Search: id:A055438
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%I A055438
%S A055438 101,402,903,1604,2505,3606,4907,6408,8109,10010,12111,14412,16913,
%T A055438 19614,22515,25616,28917,32418,36119,40020,44121,48422,52923,57624,
%U A055438 62525,67626,72927,78428,84129,90030,96131,102432,108933,115634,122535
%N A055438 100*n^2+n.
%C A055438 For all terms, the first digit is the square of the last digit. Example: 
               101, 1^2=1; 402, 2^2=4; 903, 3^2=9; 1604, 4^2=16; 44121, 21^2=441 
               [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Mar 10 2009]
%C A055438 If A=[A055438] 100*n.^2+n (101, 402, 903,. ,.,); Y=[A010859] 20 (20, 
               20, 20,. ,.,); X=[A157956] 200*n+1 (201, 401, 601, ,. .,), we have, 
               for all terms, Pell's equation X^2-A*Y^2=1. Example: 201^2-101 *20^2=1; 
               401^2-402*20^2=1; 601^2-903*20^2=1. [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), 
               Mar 10 2009]
%C A055438 If A=[A055438] 100*n.^2+n (101, 402, 903, ,..,); Y=[A157663] 8000*n+40 
               (8040, 16040, 24040,..,); X=[A157664] 80000*n^2+800*n+1 (80801, 321601, 
               722401,..,), we have, for all terms, Pell's equation X^2-A*Y^2=1. 
               Example: 80801^2-101*8040^2=1; 321601^2-402*16040^2=1; 722401^2-903*24040^2=1. 
               [From Vincenzo Librandi (vincenzo.librandi(AT)tinit), Mar 04 2009]
%H A055438 Vincenzo Librandi, <a href="http://mathforum.org/kb/message.jspa?messageID=5773864&tstart=0">
               X^2-AY^2=1</a> [From Vincenzo Librandi (vincenzo.librandi(AT)tinit), 
               Mar 04 2009]
%Y A055438 Cf. A002378, A055437, a(n)=A055436(n) if 10<=n<100.
%Y A055438 Different from A031698.
%Y A055438 Cf. A157956, A010859 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), 
               Mar 10 2009]
%Y A055438 Cf. A157663, A157664 [From Vincenzo Librandi (vincenzo.librandi(AT)tinit), 
               Mar 04 2009]
%Y A055438 Sequence in context: A158192 A062800 A031698 this_sequence A142692 A060012 
               A142507
%Y A055438 Adjacent sequences: A055435 A055436 A055437 this_sequence A055439 A055440 
               A055441
%K A055438 easy,nonn
%O A055438 1,1
%A A055438 Henry Bottomley (se16(AT)btinternet.com), May 18 2000

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