cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A272036 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is equal to 1.

Original entry on oeis.org

1, 38, 2090, 16902, 18954, 18988, 131334, 133386, 133420, 148258, 150284, 524314, 524348, 526386, 541212, 543250, 543284, 655644, 657682, 657716, 672568, 674580, 8388742, 8390794, 8390828, 8405666, 8407692, 8520098, 8522124, 8536962, 8536996, 8539048, 8913052, 8915090
Offset: 1

Views

Author

Michel Marcus, Apr 18 2016

Keywords

Comments

That is, numbers such that both A116416(n) and A116417(n) are equal to 1.
Intersection of A272034 and A272035.
A number m with an exponent k in the binary sum must have another power of 2 having an exponent at least A275288(k). - David A. Corneth, Apr 01 2017

Examples

			For n=38, 2*38_10 = 2^6 + 2^3 + 2^2 = 1001100_2, and 1/2 + 1/3 + 1/6 = 1.

Links

Crossrefs

Cf. A116416, A116417, A272034, A272035, A272083, A275288.

Programs

  • Mathematica
    Select[Range[2^20], Total[1/Flatten@ Position[Reverse@ IntegerDigits[#, 2], 1]] == 1 &] (* Michael De Vlieger, Apr 18 2016 *)
  • PARI
    is(n) = my(b = Vecrev(binary(n))); sum(k=1, #b, b[k]/k) == 1;

A272081 Irregular triangle read by rows: strictly decreasing positive integer sequences in lexicographic order with the property that the sum of inverses is the inverse of an integer.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 6, 3, 6, 3, 2, 7, 8, 9, 10, 11, 12, 12, 4, 12, 6, 12, 6, 4, 12, 6, 4, 2, 13, 14, 15, 15, 10, 15, 10, 3, 15, 10, 3, 2, 15, 10, 6, 15, 12, 10, 15, 12, 10, 4, 15, 12, 10, 4, 2, 15, 12, 10, 6, 4, 3, 16, 17, 18, 18, 9, 18, 9, 3, 18, 9, 3, 2, 18, 9
Offset: 1

Views

Author

Peter Kagey, Apr 19 2016

Keywords

Examples

			First 18 rows:
  [1]           because 1 is self-inverse.
  [2]           because 1/2 is the inverse of an integer.
  [3]
  [4]           (...)
  [5]
  [6]
  [6, 3]        because 1/6 + 1/3              = 1/2.
  [6, 3, 2]     because 1/6 + 1/3 + 1/2        = 1/1.
  [7]
  [8]
  [9]           (...)
  [10]
  [11]
  [12]
  [12, 4]       because 1/12 + 1/4             = 1/3.
  [12, 6]       because 1/12 + 1/6             = 1/4.
  [12, 6, 4]    because 1/12 + 1/6 + 1/4       = 1/2.
  [12, 6, 4, 2] because 1/12 + 1/6 + 1/4 + 1/2 = 1/1.

Links

Crossrefs

Cf. A272020, A272034, A272082, A272083.

A272082 Irregular triangle read by rows: strictly decreasing positive integer sequences in lexicographic order with the property that the sum of inverses is an integer.

Original entry on oeis.org

1, 6, 3, 2, 6, 3, 2, 1, 12, 6, 4, 2, 12, 6, 4, 2, 1, 15, 10, 3, 2, 15, 10, 3, 2, 1, 15, 12, 10, 4, 2, 15, 12, 10, 4, 2, 1, 15, 12, 10, 6, 4, 3, 15, 12, 10, 6, 4, 3, 1, 18, 9, 3, 2, 18, 9, 3, 2, 1, 18, 12, 9, 4, 2, 18, 12, 9, 4, 2, 1, 18, 12, 9, 6, 4, 3, 18, 12
Offset: 1

