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A132918
Identity matrix with interpolated zeros.
5
1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
OFFSET
1,1
COMMENTS
Identity matrix with interpolated zeros, such that the diagonal of an infinite lower triangular matrix = (1, 0, 1, 0, 1, ...) with the rest zeros.
Given A132918 = matrix M, then M * any sequence as a vector replaces alternate terms with zeros; e.g., M * [1,2,3,...] = [1,0,3,0,5,...]. M * (any infinite lower triangular matrix) replaces alternate rows with zeros; e.g., M * A007318 = (1; 0,0; 1,2,1; 0,0,0,0; ...).
a(n) = 1 if n is a positive hexagonal number, A000384: (1, 6, 15, 28, 45, 66, ...); zero otherwise.
FORMULA
a(n) = lim_{x -> n} sqrt(2) * sin(Pi*x)/(u*(sin(Pi*u/4) + cos(Pi*u/4))), where u = sqrt(1 + 8*x). (Put y = x + n. If n is not of the form N*(2*N - 1) then the limit as y -> 0 is easily seen to be 0; if n = N*(2*N - 1) then the limit has the indeterminate form 0/0. Apply L'Hôpital's rule.) - Peter Bala, Sep 13 2025
EXAMPLE
First few rows of the triangle:
1;
0, 0;
0, 0, 1;
0, 0, 0, 0;
0, 0, 0, 0, 1;
...
MATHEMATICA
A132918[n_] := Boole[IntegerQ[(Sqrt[8*n + 1] + 1)/4]]; Array[A132918, 120] (* or *)
Normal[SparseArray[Map[# -> 1 &, PolygonalNumber[6, Range[8]]]]] (* or *)
A132918row[n_] := PadLeft[{Boole[OddQ[n]]}, n];
Array[A132918row, 15] (* Paolo Xausa, Oct 31 2025 *)
PROG
(PARI) a(n) = ispolygonal(n, 6); \\ Michel Marcus, Aug 04 2023
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Gary W. Adamson, Sep 05 2007
EXTENSIONS
More terms from Michel Marcus, Mar 06 2022
STATUS
approved