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 1-4 of 4 results.

A216975 Triangle read by rows in which row n gives the lexicographically earliest minimal sum denominators among all possible n-term Egyptian fractions with unit sum.

Original entry on oeis.org

1, 0, 0, 2, 3, 6, 2, 4, 6, 12, 3, 4, 5, 6, 20, 3, 4, 6, 10, 12, 15, 3, 4, 9, 10, 12, 15, 18, 4, 5, 6, 9, 10, 15, 18, 20, 4, 6, 8, 9, 10, 12, 15, 18, 24, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 6, 7, 8, 9, 10, 12, 14, 15, 18, 24, 28, 6, 7, 9, 10, 11, 12, 14, 15, 18, 22, 28, 33, 7, 8, 9, 10, 11, 12, 14, 15, 18, 22, 24, 28, 33
Offset: 1

Views

Author

Robert Price, Sep 21 2012

Keywords

Comments

This sequence is the lexicographically earliest Egyptian fraction (denominators only) describing the minimal sum given in A213062.
Row 2 = [0,0] corresponds to the fact that 1 cannot be written as Egyptian fraction with 2 (distinct) terms.

Examples

			Row 5 = [3,4,5,6,20]: lexicographically earliest minimal sum (38) denominators among 72 possible 5-term Egyptian fractions with unit sum.
1 = 1/3 + 1/4 + 1/5 + 1/6 + 1/20.
Triangle begins:
1;
0, 0;
2, 3, 6;
2, 4, 6, 12;
3, 4, 5,  6, 20;
3, 4, 6, 10, 12, 15;

References

  • Mohammad K. Azarian, Sylvester's Sequence and the Infinite Egyptian Fraction Decomposition of 1, Problem 958, College Mathematics Journal, Vol. 42, No. 4, September 2011, p. 330. Solution published in Vol. 43, No. 4, September 2012, pp. 340-342

Links

Crossrefs

Cf. A030659, A073546, A213062, A216993.

A216993 Triangle read by rows in which row n gives the lexicographically earliest denominators with the least possible maximum value among all n-term Egyptian fractions with unit sum.

Original entry on oeis.org

1, 0, 0, 2, 3, 6, 2, 4, 6, 12, 2, 4, 10, 12, 15, 3, 4, 6, 10, 12, 15, 3, 4, 9, 10, 12, 15, 18, 3, 5, 9, 10, 12, 15, 18, 20, 4, 5, 8, 9, 10, 15, 18, 20, 24, 5, 6, 8, 9, 10, 12, 15, 18, 20, 24, 5, 6, 8, 9, 10, 15, 18, 20, 21, 24, 28, 4, 8, 9, 10, 12, 15, 18, 20, 21, 24, 28, 30, 4, 8, 9, 11, 12, 18, 20, 21, 22, 24, 28, 30, 33
Offset: 1

Views

Author

Robert Price, Sep 21 2012

Keywords

Comments

This sequence is the lexicographically earliest Egyptian fraction (denominators only) describing the minimum largest denominator given in A030659.
Row 2 = [0,0] corresponds to the fact that 1 cannot be written as an Egyptian fraction with 2 (distinct) terms.

Examples

			Row 5 = [2,4,10,12,15]: lexicographically earliest denominators with the least possible maximum value (15) among 72 possible 5-term Egyptian fractions equal to 1. 1 = 1/2 + 1/4 + 1/10 + 1/12 + 1/15.
Triangle begins:
  1;
  0, 0;
  2, 3,  6;
  2, 4,  6, 12;
  2, 4, 10, 12, 15;
  3, 4,  6, 10, 12, 15;

Links

Crossrefs

Cf. A030659, A073546, A213062, A216975, A378723.

A213062 Minimal sum x(1) +...+ x(n) such that 1/x(1) +...+ 1/x(n) = 1, the x(i) being n distinct positive integers.

Original entry on oeis.org

1, 0, 11, 24, 38, 50, 71, 87, 106, 127, 151, 185, 211, 249, 288, 325, 364, 406, 459, 508, 550, 613, 676, 728
Offset: 1

Views

Author

M. F. Hasler, Jun 03 2012

Keywords

Comments

The term a(2)=0 corresponds to the fact that 1 cannot be written as Egyptian fraction with 2 (distinct) terms.

Examples

			a(3) = 11 = 2 + 3 + 6, because 1/2+1/3+1/6 is the only Egyptian fraction with 3 terms having 1 as sum.
a(4) = 24 = 2 + 4 + 6 + 12 is the smallest sum of denominators among the six 4-term Egyptian fractions equal to 1.
a(5) = 38 = 3 + 4 + 5 + 6 + 20, least sum of denominators among 72 possible 5-term Egyptian fractions equal to 1.
a(6) = 50 = 3 + 4 + 6 + 10 + 12 + 15, least sum of denominators among 2320 possible 6-term Egyptian fractions equal to 1.
a(7) <= 71 = 3 + 5 + 20 + 6 + 10 + 12 + 15 (obtained from n=6 using 1/4 = 1/5 + 1/20).
a(8) <= 114 = 3 + 5 + 20 + 7 + 42 + 10 + 12 + 15 (obtained using 1/6 = 1/7 + 1/42).
a(9) <= 145 = 3 + 6 + 30 + 20 + 7 + 42 + 10 + 12 + 15 (obtained using 1/5 = 1/6 + 1/30).
a(10) <= 202 = 3 + 6 + 30 + 20 + 8 + 56 + 42 + 10 + 12 + 15 (obtained using 1/7 = 1/8 + 1/56).

Links

Crossrefs

Cf. A030659. - Alois P. Heinz, Sep 21 2012

Programs

  • PARI
    a(n,M=9e9,s=1,m=2)={ n==1 & return((numerator(s)==1 & 1 >= m*s || s==1)/s); sum( k=m,m+n-1,1/k ) < s & return; for(x=max(m,1\s+1),n\s, n*(x+(n-1)/2)>=M & break; (m=a(n-1, M-x, s-1/x, x+1)) & M=min(M,x+m)); M} /* For n>6, a good upper bound must be given as 2nd (optional) argument. Such a bound can be obtained using 1/x = 1/(x+1) + 1/x(x+1) in solutions for n-1, cf. Examples. */

Extensions

a(11)-a(24) from Robert Price, Aug 26 2012 - Sep 21 2012

A224910 Positive integer solutions to 1/p + 1/q + 1/r = 1 by lexicographic order.

Original entry on oeis.org

2, 3, 6, 2, 4, 4, 3, 3, 3
Offset: 0

Views

Author

Jean-François Alcover, Apr 19 2013

Keywords

Examples

			The sequence consists of the three triples {2, 3, 6}, {2, 4, 4}, {3, 3, 3}.

Links

Crossrefs

Cf. A030659.

Programs

  • Mathematica
    {p, q, r} /. {ToRules @ Reduce[0 < p <= q <= r <= 6 && 1/p + 1/q + 1/r == 1, {p, q, r}, Integers]} // Flatten
Showing 1-4 of 4 results.

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