A092666 -id:A092666 - cp's OEIS frontend

A020473 Egyptian fractions: number of partitions of 1 into reciprocals of positive integers <= n.

1, 2, 3, 5, 6, 13, 14, 24, 34, 60, 61, 168, 169, 252, 627, 1011, 1012, 2430, 2431, 7212, 15024, 16553, 16554, 50219, 60008, 64284, 92071, 260178, 260179, 844846, 844847, 1431187, 2608883, 2661217, 7946814, 22692855, 22692856, 22911815, 36004488, 120859171

Offset: 1

David W. Wilson

nonn

Number of ways to represent 1 = Sum_{k=1..n} b(k)/k, where the b(k) >= 0. - Franklin T. Adams-Watters, Aug 01 2006

Toshitaka Suzuki, Table of n, a(n) for n = 1..390
Index entries for sequences related to Egyptian fractions

Cf. A038034, A092666, A092670, A208480.

Table[Length[IntegerPartitions[1, All, 1/Range[n]]], {n, 1, 20}] (* Ben Branman, Apr 21 2012 *)

a(n) = Sum(A092666(i), i=1..n).

For prime p, a(p) = a(p-1) + 1. - Max Alekseyev, May 07 2012

A092669 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), 0

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 3, 0, 0, 5, 0, 11, 0, 0, 0, 19, 0, 0, 0, 73, 0, 86, 0, 0, 163, 0, 203, 286, 0, 0, 0, 803, 0, 1399, 0, 0, 2723, 0, 0, 4870, 0, 0, 0, 8789, 0, 13937, 14987, 42081, 0, 0, 0, 85577, 0, 0, 159982, 0, 117889, 437874, 0, 0, 0, 818640, 0

Offset: 1

Max Alekseyev, Mar 02 2004

nonn

For a given n, the Mathematica program uses backtracking to count the solutions. The solutions can be printed by uncommenting the print statement. It is very time-consuming for large n. A092671 gives the n that yield a(n) > 0. - T. D. Noe, Mar 26 2004

			a(6) = 1 since there is the only fraction 1 = 1/2+1/3+1/6.

Toshitaka Suzuki, Table of n, a(n) for n = 1..610
Harry Ruderman and Paul Erdős, Problem E2427: Bounds of Egyptian fraction partitions of unity, Amer. Math. Monthly, Vol. 81, No. 7 (1974), 780-782.
Index entries for sequences related to Egyptian fractions

Cf. A092666, A092667, A092670, A092671, A092672, A006585.

n=20; try2[lev_, s_] := Module[{nmim, nmax, si, i}, AppendTo[soln, 0]; If[lev==1, nmin=2, nmin=1+soln[[ -2]]]; nmax=n-1; Do[If[iT. D. Noe, Mar 26 2004 *)

a(n) = A092670(n) - A092670(n-1).

More terms from T. D. Noe, Mar 26 2004

More terms from T. Suzuki (suzuki(AT)scio.co.jp), Nov 24 2006

A092667 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), with max{x_i}=n.

Original entry on oeis.org

1, 1, 1, 4, 1, 44, 1, 235, 921, 4038, 1, 66147, 1, 304383, 6581754, 45353329, 1, 1100311690, 1, 44423279911, 1250831952086, 284120133400, 1, 71664788693247, 511162204140999, 55479698795314, 10715917223431762, 505603414069366830, 1, 28696102343693431631, 1, 857699266471525509621, 30399386408588668316839, 63063040603038091480

Offset: 1

Max Alekseyev, Mar 02 2004

nonn

			a(4) = 4 since there are four fractions 1=1/2+1/4+1/4, 1=1/4+1/2+1/4, 1=1/4+1/4+1/2 and 1=1/4+1/4+1/4+1/4.

Index entries for sequences related to Egyptian fractions

Cf. A092666, A038034, A092669, A002967, A000058.

a(n) = A038034(n) - A038034(n-1).

a(n) = 1 if n is prime.

A259633 a(n) = number of inequivalent necklaces with beads labeled 1/i (1 <= i <= n) such that the sum of the beads is 1 and the smallest bead is 1/n.

Original entry on oeis.org

1, 1, 1, 2, 1, 12, 1, 43, 132, 547, 1, 7834, 1, 30442, 608887, 3834978, 1, 84536629, 1, 3030450058, 79538220753, 16701983083, 1, 4136127573912, 26625599501697, 2730194738935

Offset: 1

Gordon Hamilton, Jul 02 2015

"Equivalence" refers to the cyclic group. Turning over is not allowed.

The original definition referred to slices of pie with slices of size 1/i, which add to a full pie.

			a(6) = 12 because a pie can be made in the following twelve ways (moving clockwise from a 1/6):
1 = 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6,
1 = 1/6 + 1/6 + 1/6 + 1/4 + 1/4,
1 = 1/6 + 1/6 + 1/4 + 1/6 + 1/4,
1 = 1/6 + 1/4 + 1/4 + 1/3,
1 = 1/6 + 1/4 + 1/3 + 1/4,
1 = 1/6 + 1/3 + 1/4 + 1/4,
1 = 1/6 + 1/6 + 1/6 + 1/2,
1 = 1/6 + 1/6 + 1/6 + 1/6 + 1/3,
1 = 1/6 + 1/6 + 1/3 + 1/3,
1 = 1/6 + 1/3 + 1/6 + 1/3,
1 = 1/6 + 1/3 + 1/2,
1 = 1/6 + 1/2 + 1/3.
Notice that the bottom two pies are chiral copies of one another.

