cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A244419 Coefficient triangle of polynomials related to the Dirichlet kernel. Rising powers. Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)).

Original entry on oeis.org

1, 1, 2, -1, 2, 4, -1, -4, 4, 8, 1, -4, -12, 8, 16, 1, 6, -12, -32, 16, 32, -1, 6, 24, -32, -80, 32, 64, -1, -8, 24, 80, -80, -192, 64, 128, 1, -8, -40, 80, 240, -192, -448, 128, 256, 1, 10, -40, -160, 240, 672, -448, -1024, 256, 512, -1, 10, 60, -160, -560, 672, 1792, -1024, -2304, 512, 1024
Offset: 0

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Author

Wolfdieter Lang, Jul 29 2014

Keywords

Comments

This is the row reversed version of A180870. See also A157751 and A228565.
The Dirichlet kernel is D(n,x) = Sum_{k=-n..n} exp(i*k*x) = 1 + 2*Sum_{k=1..n} T(n,x) = S(n, 2*y) + S(n-1, 2*y) = S(2*n, sqrt(2*(1+y))) with y = cos(x), n >= 0, with the Chebyshev polynomials T (A053120) and S (A049310). This triangle T(n, k) gives in row n the coefficients of the polynomial Dir(n,y) = D(n,x=arccos(y)) = Sum_{m=0..n} T(n,m)*y^m. See A180870, especially the Peter Bala comments and formulas.
This is the Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2)) due to the o.g.f. for Dir(n,y) given by (1+z)/(1 - 2*y*z + z^2) = G(z)/(1 - y*F(z)) with G(z) = (1+z)/(1+z^2) and F(z) = 2*z/(1+z^2) (see the Peter Bala formula under A180870). For Riordan triangles and references see the W. Lang link 'Sheffer a- and z- sequences' under A006232.
The A- and Z- sequences of this Riordan triangle are (see the mentioned W. Lang link in the preceding comment also for the references): The A-sequence has o.g.f. 1+sqrt(1-x^2) and is given by A(2*k+1) = 0 and A(2*k) [2, -1/2, -1/8, -1/16, -5/128, -7/256, -21/1024, -33/2048, -429/32768, -715/65536, ...], k >= 0. (See A098597 and A046161.)
The Z-sequence has o.g.f. sqrt((1-x)/(1+x)) and is given by
[1, -1, 1/2, -1/2, 3/8, -3/8, 5/16, -5/16, 35/128, -35/128, ...]. (See A001790 and A046161.)
The column sequences are A057077, 2*(A004526 with even numbers signed), 4*A008805 (signed), 8*A058187 (signed), 16*A189976 (signed), 32*A189980 (signed) for m = 0, 1, ..., 5.
The row sums give A005408 (from the o.g.f. due to the Riordan property), and the alternating row sums give A033999.
The row polynomials Dir(n, x), n >= 0, give solutions to the diophantine equation (a + 1)*X^2 - (a - 1)*Y^2 = 2 by virtue of the identity (a + 1)*Dir(n, -a)^2 - (a - 1)*Dir(n, a)^2 = 2, which is easily proved inductively using the recurrence Dir(n, a) = (1 + a)*(-1)^(n-1)*Dir(n-1, -a) + a*Dir(n-1, a) given below by Wolfdieter Lang. - Peter Bala, May 08 2025

Examples

			The triangle T(n,m) begins:
  n\m  0   1   2    3    4    5    6     7     8    9    10 ...
  0:   1
  1:   1   2
  2:  -1   2   4
  3:  -1  -4   4    8
  4:   1  -4 -12    8   16
  5:   1   6 -12  -32   16   32
  6:  -1   6  24  -32  -80   32   64
  7:  -1  -8  24   80  -80 -192   64   128
  8:   1  -8 -40   80  240 -192 -448   128   256
  9:   1  10 -40 -160  240  672 -448 -1024   256  512
  10: -1  10  60 -160 -560  672 1792 -1024 -2304  512  1024
  ...
Example for A-sequence recurrence: T(3,1) = Sum_{j=0..2} A(j)*T(2,j) = 2*(-1) + 0*2 + (-1/2)*4 = -4. Example for Z-sequence recurrence: T(4,0) = Sum_{j=0..3} Z(j)*T(3,j) = 1*(-1) + (-1)*(-4) + (1/2)*4 + (-1/2)*8 = +1. (For the A- and Z-sequences see a comment above.)
Example for the alternate recurrence: T(4,2) = 2*T(3,1) - T(3,2) = 2*(-4) - 4 = -12. T(4,3) = 0*T(3,2) + T(3,3) = T(3,3) = 8. - _Wolfdieter Lang_, Jul 30 2014
		

