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A323019
a(n) is the smallest k such that A316506(k) = n.
0
1, 2, 4, 8, 20, 40, 120, 520, 1560, 8840, 26520, 185640, 769080, 5383560, 28455960, 199191720, 1166694360, 8166860520, 61834801080, 432843607560, 3771922865880, 26403460061160, 275350369209240, 1927452584464680, 21201978429111480, 171543280017356520
OFFSET
0,2
COMMENTS
a(n) is the smallest k such that the rank of the multiplicative group of Gaussian integers modulo k is n.
FORMULA
a(0) = 1, a(1) = 2, a(2) = 4, a(3) = 8. Let p(n) be the n-th prime congruent to 1 modulo 4, q(n) be the n-th prime congruent to 3 modulo 4. Then there exists {i(n)} and {j(n)} such that i(2) = j(2) = i(3) = j(3) = 0; for n >= 4, if a(n-2)*p(i(n-2)+1) < a(n-1)*q(j(n-1)+1), then a(n) = a(n-2)*p(i(n-2)+1), i(n) = i(n-2) + 1, j(n) = j(n-2), or a(n) = a(n-1)*q(j(n-1)+1), i(n) = i(n-1), j(n) = j(n-1) + 1.
EXAMPLE
a(2) = 4, i(2) = 0, j(2) = 0;
a(3) = 8, i(3) = 0, j(3) = 0;
For n = 4, a(n-2)*p(i(n-2)+1) = a(2)*p(1) = 4*5 = 20, a(n-1)*q(j(n-1)+1) = a(3)*q(1) = 8*3 = 24. So a(4) = 20, i(4) = i(2) + 1 = 1, j(4) = j(2) = 0.
For n = 5, a(n-2)*p(i(n-2)+1) = a(3)*p(1) = 8*5 = 40, a(n-1)*q(j(n-1)+1) = a(4)*q(1) = 20*3 = 60. So a(5) = 40, i(5) = i(3) + 1 = 1, j(5) = j(3) = 0.
For n = 6, a(n-2)*p(i(n-2)+1) = a(4)*p(2) = 20*13 = 260, a(n-1)*q(j(n-1)+1) = a(5)*q(1) = 40*3 = 120. So a(6) = 120, i(6) = i(5) = 1, j(6) = j(5) + 1 = 1.
...
List of the multiplicative groups of Gaussian integers modulo members of this sequence:
a(0) = 1: the trivial group;
a(1) = 2: C_2;
a(2) = 4: C_2 X C_4;
a(3) = 8: C_2 X C_4 X C_4;
a(4) = 20: C_2 X C_4 X C_4 X C_4;
a(5) = 40: C_2 X C_4 X C_4 X C_4 X C_4;
a(6) = 120: C_2 X C_4 X C_4 X C_4 X C_4 X C_8;
a(7) = 520: C_2 X C_4 X C_4 X C_4 X C_4 X C_12 X C_12;
a(8) = 1560: C_2 X C_4 X C_4 X C_4 X C_4 X C_4 X C_12 X C_24;
a(9) = 8840: C_2 X C_4 X C_4 X C_4 X C_4 X C_4 X C_4 X C_48 X C_48;
a(10) = 26520: C_2 X C_4 X C_4 X C_4 X C_4 X C_4 X C_4 X C_8 X C_48 X C_48;
...
PROG
(PARI)
p(n) = my(i=0, k=0); while(i<n, k++; if(prime(k)%4==1, i++)); prime(k)
q(n) = my(i=0, k=0); while(i<n, k++; if(prime(k)%4==3, i++)); prime(k)
a(n) = if(n<3, 2^n, my(v=vector(n), s=vector(n), t=vector(n)); [v[2], v[3]]=[4, 8]; for(i=4, n, my(a=v[i-2]*p(s[i-2]+1), b=v[i-1]*q(t[i-1]+1)); if(a<b, [v[i], s[i], t[i]] = [a, s[i-2]+1, t[i-2]], [v[i], s[i], t[i]] = [b, s[i-1], t[i-1]+1])); v[n])
CROSSREFS
Cf. A316506.
Sequence in context: A283047 A236397 A272122 * A105319 A051389 A078006
KEYWORD
nonn
AUTHOR
Jianing Song, Jan 10 2019
STATUS
approved