Cf. A092666.

a(p) = 1 for all primes.

a(6) corrected, a(8) confirmed, a(9)-a(17) added by Alois P. Heinz, Jul 28 2015
a(18)-a(23) from Alois P. Heinz, Jul 30 2015
a(24)-a(26) from Alois P. Heinz, Sep 01 2015

A333437 Triangle read by rows: T(n,k) is the number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k , with 0 < x_1 <= ... <= x_k = n.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 1, 3, 2, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 3, 3, 2, 1, 0, 0, 0, 0, 2, 2, 3, 2, 1, 0, 0, 0, 1, 3, 6, 7, 5, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 3, 8, 15, 21, 24, 20, 11, 4, 1

Offset: 1

Seiichi Manyama, Mar 24 2020

			1 = 1/2 + 1/6 + 1/6 + 1/6 = 1/3 + 1/3 + 1/6 + 1/6 = 1/3 + 1/4 + 1/4 + 1/6. So T(6,4) = 3.
Triangle begins:
n\k  | 1  2  3  4  5   6   7   8   9  10 11 12
-----+----------------------------------------
   1 | 1;
   2 | 0, 1;
   3 | 0, 0, 1;
   4 | 0, 0, 1, 1;
   5 | 0, 0, 0, 0, 1;
   6 | 0, 0, 1, 3, 2,  1;
   7 | 0, 0, 0, 0, 0,  0,  1;
   8 | 0, 0, 0, 1, 3,  3,  2,  1;
   9 | 0, 0, 0, 0, 2,  2,  3,  2,  1;
  10 | 0, 0, 0, 1, 3,  6,  7,  5,  3,  1;
  11 | 0, 0, 0, 0, 0,  0,  0,  0,  0,  0, 1;
  12 | 0, 0, 0, 3, 8, 15, 21, 24, 20, 11, 4, 1;

Index entries for sequences related to Egyptian fractions

Row sums give A092666.

Cf. A020473, A333496.

T(n,n) = 1.

If n is prime, T(n,k) = 0 for 1 <= k < n.

A333496 Least k of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k , with 0 < x_1 <= ... <= x_k = n.

Original entry on oeis.org

1, 2, 3, 3, 5, 3, 7, 4, 5, 4, 11, 4, 13, 5, 4, 5, 17, 4, 19, 4, 5, 7, 23, 4, 8, 8, 6, 5, 29, 5, 31, 6, 5, 10, 5, 5, 37

Offset: 1

Seiichi Manyama, Mar 24 2020

			One of solutions
  n  |  [x_1, x_2, ... , x_a(n)]
-----+--------------------------------------
   4 | [2,  4,  4]
   6 | [2,  3,  6]
   8 | [2,  4,  8,  8]
   9 | [2,  6,  9,  9,  9]
  10 | [2,  5,  5, 10]
  12 | [2,  3, 12, 12]
  14 | [2,  7,  7,  7, 14]
  15 | [2,  3, 10, 15]
  16 | [2,  4,  8, 16, 16]
  18 | [2,  3,  9, 18]
  20 | [2,  4,  5, 20]
  21 | [2,  3, 14, 21, 21]
  22 | [2, 11, 11, 11, 11, 11, 22]
  24 | [2,  3,  8, 24]
  25 | [2,  4, 20, 25, 25, 25, 25, 25]
  26 | [2, 13, 13, 13, 13, 13, 13, 26]
  27 | [2,  3, 18, 27, 27, 27]
  28 | [2,  3, 12, 21, 28]
  30 | [2,  3, 10, 30, 30]
  32 | [2,  3, 16, 24, 32, 32]
  33 | [2,  3, 11, 22, 33]
  34 | [2, 17, 17, 17, 17, 17, 17, 17, 17, 34]
  35 | [2,  3, 14, 15, 35]
  36 | [2,  3,  9, 36, 36]

Index entries for sequences related to Egyptian fractions

Cf. A020473, A092666, A333437.

a(n) <= n.
a(m * n) <= a(n) + m - 1.
If p is prime, a(p) = p.
If m is odd, a(2 * m) <= (m - 1)/2 + 2 because 1 = 1/2 + (m - 1)/2 * 1/m + 1/(2 * m).

cp's OEIS Frontend

A020473 Egyptian fractions: number of partitions of 1 into reciprocals of positive integers <= n.

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A092669 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), 0

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A092667 a(n) = number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k (for any k), with max{x_i}=n.

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A259633 a(n) = number of inequivalent necklaces with beads labeled 1/i (1 <= i <= n) such that the sum of the beads is 1 and the smallest bead is 1/n.

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A333437 Triangle read by rows: T(n,k) is the number of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k , with 0 < x_1 <= ... <= x_k = n.

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A333496 Least k of Egyptian fractions 1 = 1/x_1 + ... + 1/x_k , with 0 < x_1 <= ... <= x_k = n.

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