A272036 -id:A272036 - cp's OEIS frontend

A272083 Irregular triangle read by rows: strictly decreasing positive integer sequences in lexicographic order with the property that the sum of inverses equals one.

1, 6, 3, 2, 12, 6, 4, 2, 15, 10, 3, 2, 15, 12, 10, 4, 2, 15, 12, 10, 6, 4, 3, 18, 9, 3, 2, 18, 12, 9, 4, 2, 18, 12, 9, 6, 4, 3, 18, 15, 10, 9, 6, 2, 18, 15, 12, 10, 9, 4, 3, 20, 5, 4, 2, 20, 6, 5, 4, 3, 20, 12, 6, 5, 2, 20, 15, 10, 5, 4, 3, 20, 15, 12, 10, 5

Offset: 1

Peter Kagey, Apr 19 2016

			First six rows:
[1]                   because 1/1 = 1.
[6, 3, 2]             because 1/6 + 1/3 + 1/2 = 1.
[12, 6, 4, 2]         because 1/12 + 1/6 + 1/4 + 1/2 = 1.
[15, 10, 3, 2]        because 1/15 + 1/10 + 1/3 + 1/2 = 1.
[15, 12, 10, 4, 2]    because 1/15 + 1/12 + 1/10 + 1/4 + 1/2 = 1.
[15, 12, 10, 6, 4, 3] because 1/15 + 1/12 + 1/10 + 1/6 + 1/4 + 1/3 = 1.

Peter Kagey, Table of n, a(n) for n = 1..1578 (All rows with first term less than or equal to 30.)
Wikipedia, Erdős-Graham problem.

Cf. A073546, A216975, A216993, A272020, A272036, A272081, A272082.

A272034 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is the inverse of an integer.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 36, 38, 64, 128, 256, 512, 1024, 2048, 2056, 2080, 2088, 2090, 4096, 8192, 16384, 16896, 16900, 16902, 16928, 18944, 18952, 18954, 18988, 32768, 65536, 131072, 131328, 131332, 131334, 131360, 133376, 133384, 133386, 133420, 148224, 148256, 148258, 150284

Offset: 1

Michel Marcus, Apr 18 2016

nonn

That is, numbers such that A116416(n) is equal to 1.

The powers of 2 (A000079) form a subsequence.

			For n=36, 38_10=100100_2, and 1/3 + 1/6 = 1/2, the inverse of an integer.

Peter Kagey, Table of n, a(n) for n = 1..354

Cf. A000079, A116416, A116417, A272035, A272036, A272081.

Select[Range[2^18], IntegerQ[1/Total[1/# & /@ Flatten@ Position[Reverse@ IntegerDigits[#, 2], 1]]] &] (* Michael De Vlieger, Apr 18 2016 *)

isok(n) = {my(b = Vecrev(binary(n))); numerator(sum(k=1, #b, b[k]/k)) == 1;}

A272035 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is an integer.

Original entry on oeis.org

0, 1, 38, 39, 2090, 2091, 16902, 16903, 18954, 18955, 18988, 18989, 131334, 131335, 133386, 133387, 133420, 133421, 148258, 148259, 150284, 150285, 524314, 524315, 524348, 524349, 526386, 526387, 541212, 541213, 543250, 543251, 543284, 543285, 655644, 655645, 657682

Offset: 1

Michel Marcus, Apr 18 2016

nonn

That is, numbers such that A116416(n) equals 1.

2k is in this sequence if and only if 2k + 1 is. Therefore n + a(n) is odd for all n. - Peter Kagey, Apr 19 2016

			For n=39, 39_10=100111_2, and 1/1 + 1/2 + 1/3 + 1/6 = 2, an integer.

Peter Kagey, Table of n, a(n) for n = 1..450 (All terms less than 2^30)

Cf. A116416, A116417, A272034, A272036, A272082.

Select[Range[2^20], IntegerQ@ Total[1/Flatten@ Position[Reverse@ IntegerDigits[#, 2], 1]] &] (* Michael De Vlieger, Apr 18 2016 *)

isok(n) = {my(b = Vecrev(binary(n))); denominator(sum(k=1, #b, b[k]/k)) == 1;}

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

Peter Kagey, Aug 18 2016

nonn

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)

			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)

Jon E. Schoenfield, Table of n, a(n) for n = 1..10000

Cf. A006255, A092671, A272036, A272083.

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 *)

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

A291256 Numbers n such that Sum_{k>=1} digits(k)/k = 1 where digits() are the digits of n in base 10, the least significant digit having index 1.

Original entry on oeis.org

1, 20, 300, 2010, 4000, 50000, 100110, 102100, 200200, 300010, 302000, 400100, 600000, 7000000, 20001010, 20003000, 20101100, 20301000, 40000010, 40002000, 40100100, 40300000, 60001000, 80000000, 300000200, 300100010, 300102000, 300200100, 300400000, 320101000, 340100000

Offset: 1

Michel Marcus, Aug 21 2017

			20 is a term since 0/1 + 2/2 = 1.
2010 is a term since 0/1 + 1/2 + 0/3 + 2/4 = 1.

Michel Marcus and Giovanni Resta, Table of n, a(n) for n = 1..10000 (first 59 terms from Michel Marcus)

Cf. A272036 (analog in base 2).

ndig[n_] := Sort[Sum[e[[2]] 10^(1/e[[1]] - 1), {e, #}] & /@ Select[Tally /@ (Join[ {1/n}, #] & /@ IntegerPartitions[1 - 1/n, All, 1/Range[n]]), Max[Flatten[#]] < 10 &]]; Join @@ (ndig /@ Range[20]) (* Giovanni Resta, Aug 21 2017 *)

isok(n) = my(d = Vecrev(digits(n))); sum(k=1, #d, d[k]/k) == 1;

cp's OEIS Frontend

A272083 Irregular triangle read by rows: strictly decreasing positive integer sequences in lexicographic order with the property that the sum of inverses equals one.

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A272034 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is the inverse of an integer.

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A272035 Numbers n such that the sum of the inverse of the exponents in the binary expansion of 2n is an integer.

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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.

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A291256 Numbers n such that Sum_{k>=1} digits(k)/k = 1 where digits() are the digits of n in base 10, the least significant digit having index 1.

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