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.

Showing 1-3 of 3 results.

A319894 Number of ordered pairs (i,j) with 0 < i < j < prime(n)/2 such that R(i^4,prime(n)) > R(j^4,prime(n)), where R(k,p) (with p an odd prime) denotes the unique integer r among 0,...,(p-1)/2 with k congruent to r or -r modulo p.

Original entry on oeis.org

0, 0, 0, 6, 3, 5, 10, 22, 51, 62, 58, 53, 100, 146, 194, 200, 185, 246, 242, 310, 374, 344, 422, 497, 540, 582, 652, 683, 768, 946, 916, 1011, 1180, 1294, 1108, 1387, 1592, 1656, 1829, 2050, 2048, 2386, 2365, 2186, 2184, 2770, 2902, 2890, 3296, 3292, 3754, 3063, 3562, 3650, 4184, 4391, 4164, 4506, 4812
Offset: 2

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Author

Zhi-Wei Sun, Sep 30 2018

Keywords

Comments

Conjecture: Let p be an odd prime, and let r(p) be the number of ordered pairs (i,j) with 0 < i < j < p/2 and R(i^4,p) > R(j^4,p), where R(k,p) denotes the unique integer r among 0,...,(p-1)/2 with k congruent to r or -r modulo p. Then r(p) is even if p == 3 (mod 4). Also, r(p) == (p-5)/8 (mod 2) if p == 5 (mod 8). When p == 1 (mod 8), r(p) is even if and only if 2 is a quartic residue modulo p.
See also A319311, A319480 and A319882 for similar conjectures.

Examples

			a(6) = 3 since prime(6) = 13, (R(1^4,13),R(2^4,13),...,R(6^4,13)) = (1,3,3,9,1,9), and (2,5), (3,5) and (4,5) are only pairs (i,j) with 0 < i < j < 13/2 and R(i^4,13) > R(j^4,13).
		

Crossrefs

Programs

  • Mathematica
    f[k_,p_]:=f[k,p]=Abs[Mod[PowerMod[k,4,p],p,-p/2]];Inv[p_]:=Inv[p]=Sum[Boole[f[i,p]>f[j,p]],{j,2,(p-1)/2},{i,1,j-1}];Table[Inv[Prime[n]],{n,2,60}]

A319903 Number of ordered pairs (i,j) with 0 < i < j < prime(n)/2 such that (i^8 mod prime(n)) > (j^8 mod prime(n)).

Original entry on oeis.org

0, 0, 1, 2, 7, 5, 10, 22, 45, 48, 68, 53, 104, 127, 146, 200, 203, 250, 288, 312, 387, 318, 450, 557, 536, 745, 664, 581, 722, 797, 986, 1011, 1082, 1474, 1294, 1317, 1608, 1684, 1893, 2096, 1898, 2297, 2333, 2090, 2467, 2652, 2836, 3352, 3698, 3326, 3380, 2981, 3778, 3902, 4165, 4743, 4350, 4652, 4240
Offset: 2

Views

Author

Zhi-Wei Sun, Oct 01 2018

Keywords

Comments

Conjecture 1: Let p be an odd prime, and let N be the number of ordered pairs (i,j) with 0 < i < j < p/2 and (i^8 mod p) > (j^8 mod p). When p == 1 (mod 8), we have 2 | N if and only if 2 is a quartic residue modulo p. Also, N is even if p == 3 (mod 8). When p == 5 (mod 8), we have N == (p-5)/8 (mod 2). If p == 7 (mod 8) then N == (h(-p)+1)/2 (mod 2), where h(-p) is the class number of the imaginary quadratic field Q(sqrt(-p)).
Conjecture 2: Let p be an odd prime, and let N' be the number of ordered pairs (i,j) with 0 < i < j < p/2 and R(i^8,p) > R(j^8,p), where R(k,p) denotes the unique integer r among 0,...,(p-1)/2 with k congruent to r or -r modulo p. When p == 9 (mod 16), we have 2 | N' if and only if 2 is a quartic residue modulo p. Also, N' == floor((p+1)/8) (mod 2) if p is not congruent to 9 modulo 16.
See also A319311, A319480, A319882 and A319894 for similar conjectures.

Examples

			a(4) = 1 since prime(4) = 7, and (R(1^8,7),R(2^8,7),R(3^8,7)) = (1,3,2) with R(2^8,7) > R(3^8,7).
a(5) = 2 since prime(5) = 11, and (R(1^8,11),...,R(5^8,11)) = (1,3,5,2,4) with R(2^8,11) > R(4^8,11), R(3^8,11) > R(4^8,11) and R(3^8,11) > R(5^8,11).
		

Crossrefs

Programs

  • Mathematica
    f[k_,p_]:=f[k,p]=PowerMod[k,8,p];Inv[p_]:=Inv[p]=Sum[Boole[f[i,p]>f[j,p]],{j,2,(p-1)/2},{i,1,j-1}];Table[Inv[Prime[n]],{n,2,60}]

A309012 Number of ordered pairs (i,j) with 0 < i < j < prime(n)/2 such that (i^16 mod prime(n)) > (j^16 mod prime(n)).

Original entry on oeis.org

0, 0, 0, 0, 3, 3, 0, 16, 21, 43, 30, 62, 77, 99, 129, 146, 203, 187, 228, 245, 252, 345, 372, 382, 402, 558, 570, 631, 663, 756, 901, 1114, 961, 1325, 1398, 1253, 1571, 1470, 1601, 1795, 2024, 1988, 2349, 2014, 2184, 2200, 2728, 3054, 3084, 3718, 3386, 3224, 3018, 3861, 3866, 4258, 4361, 4418, 5110, 4724
Offset: 1

Views

Author

Zhi-Wei Sun, Jul 06 2019

Keywords

Comments

Conjecture : Let p be an odd prime, and let N be the number of ordered pairs (i,j) with 0 < i < j < p/2 and (i^16 mod p) > (j^16 mod p). When p == 1 (mod 16), we have 2 | N. Also, N == |{0

Examples

			a(5) = 3 with prime(5) = 11, and (2^16 mod 11) = 9 greater than (3^16 mod 11) = 3, (4^16 mod 11) = 4 and (5^16 mod 11) = 5.
		

Programs

  • Mathematica
    r[p_]:=r[p]=Sum[Boole[PowerMod[j,16,p]>PowerMod[k,16,p]],{k,2,p/2},{j,1,k-1}];
    Print[Table[r[Prime[n]],{n,1,60}]]
Showing 1-3 of 3 results.