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Smallest prime p == 1 (mod 8) such that Q(sqrt(p)) has class number 2n+1.
3

%I #16 Jul 20 2022 08:54:12

%S 17,257,401,577,1129,1297,13033,11321,11257,38569,7057,23593,27689,

%T 8761,56857,284561,63361,25601,24337,55441,458929,14401,32401,78401,

%U 70969,69697,376897,106537,41617,160001,193601,57601,197137,367721,414433,1506473,444089,331777,156817

%N Smallest prime p == 1 (mod 8) such that Q(sqrt(p)) has class number 2n+1.

%C It seems that a(n) < A355877(n) for most n. a(n) > A355877(n) for n = 0, 1, 6, 9, 15, 20, 35, ...

%H Jianing Song, <a href="/A355876/b355876.txt">Table of n, a(n) for n = 0..57</a>

%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Class_number_(number_theory)#Class_numbers_of_quadratic_fields">Class numbers of quadratic fields</a>

%H <a href="/index/Qua#quadfield">Index entries for sequences related to quadratic fields</a>

%e p = 257 is the smallest prime congruent to 1 modulo 8 such that Q(sqrt(p)) has class number 3, so a(1) = 257.

%o (PARI) a(n) = forprime(p=2, oo, if(p%8==1 && qfbclassno(p)==2*n+1, return(p)))

%Y Cf. A355878.

%Y Similar sequences: A002148 (p == 3 (mod 8)), A355877 (p == 5 (mod 8)), A002146 (p == 7 (mod 8)).

%K nonn

%O 0,1

%A _Jianing Song_, Jul 20 2022