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.

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Original entry on oeis.org

1, 25, 841, 28561, 970225, 32959081, 1119638521, 38034750625, 1292061882721, 43892069261881, 1491038293021225, 50651409893459761, 1720656898084610641, 58451683124983302025, 1985636569351347658201, 67453191674820837076801, 2291422880374557112953025
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Numbers simultaneously square and centered square. E.g., a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker, Apr 24 2007
Solutions to A007913(x)=A007913(2x-1). - Benoit Cloitre, Apr 07 2002
From Ant King, Nov 09 2011: (Start)
Indices of positive hexagonal numbers that are also perfect squares.
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12 * sqrt(2).
(End)
Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015
Also positive integers x in the solutions to 4*x^2 - 8*y^2 - 2*x + 8*y - 2 = 0, the corresponding values of y being A253826. - Colin Barker, Jan 25 2015
Squares that are sum of two consecutive squares: y^2 = (k + 1)^2 + k^2 is equivalent to x^2 - 2*y^2 = -1 with x = 2*k + 1. - Jean-Christophe Hervé, Nov 11 2015
Squares in the main diagonal of the natural number array, A000027. - Clark Kimberling, Mar 12 2023

Examples

			From _Ravi Kumar Davala_, May 26 2013: (Start)
A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly
a(n+m)=A001333(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)), with m=1,2 is true.
A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly
a(n+m)=(1/2)*A002203(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)) is true for m=1,2
a(n+1)*a(n-1) = (a(n)+4)^2 , with n=1 is 841*1=(25+4)^2, for n=2 , 28561*25=(841+4)^2.
(End)
1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Links

Crossrefs

Cf. A000290, A001844, A007913, A000384, A046177, A016754, A253826.

Programs

  • GAP
    a := [1, 25, 841];; for i in [4..10^2] do a[i] := 35*a[i-1] - 35*a[i-2] + a[i-3]; od; a;  # Muniru A Asiru, Jan 17 2018
  • Magma
    I:=[1,25,841]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jan 20 2018
  • Maple
    CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007
  • Mathematica
    LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* Ant King, Nov 09 2011 *)
CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 20 2018 *)
  • PARI
    a(n)=if(n<0,0,sqr(subst(poltchebi(n+1)+poltchebi(n),x,3)/4))
  • PARI
    vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ Altug Alkan, Nov 11 2015

Formula

From Benoit Cloitre, Jan 19 2003: (Start)
a(n) = A078522(n) + 1.
a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)
G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).
a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000
a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - Richard Choulet, Sep 14 2007
Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2 - s(n))). - Steven Schlicker, Apr 24 2007
From Ant King, Nov 09 2011: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 8.
a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).
a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)). (End)
From Ravi Kumar Davala, May 26 2013: (Start)
a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).
a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+1)*a(n-1) = (a(n)+4)^2. (End)
a(n) = A001652(n)^2 + A046090(n)^2. - César Aguilera, Jan 15 2018
Limit_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jan 28 2018
sqrt(2*a(n))-1 = A002315(n). - Ezhilarasu Velayutham, Apr 05 2019
4*a(n) = 1 +3*A077420(n). - R. J. Mathar, Mar 05 2024
Product_{n>=0} (1 + 4/a(n)) = 2*sqrt(2) + 3 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Extensions

Entry edited by N. J. A. Sloane, Sep 14 2007

A008843 Squares of NSW numbers (A002315): x^2 such that x^2 - 2y^2 = -1 for some y.

Original entry on oeis.org

1, 49, 1681, 57121, 1940449, 65918161, 2239277041, 76069501249, 2584123765441, 87784138523761, 2982076586042449, 101302819786919521, 3441313796169221281, 116903366249966604049, 3971273138702695316401, 134906383349641674153601, 4582845760749114225906049, 155681849482120242006652081
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Also indices of triangular numbers (A000217) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 16 2015
a(n)-th triangular number is a square; subsequence of A001108. - Jaroslav Krizek, Aug 05 2016

References

  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.
  • P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 288.
  • P. F. Teilhet, Note #2094, L'Intermédiaire des Mathématiciens, 10 (1903), pp. 235-238.

Links

Crossrefs

Cf. A000217, A002315, A016754, A146313, A253826.

Programs

  • Mathematica
    LinearRecurrence[{35,-35,1},{1,49,1681},17] (* Stefano Spezia, Aug 17 2024 *)

Formula

a(n) = 34*a(n-1) - a(n-2) + 16 = A002315(n)^2 = 2*A001653(n)^2 - 1 = 2*A008844(n) - 1 = floor(A046176(n)*sqrt(2)) = 6*A055792(n+1) - a(n-1) + 4 = (6*A055792(n+2) + a(n-1) - 20)/35. - Henry Bottomley, Nov 13 2001
a(n) = A001108(2n+1). - Ira M. Gessel, Nov 05 2014
a(n) = Sum_{k=1..2*n+1} 2^(k-1)*binomial(4*n+2, 2*k). - Zoltan Zachar (zachar(AT)fellner.sulinet.hu), Oct 03 2003
O.g.f.: -(1+14*x+x^2)/((-1+x)*(1-34*x+x^2)). - R. J. Mathar, Nov 23 2007
a(n) = -(cosh((2*n - 1)*arctanh(sqrt(2))))^2 = -1 - (sinh((2*n - 1)*arctanh(sqrt(2))))^2. - Artur Jasinski, Oct 30 2008
a(n) = Sum_{k=0..4n+1} A000129(k), see Santana and Diaz-Barrero link at A002315. - Ivan N. Ianakiev, Jul 15 2022

Extensions

a(14)-a(17) from Stefano Spezia, Aug 17 2024

A046177 Squares (A000290) which are also hexagonal numbers (A000384).

Original entry on oeis.org

1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, 4446390382511295358038307980025, 5131130648390546663702275158894481
Offset: 1

Views

Author

Eric W. Weisstein

Keywords

Comments

Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - Alexander Adamchuk, Nov 06 2007
Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - Richard Choulet, May 01 2009
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - Ant King Nov 08 2011
Also centered octagonal numbers (A016754) which are also triangular numbers (A000217). - Colin Barker, Jan 16 2015
Also hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015

References

  • M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Links

Crossrefs

Cf. A008844, A046176, A253826.
Cf. A001110 (Numbers that are both triangular and square).
Cf. A000290, A000384, A016754, A253826.

Programs

  • Mathematica
    LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* Ant King, Nov 08 2011 *)
  • PARI
    Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 16 2015

Formula

a(n) = A001110(2n-1). - Alexander Adamchuk, Nov 06 2007
a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - Richard Choulet, May 01 2009
a(n+2) = 1154*a(n+1) - a(n) + 72 for n>=0. - Richard Choulet, May 01 2009
From Ant King, Nov 07 2011: (Start)
a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).
a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).
a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).
a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).
a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).
G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).
(End)
a(n) = A096979(4*n - 3). - Ivan N. Ianakiev, Sep 05 2016
a(n) = (1/2) * (A002315(n))^2 * ((A002315(n))^2 + 1) = ((2*x + 1)*sqrt(x^2 + (x+1)^2))^2, where x = (1/2)*(A002315(n)-1). - Ivan N. Ianakiev, Sep 05 2016
Showing 1-3 of 3 results.

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

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