# 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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### 2 responses to “3, 7, 15, 23, 39, 47, 71, 87, 111, 127, …”

1. m-k

Jacob A. Siehler tells about 2×2 ZZ-matrices with det=1,.

how?

by Jordan canonical forms?

2. k-mat