Christian G. Bower wrote:
The 1's had not been tallied yet, so I entered them as A114913. Interesting thing is that the sequence is very similar to A111174
A114913 Values such that A114912(a(n))=1. Values such that A000009(a(n))==2 (mod 4).
That last line should be: A000009(n)==2 (mod 4).
3, 4, 8, 10, 13, 14, 17, 18, 19, 24, 25, 28, 32, 39, 42, 43, 47, 48, 50, 52, 54, 55, 62, 67, 69, 73, 74, 75, 76, 78, 83, 84, 87, 88, 89, 90, 95, 99, 101, 103, 105, 108, 109, 112, 113, 118, 119, 123, 125, 127, 130, 132, 134, 138, 140, 143, 144, 147, 149, 153, 154, 157
A111174 Numbers n such that 24*n + 1 is prime.
3, 4, 8, 10, 13, 14, 17, 18, 19, 24, 25, 28, 32, 39, 42, 43, 47, 48, 50, 52, 54, 55, 62, 67, 69, 73, 74, 75, 78, 83, 84, 87, 88, 89, 90, 95, 99, 103, 105, 108, 109, 112, 113, 118, 119, 123, 125, 127, 130, 132, 134, 138, 140, 143, 144, 147, 153, 154, 157, 158, 162 (list)
matching in the first 28 terms. A little more searching and it appears that every number in A111174 is in A114913, but there are probably infinitely many numbers in the latter not in the former.
Yes, for example if 24*n+1 = 25p for some prime p, then n will be in A114913 but not in A111174; see below.
I looked a little at Dean Hickerson's 8/4/2005 posting about q(n) mod 64.
I didn't fully understand it, but it does not appear to have an obvious connection to prime number theory.
Here's what I wrote about Q(n) mod 64:
I looked at Q(n) mod small powers of 2 a few years ago. I found that the value of Q(n) mod 64 is determined by representations of 24n+1 by the binary quadratic form x^2 + 24 y^2. Specifically:
Let f(x,y) be determined by the values of x mod 192 and y mod 2:
x mod 192 1 5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 f(x,even) 1 51 55 29 5 15 27 25 41 11 31 53 45 39 3 49 f(x,odd) 7 5 1 11 3 9 29 15 31 13 25 19 27 17 21 23
x mod 192 49 53 55 59 61 65 67 71 73 77 79 83 85 89 91 95 f(x,even) 17 35 7 13 21 63 43 9 57 59 47 37 61 23 19 33 f(x,odd) 23 21 17 27 19 25 13 31 15 29 9 3 11 1 5 7
x mod 192 97 101 103 107 109 113 115 119 121 125 127 131 133 137 139 143 f(x,even) 33 19 23 61 37 47 59 57 9 43 63 21 13 7 35 17 f(x,odd) 7 5 1 11 3 9 29 15 31 13 25 19 27 17 21 23
x mod 192 145 149 151 155 157 161 163 167 169 173 175 179 181 185 187 191 f(x,even) 49 3 39 45 53 31 11 41 25 27 15 5 29 55 51 1 f(x,odd) 23 21 17 27 19 25 13 31 15 29 9 3 11 1 5 7
Let b(n) be the sum of f(x,y) over all solutions of
24n+1 = x^2 + 24 y^2, x>0. (0)
Then
Q(n) == b(n) (mod 64). (1)
If we're only interested in Q(n) mod 4, we can simplify this to: Let c(n) be the number of solutions of (0). If 24n+1 is not a square or if sqrt(24n+1) == 1 or 11 (mod 12), then Q(n) == c(n) (mod 4). If sqrt(24n+1) == 5 or 7 (mod 12) then Q(n) == c(n) + 2 (mod 4). For example, if n=51, then the solutions of (0) are (x,y) = (7,#7), (19,#6), (25,#5), (29,#4), (35,0) (where "#" means plus or minus), so c(51)=9. Since sqrt(24n+1) = 35 == 11 (mod 12), we should have Q(51) == c(51) == 1 (mod 4). And in fact Q(51) = 4097 == 1 (mod 4). What follows is implied by the arithmetic of Q(sqrt(-6)); I'll skip the details. If 24n+1 is prime, then there's exactly one solution of (0), so Q(n) == 2 (mod 4). More generally, we can determine c(n) from the prime factorization of 24n+1: Let 24n+1 = p_1^e_1 * ... * p_r^e_r * q_1^f_1 * ... * q_s^f_s, where the p_i's are distinct primes == 1, 5, 7, or 11 (mod 24) and the q_i's are distinct primes == 13, 17, 19, or 23 (mod 24). If some f_i is odd, then c(n) = 0. Otherwise, c(n) = (e_1 + 1) * ... * (e_r + 1). So c(n) == 2 (mod 4) iff all of the f_i's are even and all but one of the e_i's are even and the one e_i which is odd is == 1 (mod 4). Since Q(n) and c(n) are both odd if 24n+1 is a square, we can replace c by Q in this: Q(n) == 2 (mod 4) iff all of the f_i's are even and all but one of the e_i's are even and the one e_i which is odd is == 1 (mod 4). In particular, if 24n+1 = 25p for some prime p, then c(n) == 2 (mod 4), so there are infinitely many n in A114913 but not in A111174. Dean Hickerson dean@math.ucdavis.edu