Q: prime number congruences and quadratic forms: A141778=A038977?

[Richard Mathar]

sequences, let me elaborate on what I know about these quadratic
statement  4ap=z^2 (mod D) which can be rephrased that 4ap is a square
symbol and the fact that the (4a/D)=+1 in the cases considered, we also
  The equations p=z^2-Dy^2 are solvable with D=5,13 or 17 according to
solvable (with y=1,z=5 or y=29,z=7 etc), such that the solutions for 8p=z^2-Dy^2
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Date: Tue, 22 Jul 2008 19:57:40 +0200
From: "=?ISO-8859-1?Q?Beno=EEt_Jubin?=" <benoit.jubin at gmail.com>
To: seqfan at ext.jussieu.fr
Subject: A014221: broken link and shifted sequence
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Dear SeqFans,

First, there is a broken link in A014221 (an alternate link is already given):
%H A014221 H. M. Friedman, <a
href="http://www.simple-solutions.cc/testbed12/pdf/finiteseq10_8_98.pdf">Long
finite sequences</a>, J. Comb. Theory, A 95 (2001), 102-144. [A faster
site]

Also, the sequence is defined by "a(0) = 0; for n >= 0, a(n+1) =
2^a(n)", so reads 0,1,2,4,16...
I suggest the same definition with a(0)=1 instead, which shifts the
sequence.  Then:
-we would have a(n)=2^^n, a power tower of n two's
-we would have a(n)=A_3(n) (as defined in the comments of A014221)
-some statements in A094358 would become true
-a comment in A062860 would read "A014221(n)"
-the first term of A038081 may (or may not) be deleted and the offset
set to 1 (are there rooted trees of height 0... I don't think so)
-the formula in A048830 would become true
-a comment in A115658 would read "A_3(n)"
-the formula in A48828 would... have to be checked
[this list of changes is exhaustive]

The 3-ish analog A014222 is correct but lacks in the definition the
initial value a(0)=1.
The definition of the 4-ish analog A114561 is unnecessarily involved
(and shifted).
The definition of the 3/2-ish analog A081651 is correct but with
nonstandard notation; I suggest: floor((3/2)^^n), where x^^n is the
power tower x^(x^(...^x).) with n x's.
(and the current statement in "example" is false)

For the 3 sequences corresponding to b=2,3,4, I suggest to be consistent:
either
a(n) = b^^n, the power tower b^(b^(...^b).) with n b's.
or
a(0) = 1 and for n >= 0, a(n+1) = b^a(n).

What do you think?

Benoit Jubin