Tag Archives: OEIS

words-length and words in a group

the group is \mathbb{Z}_2*\mathbb{Z}_3=\langle a,\ b\mid a^2,\ b^3\rangle, and the picture:

Correctly predicts the next one. It is A164001 in the OEIS data-base. Dubbed “Spiral of triangles around a hexagon“. It has the generating function -(x+1)+\frac{x^2+2x+1}{1-x^2-x^3}, Why would it be? :) . This another is A000931.

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another awesome crescent sequence to try its sum-series of reciprocals. Check at A000127 of the Oeis. It has a beautiful connection with the Pascal’s triangle. Its generating function is


What is the g.f. of the reciprocals?regionsSAMSUNG

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3, 7, 15, 23, 39, 47, 71, 87, 111, 127, …

10 Siehler’s numbers

two consecutive reciprocal partial sums are


\approx 0.722936



1/1111 \approx 0.723836


by the way: f(n)=\#{\mathbb{Z}_n}^* (cardinal of multiplicatively invertibles elements in the ring {\mathbb{Z}_n}) satisfy:

  • f(1)=\#{\mathbb{Z}_1}^*=0, since {\mathbb{Z}_1}=\{0\} and {\mathbb{Z}_1}^*=\varnothing
  • f(2)=\#{\mathbb{Z}_2}^*=1, since {\mathbb{Z}_2}=\{0,1\} and {\mathbb{Z}_2}^*=\{1\}
  • f(3)=\#{\mathbb{Z}_3}^*=2, since {\mathbb{Z}_3}=\{0,1,2\} and {\mathbb{Z}_3}^*=\{1,2\}
  • f(4)=\#{\mathbb{Z}_4}^*=2, since {\mathbb{Z}_4}=\{0,1,2,3\} and {\mathbb{Z}_4}^*=\{1,3\}
  • f(5)=\#{\mathbb{Z}_4}^*=4, since {\mathbb{Z}_5}=\{0,1,2,3,4\} and {\mathbb{Z}_4}^*=\{1,2,3,4\}









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another reciprocal sum problem

ok we already know how to calculate how many rationals, \mathbb{Q}, exists in terms of the height function

h(a/b)=\max (|a|,|b|)


It turns out that the A171503  entry of the OEIS tells more…

meanwhile it is natural to ask for convergence of

1/3 + 1/7 + 1/15 + 1/23 + 1/31 + 1/47 + 1/71 + 1/87 + 1/111 + 1/127+...

and for (non-) existence of a generating function…

The crescent sequence could be called Siehler’s numbers


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whole sum of the reciprocal Catalan numbers

In this little post I complete the details of the calculation

2+\frac{4\sqrt{3}\pi}{27}=\sum_{n=0}^{\infty}\frac{n+1}{{2n\choose n}}=1+1+\frac{1}{2}+\frac{1}{5}+\frac{1}{14}+\frac{1}{42}+\cdots

for the reciprocals C_n=\frac{1}{n+1}{2n\choose n}, the famous so called Catalan numbers. It seems this is well known but it is scarcely quoted anywhere: Cf1, Cf2

We begin by recalling the relation {C_n}^{-1}=(2n+1)(n+1)\int_0^1t^n(1-t)^n{\rm{d}}t, that is,  in terms of the Beta function. So



=\int_0^1\frac{1+3t - 3t^2}{{(1-t+t^2)}^3}{\rm{d}}t


This was hinted to me, thanks to, by the professor Qiaochu Yuan and I am not ashamed to confess that the calculations were executed in mathematica-v5.  Why? well, the risk of introducing errors is a little-big.

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