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 1-2 of 2 results.

A378880 a(n) = A378879(n) - A083025(n).

Original entry on oeis.org

0, 1, 1, 2, -1, 2, 1, 3, 2, 0, 1, 3, -1, 2, 0, 4, -1, 3, 1, 1, 2, 2, 1, 4, -2, 0, 3, 3, -1, 1, 1, 5, 2, 0, 0, 4, -1, 2, 0, 2, -1, 3, 1, 3, 1, 2, 1, 5, 2, -1, 0, 1, -1, 4, 0, 4, 2, 0, 1, 2, -1, 2, 3, 6, -2, 3, 1, 1, 2, 1, 1, 5, -1, 0, -1, 3, 2, 1, 1, 3, 4, 0
Offset: 1

Views

Author

Clark Kimberling, Jan 14 2025

Keywords

Examples

			a(10) = 0 because the factorization 2*5 has 1 each of a Pythagorean prime, 5, and a non-Pythagorean prime, 2.
		

Crossrefs

Programs

  • Mathematica
    f[{x_, y_}] := If[Mod[x, 4] == 1, y, -y];
    s[n_] := Map[f, FactorInteger[n]];
    p[n_] := {Total[Select[s[n], # > 0 &]], -Total[Select[s[n], # < 0 &]]};
    p[1] = {0, 0};
    t = Table[p[n], {n, 1, 135}]
    u = Map[First, t]   (* A083025 *)
    v = Map[Last, t]   (* A378879 *)
    v - u (* A377625 *)

Formula

Totally additive with a(p) = 1 if p = 2 or p == 3 (mod 4), and a(p) = -1 if p == 1 (mod 4). - Amiram Eldar, Jun 09 2025

A380657 Numbers whose prime factorization has more Pythagorean prime factors than non-Pythagorean prime factors (including multiplicities).

Original entry on oeis.org

5, 13, 17, 25, 29, 37, 41, 50, 53, 61, 65, 73, 75, 85, 89, 97, 101, 109, 113, 125, 130, 137, 145, 149, 157, 169, 170, 173, 175, 181, 185, 193, 195, 197, 205, 221, 229, 233, 241, 250, 255, 257, 265, 269, 275, 277, 281, 289, 290, 293, 305, 313, 317, 325, 337
Offset: 1

Views

Author

Clark Kimberling, Jan 30 2025

Keywords

Examples

			50 appears because 2*5*5 has 2 Pythagorean prime factors but only 1 non-Pythagorean prime factor.
		

Crossrefs

Programs

  • Mathematica
    f[{x_, y_}] := If[Mod[x, 4] == 1, y, -y];
    s[n_] := Map[f, FactorInteger[n]];
    p[n_] := {Total[Select[s[n], # > 0 &]], -Total[Select[s[n], # < 0 &]]};
    p[1] = {0, 0};
    t = Table[p[n], {n, 1, 500}];
    u = Map[First, t];  (* A083025 *)
    v = Map[Last, t] ;  (* A376961 *)
    v - u (* A377625 *);
    Flatten[Position[v - u, -1]]  (* this sequence *)
Showing 1-2 of 2 results.