Search: id:A130282
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%I A130282
%S A130282 11,23,39,41,59,64,83,111,134,143,153,179,181,219,263,307,311,363,373,
%T A130282 386,419,479,543,571,584,611,683,703,759,781,839,900,923,989,1011,1103,
%U A130282 1156,1199,1299,1403,1405,1425,1511,1546,1623,1739,1769,1859,1983,2111
%N A130282 Numbers n such that A130280(n^2-1) < n-1, i.e. there is a k, 1 < k < 
               n-1, such that (n^2-1)(k^2-1)+1 is a perfect square.
%C A130282 For any n>1, the number (n^2-1)(k^2-1)+1 is a square for k = n-1 ; this 
               sequence lists those n>1 for which there is a smaller k>1 having 
               this property. This sequence contains the subsequence b(k) = 2k(k+1)-1, 
               k>1, for which A130280(b(k)^2-1) <= k < b(k)-1, since (b(k)^2-1)(k^2-1)+1 
               = (2k^3+2k^2-2k-1)^2. We have n=b(k) whenever 2n+3 is a square, the 
               square root of which is then 2k+1. (See also formula.)
%C A130282 The only elements of this sequence not of the form |P[m](k)| (see formula) 
               are seem to be non-minimal n>k+1 such that (k^2-1)(n^2-1)+1 is a 
               square, for some k occuring earlier in this sequence (thus having 
               A130280(n^2-1)=k): { 900, 1405, 19759...} with k=11; { 6161, 8322,
               ... } with k=23, ...
%F A130282 If 2n+3 is a square, then n = b(k)= 2k(k+1)-1, k = (sqrt(n/2+3/4)-1)/
               2 = floor(sqrt(n/2)) >= A130280(n^2-1). (For all k>1, b(k) is in 
               this sequence.)
%F A130282 Most terms of this sequence are in the set { P[m](k), |P[m](-k)| ; m=2,
               3,4..., k=2,3,4,... } with P[m] = 2 X P[m-1] - P[m-2], P[1]=X-1, 
               P[0]=1. Whenever a(n) = P[m](k) or a(n) = |P[m](-k)| (m,k>1), then 
               A130280(a(n)^2-1) <= k (resp. k-1 for m=2) < a(n). (No case where 
               equality does not hold is known so far.) We have P[2] = P[2](1-X) 
               and for all integers m>2,x>0: P[m](x) < (-1)^m P[m](-x) <= |P[m+1](x)| 
               with equality iff x=2. We have P[m](-1)=(-1)^m (m+1), P[m](0)=(-1)^(m(m+1)/
               2), P[m](1)=1-m, P[m](x)>0 for all x >=2 ; P[m](x) ~ 2^(m-1) x^m.
%e A130282 a(1) = 11 since n=11 is the smallest integer > 1 such that (n^2-1)(k^2-1)+1 
               is a square for 1 < k < n-1, namely for k=2.
%e A130282 Values of P[2](k+1) = 2 k^2 + 2 k - 1 for k=2,3,... are { 11,23,39,... 
               } and A130280(11^2-1)=2, A130280(23^2-1)=3, A130280(39^2-1)=4,...
%e A130282 Values of P[3](k) = 4 k^3 - 4 k^2 - 3 k + 1 for k=2,3,4... are { 11,64,
               181,... } and A130280(64^2-1)=3, A130280(181^2-1)=4,...
%e A130282 Values of -P[3](-k) = 4 k^3 + 4 k^2 - 3 k - 1 for k=2,3,4... are { 41,
               134,307,... } and A130280(134^2-1)=3, A130280(307^2-1)=4,...
%o A130282 (PARI) check(n) = { local( m = n^2-1 ); for( i=2, n-2, if( issquare( 
               m*(i^2-1)+1), return(i))) } t=0;A130282=vector(100,i,until(check(t++),
               );t)
%o A130282 (PARI) P(m,x=x)=if(m>1,2*x*P(m-1,x)-P(m-2,x),m*(x-2)+1)
%Y A130282 Cf. A084702, A094357, A130280.
%Y A130282 Sequence in context: A135978 A139493 A077345 this_sequence A019356 A046440 
               A119890
%Y A130282 Adjacent sequences: A130279 A130280 A130281 this_sequence A130283 A130284 
               A130285
%K A130282 easy,nonn
%O A130282 1,1
%A A130282 M. F. Hasler (Maximilian.Hasler(AT)gmail.com), May 20 2007, May 24 2007, 
               May 31 2007

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