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)

2 Comments

Filed under math

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?

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