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 1-8 of 8 results.

A084703 Squares k such that 2*k+1 is also a square.

Original entry on oeis.org

0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844
Offset: 0

Views

Author

Amarnath Murthy, Jun 08 2003

Keywords

Comments

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008
Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011
From M. F. Hasler, Jan 17 2012: (Start)
Bisection of A079291. The squares 2*k+1 are given in A055792.
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

Links

Crossrefs

Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.
Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.
Cf. A089928, A204576.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

Programs

  • Magma
    [4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022
  • Mathematica
    b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)
  • SageMath
    [4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

Formula

a(n) = 4*A001110(n) = A001542(n)^2.
a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - Charlie Marion, Jul 01 2003
a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - Charlie Marion, Jul 16 2003
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008
a(n) = A079291(2n). - M. F. Hasler, Jan 16 2012
From George F. Johnson, Nov 02 2012: (Start)
a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
2*a(n) + 1 = (A001541(n))^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n)*a(n+1) = (4*A029549(n))^2.
a(n+1) - a(n) = 4*A046176(n).
a(n) + a(n+1) = 4*(6*A029549(n) + 1).
a(n) = (2*A001333(n)*A000129(n))^2.
Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Empirical: a(n) = A089928(4*n-2), for n > 0. - Alex Ratushnyak, Apr 12 2013
a(n) = 4*A001109(n)^2. - G. C. Greubel, Aug 18 2022
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Extensions

Edited and extended by Robert G. Wilson v, Jun 15 2003

A130280 a(n) = smallest integer k>1 such that n(k^2-1)+1 is a perfect square, or 0 if no such number exists.

Original entry on oeis.org

2, 5, 3, 0, 2, 3, 5, 2, 0, 3, 7, 5, 4, 11, 3, 2, 4, 13, 9, 7, 2, 5, 19, 4, 0, 5, 21, 3, 11, 9, 11, 14, 2, 29, 5, 3, 6, 31, 21, 2, 13, 11, 13, 169, 3, 7, 41, 6, 0, 7, 5, 11, 22, 419, 3, 2, 5, 23, 461, 27, 8, 55, 7, 4, 2, 3, 49, 29
Offset: 1

Views

Author

M. F. Hasler, May 20 2007, May 25 2007

Keywords

Comments

A084702(n) = a(n)^2-1, resp. a(n) = sqrt(A084702(n)+1). See A130283 for values where A130280(n)=0.

Examples

			a( (2k)^2 ) <= k since (2k)^2(k^2-1)+1 = (2k^2-1)^2 (but k=1 is excluded since with k^2-1=0 this would be a trivial solution for any n).

Links

Crossrefs

Cf. A084702, A094357, A130281, A130282, A130283, A130284.
See also A306767.

Programs

  • Maple
    A130280:=proc(n) local x,y,z; if n=1 then return 2 fi; isolve(n*(x^2-1)+1=y^2,z); select(has,`union`(%),x); map(rhs,%); simplify(eval(%,z=1) union eval(%,z=0)) minus {-1,1}; if %={} then 0 else (min@op@map)(abs,%) fi end;
  • Mathematica
    $MaxExtraPrecision = 100;
r[n_, c_] := Reduce[k > 1 && j > 1 && n*(k^2 - 1) + 1 == j^2, {j, k}, Integers] /. C[1] -> c // Simplify;
a[n_] := If[rn = r[n,0] || r[n,1] || r[n,2]; rn === False, 0, k /. {ToRules[rn]} // Min];
Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 800}] (* Jean-François Alcover, May 12 2017 *)
  • PARI
    {A130280(n,L=10^15)=if(issquare(n),L=2+sqrtint(n>>2)); for( k=2, L, if( issquare(n*(k^2-1)+1),return(k)))}

Formula

If n=(2k)^2, then A130280(n) <= k, since (2k)^2(k^2-1)+1 = (2k^2-1)^2. See A130281 for the cases where equality does not hold. If n=k^2-1, then A130280(n) <= k-1 since (k^2-1)((k-1)^2-1)+1 = (k^2-k-1)^2. See A130282 for the cases where equality does not hold.

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

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.

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)}

A094357 Numbers of the form k^2 -1 such that every partial product is also of the form k^2-1.

Original entry on oeis.org

3, 8, 15, 323, 115599, 13441619843, 180680260779332208399
Offset: 1

Views

Author

Amarnath Murthy, May 22 2004

Keywords

Comments

Next term <= 32645356640144805339103579127542660095683 : The number (sqrt( product( a[j], j=1 .. n )+1)-1)^2-1 does satisfy the requirements for the next term, but is this always the smallest solution ? - M. F. Hasler, May 15 2007

Examples

			3, 8 and 15 are 1 less than a square and so are the numbers 3, 3*8, 3*8*15.

Crossrefs

Cf. A084702, A093959.

Formula

a(n+1) = A084702(A093959(n)-1). - David Wasserman, May 03 2007
For n>1, a[n+1] <= floor( sqrt( product( a[j], j=1 .. n )))^2-1 - M. F. Hasler, May 15 2007

Extensions

More terms from David Wasserman, May 03 2007

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.

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 )}
Showing 1-8 of 8 results.

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