Views

Author

Peter Kagey, Apr 19 2016

Keywords

Examples

			First 8 rows:
[1]                because 1/1 is an integer
[6, 3, 2]          because 1/6 + 1/3 + 1/2 = 1.
[6, 3, 2, 1]       because 1/6 + 1/3 + 1/2 + 1/1 = 2.
[12, 6, 4, 2]      because 1/12 + 1/6 + 1/4 + 1/2 = 1.
[12, 6, 4, 2, 1]   because 1/12 + 1/6 + 1/4 + 1/2 + 1/1 = 2.
[15, 10, 3, 2]     because 1/15 + 1/10 + 1/3 + 1/2 = 1.
[15, 10, 3, 2, 1]  because 1/15 + 1/10 + 1/3 + 1/2 + 1/1 = 2.
[15, 12, 10, 4, 2] because 1/15 + 1/12 + 1/10 + 1/4 + 1/2 = 1.

Links

Crossrefs

Cf. A272020, A272035, A272081, A272083.

A275288 Least k such that there exists a sequence b_1 < b_2 < ... < b_t = k that includes n and has a reciprocal sum of 1.

Original entry on oeis.org

1, 6, 6, 12, 20, 6, 28, 24, 18, 15, 33, 12, 65, 28, 15, 48, 85, 18, 76, 20, 28, 33, 115, 24, 100, 52, 54, 28, 145, 30, 217, 96, 33, 85, 35, 36, 296, 95, 52, 40, 246, 42, 301, 55, 45, 138, 329, 48, 196, 75, 102, 52, 371, 54, 55, 56, 76, 174, 531, 60, 305, 155
Offset: 1

Views

Author

Peter Kagey, Aug 18 2016

Keywords

Comments

From Robert Price, Jan 04 2017: (Start)
a(11) = 33 [2,3,11,22,33]
65 >= a(13) > 26 [2,3,13,26,52,60,65]; no better solution with fewer than 15 terms.
48 >= a(16) > 32 [2,3,16,18,36,48]; no better solution with fewer than 24 terms.
85 >= a(17) > 34 [2,3,15,17,34,85]; no better solution with fewer than 12 terms.
76 >= a(19) > 19 [2,3,12,19,57,76]; no better solution with fewer than 12 terms.
a(20) = 20 [2,4,5,20]
a(21) = 28 [2,4,8,21,24,28]
a(22) = 33 [2,4,11,20,22,30,33]
115 >= a(23) > 23 [2,3,10,23,69,115]; no better solution with fewer than 11 terms.
a(24) = 24 [2,3,8,24]
100 >= a(25) > 25 [2,3,10,25,60,100]; no better solution with fewer than 11 terms.
52 >= a(26) > 26 [2,3,12,26,39,52]; no better solution with fewer than 16 terms.
54 >= a(27) > 27 [2,3,12,27,36,54]; no better solution with fewer than 9 terms.
a(28) = 28 [2,3,12,21,28]
145 >= a(29) > 29 [2,4,5,29,116,145]; no better solution with fewer than 9 terms.
a(30) = 30 [2,3,12,20,30]
217 >= a(31) > 31 [2,3,9,31,93,126,217]; no better solution with fewer than 9 terms.
96 >= a(32) > 32 [2,3,9,32,72,96]; no better solution with fewer than 11 terms.
a(33) = 33 [2,3,11,22,33]
85 >= a(34) > 34 [2,3,17,20,34,60,85]; no better solution with fewer than 9 terms.
a(35) = 35 [2,3,14,15,35]
a(36) = 36 [2,3,12,18,36]
296 >= a(37) > 37 [2,3,8,37,148,222,296]; no better solution with fewer than 8 terms.
95 >= a(38) > 38 [2,4,5,38,76,95]; no better solution with fewer than 11 terms.
52 >= a(39) > 39 [2,4,6,26,39,52]; no better solution with fewer than 15 terms.
a(40) = 40 [2,3,10,24,40]
246 >= a(41) > 41 [2,3,8,41,120,205,246]; no better solution with fewer than 9 terms.
a(42) = 42 [2,3,7,42]
192 >= a(64) [2,3,8,48,64,192]; no better solution with fewer than 9 terms.
162 >= a(81) [2,3,8,72,81,108,162]; no better solution with fewer than 9 terms.
384 >= a(128) [2,3,7,96,128,336,384]; no better solution with fewer than 8 terms.
486 >= a(243) [2,3,7,81,243,336,432,486]; no better solution with fewer than 9 terms.
a(216) = 216 [2,3,8,27,216]
196 >= a(49) [2,3,8,49,98,168,196]; no better solution with fewer than 8 terms.
a(100) = 100 [2,4,5,25,100]
363 >= a(121) [2,3,7,121,176,242,336,363]; no better solution with fewer than 8 terms.
a(144) = 144 [2,3,7,112,126,144]
a(196) = 196 [2 ,3,7,84,147,196]
a(225) = 225 [2,3,9,25,90,225]
a(500) = 500 [2,4,5,25,125,500]
It appears that in most cases a(n) is a small multiple of n. For example: a(8)=3*8, a(11)=3*11, a(35)=1*35.
If not a small multiple of n, then a small rational times n. For example: a(10)=3/2*10, a(21)=4/3*21, a(22)=3/2*22.
Conjectures:
a(2^n) = 3*n
a(3^n) = 2*n
a(5^n) = 4*n
a(6^n) = n
a(7^n) = 4*n
(End)
From Peter Kagey, Jul 20 2017: (Start)
a(n) = n if and only if n is in A092671.
Every term in this sequence is in A092671.
a(a(n)) = a(n); that is, this sequence is idempotent.
(End)
From Jon E. Schoenfield, Feb 15 2020: (Start)
For any n > 1, let P be the largest divisor of n that is either a prime (p) or prime power (p^e, where e > 1). Then a(n) >= m*P where m is the smallest integer such that p divides the numerator of the sum of some subset of the set of unit fractions {1/1, 1/2, 1/3, ..., 1/m} that includes 1/(n/P).
Conjecture (confirmed for all n <= 40000): for all n > 1, the lower bound given above is tight, i.e., a(n) = m*P where m and P are as defined above. (See Example section.) (End)