Crossrefs

Dir(n, x) : A005408 (x = 1), A002878 (x = 3/2), A001834 (x = 2), A030221 (x = 5/2), A002315 (x = 3), A033890 (x = 7/2), A057080 (x = 4), A057081 (x = 9/2), A054320 (x = 5), A077416 (x = 6), A028230 (x = 7), A159678 (x = 8), A049629 (x = 9), A083043 (x = 10),
(-1)^n * Dir(n, x): A122367 (x = -3/2); A079935 (x = -2), A004253 (x = -5/2), A001653 (x = -3), A049685 (x = -7/2), A070997 (x = -4), A070998 (x = -9/2), A072256(n+1) (x = -5).

Programs

  • Mathematica
    T[n_, k_] := T[n, k] = Which[k == 0, (-1)^Quotient[n, 2], (0 <= n && n < k) || (n == -1 && k == 1), 0, True, 2 T[n-1, k-1] - T[n-2, k]];
    Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 28 2019, from Sage *)
  • Sage
    def T(n, k):
        if k == 0: return (-1)^(n//2)
        if (0 <= n and n < k) or (n == -1 and k == 1): return 0
        return 2*T(n-1, k-1) - T(n-2, k)
    for n in range(11): [T(n,k) for k in (0..n)] # Peter Luschny, Jul 29 2014

Formula

T(n, m) = [y^m] Dir(n,y) for n >= m >= 0 and 0 otherwise, with the polynomials Dir(y) defined in a comment above.
T(n, m) = 2^m*(S(n,m) + S(n-1,m)) with the entries S(n,m) of A049310 given there explicitly.
O.g.f. for polynomials Dir(y) see a comment above (Riordan triangle ((1+z)/(1+z^2), 2*z/(1+z^2))).
O.g.f. for column m: ((1 + x)/(1 + x^2))*(2*x/(1 + x^2))^m, m >= 0, (Riordan property).
Recurrence for the polynomials: Dir(n, y) = 2*y*Dir(n-1, y) - Dir(n-2, y), n >= 1, with input D(-1, y) = -1 and D(0, y) = 1.
Triangle three-term recurrence: T(n,m) = 2*T(n-1,m-1) - T(n-2,m) for n >= m >= 1 with T(n,m) = 0 if 0 <= n < m, T(0,0) = 1, T(-1,1) = 0 and T(n,0) = A057077(n) = (-1)^(floor(n/2)).
From Wolfdieter Lang, Jul 30 2014: (Start)
In analogy to A157751 one can derive a recurrence for the row polynomials Dir(n, y) = Sum_{m=0..n} T(n,m)*y^m also using a negative argument but only one recursive step: Dir(n,y) = (1+y)*(-1)^(n-1)*Dir(n-1,-y) + y*Dir(n-1,y), n >= 1, Dir(0,y) = 1 (Dir(-1,y) = -1). See also A180870 from where this formula can be derived by row reversion.
This entails another triangle recurrence T(n,m) = (1 + (-1)^(n-m))*T(n-1,m-1) - (-1)^(n-m)*T(n-1,m), for n >= m >= 1 with T(n,m) = 0 if n < m and T(n,0) = (-1)^floor(n/2). (End)
From Peter Bala, Aug 14 2022: (Start)
The row polynomials Dir(n,x), n >= 0, are related to the Chebyshev polynomials of the first kind T(n,x) by the binomial transform as follows:
(2^n)*(x - 1)^(n+1)*Dir(n,x) = (-1) * Sum_{k = 0..2*n+1} binomial(2*n+1,k)*T(k,-x).
Note that Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x). (End)
From Peter Bala, May 04 2025: (Start)
For n >= 1, the n-th row polynomial Dir(n, x) = (-1)^n * (U(n, -x) - U(n-1, -x)) = U(2*n, sqrt((1+x)/2)), where U(n, x) denotes the n-th Chebyshev polynomial of the second kind.
For n >= 1 and x < 1, Dir(n, x) = (-1)^n * sqrt(2/(1 - x )) * T(2*n+1, sqrt((1 - x)/2)), where T(n, x) denotes the n-th Chebyshev polynomial of the first kind.
Dir(n, x)^2 - 2*x*Dir(n, x)*Dir(n+1, x) + Dir(n+1, x)^2 = 2*(1 + x).
Dir(n, x) = (-1)^n * R(n, -2*(x+1)), where R(n, x) is the n-th row polynomial of the triangle A085478.
Dir(n, x) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n+k, 2*k) * (2*x + 2)^k. (End)

