A065100 -id:A065100 - cp's OEIS frontend

A154021 a(n+2) = 16*a(n+1) - a(n), with a(1)=0, a(2)=4.

0, 4, 64, 1020, 16256, 259076, 4128960, 65804284, 1048739584, 16714029060, 266375725376, 4245297576956, 67658385505920, 1078288870517764, 17184963542778304, 273881127813935100, 4364913081480183296

Offset: 1

Vincenzo Librandi, Jan 04 2009

If a(n)=x and a(n+1)=y, then 16=(x^2+y^2)/(xy+1).

In general, the sequence a(1)=0, a(2)=U; a(n+2)=U^2*a(n+1)-a(n) has the property that "If a(n)=x and a(n+1)=y then (x^2+y^2)/(xy+1)=U^2".

Vincenzo Librandi, Table of n, a(n) for n = 1..800
Index entries for linear recurrences with constant coefficients, signature (16,-1).

Cf. A065100, A154022-A154027.

I:=[0,4]; [n le 2 select I[n] else 16*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 25 2012

Nest[Append[#,16Last[#]-#[[-2]]]&,{0,4},20]  (* or *) Rest[CoefficientList[Series[4x^2/(1-16x+x^2), {x,0,22}], x]]  (* Harvey P. Dale, Apr 17 2011 *)
LinearRecurrence[{16, -1}, {0, 4}, 20] (* T. D. Noe, Apr 17 2011 *)

From R. J. Mathar, Jan 05 2011: (Start)
G.f.: 4*x^2/(1 -16*x +x^2).
a(n) = 4*A077412(n-2). (End)

375725376 replaced by 266375725376 - R. J. Mathar, Jan 07 2009

Edited by N. J. A. Sloane, Jun 23 2010 at the suggestion of Joerg Arndt.

A154027 a(n+2) = 100*a(n+1) - a(n), a(1)=0, a(2)=10.

Original entry on oeis.org

0, 10, 1000, 99990, 9998000, 999700010, 99960003000, 9995000599990, 999400099996000, 99930014999000010, 9992002099800005000, 999100279965001499990, 99900035994400349994000, 9989004499160069997900010

Offset: 1

Vincenzo Librandi, Jan 04 2009

If a(n)=x and a(n+1)=y then (x^2+y^2)/(xy+1)=100.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (100,-1).

Cf. A065100, A154021-A154026.

I:=[0, 10]; [n le 2 select I[n] else 100*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 26 2012

LinearRecurrence[{100,-1},{0,10},20] (* Vincenzo Librandi, Feb 26 2012 *)
CoefficientList[Series[(10*x)/(1-100*x+x^2),{x,0,20}],x] (* Harvey P. Dale, Aug 15 2012 *)

a(n)=([0,1;-1,100]^n*[0;10])[1,1] \\ Charles R Greathouse IV, Sep 01 2016

G.f.: (10*x)/(1 -100*x +x^2). - Harvey P. Dale, Aug 15 2012

A115169 Integers b > 0 for which there exists a positive integer a <= b such that (a^2 + b^2)/(1 + ab) is an integer.

Original entry on oeis.org

1, 8, 27, 30, 64, 112, 125, 216, 240, 343, 418, 512, 729, 1000, 1020, 1331, 1560, 1728, 2133, 2197, 2744, 3120, 3375, 4096, 4913, 5822, 5832, 6859, 7770, 8000, 9261, 10648, 12167, 13824, 15625, 16256, 16800, 17576, 18957, 19683

Offset: 1

John W. Layman, Mar 03 2006

nonn

All positive cubes are in this sequence.

Indeed, if b = k^3, then for a = k, we have a^2 + b^2 = k^2 + k^6 = (1 + k^4)*k^2 = (a*b + 1)*a^2. More generally, if the ratio (a^2 + b^2)/(a*b + 1) is an integer, it is equal to gcd(a,b)^2, thus in particular a perfect square. (This was Question 6 in the 1988 IMO.) All solutions (a,b) are member of a sequence {(x(n), x(n+1)); n = 1,2,...} where x = (0, k, k^3, k^5 - k, ...) with x(n+1) = k^2*x(n) - x(n-1) and some k >= 2, cf. A052530 for k = 2, A065100 for k = 3. (One might consider >= 0 instead > 0 in the definition, but a = 0 yields a solution for any b.) - M. F. Hasler, Jun 12 2019

