A055792 -id:A055792 - cp's OEIS frontend

A204503 Squares n^2 such that floor(n^2/9) is again a square.

0, 1, 4, 9, 16, 36, 81, 144, 225, 324, 441, 576, 729, 900, 1089, 1296, 1521, 1764, 2025, 2304, 2601, 2916, 3249, 3600, 3969, 4356, 4761, 5184, 5625, 6084, 6561, 7056, 7569, 8100, 8649, 9216, 9801, 10404, 11025, 11664, 12321, 12996, 13689, 14400, 15129, 15876

Offset: 1

M. F. Hasler, Jan 15 2012

Or: Squares which remain squares when their last base-9 digit is dropped.
(For the first three terms, which have only 1 digit in base 9, dropping that digit is meant to yield zero.)
Base-9 analog of A055792 (base 2), A055793 (base 3), A055808 (base 4), A055812 (base 5), A055851 (base 6), A055859 (base 7), A055872(base 8) and A023110 (base 10).

M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.

Select[Range[0,200]^2,IntegerQ[Sqrt[Floor[#/9]]]&] (* Harvey P. Dale, Jan 27 2012 *)

b=9;for(n=1,200,issquare(n^2\b) & print1(n^2,","))

a(n) = A204502(n)^2.

Conjectures: a(n) = 9*(n-4)^2 for n>5. G.f.: x^2*(7*x^6-12*x^5-11*x^4-x-1) / (x-1)^3. - Colin Barker, Sep 15 2014

A204514 Numbers such that floor(a(n)^2 / 8) is again a square.

Original entry on oeis.org

0, 1, 2, 3, 6, 17, 34, 99, 198, 577, 1154, 3363, 6726, 19601, 39202, 114243, 228486, 665857, 1331714, 3880899, 7761798, 22619537, 45239074, 131836323, 263672646, 768398401, 1536796802, 4478554083, 8957108166, 26102926097, 52205852194, 152139002499, 304278004998, 886731088897

Offset: 1

M. F. Hasler, Jan 15 2012

Or: Numbers whose square, with its last base-8 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 8.)

See A204504 for the squares resulting from truncation of a(n)^2, and A204512 for their square roots. - M. F. Hasler, Sep 28 2014

M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A031149=sqrt(A023110) (base 10), A204502=sqrt(A204503) (base 9), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

Cf. A003499, A055872, A204504, A204512.

A204514 := proc(n) coeftayl((x^2+2*x^3-3*x^4-6*x^5)/(1-6*x^2+x^4), x=0, n); end proc: seq(A204514(n), n=1..30); # Wesley Ivan Hurt, Sep 28 2014

CoefficientList[Series[(x^2 + 2*x^3 - 3*x^4 - 6*x^5)/(x (1 - 6*x^2 + x^4)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 28 2014 *)
LinearRecurrence[{0,6,0,-1},{0,1,2,3,6},40] (* Harvey P. Dale, Nov 23 2022 *)

b=8;for(n=0,1e7,issquare(n^2\b) & print1(n","))

A204514(n)=polcoeff((x + 2*x^2 - 3*x^3 - 6*x^4)/(1 - 6*x^2 + x^4+O(x^(n+!n))),n-1,x)

G.f. = (x^2 + 2*x^3 - 3*x^4 - 6*x^5)/(1 - 6*x^2 + x^4).
a(n) = sqrt(A055872(n)). - M. F. Hasler, Sep 28 2014
a(2n) = A001541(n-1). a(2n+1) = A003499(n-1). - R. J. Mathar, Feb 05 2020

A204516 Numbers such that floor(a(n)^2 / 7) is a square.

Original entry on oeis.org

0, 1, 2, 3, 8, 16, 45, 127, 254, 717, 2024, 4048, 11427, 32257, 64514, 182115, 514088, 1028176, 2902413, 8193151, 16386302, 46256493, 130576328, 261152656, 737201475, 2081028097, 4162056194, 11748967107, 33165873224, 66331746448

Offset: 1

M. F. Hasler, Jan 15 2012

Or: Numbers whose square, with its last base-7 digit dropped, is again a square (where for the first 3 terms, dropping the digit is meant to yield zero).