A103200 a(1)=1, a(2)=2, a(3)=11, a(4)=19; a(n) = a(n-4) + sqrt(60*a(n-2)^2 + 60*a(n-2) + 1) for n >= 5.

Original entry on oeis.org

1, 2, 11, 19, 90, 153, 712, 1208, 5609, 9514, 44163, 74907, 347698, 589745, 2737424, 4643056, 21551697, 36554706, 169676155, 287794595, 1335857546, 2265802057, 10517184216, 17838621864, 82801616185, 140443172858, 651895745267, 1105706761003, 5132364345954
Offset: 1

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Author

K. S. Bhanu and M. N. Deshpande, Mar 24 2005

Keywords

Comments

The original version of this question was as follows: Let a(1) = 1, a(2) = 2, a(3) = 11, a(4) = 19; for n = 1..4 let b(n) = sqrt(60 a(n)^2 + 60 a(n) + 1); for n >= 5 let a(n) = a(n-4) + b(n-2), b(n) = sqrt(60 a(n)^2 + 60 a(n) +1). Bhanu and Deshpande ask for a proof that a(n) and b(n) are always integers. The b(n) sequence is A103201.
This sequence is also the interleaving of two sequences c and d that can be extended backwards: c(0) = c(1) = 0, c(n) = sqrt(60 c(n-1)^2 + 60 c(n-1) +1) + c(n-2) giving 0,0,1,11,90,712,5609,... d(0) = 1, d(1) = 0, d(n) = sqrt(60 d(n-1)^2 + 60 d(n-1) +1) + d(n-2) giving 1,0,2,19,153,1208,9514,... and interleaved: 0,1,0,0,1,2,11,19,90,153,712,1208,5609,9514,... lim_{n->infinity} a(n)/a(n-2) = 1/(4 - sqrt(15)), (1/(4-sqrt(15)))^n approaches an integer as n -> infinity. - Gerald McGarvey, Mar 29 2005

References

  • K. S. Bhanu (bhanu_105(AT)yahoo.com) and M. N. Deshpande, An interesting sequence of quadruples and related open problems, Institute of Sciences, Nagpur, India, Preprint, 2005.

Crossrefs

Cf. A103201, A177187 (first differences).

Programs

  • Magma
    I:=[1,2,11,19,90]; [n le 5 select I[n] else Self(n-1)+8*Self(n-2)-8*Self(n-3)-Self(n-4)+Self(n-5): n in [1..30]]; // Vincenzo Librandi, Sep 28 2011
  • Maple
    a[1]:=1: a[2]:=2:a[3]:=11: a[4]:=19: for n from 5 to 31 do a[n]:=a[n-4]+sqrt(60*a[n-2]^2+60*a[n-2]+1) od:seq(a[n],n=1..31); # Emeric Deutsch, Apr 13 2005
  • Mathematica
    RecurrenceTable[{a[1]==1,a[2]==2,a[3]==11,a[4]==19,a[n]==a[n-4]+ Sqrt[60a[n-2]^2+60a[n-2]+1]},a[n],{n,40}] (* or *) LinearRecurrence[ {1,8,-8,-1,1},{1,2,11,19,90},40] (* Harvey P. Dale, Sep 27 2011 *)
    CoefficientList[Series[-x*(1 + x + x^2)/((x - 1)*(x^4 - 8*x^2 + 1)), {x, 0, 40}], x] (* T. D. Noe, Jun 04 2012 *)

