
. . . . . . .
. . . . . . .
251 502 1004 2008 4016 8032 16064
257 514 1028 2056 4112 8224 16448 32896
263 526 1052 2104 4208 8416 16832 33664
Since 2^(2^4) + 1 = 65537 is the 6543rd prime, column k = 15 starts with 2^15*(2^(2^16) + 1) = 2147516416 in row 6542 with 65537 in column k = 0.
For an image of the symmetric representations of sigma(m) for all values m <= 137 in the triangle see the link.
The first column is the sequence of odd primes, see A065091. The second column is the sequence of twice the primes starting with 10, see A001747. The third column is the sequence of four times the primes starting with 44, see A001749. For related references also see A033676 (largest divisor of n less than or equal to sqrt(n)).
MATHEMATICA
(* functions path[] and a237270[ ] are defined in A237270 *) atmostOneDiagonalsQ[n_]:=SubsetQ[{0, 1}, Union[Flatten[Drop[Drop[path[n], 1], - 1] - path[n - 1], 1]]]
(* data *)
Select[Range[200], Length[a237270[#]]==2 && atmostOneDiagonalsQ[#]&]
(* function for computing triangle in the Example section through row 55 *)
TableForm[Table[2^k Prime[n], {n, 2, 56}, {k, 0, Floor[Log[2, Prime[n]] - 1]}], TableDepth->2]
CROSSREFS
Cf. A000203, A033676, A071561, A163280, A237270, A237271, A237593, A241008, A241010, A247687, A250068, A250070, A250071, A365406.
Numbers n with the property that the symmetric representation of sigma(n) has four parts.
+0
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).
Numbers k such that the symmetric representation of sigma(k) has only two parts and they meet at the center of the Dyck path.
+0
9
3, 10, 78, 136, 666, 820, 1830, 2628, 4656, 5886, 6328, 16290, 18528, 28920, 32896, 39340, 48828, 56616, 62128, 78606, 80200, 83436, 88410, 93528, 100576, 104196, 135460, 146070, 166176, 180300, 187578, 190036
COMMENTS
For a proof of the formula see the link and also the links in A239929 and A071561. This formula allows for a fast computation of numbers in the sequence that does not require computations of Dyck paths. All terms are triangular numbers.
More precisely, all terms are second hexagonal numbers ( A014105). There are no terms with middle divisors. - Omar E. Pol, Oct 31 2018 Numbers k such that the concatenation of the widths of the symmetric representation of sigma(k) is a cyclops numbers ( A134808). - Omar E. Pol, Aug 29 2021
FORMULA
Terms are equal to q*(2*q + 1) where q is in A174973 and 2*q + 1 is prime.
EXAMPLE
q = 128 = 2^7 is the 15th term in A174973 for which 2*n+1 = 2^8 + 1 is prime so that a(15) = 2^7 * (2^8 + 1) = 32896. The two parts in the symmetric representation of sigma of a(15) have width 1 and sigma(a(15)) = 2 * a(15) - 2. q = 308 is the 32nd term in A174973 for which 2*n+1 is prime so that a(32) = 308 * 617 = 190036. The maximum width of the two regions is 2 and sigma(a(32)) = 415296. For n = 21, the symmetric representation of sigma(21) has two parts that meet at the center of the Dyck path, but 21 is not in the sequence because the symmetric representation of sigma(21) has more than two parts. - Omar E. Pol, Sep 18 2015 Illustration of initial terms (n = 1, 2):
. y
. |
. |_ _ _ _ _ _
. |_ _ _ _ _ |
. | | |_
. | |_ _|_
. | | |_ _
. | |_ _ |
. | | |
. |_ _ | |
. |_ _|_ | |
. | | | | |
. |_ _|_|_ _ _ _ _ _|_|_ _ x
. 3 10
.
The symmetric representation of sigma(3) = 2 + 2 = 4 has two parts and they meet at the point (2, 2), so a(1) = 3.
The symmetric representation of sigma(10) = 9 + 9 = 18 has two parts and they meet at the point (7, 7), so a(2) = 10.
(End)
Also 10 is in the sequence because the concatenation of the widths of the symmetric representation of sigma(10) is 1111111110111111111 and it is a cyclops number ( A134808). - Omar E. Pol, Aug 29 2021
MATHEMATICA
(* test for membership in A174973 *) a174973Q[n_]:=Module[{d=Divisors[n]}, Select[Rest[d] - 2 Most[d], #>0&]=={}]
a174973[n_]:=Select[Range[n], a174973Q]
(* compute numbers in the sequence *)
a262259[n_]:=Map[#(2#+1)&, Select[a174973[1, n], PrimeQ[2#+1]&]]
a262259[308] (* data *)
CROSSREFS
Cf. A000203, A000217, A014105, A071561, A134808, A174973, A191363, A235791, A236104, A237048, A237270, A237271, A237591, A237593, A239929, A245092, A249351 (widths), A262045, A262048, A262626.