Harvey P. Dale, Table of n, a(n) for n = 1..1000
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (0,0,16,0,0,-1).

Cf. A031149 (base 10), A204502 (base 9), A204514 (base 8), A204518 (base 6), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

LinearRecurrence[{0,0,16,0,0,-1},{0,1,2,3,8,16,45},30] (* or *) CoefficientList[Series[ (x+2x^2+3x^3-8x^4-16x^5-3x^6)/(1-16x^3+x^6),{x,0,30}],x] (* Harvey P. Dale, Apr 22 2023 *)

b=7;for(n=0,2e9,issquare(n^2\b) & print1(n","))

G.f. = (x + 2*x^2 + 3*x^3 - 8*x^4 - 16*x^5 - 3*x^6 )/(1 - 16*x^3 + x^6).

floor(a(n)^2 / 7) = A204517(n)^2.

A204518 Numbers such that floor(a(n)^2 / 6) is a square.

Original entry on oeis.org

0, 1, 2, 3, 5, 10, 27, 49, 98, 267, 485, 970, 2643, 4801, 9602, 26163, 47525, 95050, 258987, 470449, 940898, 2563707, 4656965, 9313930, 25378083, 46099201, 92198402, 251217123, 456335045, 912670090, 2486793147, 4517251249, 9034502498, 24616714347

Offset: 1

M. F. Hasler, Jan 15 2012

Or: Numbers whose square, with its last base-6 digit dropped, is again a square. (For the three initial terms whose square has only one digit in base 6, this is then meant to yield zero.)

Index entries for linear recurrences with constant coefficients, signature (0,0,10,0,0,-1).

Cf. A023110 (base 10), A204502 (base 9), A204514 (base 8), A204516 (base 7), A204520 (base 5), A004275 (base 4), A055793 (base 3), A055792 (base 2).

b=6;for(n=0,2e9,issquare(n^2\b) & print1(n","))

concat(0, Vec(-x^2*(x+1)*(3*x^4+7*x^3-2*x^2-x-1)/(x^6-10*x^3+1) + O(x^100))) \\ Colin Barker, Sep 18 2014

a(n) = sqrt(A055851(n)).
From Colin Barker, Sep 18 2014: (Start)
a(n) = 10*a(n-3) - a(n-6) for n > 7.
G.f.: -x^2*(x+1)*(3*x^4 + 7*x^3 - 2*x^2 - x - 1) / (x^6-10*x^3+1). (End)
a(3n+2) = A001079(n). a(3n) = A087799(n-1). - R. J. Mathar, Feb 05 2020

More terms from Colin Barker, Sep 18 2014

A055851 a(n) and floor(a(n)/6) are both squares; i.e., squares that remain squares when written in base 6 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 9, 25, 100, 729, 2401, 9604, 71289, 235225, 940900, 6985449, 23049601, 92198404, 684502569, 2258625625, 9034502500, 67074266169, 221322261601, 885289046404, 6572593581849, 21687323011225, 86749292044900

Offset: 1

Henry Bottomley, Jul 14 2000

For the first 3 terms, the above "base 6" interpretation is questionable, since they have only 1 digit in base 6. It is understood that dropping this digit yields 0. - M. F. Hasler, Jan 15 2012

Base-6 analog of A055792 (base 2), A055793 (base 3), A055808 (base 4), A055812 (base 5), A204517 (base 7), A204503 (base 9) and A023110 (base 10). - M. F. Hasler, Jan 15 2012

			a(5) = 100 because 100 = 10^2 = 244 base 6 and 24 base 6 = 16 = 4^2.

M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.

