cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-7 of 7 results.

A130283 Integers n > 0 for which A130280(n) = 0, i.e., such that there is no integer m > 1 for which n(m^2 - 1) + 1 is a square.

Original entry on oeis.org

4, 9, 25, 49, 81, 121, 169, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1369, 1521, 1681, 1849, 2025, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569, 7921, 8281, 8649, 9025, 9409
Offset: 1

Views

Author

M. F. Hasler, May 24 2007

Keywords

Comments

No term > 4 in this sequence is an even square (see formula in A130280).
A001248(k) is a term for any k. - Jinyuan Wang, Apr 14 2019

Examples

			a(1)=4 since 1(2^2-1)+1=2^2, 2(5^2-1)+1=7^2, 3(3^2-1)+1=5^2 but 4(m^2-1)+1 = 4m^2-3 can't be a square because the largest square < 4m^2 is (2m-1)^2 = 4m^2-4m+1 < 4m^2-3 for m>1.
a(2)=9 since for n=5,6,7,8 one has m=2,3,5,2, but 9(m^2-1)+1 = 9m^2-8 > 9m^2-11 >= 9m^2-6m+1 = (3m-1)^2 and therefore can't be a square.

Crossrefs

Cf. A001248, A084702, A130280, A130284, A130288.

Programs

  • Mathematica
    $MaxExtraPrecision = 200;
r[n_, c_] := Reduce[k > 1 && j > 1 && n*(k^2 - 1) + 1 == j^2, {j, k}, Integers] /. C[1] -> c // Simplify;
A130280[n_] := If[rn = r[n, 0] || r[n, 1] || r[n, 2]; rn === False, 0, k /. {ToRules[rn]} // Min];
Reap[For[n=1, n <= 2000, n++, If[A130280[n]==0, Print[n]; Sow[n]]]][[2,1]] (* Jean-François Alcover, May 12 2017 *)
  • PARI
    f(n) = for(k=2, n+1, if( issquare(n*(k^2-1)+1), return(k)))
is(n) = issquare(n) && f(n) == 0; \\ Jinyuan Wang, Apr 14 2019

Extensions

More terms from Jean-François Alcover, May 12 2017
More terms from Jinyuan Wang, Apr 14 2019

A130288 Record indices of A130280: integers n>0 for which min{ m>1 | (2n+1)^2(m^2-1)+1 is a square} < oo but bigger than for all preceding n.

Original entry on oeis.org

1, 2, 11, 14, 18, 23, 27, 34, 38, 44, 54, 59, 74, 158, 179, 284, 524
Offset: 1

Views

Author

M. F. Hasler, May 24 2007

Keywords

Comments

Most elements of this sequence seem to be 1,2 or 4 times a prime.
Corresponding values of A130280 are given in A130289. - M. F. Hasler, May 24 2007

Crossrefs

Cf. A084702, A130280, A130283, A130284, A130289.

Programs

  • PARI
    A130288(L=999,S=1)={local(R,T);for(n=S,L, if(issquare(n) || R>=T=A130280(n),next); print1(n", ");R=T)}

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.

Original entry on oeis.org

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

Views

Author

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

Keywords

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

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.

A130281 Integers n > 1 such that A130280(4n^2) < n, i.e., there is an m < n, m > 1 such that 4n^2(m^2 - 1) + 1 is a square.

Original entry on oeis.org

28, 102, 248, 390, 490, 852, 1358, 2032, 2898, 3465, 3980, 5302, 5432, 6888, 8762, 10948, 13470, 15372, 16352, 19618, 23292, 27398, 31960, 37002, 42548, 48015, 48622, 55248, 62450, 70252, 75658, 78678, 87752, 97498
Offset: 1

Views

Author

M. F. Hasler, May 20 2007

Keywords

Comments

If n>4 is an even square, n=4k^2, then A130280(n) <= k since n(k^2-1)+1 = (2k^2-1)^2. This sequence lists those k for which we have strict inequality. Most terms in this sequence belong to the subsequence b(m)=2m*(2m^2-1), m>1, for which A130280(4 b(m)^2) <= m < b(m), since 4 b(m)^2(m^2-1)+1 = (8m^4-8m^2+1)^2. For other terms k of this sequence (e.g., the subsequence 390, 3465, 15372, 48015, ...), A130280(4k^2) is even smaller.

Crossrefs

Cf. A130280.

Programs

  • PARI
    checkA130281(n)={local(m=4*n^2);for(i=2,sqrt(n),if(issquare(m*(i^2-1)+1),return(i)))}
for(n=1,99999,if(checkA130281(n),print(n", ")))

A130289 Record values in A130280: minima of { m>1 | (2n+1)^2(m^2-1)+1 is a square} bigger than for all preceding n.

Original entry on oeis.org

3, 5, 7, 11, 13, 19, 21, 29, 31, 169, 419, 461, 3269, 7127, 3877019, 22783559, 215308729
Offset: 1

Views

Author

M. F. Hasler, May 24 2007

Keywords

Comments

Most elements of this sequence seem to be 1,2 or 4 times a prime.

