Values of high degree -> Values of low degree.
Euler changed methods of computation.
So we must understand where he changed them.
The following values are correct.
02 0.452247420041065 50068 +156.50
s=36, 34, 32
[1/2^s]
s=32, 30, 28, 26, 24, 22, 20, 18, 16
[p.237]
A(2)+A(3)-A(6)-B(2)-B(3)+B(6)+1/6^s-(1/25^s+1/35^s+1/49^s)
A(a)=zeta(s)(1-1/a^s), B(a)=1-1/a^s
Round down (15)
Round down (20)
02 0.453891391902471 71689 ???
04 0.076993205698789 58289 ???
06 0.017070086856372 24580 ???
08 0.004061405366518 10701 ---
10 0.000993603574436 13435 ???
12 0.000246026470033 98293
14 0.000061244396725 99998 (-)
16 0.000015282026219
18 0.000003817278702
20 0.000000953961124 -1
22 0.000000238450446
24 0.000000059608184
26 0.000000014901555
28 0.000000003725333
30 0.000000000931326 -3
32 0.000000000232830
We can get close values by s=16.
s=14, 12, 10, 8
[p.234 middle]
Values of high degree -> Values of low degree.
d=20 (middle)
02 0.452247420041065 +157
04 0.076993139764247 +5
06 0.017070086850637 +2
08 0.004061405366518 -3
10 0.000993603574437 -804
12 0.000246026470035 -2
14 0.000061244396725
d=15 (middle)
02 0.452247420041065 +157
04 0.076993139764247 +5
06 0.017070086850637 +2
08 0.004061405366517 -2
10 0.000993603574437 -804
12 0.000246026470034 -1
14 0.000061244396725
s=10, 8, 6, 4,2
[p.234 low]
Values of high degree -> Values of low degree.
d=20 (low)
02 0.452247420041065 +157
04 0.076993139764246 +6
06 0.017070086850637 +2
08 0.004061405366518 -3
10 0.000993603574437 -804
d=15 (low)
02 0.452247420041065 +157
04 0.076993139764247 +5
06 0.017070086850637 +2
08 0.004061405366518 -3
10 0.000993603574437 -804
02 0.452247420041065 +157
04 0.076993139764246 +6
06 0.017070086850637 +2
08 0.004061405366518 -3
10 0.000993603574437 -804
12 0.000246026470035 -2
14 0.000061244396725
16 0.000015282026219
18 0.000003817278702
20 0.000000953961124 -1
22 0.000000238450446
24 0.000000059608184
26 0.000000014901555
28 0.000000003725333
30 0.000000000931326 -3
32 0.000000000232830
34 0.000000000058207
36 0.000000000014551
38 0.000000000003637
40 0.000000000000909
42 0.000000000000227
44 0.000000000000056
46 0.000000000000014
48 0.000000000000003
He might use "1/2^s" for s=32 and 34,
the method at p.237 from 16 to 30,
the method at p.234 (middle) from 8 (or 12) to 14,
the method at p.234 (low) from 2 to 6 (or 10).
Then we can express the indices 62 and 110=20*1+30*3.
p.234 (middle)