A001000 -id:A001000 - cp's OEIS frontend

A024842 a(n) = least m such that if r and s in {1/2, 1/4, 1/6, ..., 1/2n} satisfy r < s, then r < k/m < (k+2)/m < s for some integer k.

11, 29, 55, 89, 131, 181, 253, 323, 417, 505, 621, 727, 865, 989, 1149, 1291, 1473, 1633, 1837, 2053, 2243, 2481, 2731, 2949, 3221, 3505, 3751, 4057, 4375, 4649, 4989, 5341, 5643, 6017, 6403, 6733, 7141, 7561, 7993, 8363, 8817, 9283, 9761, 10169, 10669, 11181, 11705

Offset: 2

Clark Kimberling

nonn

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 12 2012

Clark Kimberling, Table of n, a(n) for n = 2..100

Cf. A001000, A024841.

leastSeparatorS[seq_, s_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
s + 1 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Map[leastSeparatorS[1/(2*Range[50]), #] &, Range[5]];
t[[3]] (* A024842 *)
(* Peter J. C. Moses, Aug 06 2012 *)

A024843 a(n) = least m such that if r and s in {1/1, 1/2, 1/3, ..., 1/n} satisfy r < s, then r < k/m < (k+3)/m < s for some integer k.

Original entry on oeis.org

9, 23, 43, 69, 101, 139, 183, 233, 289, 361, 431, 518, 601, 703, 799, 916, 1025, 1157, 1279, 1426, 1561, 1723, 1871, 2048, 2209, 2401, 2601, 2783, 2998, 3221, 3423, 3661, 3907, 4129, 4390, 4659, 4901, 5185, 5477, 5739, 6046, 6361, 6643, 6973, 7311, 7613, 7966, 8327, 8649

Offset: 2

Clark Kimberling

nonn

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 08 2012

			Using the terminology introduced at A001000, the 4th separator of the set {1/3, 1/2, 1} is a(3) = 23, since 1/3 < 8/23 < 11/23 < 1/2 < 12/23 < 15/23 < 1 and 23 is the least m for which 1/3, 1/2, 1 are thus separated using numbers k/m. - _Clark Kimberling_, Aug 08 2012

Clark Kimberling, Table of n, a(n) for n = 2..100

Cf. A001000.

leastSeparatorS[seq_, s_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
s + 1 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Map[leastSeparatorS[1/Range[50], #] &, Range[5]];
TableForm[t]
t[[4]] (* Peter J. C. Moses, Aug 08 2012 *)

A024844 a(n) = least m such that if r and s in {1/1, 1/3, 1/5, ..., 1/(2n-1)} satisfy r < s, then r < k/m < (k+3)/m < s for some integer k.

Original entry on oeis.org

7, 28, 61, 106, 163, 232, 313, 406, 511, 647, 780, 946, 1105, 1301, 1486, 1712, 1923, 2179, 2416, 2702, 2965, 3281, 3570, 3916, 4231, 4607, 4999, 5356, 5778, 6216, 6613, 7081, 7565, 8002, 8516, 9046, 9523, 10083, 10659, 11176, 11782, 12404, 12961, 13613, 14281, 14878

Offset: 2

Clark Kimberling

nonn

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 12 2012
From Jianing Song, Aug 31 2022: (Start)
Smallest m such that ceiling(m/(2*j-1)) - floor(m/(2*j+1)) = 5 for 1 <= j <= n-1.
Obviously we have a(n) > 3/(1/(2*n-3) - 1/(2*n-1)) => a(n) >= 6*n^2 - 12*n + 5. On the other hand, a(n) <= 4/(1/(2*n-3) - 1/(2*n-1)) + 1 = 2*(2*n-1)*(2*n-3) + 1: if m >= 2*(2*n-1)*(2*n-3) + 1, then m/(2*j-1) - m/(2*j+1) > 4 => ceiling(m/(2*j-1)) - floor(m/(2*j+1)) = ceiling(m/(2*j-1)-floor(m/(2*j+1))) >= ceiling(m/(2*j-1) - m/(2*j+1)) >= 5 for 1 <= j <= n-1. (End)

Clark Kimberling, Table of n, a(n) for n = 2..100

Cf. A001000, A024845.

leastSeparatorS[seq_, s_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
s + 1 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Map[leastSeparatorS[1/(2*Range[50]-1), #] &, Range[5]];
t[[4]] (* A024844 *)
(* Peter J. C. Moses, Aug 06 2012 *)

a(n) = for(m=6*n^2-12*n+5, 8*n^2-16*n+7, forstep(j=n-1, 1, -1, if(-((-m)\(2*j-1)) - m\(2*j+1) < 5, break(), if(j==1, return(m))))) \\ Jianing Song, Aug 31 2022

A024845 a(n) = least m such that if r and s in {1/2, 1/4, 1/6, ..., 1/(2*n)} satisfy r < s, then r < k/m < (k+3)/m < s for some integer k.