Cf. A023110.

b=6;for(n=1,2e9,issquare(n^2\b) & print1(n^2,",")) \\ M. F. Hasler, Jan 15 2012

a(n) = A204518(n)^2. - M. F. Hasler, Jan 15 2012

Empirical g.f.: -x^2*(9*x^8+100*x^7+25*x^6-162*x^5-296*x^4-74*x^3+9*x^2+4*x+1) / ((x-1)*(x^2+x+1)*(x^6-98*x^3+1)). - Colin Barker, Sep 15 2014

More terms added and offset changed to 1 by M. F. Hasler, Jan 16 2012

A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).

Original entry on oeis.org

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, 8119, 13860, 47321, 80782, 275807, 470832, 1607521, 2744210, 9369319, 15994428, 54608393, 93222358, 318281039, 543339720, 1855077841, 3166815962, 10812186007, 18457556052, 63018038201, 107578520350

Offset: 1

Benoit Cloitre, May 10 2003

The upper principal and intermediate convergents to 2^(1/2), beginning with 2/1, 3/2, 10/7, 17/12, 58/41, form a strictly decreasing sequence; essentially, numerators=A143609 and denominators=A084068. - Clark Kimberling, Aug 27 2008
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. We have
  a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
  a(2*n) = (1/sqrt(2))*(1 o 1 o ... o 1) (2*n terms). Cf. A049629, A108412 and A143608.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = U(2*n)/sqrt(2) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ).
It appears that this sequence consists of those numbers m such that 2*m^2 = floor( m*sqrt(2) * ceiling(m*sqrt(2)) ). Cf. A084069. (End)
Conjecture: a(n) is the earliest occurrence of n in A348295, which is to say, a(n) is the least m such that Sum_{k=1..m} (-1)^(floor(k*(sqrt(2)-1))) = Sum_{k=1..m} (-1)^A097508(k) = n. This has been confirmed for the first 32 terms by Chai Wah Wu, Oct 21 2021. - Jianing Song, Jul 16 2022

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Indranil Ghosh, Table of n, a(n) for n = 1..2608
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Bisections are A001542 and A002315.

Cf. A084069, A084070, A133566, A079496, A001541, A001653, A008844, A055792, A049629, A108412, A143608.

a := proc (n) if `mod`(n, 2) = 1 then (1/2)*(sqrt(2) + 1)^n - (1/2)*(sqrt(2) - 1)^n else (1/2)*((sqrt(2) + 1)^n - (sqrt(2) - 1)^n)/sqrt(2) end if;
end proc:
seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018

a[n_] := ((Sqrt[2]+1)^n - (Sqrt[2]-1)^n) ((-1)^n(Sqrt[2]-2) + (Sqrt[2]+2))/8;
Table[Simplify[a[n]], {n, 30}] (* after Paul Barry, Peter Luschny, Mar 29 2018 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^(n-1)*[1;2;7;12])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

