Displaying 1-10 of 23 results found.
Records in A110566 (lcm{1,2,...,n}/denominator of harmonic number H(n)).
+20
3
1, 3, 15, 45, 77, 275, 931, 1725, 1935, 5805, 29025, 41175, 166803, 1039533, 1162047, 91801713, 419498967, 2183383175, 19691916585, 216611082435, 2382721906785, 113804487945521, 22211221792244703, 422013214052649357, 425137351586922079, 936039253001457601
MATHEMATICA
c = 0; a = h = 1; t = {}; Do[a = LCM[a, n]; h = h + 1/n; b = a/Denominator[h]; If[b > c, c = b; AppendTo[t, b]], {n, 10^6}]; t
PROG
(PARI) lista(nn) = {rec = 0; for (n=1, nn, new = lcm(vector(n, k, k))/denominator(sum(k=1, n, 1/k)); if (new > rec, print1(new, ", "); rec = new); ); } \\ Michel Marcus, Mar 07 2018
1, 3, 1, 3, 15, 45, 15, 3, 1, 11, 77, 7, 1, 3, 9, 27, 9, 3, 33, 11, 1, 25, 5, 55, 275, 25, 1, 13, 39, 3, 9, 27, 9, 3, 1, 17, 1, 49, 7, 49, 931, 19, 1, 11, 319, 11, 319, 11, 1, 3, 75, 1725, 345, 15, 645, 1935, 5805, 29025, 675, 41175, 13725, 549, 20313, 6771, 183, 3, 411, 15207
COMMENTS
A110566: LCM{1,2,...,n}/denominator of harmonic number H(n).
The factor of change from a(n) to a(n+1) is: 3,3,3,5,3,3,5,3,11,7,11,7,3,3,3,3,3,11,3,11,25,5,11,5,11,25,13,3,13,3,3,3,3,..., . see A110268.
MATHEMATICA
f[n_] := LCM @@ Range[n]/Denominator[HarmonicNumber[n]]; Flatten[Union /@ Split[Table[f[n], {n, 703}]]]
Consider the sequence A110566: lcm{1,2,...,n}/denominator of harmonic number H(n). a(n) is the factor that is changed going from A110566(n) to A110566(n+1).
+20
1
1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 5, 3, 1, 1, 3, 5, 1, 3, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 11, 1, 1, 1, 1, 7, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 3, 1, 1, 1, 1, 11, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 25, 1, 1, 1, 1, 5
COMMENTS
a(n) is always an odd prime power, A061345.
EXAMPLE
A110566(4) through A110566(10) are {1,1,3,3,3,1,1}, therefore the factors are 1,3,1,1,3,1.
MATHEMATICA
f[n_] := LCM @@ Range[n]/Denominator[HarmonicNumber[n]]; Table[ LCM[f[n], f[n + 1]]/GCD[f[n], f[n + 1]], {n, 104}]
PROG
(PARI) f(n) = lcm(vector(n, k, k))/denominator(sum(k=1, n, 1/k));
a(n) = my(x = f(n+1)/f(n)); if (x > 1, x, 1/x); \\ Michel Marcus, Mar 07 2018
1, 6, 20, 21, 42, 120, 342, 506, 567, 594, 600, 610, 2184, 4896, 6108, 6162, 6498, 12760, 14067, 14157, 14201, 93942, 123462, 123519, 734413, 2451397, 4591010, 11571129, 13346540, 13619348, 13619790, 46180567
MATHEMATICA
c = 0; a = h = 1; t = {}; Do[a = LCM[a, n]; h = h + 1/n; b = a/Denominator[h]; If[b > c, c = b; AppendTo[t, n]], {n, 10^6}]; t
PROG
(PARI) lista(nn) = {rec = 0; for (n=1, nn, new = lcm(vector(n, k, k))/denominator(sum(k=1, n, 1/k)); if (new > rec, print1(n, ", "); rec = new); ); } \\ Michel Marcus, Mar 07 2018
a(n) is length of n-th run in A110566.