Formula

Comments from Pierre CAMI and Gerald McGarvey, Apr 20 2005: (Start)
Sequence satisfies a(0)=0, a(1)=1, a(2)=2, a(3)=11; for n > 3, a(n) = 8*a(n-2) - a(n-4) + 3.
G.f.: -x*(1 + x + x^2) / ( (x - 1)*(x^4 - 8*x^2 + 1) ). Note that the 3 = the sum of the coefficients in the numerator of the g.f., 8 appears in the denominator of the g.f. and 8 = 2*3 + 2. Similar relationships hold for other series defined as nonnegative n such that m*n^2 + m*n + 1 is a square, here m=60. Cf. A001652, A001570, A049629, A105038, A105040, A104240, A077288, A105036, A105037. (End)
a(2n) = (A105426(n)-1)/2, a(2n+1) = (A001090(n+2) - 5*A001090(n+1) - 1)/2. - Ralf Stephan, May 18 2007
a(1)=1, a(2)=2, a(3)=11, a(4)=19, a(5)=90, a(n) = a(n-1) + 8*a(n-2) - 8*a(n-3) - a(n-4) + a(n-5). - Harvey P. Dale, Sep 27 2011

Extensions

More terms from Pierre CAMI and Emeric Deutsch, Apr 13 2005

A105038 Nonnegative n such that 6*n^2 + 6*n + 1 is a square.

Original entry on oeis.org

0, 4, 44, 440, 4360, 43164, 427284, 4229680, 41869520, 414465524, 4102785724, 40613391720, 402031131480, 3979697923084, 39394948099364, 389969783070560, 3860302882606240, 38213059042991844, 378270287547312204, 3744489816430130200, 37066627876753989800
Offset: 0

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Author

Gerald McGarvey, Apr 03 2005

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[4x/(1-11x+11x^2-x^3),{x,0,30}],x] (* or *) LinearRecurrence[{11,-11,1},{0,4,44},30] (* Harvey P. Dale, Sep 29 2013 *)
  • PARI
    for(n=0,427284,if(issquare(6*n*(n+1)+1),print1(n,",")))
    
  • PARI
    Vec(4*x/(1-11*x+11*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2012

Formula

G.f.: 4*x/(1-11*x+11*x^2-x^3).
a(0)=0, a(1)=4, a(2)=44, a(n)=11*a(n-1)-11*a(n-2)+a(n-3). - Harvey P. Dale, Sep 29 2013
a(n) = (-6-(5-2*sqrt(6))^n*(-3+sqrt(6))+(3+sqrt(6))*(5+2*sqrt(6))^n)/12. - Colin Barker, Mar 05 2016
a(n) = (A072256(n+1) - 1)/2.

Extensions

More terms from Vladeta Jovovic, Apr 05 2005
Incorrect conjectures deleted by N. J. A. Sloane, Nov 24 2010

A098532 Sum of sixth powers of first n Fibonacci numbers.

Original entry on oeis.org

0, 1, 2, 66, 795, 16420, 278564, 5105373, 90871494, 1635675910, 29316316535, 526297607496, 9442398055752, 169448124595321, 3040546683808010, 54560921044808010, 979052407236876819, 17568407254504944748
Offset: 0

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Author

Benoit Cloitre, Sep 12 2004

Keywords

Crossrefs

Programs

  • Magma
    [(Fibonacci(n)^5*Fibonacci(n+3) + Fibonacci(2*n))/4: n in [0..30]]; // G. C. Greubel, Jan 17 2018
  • Mathematica
    Table[(Fibonacci[n]^5*Fibonacci[n+3] + Fibonacci[2*n])/4, {n,0,30}] (* G. C. Greubel, Jan 17 2018 *)
  • PARI
    a(n)=sum(i=0,n,fibonacci(i)^6);
    
  • PARI
    for(n=0,30, print1((fibonacci(n)^5*fibonacci(n+3) + fibonacci(2*n))/4, ", ")) \\ G. C. Greubel, Jan 17 2018
    

