A023110 -id:A023110 - cp's OEIS frontend

A202303 Drop the last digit of A023110(n).

0, 0, 0, 0, 1, 4, 16, 25, 36, 144, 324, 1849, 6400, 23716, 36481, 51984, 207936, 467856, 2666689, 9229444, 34199104, 52606009, 74960964, 299843856, 674648676, 3845364121, 13308852496, 49315084900, 75857828929, 108093658176, 432374632704, 972842923584, 5545012396225, 19191356070436, 71112318227344, 109386936710041, 155870980128900, 623483920515600, 1402838821160100

Offset: 1

N. J. A. Sloane, Jan 12 2012

By definition, all the terms are squares.

R. K. Guy, Neg and Reg, preprint, Jan 2012.

Cf. A023110. The square roots are in A031150.

Conjecture: a(n) = 1443*a(n-7)-1443*a(n-14)+a(n-21). - Colin Barker, Sep 20 2014

Empirical g.f.: -x^5*(x +1)*(x^16 +3*x^15 +13*x^14 +12*x^13 +312*x^12 -168*x^11 +204*x^10 +202*x^9 +426*x^8 +202*x^7 +204*x^6 +120*x^5 +24*x^4 +12*x^3 +13*x^2 +3*x +1) / ((x -1)*(x^6 +x^5 +x^4 +x^3 +x^2 +x +1)*(x^14 -1442*x^7 +1)). - Colin Barker, Sep 20 2014

Fourth leading 0 inserted by Georg Fischer, Feb 21 2022

A055792 a(n) and floor(a(n)/2) are both squares; i.e., squares which remain squares when written in base 2 and last digit is removed.

Original entry on oeis.org

0, 1, 9, 289, 9801, 332929, 11309769, 384199201, 13051463049, 443365544449, 15061377048201, 511643454094369, 17380816062160329, 590436102659356801, 20057446674355970889, 681362750825443653409, 23146276081390728245001, 786292024016459316676609

Offset: 0

Henry Bottomley, Jul 14 2000

a(n) > 0 is a square such that a(n) - 1 is a product of powers. - Michel Lagneau, Feb 16 2012

			a(2) = 9 because 9 = 3^2 = 1001_2 and 100_2 = 4 = 2^2.

Charles R Greathouse IV, Table of n, a(n) for n = 0..654
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Giovanni Lucca, Circle chains inside the arbelos and integer sequences, Int'l J. Geom. (2023) Vol. 12, No. 1, 71-82.
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A023110, A247375.

LinearRecurrence[{35, -35, 1}, {0, 1, 9, 289}, 25] (* Paolo Xausa, Jul 22 2024 *)

concat(0, Vec(-x*(9*x^2-26*x+1)/((x-1)*(x^2-34*x+1)) + O(x^100))) \\ Colin Barker, Sep 15 2014

is(n)=issquare(n) && issquare(n\2) \\ Charles R Greathouse IV, May 07 2015

a(n) = 34*a(n-1) - a(n-2) - 16 = A001541(n-1)^2 = 2*A001542(n-1)^2 + 1 = 8*A001110(n-1) + 1.
From Colin Barker, Sep 15 2014: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n > 3.
G.f.: -x*(9*x^2 - 26*x + 1) / ((x-1)*(x^2 - 34*x + 1)). (End)
a(n) = c*k^n + 1/2 + o(1) with k = 17+sqrt(288) = 33.97... and c = 17/4 - sqrt(18). - Charles R Greathouse IV, May 07 2015
a(n) = (4 + 2*(17 + 12*sqrt(2))^(1-n) + (34 - 24*sqrt(2))*(17 + 12*sqrt(2))^n)/8 for n > 0. - Colin Barker, Mar 02 2016

A055793 Numbers k such that k and floor[k/3] are both squares; i.e., squares which remain squares when written in base 3 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 49, 676, 9409, 131044, 1825201, 25421764, 354079489, 4931691076, 68689595569, 956722646884, 13325427460801, 185599261804324, 2585064237799729, 36005300067391876, 501489136705686529, 6984842613812219524, 97286307456665386801, 1355023461779503195684, 18873042157456379352769

Offset: 1

Henry Bottomley, Jul 14 2000

Or, squares of the form 3k^2+1.

