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A120870 a(n) is the number k for which there exists a unique pair (j,k) of positive integers such that (j+k+1)^2-4*k=13*n^2. 3
3, 3, 1, 12, 9, 4, 23, 17, 9, 36, 27, 16, 3, 39, 25, 9, 53, 36, 17, 69, 49, 27, 3, 64, 39, 12, 81, 53, 23, 100, 69, 36, 1, 87, 51, 13, 107, 68, 27, 129, 87, 43, 153, 108, 61, 12, 131, 81, 29, 156, 103, 48, 183, 127, 69, 9, 153, 92, 29, 181, 117, 51, 211, 144, 75, 4, 173, 101 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

The j's that match these j's comprise A120869.

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)=([n*r+1/2] + 1/2)^2-(13/4)*n/4; if n is even, then a(n)=(1+[n*r])^2-(13/4)*n^2, where [ ]=Floor.

EXAMPLE

3=([1*r+1/2] + 1/2)^2-(13/4)*1/4,

3=(1+[2*r])^2-(13/4)*2^2,

1=([3*r+1/2] + 1/2)^2-(13/4)*3/4, 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. A120869.

Sequence in context: A186826 A185418 A050609 * A010029 A143603 A094021

Adjacent sequences:  A120867 A120868 A120869 * A120871 A120872 A120873

KEYWORD

nonn

AUTHOR

Clark Kimberling, Jul 09 2006

STATUS

approved

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Last modified June 6 10:27 EDT 2020. Contains 334849 sequences. (Running on oeis4.)