Original entry on oeis.org

15, 41, 79, 129, 191, 265, 351, 449, 577, 703, 861, 1013, 1201, 1379, 1597, 1801, 2049, 2279, 2557, 2813, 3121, 3403, 3741, 4049, 4417, 4801, 5151, 5565, 5995, 6385, 6845, 7321, 7751, 8257, 8779, 9249, 9801, 10369, 10879, 11477, 12091, 12641, 13285, 13945

Offset: 2

Clark Kimberling

nonn

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 12 2012
From Jianing Song, Aug 31 2022: (Start)
Smallest m such that ceiling(m/(2*j)) - floor(m/(2*j+2)) >= 5 for 1 <= j <= n-1.
Obviously we have a(n) > 3/(1/(2*n-2) - 1/(2*n)) = 6*n*(n-1). On the other hand, a(n) <= 4/(1/(2*n-2) - 1/(2*n)) + 1 = 8*n*(n-1) + 1: if m >= 8*n*(n-1) + 1, then m/(2*j) - m/(2*j+2) > 4 => ceiling(m/(2*j)) - floor(m/(2*j+2)) = ceiling(m/(2*j)-floor(m/(2*j+2))) >= ceiling(m/(2*j) - m/(2*j+2)) >= 5 for 1 <= j <= n-1.
a(n) is odd for all n: for even m, we have ceiling(m/(2*j)) = ceiling((m-1)/(2*j)) (otherwise (m-1)/(2*j) would be an integer, which is impossible), so ceiling(m/(2*j)) - floor(m/(2*j+2)) >= 5 implies ceiling((m-1)/(2*j)) - floor((m-1)/(2*j+2)) >= 5. (End)

Clark Kimberling, Table of n, a(n) for n = 2..100

Cf. A001000, A024844.

leastSeparatorS[seq_, s_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
s + 1 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Map[leastSeparatorS[1/(2*Range[50]), #] &, Range[5]];
t[[4]] (* A024845 *)
(* Peter J. C. Moses, Aug 06 2012 *)

a(n) = for(m=6*n^2-6*n+1, 8*n^2-8*n+1, forstep(j=n-1, 1, -1, if(-((-m)\(2*j)) - m\(2*j+2) < 5, break(), if(j==1, return(m))))) \\ Jianing Song, Aug 31 2022

A024846 a(n) = least m such that if r and s in {1/1, 1/2, 1/3, ..., 1/n} satisfy r < s, then r < k/m < (k+4)/m < s for some integer k.

Original entry on oeis.org

11, 29, 55, 89, 131, 181, 239, 305, 379, 461, 551, 661, 769, 898, 1023, 1171, 1313, 1480, 1639, 1825, 2001, 2206, 2399, 2623, 2833, 3076, 3303, 3565, 3809, 4090, 4351, 4651, 4961, 5249, 5578, 5917, 6231, 6589, 6957, 7297, 7684, 8081, 8447, 8863, 9289, 9681, 10126

Offset: 2

Clark Kimberling

nonn

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 08 2012

			Using the terminology introduced at A001000, the 5th separator of the set {1/3, 1/2, 1} is a(3) = 29, since 1/3 < 10/29 < 14/29 < 1/2 < 15/29 < 19/29 < 1, and 29 is the least m for which 1/3, 1/2, 1 are thus separated using numbers k/m. - _Clark Kimberling_, Aug 08 2012

Clark Kimberling, Table of n, a(n) for n = 2..100

Cf. A001000.

leastSeparatorS[seq_, s_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
s + 1 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Map[leastSeparatorS[1/Range[50], #] &, Range[5]];
TableForm[t]
t[[5]] (* Peter J. C. Moses, Aug 08 2012 *)

A024851 Least m such that if r and s in {-F(2h) + tauF(2*h-1): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers) and tau = (1+sqrt(5))/2 (golden ratio).

Original entry on oeis.org

2, 5, 12, 30, 77, 200, 522, 1365, 3572, 9350, 24477, 64080, 167762, 439205, 1149852, 3010350, 7881197, 20633240

Offset: 2

Clark Kimberling

Possibly a duplicate of A188378. For a guide to related sequences, see A001000. - Clark Kimberling, Aug 09 2012

			Referring to the terminology introduced at A001000, m=5 is the (1st) separator of the set S = {f(1),f(2),f(3)}, where f(h) = - F(2*h) + tau*F(2*h-1).  That is, a(3) = 5, since 1/5 < f(3) < 2/5 < f(2) < 3/5 < f(1), whereas fractions k/m for m<5 do not separate the elements of S in this manner.

