Tag Archives: sum of reciprocals

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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pascal key


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allocated quizz on the MO


\sum_{k=0}^{n}\frac{4^k(k+1)}{{2k\choose k}}=\frac{4^{n+1}(2n^2+5n+2)}{5{2(n+1)\choose n+1}}+\frac{1}{5}

found!

Follow the soap opera:

 

 

 

http://mathoverflow.net/questions/67784/finite-sum-of-integers-inverses

or

http://mathoverflow.net/questions/68090/finite-sums-with-binomial-and-catalan-inverses

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Catalanization


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π day, π jour, π día, π Tag


limit of the inverse of Catalan numbers infinite serie

some relatives come to celebrate!

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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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