Numbers n having three parts in the symmetric representation of sigma(n).
+0
9
9, 15, 25, 35, 45, 49, 50, 70, 77, 91, 98, 110, 121, 130, 135, 143, 154, 169, 170, 182, 187, 190, 209, 221, 225, 238, 242, 247, 266, 286, 289, 299, 315, 322, 323, 338, 350, 361, 374, 391, 405, 418, 437, 442, 484, 493, 494, 506, 527, 529, 550, 551, 572, 578, 589, 598, 638, 646, 650, 667, 675, 676, 682
COMMENTS
Let n = 2^m * q with m >= 0 and q odd, let row(n) = floor(sqrt(8*n+1) - 1)/2), and let 1 = d_1 < ... < d_h <= row(n) < d_(h+1) < ... < d_k = q be the k odd divisors of n.
The symmetric representation of sigma(n) consists of 3 parts precisely when there is a unique i, 1 <= i < h, such that 2^(m+1) * d_i < d_(i+1) and d_h <= row(n) < 2^(m+1) * d_h.
This property of the odd divisors of n is equivalent to the n-th row of the irregular triangle of A249223 consisting of a block of positive numbers, followed by a block of zeros, followed in turn by a block of positive numbers, i.e., determining the first part and the left half of the center part of the symmetric representation of sigma(n), resulting in 3 parts. Let n be the product of two primes p and q satisfying 2 < p < q < 2*p. Then n satisfies the property above so that the odd numbers in A087718 form a subsequence.
EXAMPLE
a(4) = 35 = 5*7 is in the sequence since 1 < 2 < 5 < row(35) = 7 < 10;
a(8) = 70 = 2*5*7 is in the sequence since 1 < 4 < 5 < row(70) = 11 < 20;
140 = 4*5*7 is not in the sequence since 1 < 5 < 7 < 8 < row(140) = 16 < 20;
a(506) = 5950 = 2*25*7*17 is in the sequence since 1*4 < 5 is the only pair of odd divisors 1 < 5 < 7 < 17 < 25 < 35 < 85 < row(5950) = 108 satisfying the property (see A251820).
MATHEMATICA
segmentsSigma[n_] := Length[Select[SplitBy[a262045[n], #!=0&], First[#]!=0&]]
a279102[m_, n_] := Select[Range[m, n], segmentsSigma[#]==3&]
a279102[1, 700] (* sequence data *)
(* An equivalent, but slower computation is based on A237271 *) a279102[m_, n_] := Select[Range[m, n], a237271[#]==3&]
a279102[1, 700] (* sequence data *)
CROSSREFS
Cf. A087718, A174973 (column 1), A237048, A237270, A237271, A237593, A239929 (column 2), A249223, A251820, A262045, A279102.
Sigma(2p)/2, for odd primes p.
+0
7
6, 9, 12, 18, 21, 27, 30, 36, 45, 48, 57, 63, 66, 72, 81, 90, 93, 102, 108, 111, 120, 126, 135, 147, 153, 156, 162, 165, 171, 192, 198, 207, 210, 225, 228, 237, 246, 252, 261, 270, 273, 288, 291, 297, 300, 318, 336, 342, 345, 351, 360, 363, 378, 387, 396, 405
COMMENTS
The symmetric representation of sigma(2*p), p > 3 prime, consists of two sections each with three contiguous legs of width one (for a proof see the link).
The two ratios of successive legs in the symmetric representation of sigma(2*p) are integers 3 and 2, respectively, for all primes p > 3 satisfying p = -1(mod 6); see also A003627. If one ratio is an integer then so is the other. The sequence 2*p for primes p > 3 is a subsequence of A239929, numbers n whose symmetric representation of sigma(n) has two parts. Since sigma(2*p) = 3*(p+1), each element of the sequence is a multiple of 3; furthermore, a(n)/3 = A006254(n) = A111333(n+1).
FORMULA
a(n) = T(2*prime(n+1), 1) - T(2*prime(n+1), 4) = 3*(prime(n+1)+1)/2 = sigma(2*prime(n+1))/2 where T(n,k) is defined in A235791.
EXAMPLE
a(4) = T(22, 1) - T(22, 4) = 22 - 4 = 18 = sigma(22)/2
The last image in the Example section of A237593 includes the first four symmetric representations for this sequence, i.e., when 2*p = 10, 14, 22 & 26; see also the link for an image of the first 10 symmetric representations.