Formula

a(n) = (1/500)*(F(6*n+1) +3*F(6*n+2) -(-1)^n*(16*F(4*n+1)+8*F(4*n+2))-60*F(2*n+1) +120*F(2*n+2) -(-1)^n*40 ) where F(n)=A000045(n).
G.f.: x*(1-11*x-64*x^2-11*x^3+x^4)/((x+1)*(1-18*x+x^2)*(1-3*x+x^2)*(1+7*x+x^2)). - R. J. Mathar, Feb 26 2012
a(n) = -6*(-1)^n*A049685(n)/125 +3*A002878(n)/25 +A049629(n)/125 -2*(-1)^n/25. - R. J. Mathar, Feb 26 2012
a(n)= (F(n)^5 * F(n+3) + F(2*n))/4. - Gary Detlefs, Jan 05 2013

A131557 Triangular numbers that are the sums of five consecutive triangular numbers.

Original entry on oeis.org

55, 2485, 17020, 799480, 5479705, 257429395, 1764447310, 82891465030, 568146553435, 26690794309585, 182941425758080, 8594352876220660, 58906570947547645, 2767354935348742255, 18967732903684582930, 891079694829418784770, 6107551088415488155135
Offset: 1

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Author

Richard Choulet, Oct 06 2007

Keywords

Examples

			a(1) = 55 = 3+6+10+15+21.
		

Crossrefs

Cf. A129803.

Programs

  • Maple
    a:= n-> `if`(n<2, [0, 55][n+1], (<<0|1|0>, <0|0|1>, <1|-323|323>>^iquo(n-2, 2, 'r'). `if`(r=0, <<2485, 799480, 257429395>>, <<17020, 5479705, 1764447310>>))[1, 1]): seq(a(n), n=1..20); # Alois P. Heinz, Sep 25 2008, revised Dec 15 2011
  • Mathematica
    LinearRecurrence[{1, 322, -322, -1, 1}, {55, 2485, 17020, 799480, 5479705}, 20] (* Jean-François Alcover, Oct 05 2019 *)

Formula

The subsequences with odd indices and even indices satisfy the same recurrence relations: a(n+2) = 322*a(n+1) - a(n) - 680 and a(n+1) = 161*a(n) - 340 + 9*sqrt(320*a(n)^2 - 1360*a(n) - 175).
G.f.: -5*x*(11+486*x-635*x^2+2*x^4) / ( (x-1)*(x^2+18*x+1)*(x^2-18*x+1) ).
8*a(n) = 17 + 45*A007805(n) + 18*(-1)^n*A049629(n). - R. J. Mathar, Apr 28 2020

Extensions

More terms from Alois P. Heinz, Sep 25 2008
a(6) and a(8) corrected by Harvey P. Dale, Oct 02 2011
a(10), a(12), a(14) corrected at the suggestion of Harvey P. Dale by D. S. McNeil, Oct 02 2011

A221762 Numbers m such that 11*m^2 + 5 is a square.

Original entry on oeis.org

1, 2, 22, 41, 439, 818, 8758, 16319, 174721, 325562, 3485662, 6494921, 69538519, 129572858, 1387284718, 2584962239, 27676155841, 51569671922, 552135832102, 1028808476201, 11015040486199, 20524599852098, 219748673891878, 409463188565759
Offset: 1

Views

Author

Bruno Berselli, Jan 24 2013

Keywords

Comments

Corresponding squares are: 16, 49, 5329, 18496, 2119936, 7360369, 843728209, 2929407376, ... (subsequence of A016778).
The Diophantine equation 11*x^2+k = y^2, for |k|<11, has integer solutions with the following k values:
k = -10, the nonnegative x values are in A198947;
k = -8, " 2*A075839;
k = -7, " A221763;
k = -2, " A075839;
k = 1, " A001084;
k = 4, " A075844;
k = 5, " this sequence;
k = 9, " 3*A001084.
Also, the Diophantine equation h*x^2+5 = y^2 has infinitely many integer solutions for h = 5, 11, 19, 20, 29, 31, 41, 44, 55, 59, ...
a(n+1)/a(n) tends alternately to (1+sqrt(11))^2/10 and (4+sqrt(11))^2/5.
a(n+2)/a(n) tends to A176395^2/2.