"A Diofloortin equation": n such that 2*n^2=floor(n*sqrt(2)*ceiling(n*sqrt(2))).
a(n)*a(n+3) = -2 + a(n+1)*a(n+2).
From Paul Barry, Jun 06 2006: (Start)
G.f.: x*(1+x)^2/(1-6*x^2+x^4);
a(n) = ((sqrt(2)+1)^n-(sqrt(2)-1)^n)*((sqrt(2)/8-1/4)*(-1)^n+sqrt(2)/8+1/4);
a(n) = Sum_{k=0..floor(n/2)} 2^k*(C(n,2*k)-C(n-1,2*k+1)*(1+(-1)^n)/2). (End)
A000129(n+1) = A079496(n) + a(n). - Gary W. Adamson, Sep 18 2007
Equals A133566 * A000129, where A000129 = the Pell sequence. - Gary W. Adamson, Sep 18 2007
From Peter Bala, Mar 23 2018: (Start)
a(2*n + 2) = a(2*n + 1) + sqrt( (1 + a(2*n + 1)^2)/2 ).
a(2*n + 1) = 2*a(2*n) + sqrt( (1 + 2*a(2*n)^2) ).
More generally,
a(2*n+2*m+1) = sqrt(2)*a(2*n) o a(2*m+1), where o is the binary operation defined above, that is,
a(2*n+2*m+1) = sqrt(2)*a(2*n)*sqrt(1 + a(2*m+1)^2) + a(2*m+1)*sqrt(1 + 2*a(2*n)^2).
sqrt(2)*a(2*(n + m)) = (sqrt(2)*a(2*n)) o (sqrt(2)*a(2*m)), that is,
a(2*n+2*m) = a(2*n)*sqrt(1 + 2*a(2*m)^2) + a(2*m)*sqrt(1 + 2*a(2*n)^2).
sqrt(1 + 2*a(2*n)^2) = A001541(n).
1 + 2*a(2*n)^2 = A055792(n+1).
a(2*n) - a(2*n-1) = A001653(n).
(1 + a(2*n+1)^2)/2 = A008844(n). (End)
a(n) = A000129(n) for even n and A001333(n) for odd n. - R. J. Mathar, Oct 15 2021

A055997 Numbers k such that k*(k - 1)/2 is a square.

Original entry on oeis.org

1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522

Offset: 1

Barry E. Williams, Jun 14 2000

Numbers k such that (k-th triangular number - k) is a square.
Gives solutions to A007913(2x)=A007913(x-1). - Benoit Cloitre, Apr 07 2002
Number of closed walks of length 2k on the grid graph P_2 X P_3. - Mitch Harris, Mar 06 2004
If x = A001109(n - 1), y = a(n) and z = x^2 + y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
The product of any term a(n) with an even successor a(n + 2k) is always a square number. The product of any term a(n) with an odd successor a(n + 2k + 1) is always twice a square number. - Bradley Klee & Bill Gosper, Jul 22 2015
It appears that dividing even terms by two and taking the square root gives sequence A079496. - Bradley Klee, Jul 25 2015
The bisections of this sequence are a(2n - 1) = A055792(n) and a(2n) = A088920(n). - Bernard Schott, Apr 19 2020

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.
P. Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 35 pages 346-347.

Colin Barker, Table of n, a(n) for n = 1..100
Dario Alpern, a^4+b^3=c^2.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 46, 56.
Phil Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Kenneth Ramsey, Generalized Proof re Square Triangular Numbers, message 62 in Triangular_and_Fibonacci_Numbers Yahoo group, Oct 10, 2011.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A007913, A001541.
A001109(n-1) = sqrt{[(a(n))^2 - (a(n))]/2}.
a(n) = A001108(n-1)+1.
A001110(n-1)=a(n)*(a(n)-1)/2.
Cf. A001652, A001653, A046090.
Identical to A115599, but with additional leading term.

I:=[1,2,9]; [n le 3 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

A:= gfun:-rectoproc({a(n) = 6*a(n-1)-a(n-2)-2, a(1) = 1, a(2) = 2}, a(n), remember):
map(A,[$1..100]); # Robert Israel, Jul 22 2015

Table[ 1/4*(2 + (3 - 2*Sqrt[2])^k + (3 + 2*Sqrt[2])^k ) // Simplify, {k, 0, 20}] (* Jean-François Alcover, Mar 06 2013 *)
CoefficientList[Series[(1 - 5 x + 2 x^2) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
(1 + ChebyshevT[#, 3])/2 & /@ Range[0, 20] (* Bill Gosper, Jul 20 2015 *)
a[1]=1;a[2]=2;a[n_]:=(a[n-1]+1)^2/a[n-2];a/@Range[25] (* Bradley Klee, Jul 25 2015 *)
LinearRecurrence[{7,-7,1},{1,2,9},30] (* Harvey P. Dale, Dec 06 2015 *)