Cf. A001000.

f[n_] := f[n] = -Fibonacci[2 n] + GoldenRatio*Fibonacci[2 n - 1]
leastSeparator[seq_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
2 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Table[N[f[h], 40], {h, 1, 18}] (* A024851 *)
t1 = leastSeparator[t]
(* Peter J. C. Moses, Aug 01 2012 *)

Extended, corrected, and edited by Clark Kimberling, Aug 09 2012

a(19) from Sean A. Irvine, Jul 26 2019

A071111 a(n) is the least integer x such that there exists an integer in the open interval (x/(i+1), x/i) for i= n-1, n-2 ..., 3, 2, 1.

Original entry on oeis.org

3, 5, 7, 13, 17, 26, 31, 43, 57, 65, 82, 101, 111, 133, 157, 183, 197, 226, 257, 290, 307, 343, 381, 421, 463, 485, 530, 577, 626, 677, 703, 757, 813, 871, 931, 993, 1025, 1090, 1157, 1226, 1297, 1370, 1407, 1483, 1561, 1641, 1723, 1807, 1893, 1937, 2026, 2117

Offset: 2

Fernando Delgado, Paul Monasterios, and Adolfo Rodriguez (misterioso53(AT)hotmail.com), May 27 2002

a(10) = 57, a(100) = 8191, a(1000) = 937993.
a(n) is the least integer that for k=1, 2, ..., (n-1) can be expressed as: a(n)=p*k + b for some positive integers p and b such that p>1 and p>b>0.
This is the same sequence (apart from the initial term) as A001000. The identity of these two sequences was first proved by Rustem Aidagulov and a detailed version of the proof can be found in the Alekseyev link below.
Comments from Christopher Carl Heckman, May 23 2004: "This problem was given in Crux Mathematicorum, Vol. 23 #6 (October 1997) as Problem #2272. A solution, which includes a general formula, can be found in Crux Mathematicorum, Vol. 24 #7 (November 1998): a(n) = floor (((n + x_n) / 2)^2 + 1), where x_n = floor (n + 1 - 2 sqrt (n - 1)).
"This formula was found by Florian Herzig (then a student at Cambridge, UK), who also proved that the proposer's conjecture that a(n) = cases (1 + (n-m)^2, if m^2 <= n - 2, 1 + (n-m)^2 + (n - m), otherwise) where m = floor ((1 + sqrt (4 n - 7)) / 2) also is true although 'the proof of this fact is quite challenging'.
"The problem was also solved by Peter Tingley (then an undergraduate student at the University of Waterloo, Waterloo, Ontario), who gave the alternate formula: a(n) = n y_n + floor ((n - y_n)^2 / 4 + 1), where y_n = floor (n - 2 sqrt(n - 1) + 1), which 'is readily seen to be the same as the one obtained by Herzig.'"

			a(4)=7 because 2 lies in (7/4,7/3), 3 lies in (7/3,7/2) and 4, 5 and 6 lie in (7/2,7) and for x<7 the definition doesn't hold.

T. D. Noe, Table of n, a(n) for n=2..1000
Max Alekseyev, Proof that A001000 and A071111 are essentially the same sequence
Florian Herzig and Peter Tingley, Solutions to Problem 2272, Crux Mathematicorum 24:7 (1998), pp. 438-441.

test[x_, n_] := Module[{k}, For[k=n, x<=k(k-1), k--, If[Ceiling[x/(k-1)]-Floor[x/k]<2, Return[False]]]; True]; a[n_] := For[x=1, True, x++, If[test[x, n], Return[x]]]

For 2<=n<=200, a(n) = n^2 - n*c(n) + floor(c(n)^2/4) + 1, where c(n) = floor(sqrt(4n-5)). Is this true for all n>=2?

Edited by Dean Hickerson and Robert G. Wilson v, Jun 04 2002

A024819 a(n) = least m such that if r and s in {1/1, 1/3, 1/5,..., 1/(2n-1)} satisfy r < s, then r < k/m < s for some integer k.