Crossrefs

Cf. A049629 (numbers m such that 20*m^2 + 5 is a square), A075796 (numbers m such that 5*m^2 + 5 is a square).

Programs

  • Magma
    m:=24; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+2*x^2+x^3)/(1-20*x^2+x^4)));
    
  • Magma
    I:=[1,2,22,41]; [n le 4 select I[n] else 20*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Aug 18 2013
  • Maple
    A221762:=proc(q)
    local n;
    for n from 1 to q do if type(sqrt(11*n^2+5), integer) then print(n);
    fi; od; end:
    A221762(1000); # Paolo P. Lava, Feb 19 2013
  • Mathematica
    LinearRecurrence[{0, 20, 0, -1}, {1, 2, 22, 41}, 24]
    CoefficientList[Series[(1 + 2 x + 2 x^2 + x^3)/(1 - 20 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 18 2013 *)
  • Maxima
    makelist(expand(((-11*(-1)^n+4*sqrt(11))*(10+3*sqrt(11))^floor(n/2)-(11*(-1)^n+4*sqrt(11))*(10-3*sqrt(11))^floor(n/2))/22), n, 1, 24);
    

Formula

G.f.: x*(1+2*x+2*x^2+x^3)/(1-20*x^2+x^4).
a(n) = -a(1-n) = ((-11*(-1)^n+4*t)*(10+3*t)^floor(n/2)-(11*(-1)^n+4*t)*(10-3*t)^floor(n/2))/22, where t=sqrt(11).
a(n) = 20*a(n-2) - a(n-4) for n>4, a(1)=1, a(2)=2, a(3)=22, a(4)=41.
a(n)*a(n-3)-a(n-1)*a(n-2) = -(3/2)*(9-7*(-1)^n).
a(n+1) + a(n-1) = A198949(n), with a(0)=-1.
2*a(n-1) - a(n) = A001084(n/2-1) for even n.

A153111 Solutions of the Pell-like equation 1 + 6*A*A = 7*B*B, with A, B integers.

Original entry on oeis.org

1, 25, 649, 16849, 437425, 11356201, 294823801, 7654062625, 198710804449, 5158826853049, 133930787374825, 3477041644892401, 90269151979827601, 2343520909830625225, 60841274503616428249, 1579529616184196509249, 41006928746285492812225
Offset: 1

Views

Author

Ctibor O. Zizka, Dec 18 2008

Keywords

Comments

B is of the form B(i) = 26*B(i-1) - B(i-2) for B(0) = 1, B(1) = 25 (this sequence).
A is of the form A(i) = 26*A(i-1) - A(i-2) for A(0) = 1, A(1) = 27.
In general a Pell-like equation of the form 1 + X*A*A = (X + 1)*B*B has the solution A(i) = (4*X + 2)*A(i-1) - A(i-2), for A(0) = 1 and A(1) = (4*X + 3), and B(i) = (4*X + 2)*B(i-1) - B(i-2) for B(0) = 1 and B(1) = (4*X + 1).
Examples in the OEIS:
X = 1 gives A002315 for A(i) and A001653 for B(i);
X = 2 gives A054320 for A(i) and A072256 for B(i);
X = 3 gives A028230 for A(i) and A001570 for B(i);
X = 4 gives A049629 for A(i) and A007805 for B(i);
X = 5 gives A133283 for A(i) and A157014 for B(i);
X = 6 gives A157461 for A(i) and this sequence for B(i).
Positive values of x (or y) satisfying x^2 - 26*x*y + y^2 + 24 = 0. - Colin Barker, Feb 20 2014

Crossrefs

Cf. similar sequences listed in A238379.