			(2^2+8^2)/(1+2*8) = 68/17 = 4, an integer, so 8 is a term of the series.
From _M. F. Hasler_, Jun 12 2019: (Start)
The list of solutions starts:
     a      b     a^2+b^2   a*b+1   ratio
   ----------------------------------------
     1      1          2       2       1
     8      2         68      17       4
    27      3        738      82       9
    30      8        964     241       4
    64      4       4112     257      16
   112     30      13444    3361       4
   125      5      15650     626      25
   216      6      46692    1297      36
   240     27      58329    6481       9
   343      7     117698    2402      49
   418    112     187268   46817       4
   512      8     262208    4097      64
   729      9     531522    6562      81
  1000     10    1000100   10001     100
  1020     64    1044496   65281      16
(End)

Charles R Greathouse IV, Table of n, a(n) for n = 1..500
Steve Chow, You, Me and The Legend of Question Six, BlackPenRedPen, YouTube, Apr 20 2020.
I. Lauko, G. Pinter and L. Pinter, Another Step Further... On a Problem of the 1988 IMO, Math. Mag. 79 (2006), 45-53.
Simon Pampena, The Legend of Question Six, Numberphile, YouTube, Aug 16 2016.

Cf. A000578 (cubes), A052530 (subsequence of terms for ratio 2^2, for n >= 2), A065100 (subsequence of terms for ratio 3^2).

isok(n) = for(m=0, n, if (denominator((m^2+n^2)/(1+m*n))==1, return(1))); return (0); \\ Michel Marcus, Sep 18 2017

is_A115169(n)=for(a=1,n\3+1,(a^2+n^2)%(1+a*n)||return(1)) \\ M. F. Hasler, Jun 12 2019

is(n)=my(s=sqrtnint(n,3),n2=n^2); for(b=1,s, if((n2+b^2)%(n*b+1)==0, return(1))); for(K=2,sqrtint((n2+(s+1)^2)\(n*s+n+1)), my(k=K^2); if(issquare(k^2*n2-4*n2+4*k), return(1))); 0 \\ Charles R Greathouse IV, Nov 08 2021

Edited by M. F. Hasler, Jun 12 2019

A154022 a(n) = 5*A097780(n-2).

Original entry on oeis.org

0, 5, 125, 3120, 77875, 1943755, 48516000, 1210956245, 30225390125, 754423796880, 18830369531875, 470004814499995, 11731289992968000, 292812245009700005, 7308574835249532125, 182421558636228603120

Offset: 1

Vincenzo Librandi, Jan 04 2009

If a(n)=x and a(n+1)=y, then (x^2+y^2)/(xy+1)=25.

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (25,-1).

Cf. A065100, A154021-A154027.

I:=[0, 5]; [n le 2 select I[n] else 25*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Jan 29 2012

CoefficientList[Series[(5*z)/(z^2-25*z+1),{z,0,25}],z] (* Vincenzo Librandi, Jan 29 2012 *)
LinearRecurrence[{25,-1},{0,5},20] (* Harvey P. Dale, Mar 06 2018 *)

concat(0,Vec(5/(1-25*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Dec 27 2011

a(n) = +25*a(n-1) -a(n-2).

G.f.: 5*x^2/(1 -25*x +x^2). - R. J. Mathar, Jan 05 2011

Edited by N. J. A. Sloane, Jun 23 2010 at the suggestion of Joerg Arndt.

A154026 a(n+2) = 81*a(n+1) - a(n), a(1)=0, a(2)=9.

Original entry on oeis.org

0, 9, 729, 59040, 4781511, 387243351, 31361929920, 2539929080169, 205702893563769, 16659394449585120, 1349205247522830951, 109268965654899721911, 8849437012799354643840, 716695129071092826429129, 58043456017745719586115609, 4700803242308332193648935200

Offset: 1

Vincenzo Librandi, Jan 04 2009

If a(n)=x and a(n+1)=y then (x^2+y^2)/(xy+1)=81.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (81,-1).

Cf. A065100, A154021-A154027.

I:=[0,9]; [n le 2 select I[n] else 81*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Feb 25 2012

LinearRecurrence[{81,-1},{0,9},20] (* Harvey P. Dale, Sep 15 2011 *)

concat(0,Vec(9/(x^2-81*x+1)+O(x^98))) \\ Charles R Greathouse IV, Dec 27 2011

G.f.: (9*x)/(1 -81*x +x^2). - Harvey P. Dale, Sep 15 2011

More terms from Harvey P. Dale, Sep 15 2011

A154023 a(n+2) = 36*a(n+1) - a(n), a(1)=0, a(2)=6.

