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A117484 Number of triangular numbers mod n. 3
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 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

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)

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

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).

FORMULA

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

EXAMPLE

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

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 *)

CROSSREFS

Cf. A000224.

Sequence in context: A094950 A087874 A166267 * A086835 A046701 A140472

Adjacent sequences:  A117481 A117482 A117483 * A117485 A117486 A117487

KEYWORD

mult,easy,nonn

AUTHOR

Franklin T. Adams-Watters, Apr 25 2006

STATUS

approved

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Last modified July 5 03:37 EDT 2020. Contains 335459 sequences. (Running on oeis4.)