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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.
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%I #14 Sep 28 2024 19:06:07

%S 11,23,39,41,59,64,83,111,134,143,153,179,181,219,263,307,311,363,373,

%T 386,419,479,543,571,584,611,683,703,759,781,839,900,923,989,1011,

%U 1103,1156,1199,1299,1403,1405,1425,1511,1546,1623,1739,1769,1859,1983,2111

%N 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 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 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 occurring earlier in this sequence (thus having A130280(n^2-1)=k): { 900, 1405, 19759...} with k=11; { 6161, 8322,... } with k=23, ...

%H Michael Usher, <a href="https://arxiv.org/abs/1801.06762">Infinite staircases in the symplectic embedding problem for four-dimensional ellipsoids into polydisks</a>, arXiv:1801.06762 [math.SG], 2018.

%F 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 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 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 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 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 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 (PARI) check(n) = { local( m = n^2-1 ); for( i=2, n-2, if( issquare( m*(i^2-1)+1), return(i))) }

%o t=0;A130282=vector(100,i,until(check(t++),);t)

%o (PARI) P(m,x=x)=if(m>1,2*x*P(m-1,x)-P(m-2,x),m*(x-2)+1)

%Y Cf. A084702, A094357, A130280.

%K easy,nonn

%O 1,1

%A _M. F. Hasler_, May 20 2007, May 24 2007, May 31 2007