Programs

  • Magma
    I:=[1,25]; [n le 2 select I[n] else 26*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 22 2014
  • Mathematica
    CoefficientList[Series[(1 - x)/(x^2 - 26 x + 1), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 22 2014 *)
    LinearRecurrence[{26, -1}, {1, 25}, 20] (* Jean-François Alcover, Jan 07 2019 *)
  • PARI
    Vec(-x*(x-1)/(x^2-26*x+1) + O(x^100)) \\ Colin Barker, Feb 20 2014
    

Formula

a(n) = 26*a(n-1) - a(n-2). - Colin Barker, Feb 20 2014
G.f.: -x*(x - 1) / (x^2 - 26*x + 1). - Colin Barker, Feb 20 2014
a(n) = (1/14)*(7 - sqrt(42))*(1 + (13 + 2*sqrt(42))^(2*n - 1))/(13 + 2*sqrt(42))^(n - 1). - Bruno Berselli, Feb 25 2014
E.g.f.: (1/7)*(7*cosh(2*sqrt(42)*x) - sqrt(42)*sinh(2*sqrt(42)*x))*exp(13*x) - 1. - Franck Maminirina Ramaharo, Jan 07 2019

Extensions

More terms from Philippe Deléham, Sep 19 2009; corrected by N. J. A. Sloane, Sep 20 2009
Additional term from Colin Barker, Feb 20 2014

A157459 Expansion of 72*x^2 / (1 - 323*x + 323*x^2 - x^3).

Original entry on oeis.org

0, 72, 23256, 7488432, 2411251920, 776415629880, 250003421569512, 80500325329753056, 25920854752758914592, 8346434730063040745640, 2687526062225546361181560, 865375045601895865259716752, 278648077157748243067267612656, 89723815469749332371794911558552
Offset: 1

Views

Author

Paul Weisenhorn, Mar 01 2009

Keywords

Comments

This sequence is part of a solution of a more general problem involving two equations, three sequences a(n), b(n), c(n) and a constant A:
A * c(n) + 1 = a(n)^2,
(A+1) * c(n) + 1 = b(n)^2; for details see comment in A157014.
A157459 is the c(n) sequence for A=4.

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{323,-323,1},{0,72,23256},20] (* Harvey P. Dale, Feb 28 2021 *)
  • PARI
    concat(0, Vec(72*x^2/(1-323*x+323*x^2-x^3)+O(x^20))) \\ Charles R Greathouse IV, Sep 26 2012
    
  • PARI
    a(n) = -round((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)*(9+4*sqrt(5)+(-9+4*sqrt(5))*(161+72*sqrt(5))^n))/80 \\ Colin Barker, Jul 25 2016

Formula

4*a(n) + 1 = A007805(n-1)^2.
5*a(n) + 1 = A049629(n-1)^2.
G.f.: 72*x^2/(1 - 323*x + 323*x^2 - x^3).
c(1) = 0, c(2) = 72, c(3) = 323*c(2), c(n) = 323*(c(n-1) - c(n-2)) + c(n-3) for n>3.
a(n) = -((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)*(9+4*sqrt(5)+(-9+4*sqrt(5))*(161+72*sqrt(5))^n))/80. - Colin Barker, Jul 25 2016
a(n) = 72*A298271(n-1). - Greg Dresden, Dec 02 2021
a(n) = 2*A201003(n-1). - Amiram Eldar, Dec 01 2024

Extensions

Edited by Alois P. Heinz, Sep 09 2011

A195615 Numerators b(n) of Pythagorean approximations b(n)/a(n) to 2.

Original entry on oeis.org

15, 273, 4895, 87841, 1576239, 28284465, 507544127, 9107509825, 163427632719, 2932589879121, 52623190191455, 944284833567073, 16944503814015855, 304056783818718321, 5456077604922913919, 97905340104793732225, 1756840044281364266127, 31525215456959763058065
Offset: 1

Views

Author

Clark Kimberling, Sep 22 2011

Keywords

Comments

See A195500 for discussion and list of related sequences; see A195614 for Mathematica program.