MATHEMATICA
a[n_]:=3(Prime[n+1]+1)/2
Map[a, Range[55]] (* data *)
DivisorSigma[1, 2#]/2&/@Prime[Range[2, 60]] (* Harvey P. Dale, Jan 07 2023 *)
PROG
(PARI) vector(100, n, 3*(prime(n+1)+1)/2) \\ Derek Orr, Sep 19 2014 (PARI) vector(60, n, sigma(2*prime(n+1))/2) \\ Michel Marcus, Nov 25 2014
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.
+0
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.
+0
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 n with the property that the number of parts in the symmetric representation of sigma(n) equals the number of divisors of n.
+0
5
1, 3, 5, 7, 9, 11, 13, 17, 19, 21, 23, 25, 27, 29, 31, 33, 37, 39, 41, 43, 47, 49, 51, 53, 55, 57, 59, 61, 65, 67, 69, 71, 73, 79, 81, 83, 85, 87, 89, 93, 95, 97, 101, 103, 107, 109, 111, 113, 115, 119, 121, 123, 125, 127, 129, 131, 133, 137, 139, 141, 145
COMMENTS
Numbers n such that A243982(n) = 0. First differs from A151991 at a(25). Let n = 2^m * q with m >= 0 and q odd. Let c_n denote the count of regions in the symmetric representation of sigma(n), which is determined by the positions of 1's in the n-th row of A237048. The maximum of c_n occurs when odd and even positions of 1's alternate implying that all regions have width 1, denoted by w_n = 1. When m > 0 then sigma_0(n) > sigma_0(q) and c_n = sigma_0(n) is impossible. Therefore, exactly those odd n with w_n = 1 are in this sequence. Furthermore, since the 1's in A237048 represent the odd divisors of n, their odd-even alternation expresses the property 2*f < g for any two adjacent divisors f < g of odd number n; in other words, this sequence is also the complement of A090196 relative to the odd numbers. This last property permits computations of elements in this sequence faster than with function a244579, which is based on Dyck paths. - Hartmut F. W. Hoft, Oct 11 2015 Also, integers n such that for any pair a < b of divisors of n the inequality 2*a < b holds (hence n is odd).
Let 1 = d_1 < ... < d_k = n be all (odd) divisors of n. The property 2*d_i < d_(i+1), for 1 <= i < k, is equivalent for the 1's in the n-th row of A249223 to be in positions 1 = d_1 < 2 < d_2 < 2*d_2 < ... < d_i <2*d_i < d_(i+1) < ... where 2*d_i represents the odd divisor e_i with d_i * e_i = n. In other words, the odd divisors are the number of parts in the symmetric representation of sigma(n). The rightmost 1 in the n-th row occurs in an odd (even) position when k is odd (even). As a consequence this sequence is also the complement of A090196 in the set of odd numbers. (End)
EXAMPLE
9 is in the sequence because the parts of the symmetric representation of sigma(9) are [5, 3, 5] and the divisors of 9 are [1, 3, 9] and in both cases there is the same number of elements: A237271(9) = A000005(9) = 3. See the link for a diagram of the symmetric representations of sigma for sequence data listed above. The symmetric representations of sigma(a(35)) = sigma(81) = sigma(3^4) consists of 5 regions whose areas are [41, 15, 9, 15, 41] and computed as 41 = (3^4+3^0)/2, 15 = (3^3+3^1)/2, and 9 = 3^2 for the central area. Observe also that the 81st row in triangle A237048 is [ 1 1 1 0 0 1 0 0 1 0 0 0 ] with the 1's in positions 1, 2, 3, 6, and 9. This is the largest count for the symmetric regions of sigma shown in the diagram. - Hartmut F. W. Hoft, Oct 11 2015
MATHEMATICA
(* Function a237270[] is defined in A237270 *) a244579[m_, n_] := Select[Range[m, n], Length[a237270[#]] == Length[Divisors[#]]&]
a244579[1, 104]] (* data *)
(* alternative function using the divisor property *)
divisorPairsQ[n_] := Module[{d=Divisors[n]}, Select[2*Most[d] - Rest[d], # >= 0&] == {}]
a244579Alt[m_?OddQ, n_] := Select[Range[m, n, 2], divisorPairsQ]
a244579Alt[1, 145] (* data *)
CROSSREFS
Cf. A000005, A001227, A005279, A071561, A071562, A090196, A000203, A196020, A236104, A237048, A237270, A237271, A237593, A238443, A238524, A239657, A239929, A239663, A240062, A241558, A241559, A243982, A245092.
Search completed in 0.024 seconds
|