+20
1
5, 3, 9, 2, 1, 3, 1, 2, 6, 9, 2, 5, 5, 9, 3, 3, 3, 5, 4, 7, 12, 5, 5, 10, 1, 4, 31, 6, 7, 20, 9, 9, 9, 27, 29, 17, 5, 7, 35, 6, 1, 18, 2, 14, 29, 29, 29, 20, 2, 14, 6, 19, 4, 30, 8, 27, 6, 2, 8, 11, 4, 4, 19, 18, 5, 14, 18, 26, 11, 72, 10, 19, 6, 11, 22, 11, 33, 6, 5, 22, 4, 7, 99, 97, 2, 44, 9
MATHEMATICA
f[n_] := LCM @@ Range[n]/Denominator[HarmonicNumber[n]]; Length /@ Split[Table[f[n], {n, 1220}]]
a(n) is the least index k such that the n-th odd squarefree number A056911(n) divides A110566(k).
+20
0
1, 6, 20, 42, 33, 156, 20, 272, 342, 2058, 506, 377, 930, 77, 14406, 629, 162, 1640, 559, 2162, 4624, 1166, 110, 6498, 3422, 610, 342732, 4422, 506, 4970, 5256, 42, 6162, 6806
COMMENTS
According to a theorem proven by Shiu (2016), a(n) exists for all n.
EXAMPLE
-- ---------- -------- --------------------------
1 1 1 1 = 1 * 1
2 3 6 3 = 3 * 1
3 5 20 15 = 5 * 3
4 7 42 77 = 7 * 11
5 11 33 11 = 11 * 1
6 13 156 13 = 13 * 1
7 15 20 15 = 15 * 1
8 17 272 17 = 17 * 1
9 19 342 931 = 19 * 49
10 21 2058 1911 = 21 * 91
11 23 506 1725 = 23 * 75
12 29 377 319 = 29 * 11
13 31 930 3751 = 31 * 121
14 33 77 33 = 33 * 1
15 35 14406 2430488445 = 35 * 69442527
16 37 629 20313 = 37 * 549
17 39 162 39 = 39 * 1
18 41 1640 6519 = 41 * 159
19 43 559 645 = 43 * 15
20 47 2162 12831 = 47 * 273
21 51 4624 9537 = 51 * 187
22 53 1166 53 = 53 * 1
23 55 110 55 = 55 * 1
24 57 6498 419498967 = 57 * 7359631
25 59 3422 6431 = 59 * 109
26 61 610 41175 = 61 * 675
27 65 342732 974285 = 65 * 14989
28 67 4422 2211 = 67 * 33
29 69 506 1725 = 69 * 25
30 71 4970 2343 = 71 * 33
31 73 5256 7227 = 73 * 99
32 77 42 77 = 77 * 1
33 79 6162 91801713 = 79 * 1162047
34 83 6806 1200097 = 83 * 14459
MATHEMATICA
max = 64; osf = Select[Range[1, 64, 2], SquareFreeQ]; m = Length[osf]; c = 0; s = Table[0, {m}]; h = 0; lcm = 1; n = 1; While[c < m, h += 1/n; lcm = LCM[lcm, n]; r = lcm/Denominator[h]; Do[If[s[[k]] == 0 && Divisible[r, osf[[k]]], c++; s[[k]] = n], {k, 1, m}]; n++]; s
a(n) = (1/1 + 1/2 + ... + 1/n)*lcm{1,2,...,n}.
+10
18
1, 3, 11, 25, 137, 147, 1089, 2283, 7129, 7381, 83711, 86021, 1145993, 1171733, 1195757, 2436559, 42142223, 42822903, 825887397, 837527025, 848612385, 859193865, 19994251455, 20217344325, 102157567401, 103187226801, 312536252003, 315404588903, 9227046511387
COMMENTS
By Wolstenholme's theorem, if p > 3 is a prime, then p^2 | a(p-1).
