A077417 -id:A077417 - cp's OEIS frontend

A080875 a(n)a(n+3) - a(n+1)a(n+2) = 5, given a(0)=a(1)=1, a(2)=6.

1, 1, 6, 11, 71, 131, 846, 1561, 10081, 18601, 120126, 221651, 1431431, 2641211, 17057046, 31472881, 203253121, 375033361, 2421980406, 4468927451, 28860511751, 53252096051, 343904160606, 634556225161, 4097989415521

Offset: 0

Paul D. Hanna, Feb 22 2003

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..1500
Index entries for linear recurrences with constant coefficients, signature (0, 12, 0, -1).

Cf. A080871, A080872, A080873, A080874.

Bisections are A023038 and A077417.

LinearRecurrence[{0,12,0,-1},{1,1,6,11},30] (* Harvey P. Dale, Jul 14 2024 *)

G.f.: (-x^3 - 6*x^2 + x + 1)/(x^4 - 12*x^2 + 1).
a(n+4) = 12*a(n+2)-a(n). [Richard Choulet, Dec 04 2008]
a(n) = (1/4 + ((sqrt(6 + sqrt(35)) - sqrt(6 - sqrt(35)))/(4*sqrt(35))))*(sqrt(6 + sqrt(35)))^n + (1/4 + ((sqrt(6 + sqrt(35)) - sqrt(6 - sqrt(35)))/(4*sqrt(35))))*(sqrt(6 - sqrt(35)))^n + (1/4 - ((sqrt(6 + sqrt(35)) - sqrt(6 - sqrt(35)))/(4*sqrt(35))))*( - sqrt(6 + sqrt(35)))^n + (1/4 - ((sqrt(6 + sqrt(35)) - sqrt(6 - sqrt(35)))/(4*sqrt(35))))*( - (sqrt(6 - sqrt(35))))^n. [Richard Choulet, Dec 06 2008]

A123971 Triangle T(n,k), read by rows, defined by T(n,k)=3*T(n-1,k)-T(n-1,k-1)-T(n-2,k), T(0,0)=1, T(1,0)=2, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n.

Original entry on oeis.org

1, 2, -1, 5, -5, 1, 13, -19, 8, -1, 34, -65, 42, -11, 1, 89, -210, 183, -74, 14, -1, 233, -654, 717, -394, 115, -17, 1, 610, -1985, 2622, -1825, 725, -165, 20, -1, 1597, -5911, 9134, -7703, 3885, -1203, 224, -23, 1, 4181, -17345, 30691, -30418, 18633, -7329

Offset: 0

Gary W. Adamson and Roger L. Bagula, Oct 30 2006

This entry is the result of merging two sequences, this one and a later submission by Philippe Deléham, Nov 29 2013 (with edits from Ralf Stephan, Dec 12 2013). Most of the present version is the work of Philippe Deléham, the only things remaining from the original entry are the sequence data and the Mathematica program. - N. J. A. Sloane, May 31 2014
Subtriangle of the triangle given by (0, 2, 1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, -2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.
Apart from signs, equals A126124.
Row sums = 1.
Sum_{k=0..n} T(n,k)*(-x)^k = A001519(n+1), A079935(n+1), A004253(n+1), A001653(n+1), A049685(n), A070997(n), A070998(n), A072256(n+1), A078922(n+1), A077417(n), A085260(n+1), A001570(n+1) for x=0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 respectively.

			Triangle begins:
  1
  2, -1
  5, -5, 1
  13, -19, 8, -1
  34, -65, 42, -11, 1
  89, -210, 183, -74, 14, -1
  233, -654, 717, -394, 115, -17, 1
Triangle (0, 2, 1/2, 1/2, 0, 0, ...) DELTA (1, -2, 0, 0, ...) begins:
  1
  0, 1
  0, 2, -1
  0, 5, -5, 1
  0, 13, -19, 8, -1
  0, 34, -65, 42, -11, 1
  0, 89, -210, 183, -74, 14, -1
  0, 233, -654, 717, -394, 115, -17, 1

Cf. A094954, A098495, A123971, A126124, A152063, A001519, A079935, A004253, A001653, A049685, A070997, A070998, A072256, A078922, A077417, A085260, A001570, A001870, A126124.

