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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1,2,4,8,16,31,57,99,…


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

\frac{1-3x+4x^2-2x^3+x^4}{(1-x)^5}

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

1/3+1/7+1/15+1/23+1/39+1/47+1/71+1/87+1/111+1/127+1/167+1/183+1/231+1/255+1/287+1/319+1/383+1/407+1/479+1/511+1/559+1/599+1/687+1/719+1/799+1/847+1/919+1/967+1/10794

\approx 0.722936

and

1/3+1/7+1/15+1/23+1/39+1/47+1/71+1/87+1/111+1/127+1/167+1/183+1/231+1/255+1/287+1/319+1/383+1/407+1/479+1/511+1/559+1/599+1/687+1/719+1/799+1/847+1/919+1/967+1/1079+

1/1111 \approx 0.723836

(04.nov.2011)

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\}

Hence

s(1)=3+4f(1)=3

s(2)=s(1)+4f(2)=7

s(3)=s(2)+4f(3)=15

s(4)=s(3)+4f(4)=23

s(5)=s(4)+4f(5)=39

s(n+1)=s(n)+4f(n+1)

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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|)

gcd(a,b)=1

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

http://192.20.225.10/~njas/sequences/A171503

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

\sum_{n=0}^{\infty}{C_n}^{-1}=\sum_{n=0}^{\infty}(2n+1)(n+1)\int_0^1t^n(1-t)^n{\rm{d}}t

=\int_0^1\left[\sum_{n=0}^{\infty}(2n+1)(n+1)t^n(1-t)^n\right]{\rm{d}}t

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

=2+\frac{4\sqrt{3}\pi}{27}

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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