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A120869 a(n) is the number j for which there exists a unique pair (j,k) of positive integers such that (j+k+1)^2-4*k=13*n^2. 3
1, 4, 9, 3, 9, 17, 3, 12, 23, 1, 13, 27, 43, 12, 29, 48, 9, 29, 51, 4, 27, 52, 79, 23, 51, 81, 17, 48, 81, 9, 43, 79, 117, 36, 75, 116, 27, 69, 113, 16, 61, 108, 3, 51, 101, 153, 39, 92, 147, 25, 81, 139, 9, 68, 129, 192, 53, 117, 183, 36, 103, 172, 17, 87, 159, 233, 69, 144 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

The k's that match these j's comprise A120870.

REFERENCES

Clark Kimberling, The equation (j+k+1)^2-4*k=Q*n^2 and related dispersions, Journal of Integer Sequences 10 (2007, Article 07.2.7) 1-17.

LINKS

Table of n, a(n) for n=1..68.

FORMULA

Let r=(1/2)*sqrt(13). If n is odd, then a(n)=13*n^2-([n*r+1/2]- 1/2)^2; if n is even, then a(n)=13*n^2-[n*r])^2, where [ ]=Floor.

EXAMPLE

1=13*(1/2)^2-([1*r+1/2]- 1/2)^2,

4=13*(2/2)^2-[2*r])^2,

9=13*(3/2)^2-([3*r+1/2]- 1/2)^2, etc. Moreover,

for n=1, the unique (j,k) is (1,3): (1+3+1)^2-4*3=13*1;

for n=2, the unique (j,k) is (4,3): (4+3+1)^2-4*3=13*4;

for n=3, the unique (j,k) is (9,1): (9+1+1)^2-4*1=13*9.

CROSSREFS

Cf. A120870.

Sequence in context: A292340 A329218 A224953 * A070436 A259450 A219731

Adjacent sequences:  A120866 A120867 A120868 * A120870 A120871 A120872

KEYWORD

nonn

AUTHOR

Clark Kimberling, Jul 09 2006

STATUS

approved

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Last modified June 1 19:32 EDT 2020. Contains 334762 sequences. (Running on oeis4.)