# 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.

Filed under free group, math, sum of reciprocals

## 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!

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

or

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

Filed under math

## π day, π jour, π día, π Tag

limit of the inverse of Catalan numbers infinite serie

some relatives come to celebrate!

Filed under math

## 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?

Filed under math

## 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

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

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

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

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

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

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