AW: a sequence from analysis

[Pfoertner, Hugo]

-----Ursprüngliche Nachricht-----
Von: N. J. A. Sloane [mailto:njas at research.att.com]
Gesendet am: 16 February, 2004 21:17
An: seqfan at ext.jussieu.fr
Betreff: a sequence from analysis


The following are a very nice pair of sequences
for which i could use more terms!

Thanks

Neil

%I A088403
%S A088403 0,227
%N A088403 Values m_0 = 0, m_1, m_2, ... associated with series T shown
below.
%C A088403 T = 1
%C A088403 - (1/2 + 1/4 + 1/6 + ... + 1/(2m_1))
%C A088403 + (1/3 + 1/5 + 1/7 + ... + 1/(2m_2+1))
%C A088403 - (1/(2m_1+2) + 1/(2m_1+4) + ... + 1/(2m_3)
%C A088403 + (1/(2m_2+3) + 1/(2m_2+5) + ... + 1/(2m_4+1))
%C A088403 - (1/(2m_3+2) + 1/(2m_3+4) + ... + 1/(2m_5)
%C A088403 + (1/(2m_4+3) + 1/(2m_4+5) + ... + 1/(2m_6+1))
%C A088403 - ...
%C A088403 so that the partial sums of the terms through the ends of the
rows are respectively 1, just < -2, just > 3, just < -4, just > 5, etc.
%C A088403 Every positive number appears exactly once as a denominator in T.
%C A088403 The series T is a divergent rearrangement of the conditionally
convergent series Sum_{j>=1} (-1)^j/j which has the entire real number
system as its set of limit points.
%D A088403 B. R. Gelbaum and J. M. H. Olmsted, Counterexamples in Analysis,
Holden-Day, San Francisco, 1964; see p. 55.
%e A088403 1 - (1/2 + 1/4 + 1/6 + ... + 1/454) = -2.002183354..., which is
just less than -2; so a(1) = m_1 = 227.
%O A088403 0,2
%K A088403 nonn,nice,more,bref
%Y A088403 Cf. A088507.
%A A088403 njas, Feb 16 2004



Neil, SeqFans,

I just wrote a little Fortran program using 64-bit floats to calculate the
next 2 terms in A088403:

C Neil Sloane's A088403
      implicit real*16 (p-z)
      integer*8  m1, m2, m3, m4
      s = 1.0Q0
      m1 = 0
C...+....1....+....2....+....3....+....4....+....5....+....6....+....7..
10    continue
      m1 = m1 + 1
      t = 1.0Q0/qext(m1+m1)
      s = s - t
      if ( s .lt. -2.0Q0 ) then
        write (*,*) m1, s
        m3 = m1
        goto 15
      endif
      goto 10
15    continue
C...+....1....+....2....+....3....+....4....+....5....+....6....+....7..
      m2 = 0
20    continue
      m2 = m2 + 1
      t = 1.0Q0/qext(m2+m2+1)
      s = s + t
      if ( s .gt. 3.0Q0 ) then
        write (*,*) m2, s
        m4 = m2
        goto 25
      endif
      goto 20
25    continue
C...+....1....+....2....+....3....+....4....+....5....+....6....+....7..
C Diagnosis to see progress
      target = 2.0Q0
      write (*,*) ' m3_start=', m3+1
30    continue
      m3 = m3 + 1
      t = 1.0Q0/qext(m3+m3)
      s = s - t
      if ( s .lt. target ) then
        write (*,*) m3, s
        target = target - 1.0Q0
      endif
      if ( s .lt. -4.0Q0 ) then
        write (*,*) m3, s
        goto 35
      endif
      goto 30
35    continue
C...+....1....+....2....+....3....+....4....+....5....+....6....+....7..
      write (*,*) ' m4_start=', m4+1
C Diagnosis to see progress
      target = -3.0Q0
40    continue
      m4 = m4 + 1
      t = 1.0Q0/qext(m4+m4+1)
      s = s + t
      if ( s .gt. target ) then
        write (*,*) m4, s
        target = target + 1.0Q0
      endif
      if ( s .gt. 5.0Q0 ) then
        write (*,*) m4, s
        goto 45
      endif
      goto 40
45    continue
      end

The results are:
                  227  -2.00218335417278301168821810857870
                22945   3.00002111305704094022709303989168
 m3_start=                   228
                 1681   1.99987586182063452263072644346311
                12422  0.999964322905817529888522274884975
                91784 -2.723741766763051976129695675715836E-0006
               678197  -1.00000027644030271358908625231417
              5011237  -2.00000008949500077609454932713499
             37028308  -3.00000000155465546011825940052659
            273604248  -4.00000000108863447148035343522068
            273604248  -4.00000000108863447148035343522068
 m4_start=                 22946
               169549  -2.99999788153012963960202669811616
              1252809  -1.99999966215429848913139095062578
              9257077 -0.999999952272842961021019165421510
             68401062  6.396099474124162027484980949915391E-0009
            505419285   1.00000000072408638564662275136320
           3734571452   2.00000000009008129850949083036703

....working, but without a chance to finish in a reasonable time.

 From an extrapolation the next term will be ~1.5066336*10^12.

If I made no mistakes, the first terms of A088403 will be

1, 227, 22945,  273604248, ~1.5066336*10^12

Hugo