Mathematica ( general k th center) Clear[M, T, d, a, x, k] k = 3 T[n_, m_, d_] := If[ n == m && n < d && m < d, k, If[n == m - 1 || n == m + 1, -1, If[n == m == d, k - 1, 0]]] M[d_] := Table[T[n, m, d], {n, 1, d}, {m, 1, d}] Table[M[d], {d, 1, 10}] Table[Det[M[d]], {d, 1, 10}] Table[Det[M[d] - x*IdentityMatrix[d]], {d, 1, 10}] a = Join[{M[1]}, Table[CoefficientList[ Det[M[d] - x*IdentityMatrix[d]], x], {d, 1, 10}]] Flatten[a] MatrixForm[a] Table[NSolve[Det[M[d] - x*IdentityMatrix[d]] == 0, x], {d, 1, 10}] Table[x /. NSolve[Det[M[d] - x*IdentityMatrix[d]] == 0, x][[d]], {d, 1, 10}]

T(n,k)=polcoeff(polcoeff(Ser((1-x)/(1+(y-3)*x+x^2)),n,x),n-k,y) \\ Ralf Stephan, Dec 12 2013

@CachedFunction
def A123971(n,k): # With T(0,0) = 1!
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = 2*A123971(n-1,k) if n==1 else 3*A123971(n-1,k)
    return A123971(n-1,k-1) - A123971(n-2,k) - h
for n in (0..9): [A123971(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

T(n,k) = (-1)^n*A126124(n+1,k+1).
T(n,k) = (-1)^k*Sum_{m=k..n} binomial(m,k)*binomial(m+n,2*m). - Wadim Zudilin, Jan 11 2012
G.f.: (1-x)/(1+(y-3)*x+x^2).
T(n,0) = A001519(n+1) = A000045(2*n+1).
T(n+1,1) = -A001870(n).

Edited by N. J. A. Sloane, May 31 2014

A159681 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 5n(j)+1=a(j)a(j) and 7n(j)+1=b(j)b(j) with positive integer numbers.

Original entry on oeis.org

0, 24, 3432, 487344, 69199440, 9825833160, 1395199109304, 198108447688032, 28130004372591264, 3994262512460271480, 567157146764985958920, 80532320578115545895184, 11435022364945642531157232, 1623692643501703123878431784, 230552920354876897948206156120

Offset: 1

Paul Weisenhorn, Apr 19 2009

Colin Barker, Table of n, a(n) for n = 1..450
Index entries for linear recurrences with constant coefficients, signature (143,-143,1).

Cf. A077417, A077416, A157456.

R:=PowerSeriesRing(Integers(), 30); [0] cat Coefficients( R!(24*x^2/((1-x)*(1-142*x+x^2)))); // G. C. Greubel, Jun 03 2018

for a from 1 by 2 to 100000 do b:=sqrt((7*a*a-2)/5): if (trunc(b)=b) then
n:=(a*a-1)/5: La:=[op(La),a]:Lb:=[op(Lb),b]:Ln:=[op(Ln),n]: end if: end do:
# Second program
seq((6/35)*(simplify(ChebyshevU(n,71) -141*ChebyshevU(n-1,71)) -1), n=1..30); # G. C. Greubel, Sep 27 2022

LinearRecurrence[{143,-143,1}, {0, 24, 3432}, 30] (* or *) CoefficientList[Series[24*x^2/((1-x)*(1-142*x+x^2)), {x,0,30}], x] (* G. C. Greubel, Jun 03 2018 *)

concat(0, Vec(-24*x^2/((x-1)*(x^2-142*x+1)) + O(x^20))) \\ Colin Barker, Jul 26 2016

a(n) = round((-12+(6+sqrt(35))*(71+12*sqrt(35))^(-n)-(-6+sqrt(35))*(71+12*sqrt(35))^n)/70) \\ Colin Barker, Jul 26 2016

[(6/35)*(-1 + chebyshev_U(n, 71) - 141*chebyshev_U(n-1, 71)) for n in range(1,30)] # G. C. Greubel, Sep 27 2022

The a(j) recurrence is a(1)=1, a(2)=11, a(t+2) = 12*a(t+1) - a(t) resulting in terms 1, 11, 131, 1561, ... (A077417).
The b(j) recurrence is b(1)=1, b(2)=13, b(t+2) = 12*b(t+1) - b(t) resulting in terms 1, 13, 155, 1847, ... (A077416).
The n(j) recurrence is n(0)=n(1)=0, n(2)=24, n(t+3) = 143*(n(t+2) - n(t+1)) + n(t) resulting in terms 0, 0, 24, 3432, 487344, ... (this sequence).
G.f.: 24*x^2/((1-x)*(1-142*x+x^2)). - R. J. Mathar, Apr 20 2009
a(n) = (-12+(6+sqrt(35))*(71+12*sqrt(35))^(-n)-(-6+sqrt(35))*(71+12*sqrt(35))^n)/70. - Colin Barker, Jul 26 2016
a(n) = (6/35)*(ChebyshevU(n, 71) - 141*ChebyshevU(n-1, 71) - 1). - G. C. Greubel, Sep 27 2022

More terms from R. J. Mathar, Apr 20 2009

A174227 Expansion of -(10x + sqrt((1-10x)(1-14x)))/(2*x).

