A055793 -id:A055793 - cp's OEIS frontend

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

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

A007654 Numbers k such that the standard deviation of 1,...,k is an integer.

Original entry on oeis.org

0, 3, 48, 675, 9408, 131043, 1825200, 25421763, 354079488, 4931691075, 68689595568, 956722646883, 13325427460800, 185599261804323, 2585064237799728, 36005300067391875, 501489136705686528, 6984842613812219523, 97286307456665386800, 1355023461779503195683

Offset: 1

N. J. A. Sloane

Gives solutions k to the Diophantine equation m^2 = k*(k+1)/3. - Anton Lorenz Vrba (anton(AT)a-l-v.net), Jun 28 2005
If x=a(n), y=a(n+1), z=a(n+2) are three consecutive terms, then x^2 - 16*y*x + 14*x*z + 16*y^2 - 16*z*y + z^2 = 144. The formula is symmetric in x and z, so it is also valid for x=a(n+2), y=a(n+1), z=a(n). - Alexander Samokrutov, Jul 02 2015
From Bernard Schott, Apr 09 2021 (Start):
Corresponding solutions m (of first comment) are in A011944.
Equivalently, numbers k such that k/3 and k+1 are both perfect squares. (End)

Guy Alarcon and Yves Duval, TS: Préparation au Concours Général, RMS, Collection Excellence, Paris, 2010, chapitre 13, Questions proposées aux élèves de Terminale S, Exercice 1, p. 220, p. 223.
D. A. Benaron, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 1..100
Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A001353, A007655, A011944, A055793, A098301.

I:=[0,3]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+6: n in [1..20]]; // Vincenzo Librandi, Mar 05 2016

RecurrenceTable[{a[m] == 14 a[m - 1] - a[m - 2] + 6, a[1] == 0, a[2] == 3}, a, {m, 1, 17}] (* Michael De Vlieger, Jul 02 2015 *)
CoefficientList[Series[-3 x^2*(1 + x)/(-1 + x)/(1 - 14 x + x^2), {x, 0, 17}], x] (* Michael De Vlieger, Feb 02 2016 *)

concat(0,3*Vec((1+x)/(1-x)/(1-14*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, May 14 2013

a(n) = 3*A098301(n-1).
a(m) = 14*a(m-1) - a(m-2) + 6.
G.f.: -3*x^2*(1+x)/(-1+x)/(1-14*x+x^2) = -3 + (1/2)/(-1+x) + (1/2)*(-97*x+7)/(1-14*x+x^2). - R. J. Mathar, Nov 20 2007
a(n) = (-2 + (7-4*sqrt(3))^n*(7+4*sqrt(3)) + (7-4*sqrt(3))*(7+4*sqrt(3))^n)/4. - Colin Barker, Mar 05 2016
From Bernard Schott, Apr 09 2021: (Start)
a(n) = 3 * A001353(n-1)^2.
a(n) = A055793(n+1) - 1 = A001075(n-1)^2 - 1. (End)
2*a(n) = A011943(n)-1. - R. J. Mathar, Mar 16 2023

Corrected by Keith Lloyd, Mar 15 1996

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

A098301 Member r=16 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 16, 225, 3136, 43681, 608400, 8473921, 118026496, 1643897025, 22896531856, 318907548961, 4441809153600, 61866420601441, 861688079266576, 12001766689130625, 167163045568562176, 2328280871270739841, 32428769152221795600, 451674487259834398561

Offset: 0

Wolfdieter Lang, Oct 18 2004

Also m such that (3*m^2 + m)/4 = m*(3*m + 1)/4 is a perfect square. - Ctibor O. Zizka, Oct 15 2010

Consequently A049451(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011

Colin Barker, Table of n, a(n) for n = 0..874
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A001353, A049451, A092184.

LinearRecurrence[{# - 1, -# + 1, 1}, {0, 1, #}, 20] &[16] (* Michael De Vlieger, Feb 23 2021 *)

concat(0, Vec(x*(1+x)/((1-x)*(1-14*x+x^2)) + O(x^50))) \\ Colin Barker, Jun 15 2015

a(n) = (T(n, 7)-1)/6 with Chebyshev's polynomials of the first kind evaluated at x=7: T(n, 7) = A011943(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n)/2; therefore: a(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n - 2)/12.
a(n) = A001353(n)^2 = S(n-1, 4)^2 with Chebyshev's polynomials of the second kind evaluated at x=4, S(n, 4):=U(n, 2).
a(n) = 14*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), n >= 3.
G.f.: x*(1+x)/((1-x)*(1 - 14*x + x^2)) = x*(1+x)/(1 - 15*x + 15*x^2 - x^3) (from the Stephan link, see A092184).
Conjecture: 4*A007655(n+1) + A046184(n) = A055793(n+2) + a(n+1). - Creighton Dement, Nov 01 2004
a(n) = (A001075(n)^2-1)/3. - Parker Grootenhuis, Nov 28 2017

A204517 Square root of floor[A055859(n)/7].

Original entry on oeis.org

0, 0, 0, 1, 3, 6, 17, 48, 96, 271, 765, 1530, 4319, 12192, 24384, 68833, 194307, 388614, 1097009, 3096720, 6193440, 17483311, 49353213, 98706426, 278635967, 786554688, 1573109376, 4440692161, 12535521795, 25071043590, 70772438609, 199781794032, 399563588064

Offset: 1

M. F. Hasler, Jan 15 2012

nonn

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

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=7;for(n=1,2e9,issquare(n^2\b) & print1(sqrtint(n^2\b),","))

A204517(n)=polcoeff((x^4 + 3*x^5 + 6*x^6 + x^7)/(1 - 16*x^3 + x^6+O(x^n)),n)

A204517(n) = sqrt(floor(A204516(n)^2/7)).

G.f. = (x^4 + 3*x^5 + 6*x^6 + x^7)/(1 - 16*x^3 + x^6)

A204512 Square roots of [A055872/8]: Their square written in base 8, with some digit appended, is again a square.

Original entry on oeis.org

0, 0, 0, 1, 2, 6, 12, 35, 70, 204, 408, 1189, 2378, 6930, 13860, 40391, 80782, 235416, 470832, 1372105, 2744210, 7997214, 15994428, 46611179, 93222358, 271669860, 543339720, 1583407981, 3166815962, 9228778026, 18457556052, 53789260175, 107578520350

Offset: 1

M. F. Hasler, Jan 15 2012

Base-8 analog of A031150. The square of the terms (= truncated squares A055872) are listed in A204504.

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).

CoefficientList[Series[(x^4 (1+2x))/(1-6x^2+x^4),{x,0,40}],x] (* Harvey P. Dale, Nov 30 2020 *)

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

a(n)=polcoeff((2*x^5 + x^4)/(x^4 - 6*x^2 + 1+O(x^n)),n)

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

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

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

Original entry on oeis.org

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)

cp's OEIS Frontend

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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A007654 Numbers k such that the standard deviation of 1,...,k is an integer.

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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.

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A098301 Member r=16 of the family of Chebyshev sequences S_r(n) defined in A092184.

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A204517 Square root of floor[A055859(n)/7].

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A204512 Square roots of [A055872/8]: Their square written in base 8, with some digit appended, is again a square.

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