A111233 -id:A111233 - cp's OEIS frontend

A217693 Numbers of distinct integers obtained from summing up subsets of {1, 1/2, 1/3, ..., 1/n}.

1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

Michel Marcus, Oct 11 2012

nonn

a(n) <= A111233(n).
a(n) <= floor(Sum_{k=1..n} 1/k) = A055980(n). - Joerg Arndt, Oct 13 2012
a(n) <= 4 for n <= 94, a(n) <= 5 for n <= 257, a(n) <= 6 for n <= 689. That is because if there is a term 1/a with p dividing a for a prime p, then there must be another term 1/b with p dividing b. Hence, not all terms from 1/1 to 1/n can be summed up. Cf. the "filter" function in my Sage script. - Manfred Scheucher, Aug 17 2015
a(k) = n for all k such that A101877(n) <= k < A101877(n+1). - Jon E. Schoenfield, May 12 2017

			1, 1/2 + 1/3 + 1/6 = 1 and 1 + 1/2 + 1/3 + 1/6 = 2 are integers, but only 2 of them are distinct, so a(6)=2.
a(24)=3 because 1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/8 + 1/9 + 1/10 + 1/15 + 1/18 + 1/20 + 1/24 = 3 and Sum_{k=1..n} 1/k < 4 for all n <= 30.
a(65)=4 because the sum of the reciprocals of the integers in { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 18, 20, 22, 24, 26, 27, 28, 30, 33, 35, 36, 40, 42, 45, 48, 52, 54, 56, 60, 63, 65 } is 4 and Sum_{k=1..n} 1/k < 5 for all n <= 82. - _Jon E. Schoenfield_, Apr 30 2018

P. Erdos and R. L. Graham, Old and new problems and results in combinatorial number theory, Université de Genève, 1980.

Manfred Scheucher, Sage Script
H. Yokota, On number of integers representable as sums of unit fractions, Canad. Math. Bull. Vol. 33 (2), 1990.
H. Yokota, On Number of Integers Representable as a Sum of Unit Fractions, II, Journal of Number Theory 67, 162-169, 1997.

Cf. A101877, A111233.

ufr(n) = {tab = []; for (i=1, 2^n - 1, vb = binary(i); while(length(vb) < n, vb = concat(0, vb););; val = sum(j=1, length(vb), vb[j]/j); if (denominator(val) == 1, tab = concat(tab, val); ); ); return (length(Set(tab))); }

a(25)-a(46) from Manfred Scheucher, Aug 17 2015

a(47)-a(87) from Jon E. Schoenfield, Apr 30 2018

A358332 Numbers whose prime indices have arithmetic and geometric mean differing by one.

Original entry on oeis.org

57, 228, 1064, 1150, 1159, 2405, 3249, 7991, 29785, 29999, 30153, 35378, 51984, 82211, 133931, 185193, 187039, 232471, 242592, 374599, 404225, 431457, 685207, 715129, 927288, 1132096, 1165519, 1322500, 1343281, 1555073, 1872413, 2016546, 2873687, 3468319, 4266421, 4327344

Offset: 1

Gus Wiseman, Nov 09 2022

nonn

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

			The terms together with their prime indices begin:
      57: {2,8}
     228: {1,1,2,8}
    1064: {1,1,1,4,8}
    1150: {1,3,3,9}
    1159: {8,18}
    2405: {3,6,12}
    3249: {2,2,8,8}
    7991: {18,32}
   29785: {3,4,9,12}
   29999: {32,50}
   30153: {2,8,9,9}
   35378: {1,4,4,8,8}
   51984: {1,1,1,1,2,2,8,8}
   82211: {50,72}
  133931: {4,8,8,16}
  185193: {2,2,2,8,8,8}
  187039: {72,98}
  232471: {12,18,27}

Max Alekseyev, Table of n, a(n) for n = 1..1458
Wikipedia, Geometric mean

The partitions with these Heinz numbers are counted by A358331.
A000040 lists the primes.
A001222 counts prime indices, distinct A001221.
A003963 multiplies together prime indices.
A056239 adds up prime indices.
A067538 counts partitions with integer average, ranked by A316413.
A067539 counts partitions with integer geometric mean, ranked by A326623.
A078175 lists numbers whose prime factors have integer average.
A320322 counts partitions whose product is a perfect power.
Cf. A000720, A051293, A111233, A215366, A289508, A289509, A326027, A326624, A326028, A326645, A357710.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Select[Range[10000],Mean[primeMS[#]]==1+GeometricMean[primeMS[#]]&]

isok(k) = if (k>1, my(f=factor(k), vf=List()); for (i=1, #f~, for (j=1, f[i,2], listput(vf, primepi(f[i,1])))); my(v = Vec(vf)); vecsum(v)/#v == 1 + sqrtn(vecprod(v), #v);); \\ Michel Marcus, Nov 11 2022

More terms from Michel Marcus, Nov 11 2022

cp's OEIS Frontend

A217693 Numbers of distinct integers obtained from summing up subsets of {1, 1/2, 1/3, ..., 1/n}.

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