See A023110, A204503, A204512, A204517, A204519, A055812, A055808 and A055792 for the analog in other bases.

			a(3) = 49 because 49 = 7^2 = 1211 base 3 and 121 base 3 = 16 = 4^2.

Vincenzo Librandi, Table of n, a(n) for n = 1..800
Tom C. Brown and Peter J Shiue, Squares of second-order linear recurrence sequences, Fib. Quart., 33 (1994), 352-356.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 48, 56.
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A023110, A098301.

Cf. also A023110, A204503, A204512, A204517, A204519, A055812, A055808 and A055792 for the analog in other bases.

I:=[0, 1, 4]; [n le 3 select I[n] else 14*Self(n-1) - Self(n-2) - 6: n in [1..30]]; // Vincenzo Librandi, Jan 27 2013

A055793 := proc(n) coeftayl(x*(1-11*x+4*x^2)/((1-x)*(1-14*x+x^2)), x=0, n); end proc: seq(A055793(n), n=0..20); # Wesley Ivan Hurt, Sep 28 2014

CoefficientList[Series[x*(1 - 11*x + 4*x^2)/((1 - x)*(1 - 14*x + x^2)), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 28 2014 *)
LinearRecurrence[{15,-15,1},{0,1,4,49},40] (* Harvey P. Dale, Jun 19 2021 *)

sq3nsqplus1(n) = { for(x=1,n, y = 3*x*x+1; \ print1(y" ") if(issquare(y),print1(y" ")) ) }

a(n) = 3*A098301(n-2)+1. - R. J. Mathar, Jun 11 2009
a(n) = 14*a(n-1)-a(n-2)-6, with a(0)=1, a(1)=4. (See Brown and Shiue)
a(n) = (A001075(n-2))^2. - Johannes Boot Dec 16 2011, corrected by M. F. Hasler, Jan 15 2012
G.f.: x*(1 - 11*x + 4*x^2)/((1 - x)*(1 - 14*x + x^2)). - M. F. Hasler, Jan 15 2012

More terms from Cino Hilliard, Mar 01 2003

A031149 Numbers whose square with its last digit deleted is also a square.

Original entry on oeis.org

0, 1, 2, 3, 4, 7, 13, 16, 19, 38, 57, 136, 253, 487, 604, 721, 1442, 2163, 5164, 9607, 18493, 22936, 27379, 54758, 82137, 196096, 364813, 702247, 870964, 1039681, 2079362, 3119043, 7446484, 13853287, 26666893, 33073696, 39480499, 78960998, 118441497, 282770296

Offset: 1

Patrick De Geest

Square root of A023110(n).
For the first 4 terms, the square has only one digit, but in analogy to the sequences in other bases (A204502, A204512,  A204514, A204516, A204518, A204520, A004275, A055793, A055792), it is understood that deleting this digit yields 0.
From Robert Israel, Feb 16 2016: (Start)
Solutions x to x^2 = 10*y^2 + j, j in {0,1,4,6,9}, in increasing order of x.
j=0 occurs only for x=0.
Let M be the 2 X 2 matrix [19, 60; 6, 19].
Solutions of x^2 = 10*y^2 + 1 are (x,y)^T = M^k (1,0)^T for k >= 0.
Solutions of x^2 = 10*y^2 + 4 are (x,y)^T = M^k (2,0)^T for k >= 0.
Solutions of x^2 = 10*y^2 + 6 are (x,y)^T = M^k (4,1)^T and M^k (16,5)^T for k >= 0.
Solutions of x^2 = 10*y^2 + 9 are (x,y)^T = M^k (3,0)^T, M^k (7,2)^T, M^k (13,4)^T for k >= 0.
Since (1,0)^T <= (2,0)^T <= (3,0)^T <= (4,1)^T <= (7,2)^T <= (13,4)^T <= (16,5)^T <= (19,6)^T = M (1,0)^T (element-wise) and M has positive entries, we see that the terms always occur in this order, for successive k.
The eigenvalues of M are 19 + 6*sqrt(10) and 19 - 6*sqrt(10).
From this follow my formulas below and the G.f. (End)

			364813^2 = 133088524969, 115364^2 = 13308852496.