Conjecture: for n > 3, if n^2 | a(n-1), then n is a prime.
Note that if n = p^2 with prime p > 3, then n | a(n-1).
It seems that composite numbers n such that n | a(n-1) are only the squares n = p^2 of primes p > 3.
Primes p such that p^3 | a(p-1) are the Wolstenholme primes A088164.
The n-th triangular number n(n+1)/2 | a(n) for n = 1, 2, 6, 4422, ... (End)
MAPLE
a:= n-> add(1/k, k=1..n)*ilcm($1..n):
PROG
(GAP) List([1..30], n->Sum([1..n], k->1/k)*Lcm([1..n])); # Muniru A Asiru, Apr 02 2018
(PARI) a(n) = sum(k=1, n, 1/k)*lcm([1..n]); \\ Michel Marcus, Apr 02 2018
(Magma) [HarmonicNumber(n)*Lcm([1..n]):n in [1..30]]; // Marius A. Burtea, Aug 07 2019
Numbers k such that lcm(1,2,3,...,k) equals the denominator of the k-th harmonic number H(k).
+10
18
1, 2, 3, 4, 5, 9, 10, 11, 12, 13, 14, 15, 16, 17, 27, 28, 29, 30, 31, 32, 49, 50, 51, 52, 53, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149
COMMENTS
Numbers k such that A110566(k) = 1.
Shiu (2016) conjectured that this sequence is infinite. - Amiram Eldar, Feb 02 2021
MATHEMATICA
Select[Range[250], LCM@@Range[ # ]==Denominator[HarmonicNumber[ # ]]&]
PROG
(PARI) isok(n) = lcm(vector(n, i, i)) == denominator(sum(i=1, n, 1/i)); \\ Michel Marcus, Mar 07 2018
(Python)
from fractions import Fraction
from sympy import lcm
k, l, h, A098464_list = 1, 1, Fraction(1, 1), []
while k < 10**6:
if l == h.denominator:
k += 1
l = lcm(l, k)
Least number k such that lcm{1,2,...,k}/denominator of harmonic number H(k) = 2n-1.
+10
14
1, 6, 105, 44, 63, 33, 156, 20, 272, 343, 38272753, 11881, 100, 66, 822, 28861, 77
COMMENTS
First occurrence of 2n-1 in A110566.
Sequence continues: a(18)=?, 1332, 162, 2758521, 24649, 21, a(24)=?, 294, a(26)=?, 1166, 110, 126059, 201957, 3660, 37553041, 344929, 296341, a(35)=?, 25155299, a(37)=?, 500, 42
MATHEMATICA
a = h = 1; t = Table[0, {100}]; Do[a = LCM[a, n]; h = h + 1/n; b = a/Denominator[h]; If[b < 101 && t[[(b + 1)/2]] == 0, t[[(b + 1)/2]] = n], {n, 500000}]; t
PROG
(Python)
from fractions import Fraction
from sympy import lcm
k, l, h = 1, 1, Fraction(1, 1)
while l != h.denominator*(2*n-1):
k += 1
l = lcm(l, k)
h += Fraction(1, k)
CROSSREFS
Cf. A110566, A098464, A112813, A112814, A112815, A112816, A112817, A112818, A112819, A112820, A112821.
Numbers k such that lcm(1,2,3,...,k)/3 equals the denominator of the k-th harmonic number H(k).
+10
13
6, 7, 8, 18, 19, 25, 26, 54, 55, 56, 57, 58, 59, 60, 61, 62, 72, 73, 74, 75, 76, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 216, 217, 218, 219, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231
MATHEMATICA
f[n_] := LCM @@ Range[n]/Denominator[ HarmonicNumber[n]]; Select[ Range[231], f[ # ] == 3 &]
PROG
(PARI) isok(n) = lcm(vector(n, i, i)) == 3*denominator(sum(i=1, n, 1/i)); \\ Michel Marcus, Mar 07 2018
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