Examples

			a(1)  =  1 via [1]
a(2)  =  6 via [2, 3, 6]
a(3)  =  6 via [2, 3, 6]
a(4)  = 12 via [2, 4, 6, 12]
a(5)  = 20 via [2, 4, 5, 20]
a(6)  =  6 via [2, 3, 6]
a(7)  = 28 via [2, 4, 7, 14, 28]
a(8)  = 24 via [2, 3, 8, 24]
a(9)  = 18 via [2, 3, 9, 18]
a(10) = 15 via [2, 3, 10, 15]
a(11) > 30
a(12) = 12 via [2, 4, 6, 12]
a(13) > 30
a(14) = 28 via [2, 4, 7, 14, 28]
a(15) = 15 via [2, 3, 10, 15]
a(16) > 30
a(17) > 30
a(18) = 18 via [2, 3, 9, 18]
From _Jon E. Schoenfield_, Feb 15 2020: (Start)
For n=31, the largest prime or prime power divisor of n is P=31, and the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5, 1/6} has no subset sum that includes 1/(n/P) = 1/1 and has a numerator divisible by 31, but the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7} does have such a subset sum, namely, 1/1 + 1/3 + 1/7 = 31/21, so a(31) >= 7*31 = 217. In fact, the numbers 1*31=31, 3*31=93, and 7*31=217 are elements of many sets of integers that include n=31, include no element > 217, and have a reciprocal sum of 1 (one such set is {2,3,12,28,31,93,217}), so a(31)=217.
For n=62, the largest prime or prime power divisor of n is again P=31, and the set of unit fractions {1/1, 1/2, 1/3, 1/4} has no subset sum that includes 1/(n/P) = 1/2 and has a numerator divisible by 31, but the set of unit fractions {1/1, 1/2, 1/3, 1/4, 1/5} does have such a subset sum, namely, 1/2 + 1/3 + 1/5 = 31/30, so a(62) >= 5*31 = 155. In fact, the numbers 2*31=62, 3*31=93, and 5*31=155 are elements of many sets of integers that include n=62, include no element > 155, and have a reciprocal sum of 1 (one such set is {2,3,12,20,62,93,155}), so a(62)=155.
(End)

Links

Crossrefs

Cf. A006255, A092671, A272036, A272083.

