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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.

%I A275288 #50 Mar 09 2020 02:25:14
%S A275288 1,6,6,12,20,6,28,24,18,15,33,12,65,28,15,48,85,18,76,20,28,33,115,24,
%T A275288 100,52,54,28,145,30,217,96,33,85,35,36,296,95,52,40,246,42,301,55,45,
%U A275288 138,329,48,196,75,102,52,371,54,55,56,76,174,531,60,305,155
%N 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.
%C A275288 From _Robert Price_, Jan 04 2017: (Start)
%C A275288 a(11) = 33       [2,3,11,22,33]
%C A275288 65 >= a(13) > 26  [2,3,13,26,52,60,65]; no better solution with fewer than 15 terms.
%C A275288 48 >= a(16) > 32  [2,3,16,18,36,48]; no better solution with fewer than 24 terms.
%C A275288 85 >= a(17) > 34  [2,3,15,17,34,85]; no better solution with fewer than 12 terms.
%C A275288 76 >= a(19) > 19  [2,3,12,19,57,76]; no better solution with fewer than 12 terms.
%C A275288 a(20) = 20       [2,4,5,20]
%C A275288 a(21) = 28       [2,4,8,21,24,28]
%C A275288 a(22) = 33       [2,4,11,20,22,30,33]
%C A275288 115 >= a(23) > 23  [2,3,10,23,69,115]; no better solution with fewer than 11 terms.
%C A275288 a(24) = 24       [2,3,8,24]
%C A275288 100 >= a(25) > 25  [2,3,10,25,60,100]; no better solution with fewer than 11 terms.
%C A275288 52 >= a(26) > 26  [2,3,12,26,39,52]; no better solution with fewer than 16 terms.
%C A275288 54 >= a(27) > 27  [2,3,12,27,36,54]; no better solution with fewer than 9 terms.
%C A275288 a(28) = 28       [2,3,12,21,28]
%C A275288 145 >= a(29) > 29  [2,4,5,29,116,145]; no better solution with fewer than 9 terms.
%C A275288 a(30) = 30       [2,3,12,20,30]
%C A275288 217 >= a(31) > 31  [2,3,9,31,93,126,217]; no better solution with fewer than 9 terms.
%C A275288 96 >= a(32) > 32  [2,3,9,32,72,96]; no better solution with fewer than 11 terms.
%C A275288 a(33) = 33       [2,3,11,22,33]
%C A275288 85 >= a(34) > 34  [2,3,17,20,34,60,85]; no better solution with fewer than 9 terms.
%C A275288 a(35) = 35       [2,3,14,15,35]
%C A275288 a(36) = 36       [2,3,12,18,36]
%C A275288 296 >= a(37) > 37  [2,3,8,37,148,222,296]; no better solution with fewer than 8 terms.
%C A275288 95 >= a(38) > 38  [2,4,5,38,76,95]; no better solution with fewer than 11 terms.
%C A275288 52 >= a(39) > 39  [2,4,6,26,39,52]; no better solution with fewer than 15 terms.
%C A275288 a(40) = 40       [2,3,10,24,40]
%C A275288 246 >= a(41) > 41  [2,3,8,41,120,205,246]; no better solution with fewer than 9 terms.
%C A275288 a(42) = 42       [2,3,7,42]
%C A275288 192 >= a(64)      [2,3,8,48,64,192]; no better solution with fewer than 9 terms.
%C A275288 162 >= a(81)      [2,3,8,72,81,108,162]; no better solution with fewer than 9 terms.
%C A275288 384 >= a(128)     [2,3,7,96,128,336,384]; no better solution with fewer than 8 terms.
%C A275288 486 >= a(243)     [2,3,7,81,243,336,432,486]; no better solution with fewer than 9 terms.
%C A275288 a(216) = 216     [2,3,8,27,216]
%C A275288 196 >= a(49)      [2,3,8,49,98,168,196]; no better solution with fewer than 8 terms.
%C A275288 a(100) = 100     [2,4,5,25,100]
%C A275288 363 >= a(121)     [2,3,7,121,176,242,336,363]; no better solution with fewer than 8 terms.
%C A275288 a(144) = 144     [2,3,7,112,126,144]
%C A275288 a(196) = 196     [2 ,3,7,84,147,196]
%C A275288 a(225) = 225     [2,3,9,25,90,225]
%C A275288 a(500) = 500     [2,4,5,25,125,500]
%C A275288 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.
%C A275288 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.
%C A275288 Conjectures:
%C A275288    a(2^n) = 3*n
%C A275288    a(3^n) = 2*n
%C A275288    a(5^n) = 4*n
%C A275288    a(6^n) = n
%C A275288    a(7^n) = 4*n
%C A275288 (End)
%C A275288 From _Peter Kagey_, Jul 20 2017: (Start)
%C A275288 a(n) = n if and only if n is in A092671.
%C A275288 Every term in this sequence is in A092671.
%C A275288 a(a(n)) = a(n); that is, this sequence is idempotent.
%C A275288 (End)
%C A275288 From _Jon E. Schoenfield_, Feb 15 2020: (Start)
%C A275288 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).
%C A275288 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)
%H A275288 Jon E. Schoenfield, <a href="/A275288/b275288.txt">Table of n, a(n) for n = 1..10000</a>
%e A275288 a(1)  =  1 via [1]
%e A275288 a(2)  =  6 via [2, 3, 6]
%e A275288 a(3)  =  6 via [2, 3, 6]
%e A275288 a(4)  = 12 via [2, 4, 6, 12]
%e A275288 a(5)  = 20 via [2, 4, 5, 20]
%e A275288 a(6)  =  6 via [2, 3, 6]
%e A275288 a(7)  = 28 via [2, 4, 7, 14, 28]
%e A275288 a(8)  = 24 via [2, 3, 8, 24]
%e A275288 a(9)  = 18 via [2, 3, 9, 18]
%e A275288 a(10) = 15 via [2, 3, 10, 15]
%e A275288 a(11) > 30
%e A275288 a(12) = 12 via [2, 4, 6, 12]
%e A275288 a(13) > 30
%e A275288 a(14) = 28 via [2, 4, 7, 14, 28]
%e A275288 a(15) = 15 via [2, 3, 10, 15]
%e A275288 a(16) > 30
%e A275288 a(17) > 30
%e A275288 a(18) = 18 via [2, 3, 9, 18]
%e A275288 From _Jon E. Schoenfield_, Feb 15 2020: (Start)
%e A275288 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.
%e A275288 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.
%e A275288 (End)
%t A275288 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 *)
%Y A275288 Cf. A006255, A092671, A272036, A272083.
%K A275288 nonn
%O A275288 1,2
%A A275288 _Peter Kagey_, Aug 18 2016
%E A275288 a(11)-a(12) from _Robert Price_, Jan 07 2017
%E A275288 a(13)-a(58) from _David A. Corneth_, Jul 20 2017
%E A275288 a(59)-a(62) from _Jon E. Schoenfield_, Feb 15 2020