I take that back, you did try to help, and it's my responsibility to pursue
this further. If M49* is confirmed and revealed, I'll re-tool and re-run my
old program.

This may help:
[YOUTUBE]ewS1Kkzl8mw[/YOUTUBE]
[YOUTUBE]vSdws014e4k[/YOUTUBE]
(Prof.Caffo kindly makes his course [URL="https://www.youtube.com/playlist?list=PLpl-gQkQivXjqHAJd2t-J_One_fYE55tC"]fully available[/URL] without registration.)

You can estimate the confidence interval for [TEX]\beta_1[/TEX]. And when you do, estimate how many data points you need to confidently establish that one slope fits data any better that the other. It will not matter if you have 48 data points or 49; you may rather need more than a hundred to get a decent probability of discerning these two close slopes.

Now, Wagstaff conjecture [I]has[/I] reasons whether you are capable to evaluate them or not. And "your" conjecture's reason is "it is 3/2 because it is 3/2"?

[QUOTE=Batalov;421907]This may help:...
(Prof.Caffo kindly makes his course [URL="https://www.youtube.com/playlist?list=PLpl-gQkQivXjqHAJd2t-J_One_fYE55tC"]fully available[/URL] without registration.)

You can estimate the confidence interval for [TEX]\beta_1[/TEX]. And when you do, estimate how many data points you need to confidently establish that one slope fits data any better that the other. It will not matter if you have 48 data points or 49; you may rather need more than a hundred to get a decent probability of discerning these two close slopes.

Now, Wagstaff conjecture [I]has[/I] reasons whether you are capable to evaluate them or not. And "your" conjecture's reason is "it is 3/2 because it is 3/2"?[/QUOTE]

Thanks for the video links - it'll take a little time for me to absorb them.

My conjecture has three basic parts, but the 3/2 comes from a close fit for
M1 thru M39 and a near exact match for M23. But if it's right, M50 will almost
have to be big, like over 100M, and like M51 around say 200M just to make up
for the ground it lost by the more recent finds in 30M - 55M.

The other two parts of the conjecture are:
the ratio of consecutive MP exponents is bounded (by ten),
and every number of digits in decimal is represented by
at least one MP exponent with lead digit equal to "1".

The 3/2 is very Pythagorean. I believe someday a
formula will be discovered with (3/2)^n the dominant term.