Programs

  • Mathematica
    Table[SelectFirst[Range@ 20, MemberQ[Map[Total, 1/DeleteCases[Rest@ Subsets[Range@ #, #], w_ /; FreeQ[w, n]]], 1] &] /. k_ /; MissingQ@ k -> 0, {n, 12}] (* Michael De Vlieger, Aug 18 2016, Version 10.2, values of a(n) > 20 appear as 0 *)

Extensions

a(11)-a(12) from Robert Price, Jan 07 2017
a(13)-a(58) from David A. Corneth, Jul 20 2017
a(59)-a(62) from Jon E. Schoenfield, Feb 15 2020

A294651 Least possible value for the highest denominator in the decomposition of unity as a sum of different unitary fractions the greatest of which is 1/n.

Original entry on oeis.org

1, 6, 15, 20, 24, 28, 33, 40, 48, 52, 65, 65, 75, 76, 85, 88, 91, 100, 105, 115, 115, 119, 132, 140, 144, 145, 155, 161, 162, 171, 217, 174, 182, 190, 195, 196, 296, 200, 207, 220, 246, 224, 301, 231, 238, 253, 329, 275, 280, 287, 288, 296, 371, 300, 304, 305
Offset: 1

Views

Author

Javier Múgica, Nov 06 2017

Keywords

Comments

The decompositions need not be unique. E.g., for a(7) either 1/12 or 1/20 + 1/30 may be used in the decomposition indifferently.
For prime numbers p and any fixed epsilon < 1, a(p) > epsilon*p*log(p) for all sufficiently large p.

Examples

			1 = 1/3 + 1/4 + 1/6 + 1/10 + 1/12 + 1/15, and there is no such decomposition starting at 1/3 and having a greatest denominator smaller than 15, so a(3)=15.

Links

Crossrefs

Cf. A192881, which looks at decompositions with the least possible number of terms. Those from this sequence achieve those bounds up to a(7), with exception of a(3). However, n=7 is likely the last value of n for which this holds.
Cf. A272083.

Extensions

a(18)-a(24) from Jon E. Schoenfield, Dec 22 2019
a(25)-a(56) from Jon E. Schoenfield, Jan 01 2020

A305442 Number of subsets of {1, 2, ..., n} such that the sum of the reciprocals is strictly less than 1.

Original entry on oeis.org

1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 501, 918, 1686, 3110, 5724, 10543, 19435, 35857, 66198, 122294, 226135, 418351, 774372, 1434089, 2657205, 4925796, 9135403, 16949546, 31460330, 58415177, 108502732, 201603881, 374707879, 696649896, 1295562234, 2410000999
Offset: 0

Views

Author

Peter Kagey, Jun 01 2018

Keywords

Examples

			For n = 4 the a(4) = 7 subsets are:
{}     because 0 < 1,
{2}    because 1/2 < 1,
{2, 3} because 1/2 + 1/3 = 5/6 < 1,
{2, 4} because 1/2 + 1/4 = 3/4 < 1,
{3}    because 1/3 < 1,
{3, 4} because 1/3 + 1/4 = 7/12 < 1, and
{4}    because 1/4 < 1.

Crossrefs

Cf. A092669, A092670, A212657, A272083.

Programs

  • Mathematica
    a[n_] := 1 + Length@ Select[Subsets[Range[2,n], {1, n-1}], Total[1/#] < 1  &]; Array[a, 15] (* Giovanni Resta, Jun 01 2018 *)

Formula

a(n) = A212657(n) - A092670(n).

Extensions

a(26)-a(36) from Giovanni Resta, Jun 01 2018
Showing 1-6 of 6 results.

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