How to check if a number is present in the series $1, 7, 25, 63, 129, 231...$?
The nth term of the series is $(2 n + 1) (2n^2 + 2 n + 3) / 3$.
if $N = 7$, then it is present in the series.
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1I don't know exactly what sort of answer you expect to be getting... you could expand this out as a cubic and solve the cubic and check to see if the solution is an integer. If it is, then the number is present in the series. It will be quite ugly though. – JMoravitz May 19 '20 at 14:30
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reminds me of how to check if a number is Fibonacci – J. W. Tanner May 19 '20 at 14:35
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Suppose we seek an $n$ solving $O_n=N$, with $n$th centered octahedral number$$O_n:=\frac13(2n+1)(2n^2+2n+3)\in(\tfrac16(2n)^3,\,\tfrac16(2n+1)^3).$$Since $(2n)^3\le6N\le(2n+1)^3$, there's at most one candidate $n$ to check.
J.G.
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This is the sequence A001845 giving the centered octahedral numbers (crystal ball sequence for cubic lattice). For smaller numbers you can simply check the given list there.
Of course, with a formula at hand, you can simply solve the equation $$ (2n+1)(2n^2+2n+3)/3=k $$ for $n$ and see if it has an integral solution. For example, take $k=20$. The formula shows that we need $n<2$. But for $n=0,1$ it is false. So first, estimate $n$, and then check for those $n$.
Dietrich Burde
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2Surely the OP is looking for a way other than cross referencing a table... – JMoravitz May 19 '20 at 14:27
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$$6N=8n^3+12n^2+16n+6\\=(2n+1)^3+5(2n+1)$$ Let $$M =round\left(\sqrt[3]{6N-5\sqrt[3]{6N}}\right)$$ If $6N-5M=M^3$, and $M$ is odd, then I think $N$ is in the sequence.
Empy2
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