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.

11, 23, 39, 41, 59, 64, 83, 111, 134, 143, 153, 179, 181, 219, 263, 307, 311, 363, 373, 386, 419, 479, 543, 571, 584, 611, 683, 703, 759, 781, 839, 900, 923, 989, 1011, 1103, 1156, 1199, 1299, 1403, 1405, 1425, 1511, 1546, 1623, 1739, 1769, 1859, 1983, 2111

Offset: 1

Author

M. F. Hasler, May 20 2007, May 24 2007, May 31 2007

Keywords

• easy
• nonn

Comments

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.)

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, ...

Examples

			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.
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,...
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,...
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,...
		

Links

• Michael Usher, Infinite staircases in the symplectic embedding problem for four-dimensional ellipsoids into polydisks, arXiv:1801.06762 [math.SG], 2018.

Crossrefs

Cf. A084702, A094357, A130280.

Programs

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)

PARI

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

Formula

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.)

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.