A204512 -id:A204512 - cp's OEIS frontend

A204504 A204512(n)^2 = floor[A055872(n)/8]: Squares such that appending some digit in base 8 yields another square.

0, 0, 0, 1, 4, 36, 144, 1225, 4900, 41616, 166464, 1413721, 5654884, 48024900, 192099600, 1631432881, 6525731524, 55420693056, 221682772224, 1882672131025, 7530688524100, 63955431761796, 255821727047184, 2172602007770041, 8690408031080164, 73804512832419600

Offset: 1

M. F. Hasler, Jan 15 2012

Base-8 analog of A202303.

See also A031149=sqrt(A023110) (base 10), A204502=sqrt(A204503) (base 9),
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).

b=8;for(n=1,2e9,issquare(n^2\b) & print1((n^2\b)","))

a(n)=polcoeff(x^4*(1 + 4*x + x^2 + 4*x^3)/(1 - 35*x^2 + 35*x^4 - x^6+O(x^n)), n)

a(n)=A204512(n)^2.

G.f. = x^4*(1 + 4*x + x^2 + 4*x^3)/(1 - 35*x^2 + 35*x^4 - x^6)

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

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

A031150 Appending a digit to n^2 gives another perfect square.

Original entry on oeis.org

1, 2, 4, 5, 6, 12, 18, 43, 80, 154, 191, 228, 456, 684, 1633, 3038, 5848, 7253, 8658, 17316, 25974, 62011, 115364, 222070, 275423, 328776, 657552, 986328, 2354785, 4380794, 8432812, 10458821, 12484830, 24969660, 37454490

Offset: 1

Patrick De Geest

Square root of 'Squares from A023110 with last digit removed'.

One could include an initial '0', and even list it with multiplicity 3 or 4, since 00, 01, 04 and 09 are all perfect squares: In analogy to corresponding sequences for other bases, this sequence could be defined as sqrt(floor[A023110/10]), see A204512 [base 8], A204517 (base 7), A204519 (base 6), A204521 [base 5], A001353 [base 3], A001542 [base 2]. (For bases 4 and 9, the corresponding sequence contains all integers.) - M. F. Hasler, Jan 16 2012

			5^2 = 25 and 16^2 = 256, so 5 is in the sequence.
115364^2 = 13308852496, 364813^2 = 133088524969.

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

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Joshua Stucky, Pell's Equation and Truncated Squares, Number Theory Seminar, Kansas State University, Feb 19 2018.
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,38,0,0,0,0,0,0,-1).

Cf. A023110, A030686, A030687, A053784.

See A202303 for the resulting squares.

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

CoefficientList[Series[(x^10 + 2 x^9 + 4 x^8 + 5 x^7 + 18 x^6 + 12 x^5 + 6 x^4 + 5 x^3 + 4 x^2 + 2 x + 1)/(x^14 - 38 x^7 + 1), {x, 0, 50}], x] (* Vincenzo Librandi, Oct 19 2013 *)
LinearRecurrence[{0,0,0,0,0,0,38,0,0,0,0,0,0,-1},{1,2,4,5,6,12,18,43,80,154,191,228,456,684},40] (* Harvey P. Dale, Jun 09 2017 *)

G.f.: x*(x^10+2*x^9+4*x^8+5*x^7+18*x^6+12*x^5+6*x^4+5*x^3+4*x^2+2*x+1) / (x^14-38*x^7+1). - Colin Barker, Jan 30 2013

A204519 Square root of floor(A055851(n)/6).

Original entry on oeis.org

0, 0, 0, 1, 2, 4, 11, 20, 40, 109, 198, 396, 1079, 1960, 3920, 10681, 19402, 38804, 105731, 192060, 384120, 1046629, 1901198, 3802396, 10360559, 18819920, 37639840, 102558961, 186298002

Offset: 1

M. F. Hasler, Jan 15 2012

Base-6 analog of A031150 [base 10], A204512 [base 8], A204517 (base 7), A204521 [base 5], A001353 [base 3], A001542 [base 2]. For bases 4 and 9, the corresponding sequence contains all integers.

Cf. A055792, A055793, A055812, A204517, A204512, A204503 and A023110.

Sqrt[Floor[Select[Range[100000],IntegerQ[Sqrt[Quotient[#^2,6]]]&]^2/6]] (* Vaclav Kotesovec, Nov 26 2012 *)

b=6;for(n=1,2e9,issquare(n^2\b) & print1(sqrtint(n^2\b),","))

Conjecture (for n>=8): a(n) = 10*a(n-3) - a(n-6). - Vaclav Kotesovec, Nov 26 2012

Empirical g.f.: x^4*(x^3+4*x^2+2*x+1) / (x^6-10*x^3+1). - Colin Barker, Sep 15 2014

More terms from Vaclav Kotesovec, Nov 26 2012

A204521 Square root of floor(A055812(n) / 5).

Original entry on oeis.org

0, 0, 0, 1, 3, 4, 8, 21, 55, 72, 144, 377, 987, 1292, 2584, 6765, 17711, 23184, 46368, 121393, 317811, 416020, 832040, 2178309, 5702887, 7465176

Offset: 1

M. F. Hasler, Jan 15 2012

nonn

Or: Numbers whose square yields another square when written in base 5.
(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.)
Base-5 analog of A031150 [base 10], A001353 [base 3], A001542 [base 2].
The square roots of A055812 are listed in A204520.

Cf. A055792, A055793, A204519, A204517, A204512, A204503 and A023110.

b=5;for(n=1,2e9,issquare(n^2\b) && print1(sqrtint(n^2\b),","))

Empirical g.f.: x^4*(x^5+3*x^4+8*x^3+4*x^2+3*x+1) / ((x^4-4*x^2-1)*(x^4+4*x^2-1)). - Colin Barker, Sep 15 2014

cp's OEIS Frontend

A204504 A204512(n)^2 = floor[A055872(n)/8]: Squares such that appending some digit in base 8 yields another square.

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

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A031150 Appending a digit to n^2 gives another perfect square.

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A204519 Square root of floor(A055851(n)/6).

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