Sequence & Its Numerators & Denominators
Pfoertner, Hugo
Hugo.Pfoertner at muc.mtu.de
Wed Feb 26 10:02:34 CET 2003
Leroy, SeqFans,
a few results (what I can do without BigNums):
M = 1
k n d
2 3 2
3 21 10
4 861 310
5 1275141 363010
6 2551762438701 594665194510
7 9546380157472159016030421 1871071000515058250871610
M = 2
2 4 3
3 56 33
4 9968 4785
5 294115808 118289985
6 242590126064151488 83574429584465985
7 1.582486013449121321571784280714996E+0035
4.753334809763917319511326686829158E+0034
M = 3
2 5 4
3 115 76
4 57155 31996
5 13457544835 6509938156
6 718532108172999980195 305202648617286353116
7 1.987460976488531436231264449305835E+0042
7.510423643158930656025467393343864E+0041
M = 4
2 6 5
3 204 145
4 225624 139345
5 266503910064 145175334145
6 361476924706582858606944 175835054422073316222145
7 6.497828978375256831345904988205142E+0047
2.851592252760362666671714028680662E+0047
M = 5
2 7 6
3 329 246
4 703073 465186
5 3125677649801 1851696597486
6 60424917190276611506698577 32367813634013973326521626
7 2.216749800419013098487257188463868E+0052
1.082678765177426393429230538456141E+0052
M = 6
2 8 7
3 496 385
4 1858016 1293985
5 25521830405056 16099866182785
6 4729979071868915640350500736 2724594016670394605535321985
7 1.600107574782880396587666104603987E+0056
8.474704862481751818275190171802905E+0055
M = 7
2 9 8
3 711 568
4 4346343 3149560
5 159613018433703 105743204607160
6 212090193707026209909082272423 129327569784001903386257059960
7 3.697339705210831279240508910851950E+0059
2.087293855962712766461507748054839E+0059
M = 8
2 10 9
3 980 801
4 9253160 6921441
5 813058161891920 560267953544961
6 6199569461904967962915175521440 3958143639579167493986989412481
7 3.565548649714140138596310964638573E+0062
2.119771925416881455775058496250262E+0062
M = 9
2 11 10
3 1309 1090
4 18274949 14029390
5 3518536622075629 2504301264532090
6 129043851071919331586456316579989 85574806231235220451224136679590
7 1.719566445968690086105739864765900E+0065
1.067091704188203041022890715625893E+0065
M =10
2 12 11
3 1704 1441
4 33947088 26631121
5 13332145853048736 9749706686973121
6 2037430153619986537231286982027072 1394901896233710650914056326731201
7 4.719524667825613862537115696813998E+0067
3.036590314739899779293393958023222E+0067
Fortran program:
C a[k+1] = a[k] + 1 / ( m + 1 / a[k] ), a[1] = 1
C Proposed by Leroy Quet
C Hugo Pfoertner, 26.02.2003
C
REAL*16 A, M, ONE, N, D, NN, DD, MXINT
PARAMETER ( ONE = 1.0Q0, MXINT=1.0Q34 )
INTEGER K, MM, I
DO 10 MM = 1, 10
WRITE (*,*) ' '
WRITE (*,*) ' M = ', MM
WRITE (10,*) ' M = ', MM
M = QEXT (MM)
A = ONE
N = ONE
D = ONE
DO 20 I = 2, 7
A = A + ONE / ( M + ONE / A )
NN = N * ( M*N + D + D )
DD = D * ( M*N + D )
WRITE (*,*) I, A, A - QEXT(NN)/QEXT(DD)
N = NN
D = DD
IF ( N .LE. MXINT .AND. D .LE. MXINT ) THEN
WRITE (10,1000) I, N, D
1000 FORMAT ( I3, 2F36.0 )
ELSE
WRITE (10,*) I, N, D
ENDIF
20 CONTINUE
10 CONTINUE
END
Currently I am too busy to submit the sequences to the OEIS.
Best Regards
Hugo Pfoertner
-----Ursprüngliche Nachricht-----
Von: Leroy Quet [mailto:qqquet at mindspring.com]
Gesendet am: 26 February, 2003 04:10
An: seqfan at ext.jussieu.fr
Betreff: Sequence & Its Numerators & Denominators
I just submitted this to sci.math. I already sent the m=1 example to
seq.fan a little while back. Perhaps someone could calculate the
numerator/denominator sequences for the first couple of m's and post them
to the EIS.
Thanks,
Leroy Quet
---------------
This is a generalization of the sequence discussed at:
http://mathforum.org/discuss/sci.math/t/482979
Given a fixed positive integer m;
Let a[1] = any nonzero rational where its denominator is relatively
prime to m.
Let:
1
a[k+1] = a[k] + ------------- , for all positive integers k.
m + 1 /a[k]
In linear mode:
a[k+1] = a[k] + 1/(m + 1 /a[k]) , for all positive integers k.
If n[k] = numerator of a[k] in reduced form,
and d[k] = denominator of a[k] in reduced form,
then, for k = all positive integers,
n[k+2] =
n[k+1]*(m*n[k+1] + n[k+1]^2/(2n[k]^2) -m^2 *n[k]^2/2),
and d[k+2] =
d[k+1]*(d[k+1] + d[k+1]^2/d[k]^2 -d[k]^2)
The above are from these recursions:
n[k+1] = n[k] *(m*n[k] + 2*d[k]),
d[k+1] = d[k] *(m*n[k] + d[k])
Noteworthy facts:
*d[2] involves m. But the recursion, for d[k+2] in terms of d[k] and
d[k+1], itself is independent of m.
*Each d[k+1] is divisible by d[k], and each n[k+1] is divisible by
n[k].
*(d[k+1]/d[k]) is relatively prime to (n[j+1]/n[j]) for every positive
integer k and every positive integer j.
Thanks,
Leroy Quet
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