R. K. Guy, Neg and Reg, preprint, Jan 2012. [From N. J. A. Sloane, Jan 12 2012]

Dmitry Petukhov and Robert Israel, Table of n, a(n) for n = 1..4400 (n = 1..67 from Dmitry Petukhov)
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.

Cf. A000290, A023110, A030686, A030687, A031150.

for i from 1 to 150000 do if (floor(sqrt(10 * i^2 + 9)) > floor(sqrt(10 * i^2))) then print(floor(sqrt(10 * i^2 + 9))) end if end do;

fQ[n_] := IntegerQ@ Sqrt@ Quotient[n^2, 10]; Select[ Range[ 0, 40000000], fQ] (* Harvey P. Dale, Jun 15 2011 *) (* modified by Robert G. Wilson v, Jan 16 2012 *)

s=[]; for(n=0, 1e7, if(issquare(n^2\10), s=concat(s,n))); s \\ Colin Barker, Jan 17 2014; typo fixed by Zak Seidov, Jan 31 2014

Appears to satisfy: a(n)=38a(n-7)-a(n-14) which would require a(-k) to look like 3, 2, 1, 4, 7, 13, 16, 57, 38, 19, 136, ... for k>0. - Henry Bottomley, May 08 2001
Empirical g.f.: x^2*(1 + 2*x + 3*x^2 + 4*x^3 + 7*x^4 + 13*x^5 + 16*x^6 - 19*x^7 - 38*x^8 - 57*x^9 - 16*x^10 - 13*x^11 - 7*x^12 - 4*x^13) / ((1 - 38*x^7 + x^14)). - Colin Barker, Jan 17 2014
a(n) = 38*a(n-7) - a(n-14) for n>15 (conjectured). - Colin Barker, Dec 31 2017
With e1 = 19 + 6*sqrt(10) and e2 = 19 - 6*sqrt(10),
   a(2+7k) = (e1^k + e2^k)/2,
   a(3+7k) = e1^k + e2^k,
   a(4+7k) = (3/2) (e1^k + e2^k),
   a(5+7k) = (2+sqrt(10)/2) e1^k + (2-sqrt(10)/2) e2^k,
   a(6+7k) = (7/2+sqrt(10)) e1^k + (7/2-sqrt(10)) e2^k,
   a(7+7k) = (13/2+2 sqrt(10)) e1^k + (13/2-2 sqrt(10)) e2^k,
a(8+7k) = (8+5 sqrt(10)/2) e1^k + (8-5 sqrt(10)/2) e2^k. - Robert Israel, Feb 16 2016

4 initial terms added by M. F. Hasler, Jan 15 2012

a(40) from Robert G. Wilson v, Jan 15 2012

A055812 a(n) and floor(a(n)/5) are both squares; i.e., squares which remain squares when written in base 5 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 9, 49, 81, 324, 2209, 15129, 25921, 103684, 710649, 4870849, 8346321, 33385284, 228826129, 1568397609, 2687489281, 10749957124, 73681302249, 505019158609, 865363202001, 3461452808004

Offset: 1

Henry Bottomley, Jul 14 2000

For the first 3 terms, the above "base 5" interpretation is questionable, since they have only 1 digit in base 5. It is understood that dropping this digit yields 0. - M. F. Hasler, Jan 15 2012

			a(4) = 49 because 49 = 7^2 = 144 base 5 and 14 base 5 = 9 = 3^2.

For analogs in other bases see A055792 (base 2), A055793 (base 3), A055808 (base 4), A055851 (base 6), A204517 (base 7), A204512 (base 8), A204503 (base 9) and A023110 (base 10).

Squares of A204520. The square roots of floor[a(n)/5] are given in A204521.

b=5;for(n=1,2e9,issquare(n^2\b) && print1(n^2,","))  \\ M. F. Hasler, Jan 15 2012

Empirical g.f.: -x^2*(9*x^11 +49*x^10 +324*x^9 +81*x^8 -698*x^7 -698*x^6 -968*x^5 -242*x^4 +49*x^3 +9*x^2 +4*x +1) / ((x -1)*(x +1)*(x^2 -4*x -1)*(x^2 +1)*(x^2 +4*x -1)*(x^4 +18*x^2 +1)). - Colin Barker, Sep 15 2014

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

A204502 Numbers such that floor[a(n)^2 / 9] is a square.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174, 177

Offset: 1

M. F. Hasler, Jan 15 2012

nonn

Or, numbers n such that n^2, with its last base-9 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 9.)