Crossrefs

Programs

  • Magma
    [Fibonacci(3*n+1)*Fibonacci(3*n+2): n in [1..40]]; // G. C. Greubel, Feb 13 2023
    
  • Mathematica
    LinearRecurrence[{17,17,-1}, {15,273,4895}, 40] (* G. C. Greubel, Feb 13 2023 *)
  • PARI
    Vec(x*(15+18*x-x^2)/((1+x)*(1-18*x+x^2)) + O(x^50)) \\ Colin Barker, Jun 04 2015
    
  • SageMath
    [fibonacci(3*n+1)*fibonacci(3*n+2) for n in range(1,41)] # G. C. Greubel, Feb 13 2023

Formula

From Colin Barker, Jun 04 2015: (Start)
a(n) = 17*a(n-1) + 17*a(n-2) - a(n-3).
G.f.: x*(15 + 18*x - x^2)/((1+x)*(1-18*x+x^2)). (End)
a(n) = ((-1)^n + (2+sqrt(5))*(9+4*sqrt(5))^n + (2-sqrt(5))*(9+4*sqrt(5))^(-n))/5. - Colin Barker, Mar 04 2016
From G. C. Greubel, Feb 13 2023: (Start)
a(n) = Fibonacci(3*n+1)*Fibonacci(3*n+2).
a(n) = (1/5)*(4*A049629(n) + (-1)^n - 5*[n=0]). (End)

A267797 Lucas numbers of the form (x^3 + y^3) / 2 where x and y are distinct positive integers.

Original entry on oeis.org

76, 1364, 24476, 439204, 7881196, 141422324, 2537720636, 45537549124, 817138163596, 14662949395604, 263115950957276, 4721424167835364, 84722519070079276, 1520283919093591604, 27280388024614569596, 489526700523968661124, 8784200221406821330636
Offset: 1

Views

Author

Altug Alkan, Jan 24 2016

Keywords

Comments

Lucas numbers that are the averages of 2 distinct positive cubes.
Inspired by relation between sequence A024851 and A188378.
Corresponding indices are 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, ...
6*n + 3 is the corresponding form of indices.
Corresponding y values are listed by A188378, for n > 0. Note that corresponding x values are A188378(n) - 2, for n > 0.

Examples

			Lucas number 76 is a term because 76 = (3^3 + 5^3) / 2.
Lucas number 1364 is a term because 1364 = (10^3 + 12^3) / 2.
Lucas number 24476 is a term because 24476 = (28^3 + 30^3) / 2.
Lucas number 439204 is a term because 439204 = (75^3 + 77^3) / 2.
Lucas number 7881196 is a term because 7881196 = (198^3 + 200^3) / 2.
Lucas number 141422324 is a term because 141422324 = (520^3 + 522^3) / 2.
		

Crossrefs

Programs

  • Magma
    [Fibonacci(6*n+4)+Fibonacci(6*n+2): n in [1..20]]; // Vincenzo Librandi, Jan 24 2016
  • Mathematica
    Table[Fibonacci[6 n + 4] + Fibonacci[6 n + 2], {n, 1, 20}] (* Vincenzo Librandi, Jan 24 2016 *)
    LinearRecurrence[{18,-1},{76,1364},20] (* Harvey P. Dale, Jul 23 2024 *)
  • PARI
    l(n) = fibonacci(n+1) + fibonacci(n-1);
    is(n) = for(i=ceil(sqrtn(n\2+1, 3)), sqrtn(n-.5, 3), ispower(n-i^3, 3) && return(1));
    for(n=1, 120, if(is(2*l(n)), print1(l(n), ", ")));
    
  • PARI
    a(n) = ((5*fibonacci(n)*fibonacci(n+1) + 1 + (-1)^n)^3 + (5*fibonacci(n)*fibonacci(n+1) - 1 + (-1)^n)^3) / 2;
    
  • PARI
    a(n) = (fibonacci(6*n+4) + fibonacci(6*n+2));
    
  • PARI
    Vec(4*x*(19-x)/(1-18*x+x^2) + O(x^20)) \\ Colin Barker, Jan 24 2016
    

Formula

a(n) = A000032(A016945(n)), for n > 0.
a(n) = A188378(n)^3 - 3*A188378(n)^2 + 6*A188378(n) - 4, for n > 0.
From Colin Barker, Jan 24 2016: (Start)
a(n) = (9+4*sqrt(5))^(-n)*(2-sqrt(5)+(2+sqrt(5))*(9+4*sqrt(5))^(2*n)).
a(n) = 18*a(n-1)-a(n-2) for n>2.
G.f.: 4*x*(19-x) / (1-18*x+x^2).
(End)
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