Vec((1-5*x+2*x^2)/((1-x)*(1-6*x+x^2))+O(x^66)) /* Joerg Arndt, Mar 06 2013 */

t(n)=(1+sqrt(2))^(n-1);
for(k=1,24,print1(round((1/4)*(t(k)^2 + t(k)^(-2) + 2)),", ")) \\ Hugo Pfoertner, Nov 29 2019

a(n) = (1 + polchebyshev(n-1, 1, 3))/2; \\ Michel Marcus, Apr 21 2020

a(n) = 6*a(n - 1) - a(n - 2) - 2; n >= 3, a(1) = 1, a(2) = 2.
G.f.: x*(1 - 5*x + 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) - 1 + sqrt(2*a(n)*(a(n) - 1)) = A001652(n - 1). - Charlie Marion, Jul 21 2003; corrected by Michel Marcus, Apr 20 2020
a(n) = IF(mod(n; 2)=0; (((1 - sqrt(2))^n + (1 + sqrt(2))^n)/2)^2; 2*((((1 - sqrt(2))^(n + 1) + (1 + sqrt(2))^(n + 1)) - (((1 - sqrt(2))^n + (1 + sqrt(2))^n)))/4)^2). The odd-indexed terms are a(2n + 1) = [A001333(2n)]^2; the even-indexed terms are a(2n) = [A001333(2n - 1)]^2 + 1 = 2*[A001653(n)]^2. - Antonio Alberto Olivares, Jan 31 2004; corrected by Bernard Schott, Apr 20 2020
A053141(n + 1) + a(n + 1) = A001541(n + 1) + A001109(n + 1). - Creighton Dement, Sep 16 2004
a(n) = (1/2) + (1/4)*(3+2*sqrt(2))^(n-1) + (1/4)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Feb 21 2006; corrected by Michel Marcus, Apr 20 2020
a(n) = A001653(n)-A001652(n-1). - Charlie Marion, Apr 10 2006; corrected by Michel Marcus, Apr 20 2020
a(2k) = A001541(k)^2. - Alexander Adamchuk, Nov 24 2006
a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with mA001653(m)*A011900(m-n) - A002315(m)*A046090(m-n) - A001108(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = +7*a(n-1) -7*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 06 2013
a(n) * a(n+2) = (A001108(n)-A001652(n)+3*A046090(n))^2. - Robert Israel, Jul 23 2015
sqrt(a(n+1)*a(n-1)) = a(n)+1 - Bradley Klee & Bill Gosper, Jul 25 2015
a(n) = 1 + sum{k=0..n-2} A002315(k). - David Pasino, Jul 09 2016; corrected by Michel Marcus, Apr 20 2020
E.g.f.: (2*exp(x) + exp((3-2*sqrt(2))*x) + exp((3+2*sqrt(2))*x))/4. - Ilya Gutkovskiy, Jul 09 2016
sqrt(a(n)*(a(n)-1)/2) = A001542(n)/2. - David Pasino, Jul 09 2016
Limit_{n -> infinity} a(n)/a(n-1) = A156035. - César Aguilera, Apr 07 2018
a(n) = (1/4)*(t^2 + t^(-2) + 2), where t = (1+sqrt(2))^(n-1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) + sqrt(a(n) - 1) = (1 + sqrt(2))^(n - 1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) - sqrt(a(n) - 1) = (-1 + sqrt(2))^(n - 1). - Bernard Schott, Apr 18 2020

A055872 a(n) and floor(a(n)/8) are both squares; i.e., squares that remain squares when written in base 8 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 9, 36, 289, 1156, 9801, 39204, 332929, 1331716, 11309769, 45239076, 384199201, 1536796804, 13051463049, 52205852196, 443365544449, 1773462177796, 15061377048201, 60245508192804

Offset: 1

Henry Bottomley, Jul 14 2000

For the first 3 terms which have only 1 digit in base 8, removing this digit is meant to yield 0.

