[QUOTE=3.14159;227786]Oh.

We can construct it similarly to how the Proth numbers are constructed. Should there be a k < b[sup]b[/sup] restriction? Or do we allow all the 4n + 1 primes in?[/QUOTE]

My sequences include even b = 1.

Good thought for a third way to do it, though: a sequence of all the primes of the form k*b^b+1, subject to k < b^b.

[QUOTE=CRGreathouse]Mt sequences include even b = 1.
[/QUOTE]

b can't be 1, or else it would be an exact duplicate of [URL="http://www.research.att.com/~njas/sequences/A000040"]this[/URL].

b has to be greater than 2.

[QUOTE=3.14159;227789]b can't be 1, or else it would be an exact duplicate of [URL="http://www.research.att.com/~njas/sequences/A000040"]this[/URL].[/QUOTE]

I suppose you didn't understand my suggestions, then, as all three included b = 1 and none of them are the same as A000040.

[QUOTE=CRGreathouse]I suppose you didn't understand my suggestions, then, as all three included b = 1 and none of them are the same as A000040.
[/QUOTE]

It would in fact be the same because:

(p-1) * 1[sup]1[/sup] + 1 = p

And therefore, can be any odd prime number.

[QUOTE=CRGreathouse;227785]We need to find a good sequence without arbitrary points. So no minimum of 60 on b and no maximum of 10,000 on k.

I can think of a few ways to do this, but I'll let you decide. One example: consider a two-dimensional array where each row corresponds to a b-value and the values in a row are the least k-values that produce a prime. Then read this array by antidiagonals (or some other reasonable way). The first column is A070855, while the first row is A000040.

Another way: for each prime p, give the largest b value such that there exists a k with k*b^b+1 = p. Then give the values for 2, 3, 5, ....

Maybe both of these are good.


These examples avoid not only the arbitrariness of bounds, but also make ordinary sequences or 'tabl's rather than 'tabf's.[/QUOTE]

in that case the anti-diagonals read 2,5,3,109 ? Or 2,3,5,109, neither are found in what I'm searching.

[QUOTE=science_man_88;227792]in that case the anti-diagonals read 2,5,3,109 ? Or 2,3,5,109, neither are found in what I'm searching.[/QUOTE]

I'm sure neither are in.

The canonical antidiagonal mapping functions for the OEIS are
[code]t1(n)=floor(-1/2+sqrt(2+2*n))
t2(n)=n-binomial(floor(1/2+sqrt(2+2*n)), 2)[/code]
so I get
2, 3, 5, 5, 13, 109, 7, 29,163, 257, ...
although flipping rows and columns gives the equally-valid
2, 5, 3, 109, 13, 5, 257, 163, 29, 7, ...
.

[QUOTE=CRGreathouse]That doesn't match any of my sequences, sorry.
[/QUOTE]

They are irrelevant to what I was discussing.

If there's no k < b[sup]b[/sup] restriction, b > 1.

If there is a k < b[sup]b[/sup] restriction, b ≥ 1.

[QUOTE=3.14159;227796]They are irrelevant to what I was discussing.[/QUOTE]

I was quite precise, actually. Admittedly, I didn't define how to read by antidiagonals, but there's a standard way to do that in the OEIS. But regardless, none of them could be interpreted as A000040 except by a person who thinks "antidiagonal" means "row".

[QUOTE=CRGreathouse;227795]I'm sure neither are in.

The canonical antidiagonal mapping functions for the OEIS are
[code]t1(n)=floor(-1/2+sqrt(2+2*n))
t2(n)=n-binomial(floor(1/2+sqrt(2+2*n)), 2)[/code]
so I get
2, 3, 5, 5, 13, 109, 7, 29,163, 257, ...
although flipping rows and columns gives the equally-valid
2, 5, 3, 109, 13, 5, 257, 163, 29, 7, ...
.[/QUOTE]

closest I find is [url]http://www.research.att.com/~njas/sequences/A097453[/url]

[QUOTE=CRGreathouse]I was quite precise, actually. Admittedly, I didn't define how to read by antidiagonals, but there's a standard way to do that in the OEIS. But regardless, none of them could be interpreted as A000040 except by a person who thinks "antidiagonal" means "row".
[/QUOTE]

If there is no k < b[sup]b[/sup] restriction, and b = 1 is allowed, it is precisely the same as A000040.

I will begin work on the sequence with the k < b[sup]b[/sup] restriction.