Original entry on oeis.org

2, 4, 11, 16, 29, 37, 56, 67, 92, 121, 137, 172, 211, 254, 277, 326, 379, 436, 466, 529, 596, 667, 704, 781, 862, 947, 1036, 1082, 1177, 1276, 1379, 1486, 1597, 1654, 1771, 1892, 2017, 2146, 2279, 2347, 2486, 2629, 2776, 2927, 3082, 3161, 3322, 3487, 3656, 3829, 4006, 4187

Offset: 2

Clark Kimberling

nonn

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 07 2012

Clark Kimberling, Table of n, a(n) for n = 2..300

Cf. A001000.

leastSeparator[seq_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]]
< 2 + Floor[n #1[[2]]] &) /@
Partition[Take[seq, k], 2, 1], n++];
n, {k, 2, Length[seq]}]];
t = Table[1/(2 h - 1), {h, 1, 101}];
leastSeparator[t]
(* Peter J. C. Moses, Aug 01 2012 *)

A024820 a(n) = least m such that if r and s in {1/2, 1/4, 1/6,..., 1/2n} satisfy r < s, then r < k/m < s for some integer k.

Original entry on oeis.org

3, 5, 13, 19, 33, 41, 61, 85, 99, 129, 163, 181, 221, 265, 313, 339, 393, 451, 513, 545, 613, 685, 761, 841, 883, 969, 1059, 1153, 1251, 1301, 1405, 1513, 1625, 1741, 1861, 1923, 2049, 2179, 2313, 2451, 2593, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3699, 3873, 4051, 4233

Offset: 2

Clark Kimberling

nonn

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 07 2012

Clark Kimberling, Table of n, a(n) for n = 2..300

Cf. A001000.

leastSeparator[t_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
2 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[t, k], 2, 1], n++]; n, {k, 2, Length[t]}]];
t = Flatten[{1/(2*Range[60])}]
leastSeparator[t]

A024821 Least m such that if r and s in {1/sqrt(h): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k.

Original entry on oeis.org

4, 5, 9, 11, 17, 21, 25, 32, 40, 43, 55, 61, 67, 73, 87, 94, 105, 113, 125, 137, 145, 153, 166, 179, 188, 202, 216, 226, 246, 256, 271, 281, 297, 307, 329, 340, 351, 368, 385, 403, 421, 439, 451, 469, 481, 500, 519, 538, 551, 564, 584, 604, 624, 645, 666, 687, 708, 722, 743

Offset: 2

Clark Kimberling

nonn

For a guide to related sequences, see A001000. - Clark Kimberling, Aug 07 2012

Clark Kimberling, Table of n, a(n) for n = 2..200

Cf. A001000.

leastSeparator[seq_] := Module[{n = 1},
Table[While[Or @@ (Ceiling[n #1[[1]]] <
2 + Floor[n #1[[2]]] &) /@ (Sort[#1, Greater] &) /@
Partition[Take[seq, k], 2, 1], n++]; n, {k, 2, Length[seq]}]];
t = Flatten[Table[1/Sqrt[h], {h, 1, 60}]];
leastSeparator[t]
(* Peter J. C. Moses, Aug 01 2012 *)

cp's OEIS Frontend

A024842 a(n) = least m such that if r and s in {1/2, 1/4, 1/6, ..., 1/2n} satisfy r < s, then r < k/m < (k+2)/m < s for some integer k.

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A024843 a(n) = least m such that if r and s in {1/1, 1/2, 1/3, ..., 1/n} satisfy r < s, then r < k/m < (k+3)/m < s for some integer k.

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A024844 a(n) = least m such that if r and s in {1/1, 1/3, 1/5, ..., 1/(2n-1)} satisfy r < s, then r < k/m < (k+3)/m < s for some integer k.

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A024845 a(n) = least m such that if r and s in {1/2, 1/4, 1/6, ..., 1/(2*n)} satisfy r < s, then r < k/m < (k+3)/m < s for some integer k.

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A024846 a(n) = least m such that if r and s in {1/1, 1/2, 1/3, ..., 1/n} satisfy r < s, then r < k/m < (k+4)/m < s for some integer k.

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A024851 Least m such that if r and s in {-F(2*h) + tau*F(2*h-1): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers) and tau = (1+sqrt(5))/2 (golden ratio).

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A071111 a(n) is the least integer x such that there exists an integer in the open interval (x/(i+1), x/i) for i= n-1, n-2 ..., 3, 2, 1.

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A024819 a(n) = least m such that if r and s in {1/1, 1/3, 1/5,..., 1/(2n-1)} satisfy r < s, then r < k/m < s for some integer k.

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A024851 Least m such that if r and s in {-F(2h) + tauF(2*h-1): h = 1,2,...,n} satisfy r < s, then r < k/m < s for some integer k, where F = A000045 (Fibonacci numbers) and tau = (1+sqrt(5))/2 (golden ratio).