Base-8 analog of A055792 (base 2), A055793 (base 3), A055808 (base 4), A055812 (base 5), A055851 (base 6), A055859 (base 7), A204503 (base 9) and A023110 (base 10). - M. F. Hasler, Jan 15 2012

			a(5) = 289 because 289 = 17^2 = 441 base 8 and 44 base 8 = 36 = 6^2.

M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012.
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (0,35,0,-35,0,1)

Cf. A023110, A055792 (bisection).

Select[Range[0,8*10^6]^2,IntegerQ[Sqrt[FromDigits[Most[ IntegerDigits[ #,8]], 8]]]&] (* Harvey P. Dale, Aug 02 2016 *)

b=8;for(n=1,200,issquare(n^2\b) && print1(n^2,",")) \\ M. F. Hasler, Jan 15 2012

a(n) = A204514(n)^2. - M. F. Hasler, Jan 15 2012

Empirical g.f.: -x^2*(4*x+1)*(9*x^4-26*x^2+1) / ((x-1)*(x+1)*(x^2-6*x+1)*(x^2+6*x+1)). - Colin Barker, Sep 15 2014

More terms added and offset changed to 1 by M. F. Hasler, Jan 15 2012

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

Amarnath Murthy, Jun 08 2003

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

D. W. Wilson, Table of n, a(n-1) for n = 1..100 (offset=1)
Emrah Kılıç, Yücel Türker Ulutaş, and Neşe Ömür, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011), Article 11.5.6, table 3, k=2.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

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).

[4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022

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 *)

[4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

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

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

A247375 Numbers m such that floor(m/2) is a square.

Original entry on oeis.org

0, 1, 2, 3, 8, 9, 18, 19, 32, 33, 50, 51, 72, 73, 98, 99, 128, 129, 162, 163, 200, 201, 242, 243, 288, 289, 338, 339, 392, 393, 450, 451, 512, 513, 578, 579, 648, 649, 722, 723, 800, 801, 882, 883, 968, 969, 1058, 1059, 1152, 1153, 1250, 1251, 1352, 1353

Offset: 0

Bruno Berselli, Sep 15 2014

Union of A001105 and A058331.

Squares of the sequence are listed in A055792.

Jens Kruse Andersen, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A001105, A055792, A058331, A213037.

Cf. A130404 (numbers m such that floor(m/2) is a triangular number).

[n: n in [0..1400] | IsSquare(Floor(n div 2))];

Select[Range[0, 1400], IntegerQ[Sqrt[Floor[#/2]]] &]
LinearRecurrence[{1,2,-2,-1,1},{0,1,2,3,8},70] (* Harvey P. Dale, Oct 21 2021 *)

[n for n in [0..1400] if is_square(floor(n/2))]

G.f.: x*( 1 + x - x^2 + 3*x^3 ) / ( (1 - x)^3*(1 + x)^2 ).
a(n) = 1 + ( 2*n*(n-1) + (2*n-3)*(-1)^n - 1 )/4.
a(n+1) = 1 + A213037(n).
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n >= 5. - Wesley Ivan Hurt, Dec 18 2020
Sum_{n>=1} 1/a(n) = Pi^2/12 + coth(Pi/sqrt(2))*Pi/(2*sqrt(2)) + 1/2. - Amiram Eldar, Sep 24 2022

cp's OEIS Frontend

A204503 Squares n^2 such that floor(n^2/9) is again a square.

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A204514 Numbers such that floor(a(n)^2 / 8) is again a square.

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A204516 Numbers such that floor(a(n)^2 / 7) is a square.

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A204518 Numbers such that floor(a(n)^2 / 6) is a square.

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A055851 a(n) and floor(a(n)/6) are both squares; i.e., squares that remain squares when written in base 6 and last digit is removed.

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A084068 a(1) = 1, a(2) = 2; a(2*k) = 2*a(2*k-1) - a(2*k-2), a(2*k+1) = 4*a(2*k) - a(2*k-1).

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A055997 Numbers k such that k*(k - 1)/2 is a square.

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A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).