Original entry on oeis.org

0, 6, 216, 7770, 279504, 10054374, 361677960, 13010352186, 468011000736, 16835385674310, 605605873274424, 21784976052204954, 783653532006103920, 28189742176167536166, 1014047064810025198056, 36477504590984739593850

Offset: 1

Vincenzo Librandi, Jan 04 2009

If a(n)=x and a(n+1)=y then (x^2+y^2)/(xy+1)=36.

Vincenzo Librandi, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (36,-1).

Cf. A065100, A154021-A154027.

LinearRecurrence[{36,-1},{0,6},50] (* Vincenzo Librandi, Jan 30 2012 *)

concat(0,Vec(6/(1-36*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Dec 27 2011

From R. J. Mathar, Oct 18 2010: (Start)
a(n)= +36*a(n-1) -a(n-2)
a(n) = 6*A144128(n-1).
G.f.: 6*x/(1 -36*x +x^2). (End)

Edited by N. J. A. Sloane, Jun 23 2010 at the suggestion of Joerg Arndt.

Missing digit inserted in a(8) by R. J. Mathar, Oct 18 2010

A154024 a(n+2) = 49*a(n+1) - a(n), a(1)=0, a(2)=7.

Original entry on oeis.org

0, 7, 343, 16800, 822857, 40303193, 1974033600, 96687343207, 4735705783543, 231952896050400, 11360956200686057, 556454900937566393, 27254929189740067200, 1334935075396325726407

Offset: 1

Vincenzo Librandi, Jan 04 2009

If a(n)=x and a(n+1)=y, then (x^2+y^2)/(xy+1)=49.

Vincenzo Librandi, Table of n, a(n) for n = 1..600
Index entries for linear recurrences with constant coefficients, signature (49,-1).

Cf. A065100, A154021-A154027.

I:=[0, 7]; [n le 2 select I[n] else 49*Self(n-1)-Self(n-2): n in [1..50]]; // Vincenzo Librandi, Feb 25 2012

LinearRecurrence[{49,-1},{0,7},30] (* Vincenzo Librandi, Feb 25 2012 *)
Rest@ CoefficientList[Series[7 x^2/(1 - 49 x + x^2), {x, 0, 14}], x] (* Michael De Vlieger, Sep 01 2016 *)

concat(0,Vec(7/(1-49*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Dec 27 2011

G.f.: 7*x^2/(1 -49*x +x^2). - R. J. Mathar, Jan 05 2011

A154025 a(n+2) = 64*a(n+1) - a(n), a(1)=0, a(2)=8.

Original entry on oeis.org

0, 8, 512, 32760, 2096128, 134119432, 8581547520, 549084921848, 35132853450752, 2247953535926280, 143833893445831168, 9203121226997268472, 588855924634379351040, 37677576055373281198088, 2410776011619255617326592

Offset: 1

Vincenzo Librandi, Jan 04 2009

If a(n)=x and a(n+1)=y then (x^2+y^2)/(xy+1)=64.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (64,-1).

Cf. A065100, A154021-A154027.

I:=[0, 8]; [n le 2 select I[n] else 64*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Jan 29 2012

CoefficientList[Series[(8z)/(z^2-64z+1),{z,0,20}],z] (* Vincenzo Librandi, Jan 29 2012 *)
LinearRecurrence[{64,-1},{0,8},20] (* Harvey P. Dale, Dec 19 2023 *)

a(n)=([0,1; -1,64]^(n-1)*[0;8])[1,1] \\ Charles R Greathouse IV, Sep 01 2016

G.f.: 8*x^2/(1 -64*x +x^2). - R. J. Mathar, Jan 05 2011

cp's OEIS Frontend

A154021 a(n+2) = 16*a(n+1) - a(n), with a(1)=0, a(2)=4.

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A154027 a(n+2) = 100*a(n+1) - a(n), a(1)=0, a(2)=10.

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A115169 Integers b > 0 for which there exists a positive integer a <= b such that (a^2 + b^2)/(1 + ab) is an integer.

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A154022 a(n) = 5*A097780(n-2).

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A154026 a(n+2) = 81*a(n+1) - a(n), a(1)=0, a(2)=9.

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A154023 a(n+2) = 36*a(n+1) - a(n), a(1)=0, a(2)=6.

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A154024 a(n+2) = 49*a(n+1) - a(n), a(1)=0, a(2)=7.

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