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

The squares are in A204503, the squares with last base-9 digit dropped in A204504, and the square roots of the latter in A028310.

Cf. A031149=sqrt(A023110) (base 10), A204514=sqrt(A055872) (base 8), 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).

Select[Range[0,200],IntegerQ[Sqrt[Floor[#^2/9]]]&] (* Harvey P. Dale, May 05 2018 *)

b=9;for(n=0,200,issquare(n^2\b) & print1(n","))

Conjecture: a(n) = 3*n-12 for n>5. G.f.: x^2*(x^2+x+1)*(x^3-x+1)/(x-1)^2. [Colin Barker, Nov 23 2012]

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

Original entry on oeis.org

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

A055808 a(n) and floor(a(n)/4) are both squares; i.e., squares that remain squares when written in base 4 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484, 576, 676, 784, 900, 1024, 1156, 1296, 1444, 1600, 1764, 1936, 2116, 2304, 2500, 2704, 2916, 3136, 3364, 3600, 3844, 4096, 4356, 4624, 4900, 5184, 5476, 5776, 6084, 6400, 6724, 7056, 7396, 7744, 8100

Offset: 0

Henry Bottomley, Jul 14 2000

Let A(x) = (1 + k*x + (k-1)*x^2). Then the coefficients of (A(x))^2 sum to 4*k^2, where k = (n - 1). Examples: if k = 3 we have (1 + 3*x + 2*x^2)^2 = (1 + 6*x + 13x^2 + 12*x^3 + 4*x^4), and ( 1 + 6 + 13 + 12 + 4) = 36. If k = 4 we have (1 + 4*x + 3*x^2)^2 = (1 + 8*x + 22*x^2 + 24*x^3 + 9*x^4), and (1 + 8 + 22 + 24 + 9) = 64 = a(5). - Gary W. Adamson, Aug 02 2015

For n>0, a(n) are the Engel expansion of A197036. - Benedict W. J. Irwin, Dec 15 2016

			36 is in the sequence because 36 = 6^2 = 210 base 3 and 21 base 4 = 9 = 3^2.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A023110. Essentially A016742 with one addition.

[Floor((2*n^2)/(1 + n))^2: n in [0..60]]; // Vincenzo Librandi, Aug 03 2015

Join[{0, 1}, LinearRecurrence[{3, -3, 1}, {4, 16, 36}, 50]] (* Vincenzo Librandi, Aug 03 2015 *)

concat(0, Vec(x*(x^3-7*x^2-x-1)/(x-1)^3 + O(x^100))) \\ Colin Barker, Sep 15 2014

is_ok(n)=issquare(n) && issquare(floor(n/4));
first(m)=my(v=vector(m),r=0);for(i=1,m,while(!is_ok(r),r++);v[i]=r;r++;);v; /* Anders Hellström, Aug 08 2015 */

a(n) = A004275(n)^2. - M. F. Hasler, Jan 16 2012

a(n) = 4*(-1+n)^2 for n>1; a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>4; G.f.: x*(x^3-7*x^2-x-1) / (x-1)^3. - Colin Barker, Sep 15 2014

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

cp's OEIS Frontend

A202303 Drop the last digit of A023110(n).

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A055792 a(n) and floor(a(n)/2) are both squares; i.e., squares which remain squares when written in base 2 and last digit is removed.

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A055793 Numbers k such that k and floor[k/3] are both squares; i.e., squares which remain squares when written in base 3 and last digit is removed.

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A031149 Numbers whose square with its last digit deleted is also a square.

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A055812 a(n) and floor(a(n)/5) are both squares; i.e., squares which remain squares when written in base 5 and last digit is removed.

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A204502 Numbers such that floor[a(n)^2 / 9] is a square.

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

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