cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A323019 a(n) is the smallest k such that A316506(k) = n.

Original entry on oeis.org

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

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Author

Jianing Song, Jan 10 2019

Keywords

Comments

a(n) is the smallest k such that the rank of the multiplicative group of Gaussian integers modulo k is n.

Examples

			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;
...
		

Crossrefs

Cf. A316506.

Programs

  • PARI
    p(n) = my(i=0, k=0); while(i
    				

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.