Crossrefs

Cf. A084702, A130280, A130283, A130284, A130288.

Programs

  • PARI
    A130289( L=999, S=1 )={ local( R, T ); for( n=S, L, if( issquare(n) || R >= T = A130280(n), next ); print1( T ", " ); R=T )}

A306767 Least x > 0 such that (x^2-1)/(y^2-1) = n, where y=A130280(n), or 0 if no such x exists.

Original entry on oeis.org

2, 7, 5, 0, 4, 7, 13, 5, 0, 9, 23, 17, 14, 41, 11, 7, 16, 55, 39, 31, 8, 23, 91, 19, 0, 25, 109, 15, 59, 49, 61, 79, 10, 169, 29, 17, 36, 191, 131, 11, 83, 71, 85, 1121, 19, 47, 281, 41, 0, 49, 35, 79, 160, 3079, 21, 13, 37, 175, 3541, 209, 62, 433, 55, 31, 14, 23, 401, 239
Offset: 1

Views

Author

Jinyuan Wang, Apr 13 2019

Keywords

Crossrefs

Cf. A130280, A130283.

Programs

  • PARI
    a(n, L=10^15) = if(issquare(n), L=2+n^2); for(k=2, L, if((k^2-1)%n==0,if(issquare(1+(k^2-1)/n), return(k))));

Formula

a(n) = sqrt(n*(A130280(n)^2-1) + 1).
a(A130283(n)) = 0.

A130284 Integers j > 0 such that (2j+1)^2(m^2-1) + 1 is a square for some integer m > 1.

Original entry on oeis.org

7, 17, 31, 49, 71, 97, 104, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 594, 647, 721, 799, 881, 967, 1057, 1151, 1249, 1351, 1455, 1457, 1567, 1681, 1799, 1921, 1952, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 3041, 3199, 3361, 3527, 3697, 3871, 4049
Offset: 1

Views

Author

M. F. Hasler, May 24 2007, May 29 2007

Keywords

Comments

All terms > 4 in A130283 are odd squares, but not all odd squares are in that sequence: This sequence here gives the exceptions as (2a(n)+1)^2. The sequence consists mainly of the subsequences: (1) A056220(k) = 2k^2-1 with k>1: {7,17,31,49,...}, for which m=k gives (1+2*A056220(k))^2(k^2-1)+1 = k^2(4k^2-3)^2; (2) 2*A079414(k) = 2k^2(4k^2-3) with k>1: {104,594,1952,4850,...}, for which m=k gives (1+4*A079414(k))^2(k^2-1)+1 = k^2(16k^4-20k^2+5)^2. A third subsequence starts {1455,20195,...}; up to 20195, all terms are in one of these subsequences.

Examples

			Up to k=17, a(k)=P[1](k+1) with P[1] = 2x^2 - 1, A130280(a(k)) = k+1.
a(18) = P[2](2) < P[1](19) with P[2] = 2x^2*(4x^2 - 3), A130280(a(18)) = 2.
a(106) = P[1](100) < a(107) = P[3](3) < a(108) = P[4](2) < a(109) = P[1](101).

Crossrefs

Cf. A084702, A130280, A130284, A130288.
Cf. A130280, A130283, A130281.

Programs

  • Mathematica
    r[n_] := Reduce[m>1 && k>1 && (2n+1)^2*(m^2-1)+1 == k^2, {m, k}, Integers];
Reap[For[n=1, n <= 5000, n++, If[r[n] =!= False, Print[n]; Sow[n]]]][[2,1]] (* Jean-François Alcover, May 12 2017 *)
  • PARI
    A130284( LIM=9999, START=1 )={ local(N); for( n=START, LIM, N=(2*n+1)^2; for( m=2, sqrtint(n>>1+1), if(!issquare( N*(m^2-1)+1 ), next); print1(n", "); next(2))) }
  • PARI
    {Q(k,x=x)=if(m>0,(4*x^2-2)*Q(k-1,x)-Q(k-2,x),1)} {P(k,x=x)=if(type(x=(x^2*Q(k,x)^2-1)/(x^2-1))!="t_POL",sqrtint(x)\2,((-1)^k*Pol(sqrt(x))-1)/2)}

Formula

A130284 = { P[k](m) ; k=1,2,3,..., m=2,3,4,... } where P[k] = (sqrt((X^2 Q[k]^2 - 1)/(X^2 - 1))-1)/2 and Q[0] = Q[-1] = 1, Q[k+1] = (4X^2 -2)*Q[k] - Q[k-1]. Furthermore, (2P[k](m)+1)^2 (m^2 - 1)+1 = m^2 Q[k](m)^2, thus A130280(P[k](m)) <= m. So far, no case is known where we have strict inequality.
Showing 1-7 of 7 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.