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 1-3 of 3 results.

A093959 Squares arising as partial products in A094357 + 1.

Original entry on oeis.org

4, 25, 361, 116281, 13441851721, 180680260806215679961, 32645356640144805339464939649128207984041
Offset: 0

Views

Author

Amarnath Murthy, May 22 2004

Keywords

Examples

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

Crossrefs

Cf. A094357.

Formula

a(n) = 1+prod_{i=1..n} A094357(i). - David Wasserman, May 03 2007

Extensions

More terms from David Wasserman, May 03 2007

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.

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.
Showing 1-3 of 3 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.