Original entry on oeis.org

1, 1, 12, 145, 1764, 21602, 266232, 3301349, 41178660, 516512462, 6513158376, 82542517386, 1051024082472, 13442267711940, 172638285341040, 2225824753934445, 28802104070304420, 373966734921011990

Offset: 0

Paul Barry, Mar 12 2010

Hankel transform is A077417.

The g.f. A(x) satisfies the continued fraction relation A(x) = 1/(1-x/(1-10*x-x*A(x))).

with(LREtools): with(FormalPowerSeries): # requires Maple 2022
ogf:= -(10*x + sqrt((1-10*x)*(1-14*x)))/(2*x): req:= FindRE(ogf,x,u(n));
init:= [1, 1, 12, 145]: iseq:= seq(u(i-1)=init[i],i=1..nops(init)):
rmin:= subs(n=n-2, MinimalRecurrence(req,u(n),{iseq})[1]); # Mathar's recurrence
a:= gfun:-rectoproc({rmin, iseq}, u(n), remember):
seq(a(n),n=0..17); # Georg Fischer, Nov 03 2022
# Alternative, using function FindSeq from A174403:
ogf := -(10*x + sqrt((1-10*x)*(1-14*x)))/(2*x):
a := FindSeq(ogf): seq(a(n), n=0..17); # Peter Luschny, Nov 04 2022

a(n) = sqrt(5/7) * 10^n * (6*hypergeom([1/2, n+1],[1],2/7)-7*hypergeom([1/2, n],[1],2/7)) / (n+1) for n > 0. - Mark van Hoeij, Jul 02 2010

D-finite with recurrence: (n+1)*a(n) +12*(1-2*n)*a(n-1) +140*(n-2)*a(n-2)=0. - R. J. Mathar, Sep 30 2012

Definiton corrected by Peter Luschny, Nov 05 2022

A385121 a(n+1) = 12*a(n) - a(n-1), a(0) = a(1) = 2, a(n) = a(1-n).

Original entry on oeis.org

2, 2, 22, 262, 3122, 37202, 443302, 5282422, 62945762, 750066722, 8937854902, 106504192102, 1269112450322, 15122845211762, 180205030090822, 2147337515878102, 25587845160446402, 304906804409478722, 3633293807753298262, 43294618888630100422

Offset: 0

Michael Somos, Jun 18 2025

If x = 2, y = 6, z = a(n), w = a(n+1), then x^2+y^2+z^2+w^2 = x*y*z*w.

			G.f. = 2 + 2*x + 22*x^2 + 262*x^3 + 3122*x^4 + 37202*x^5 + ...

Index entries for linear recurrences with constant coefficients, signature (12,-1).

Cf. A061292, A077417.

a[ n_] := Which[n<1, a[1-n], n==1, 2, True, 12*a[n-1] - a[n-2]];

{a(n) = if(n<1, a(1-n), n==1, 2, 12*a(n-1) - a(n-2))};

G.f.: (2 - 22*x)/(1 - 12*x + x^2).
0 = 40 + a(n)^2 - 12*a(n)*a(n+1) + a(n+1)^2 for all n in Z.
a(n) = 2 * A077417(n-1).
E.g.f.: 2*exp(6*x)*(7*cosh(sqrt(35)*x) - sqrt(35)*sinh(sqrt(35)*x))/7. - Stefano Spezia, Aug 29 2025

cp's OEIS Frontend

A080875 a(n)*a(n+3) - a(n+1)*a(n+2) = 5, given a(0)=a(1)=1, a(2)=6.

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A123971 Triangle T(n,k), read by rows, defined by T(n,k)=3*T(n-1,k)-T(n-1,k-1)-T(n-2,k), T(0,0)=1, T(1,0)=2, T(1,1)=-1, T(n,k)=0 if k<0 or if k>n.

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A159681 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 5*n(j)+1=a(j)*a(j) and 7*n(j)+1=b(j)*b(j) with positive integer numbers.

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A174227 Expansion of -(10*x + sqrt((1-10*x)*(1-14*x)))/(2*x).

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A385121 a(n+1) = 12*a(n) - a(n-1), a(0) = a(1) = 2, a(n) = a(1-n).

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A080875 a(n)a(n+3) - a(n+1)a(n+2) = 5, given a(0)=a(1)=1, a(2)=6.

A159681 The general form of the recurrences are the a(j), b(j) and n(j) solutions of the 2 equations problem: 5n(j)+1=a(j)a(j) and 7n(j)+1=b(j)b(j) with positive integer numbers.

A174227 Expansion of -(10x + sqrt((1-10x)(1-14x)))/(2*x).