A117484 Number of triangular numbers mod n.

1, 2, 2, 4, 3, 4, 4, 8, 4, 6, 6, 8, 7, 8, 6, 16, 9, 8, 10, 12, 8, 12, 12, 16, 11, 14, 11, 16, 15, 12, 16, 32, 12, 18, 12, 16, 19, 20, 14, 24, 21, 16, 22, 24, 12, 24, 24, 32, 22, 22, 18, 28, 27, 22, 18, 32, 20, 30, 30, 24, 31, 32, 16, 64, 21, 24, 34, 36, 24, 24, 36, 32, 37, 38, 22

Offset: 1

Author

Franklin T. Adams-Watters, Apr 25 2006

Keywords

• mult
• easy
• nonn
• nice

Comments

Same as A000224 (number of squares mod n) for n odd, since there we can divide by 2 and then complete the square.

From David Morales Marciel, Jul 13 2015: (Start)

a(n) is also the total number of vertices of an n-gon that are a "final vertex" of a bouncing pattern representing the modulo-n series (an image of the bouncing pattern is included in the LINKS section). It is defined by the following algorithm:

(1) Define counter c=1.

(2) Start at any desired vertex.

(3) Mark the current vertex as a "final vertex".

(4) Advance clockwise c vertices.

(5) Set c=c+1.

(6) Repeat from (3).

The pattern of "final vertices" is cyclic: after some repetitions of steps (3)-(6), the marking of vertices is repeated and it is possible to count how many vertices of the n-gon contain a "final vertex" mark.

Examples: trivial case: a(n)=1 (one vertex is always a "final vertex"). From that point following the algorithm: a(2)=2 (segment, both vertices are a "final vertex"), a(3)=2 (triangle, only two vertices are "final vertex"), etc.

(End)

Examples

			When n=3, there is no triangular number which is congruent to 2 (mod 3) but only == 0 or 1 (mod 3), so a(3) = 2. - _Robert G. Wilson v_, Sep 16 2015
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Nicholas John Bizzell-Browning, LIE scales: Composing with scales of linear intervallic expansion, Ph. D. Thesis, Brunel Univ. (UK, 2024). See p. 21.
• David Morales Marciel, Bouncing patterns of some modulo-n series (n-gons 3 to 14, 16, 17 and 27), according to the algorithm explained in the comments. Current sequence belongs to the bouncing pattern of the 3-gon sample (first at the top).

Crossrefs

Cf. A000224.

Programs

Maple

a:= proc(n) local F, f;
F:= ifactors(n)[2];
mul(seq(`if`(f[1]=2, 2^f[2], floor(f[1]^(f[2]+1)/(2*f[1]+2))+1), f=F))
end proc:
map(a, [$1..100]); # Robert Israel, Jul 13 2015

Mathematica

f[n_] := Block[{fi = FactorInteger@ n, k = t = 1}, lng = 1 + Length@ fi; While[k < lng, t = t*If[ fi[[k, 1]] == 2, 2^fi[[k, 2]], Floor[1 + fi[[k, 1]]^(fi[[k, 2]] + 1)/(2 + 2fi[[k, 1]]) ]]; k++]; t]; Array[f, 75] (* Robert G. Wilson v, Sep 16 2015, after Robert Israel *)

PARI

a(n) = {my(f = factor(n)); prod(i = 1, #f~, if(f[i,1] == 2, 2^f[i,2], (f[i,1]^(f[i,2]+1)\(2*f[i,1] + 2)) + 1));} \\ Amiram Eldar, Sep 05 2023

Formula

Multiplicative with a(2^e) = 2^e, a(p^e) = floor(p^(e+1)/(2p+2))+1 for p > 2.