[seqfan] Seqs A002061 and A034111
Christian G.Bower
bowerc at usa.net
Fri Oct 9 04:29:53 CEST 1998
Patrick DeGeest wrote:
-----
Last month I submitted the following sequence to Sloane's table :
%I A034111
%S A034111
21,31,43,57,73,91,111,133,157,183,211,241,273,307,343,381,421,463,507,
%T A034111 553,601,651,703,757,813,871,931,993,1057,1123,1191,1261,1333,1407,
%U A034111 1483,1561,1641,1723,1807,1893,1981,2071,2163,2257,2353,2451,2551,2653
%N A034111 Decimal part of square root of a(n) starts with 5 : first term of runs.
%C A034111 Is basically sequence A002061 (n^2-n+1 = central polygonal numbers) starting from the sixth term.
%O A034111 0,1
%K A034111 nonn,base
%Y A034111 Cf. A034101.
%A A034111 Patrick De Geest (Patrick.DeGeest at ping.be), September 1998.
When I checked for possible presence I came across the next sequence :
%I A002061 M2638 N1049
%S A002061 1,1,3,7,13,21,31,43,57,73,91,111,133,157,183,211,241,273,307,343,381,
%T A002061 421,463,507,553,601,651,703,757,813,871,931,993,1057,1123,1191,1261
%N A002061 Central polygonal numbers: n^2 - n + 1.
%R A002061 HO50 22. HO70 87.
%O A002061 0,3
%A A002061 njas
%K A002061 nonn,easy
I never anticipated that such different approaches can yield the same terms.
But then again, I am not a genuine mathematician. I would appreciate it very much
if someone could explain this phenomenon or give me more insight into this matter.
Thanks in advance.
------
sqrt(23) = 4.7958...
sqrt(24) = 4.8990...
sqrt(25) = 5
sqrt(26) = 5.0990...
sqrt(27) = 5.1962...
If x>=25 then sqrt(x+1)-sqrt(x) < 0.1,
Thus the start of a run of numbers x such that the first number after the
decimal point is k is of the form:
ceiling((m+k/10)^2) for some integer m.
Example, let k=2:
5.2^2=27.04, ceiling(5.2^2)=28, sqrt(28)=5.2915...
6.2^2=38.44, ceiling(6.2^2)=39, sqrt(39)=6.2450...
...
If k=5 then
ceiling((m+5/10)^2)=ceiling(m^2+m+0.25)=m^2+m+1
Thus all the sufficiently large numbers of A034111 are of the form
m^2+m+1 for integers m.
How do we get to n^2-n+1 of A002061?
Substitute m=n-1:
(n-1)^2+(n-1)+1=n^2-2n+1+n-1+1=n^2-n+1.
Christian
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