A001570 -id:A001570 - cp's OEIS frontend

A157014 Expansion of x(1-x)/(1 - 22x + x^2).

1, 21, 461, 10121, 222201, 4878301, 107100421, 2351330961, 51622180721, 1133336644901, 24881784007101, 546265911511321, 11992968269241961, 263299036011811821, 5780585823990618101, 126909589091781786401, 2786230374195208682721, 61170158643202809233461

Offset: 1

Paul Weisenhorn, Feb 21 2009

This sequence is part of a solution of a general problem involving 2 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, where solutions are given by the recurrences:
a(1) = 1, a(2) = 4*A+1, a(n) = (4*A+2)*a(n-1)-a(n-2) for n>2, resulting in a(n) terms 1, 4*A+1, 16*A^2+12*A+1, 64*A^3+80*A^2+24*A+1, ...;
b(1) = 1, b(2) = 4*A+3, b(n) = (4*A+2)*b(n-1)-b(n-2) for n>2, resulting in b(n) terms 1, 4*A+3, 16*A^2+20*A+5, 64*A^3+112*A^2+56*A+7, ...;
c(1) = 0, c(2) = 16*A+8, c(3) = (16*A^2+16*A+3)*c(2), c(n) = (16*A^2+16*A+3) * (c(n-1)-c(n-2)) + c(n-3) for n>3, resulting in c(n) terms 0, 16*A+8, 256*A^3+384*A^2+176*A+24, 4096*A^5 + 10240*A^4 + 9472*A^3 + 3968*A^2 + 736*A + 48, ... .
A157014 is the a(n) sequence for A=5.
For other A values the a(n), b(n) and c(n) sequences are in the OEIS:
  A   a-sequence    b-sequence     c-sequence
  1     A001653       A002315(n-1)   A078522
  2     A072256       A054320(n-1)   A045502(n-1)
  3     A001570       A028230        A059989(n-1)
  4     A007805       A049629(n-1)   A157459
  5  -> A157014 <-    A133283        A157460
  6     A153111       A157461        A157874
  7     A157877       A157878        A157879
  8     A077420(n-1)  A046176        A157880
  9     A097315(n-1)  A097314(n-1)   A157881
Positive values of x (or y) satisfying x^2 - 22xy + y^2 + 20 = 0. - Colin Barker, Feb 19 2014
From Klaus Purath, Apr 22 2025: (Start)
Nonnegative solutions to the Diophantine equation 5*b(n)^2 - 6*a(n)^2 = -1. The corresponding b(n) are A133283(n). Note that (b(n+1)^2 - b(n)*b(n+2))/4 = 6 and (a(n)*a(n+2) - a(n+1)^2)/4 = 5.
(a(n) + b(n))/2 = (b(n+1) - a(n+1))/2 = A077421(n-1) = Lucas U(22,1). Also b(n)*a(n+1) - b(n+1)*a(n) = -2.
a(n)=(t(i+2*n-1) + t(i))/(t(i+n) + t(i+n-1)) as long as t(i+n) + t(i+n-1) != 0 for any integer i and n >= 1 where (t) is a sequence satisfying t(i+3) = 21*t(i+2) - 21*t(i+1) + t(i) or t(i+2) = 22*t(i+1) - t(i) without regard to initial values and including this sequence itself. (End)

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (22,-1).

Cf. similar sequences listed in A238379.

a:=[1,21];; for n in [3..20] do a[n]:=22*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jan 14 2020

I:=[1,21]; [n le 2 select I[n] else 22*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 21 2014

seq( simplify(ChebyshevU(n-1,11) - ChebyshevU(n-2,11)), n=1..20); # G. C. Greubel, Jan 14 2020

CoefficientList[Series[(1-x)/(1-22x+x^2), {x,0,20}], x] (* Vincenzo Librandi, Feb 21 2014 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
] (* Complement of A041049 *)
a[30, 20] (* Gerry Martens, Jun 07 2015 *)
Table[ChebyshevU[n-1, 11] - ChebyshevU[n-2, 11], {n,20}] (* G. C. Greubel, Jan 14 2020 *)

Vec((1-x)/(1-22*x+x^2)+O(x^20)) \\ Charles R Greathouse IV, Sep 23 2012

[chebyshev_U(n-1,11) - chebyshev_U(n-2,11) for n in (1..20)] # G. C. Greubel, Jan 14 2020

G.f.: x*(1-x)/(1-22*x+x^2).
a(1) = 1, a(2) = 21, a(n) = 22*a(n-1) - a(n-2) for n>2.
5*A157460(n)+1 = a(n)^2 for n>=1.
6*A157460(n)+1 = A133283(n)^2 for n>=1.
a(n) = (6+sqrt(30)-(-6+sqrt(30))*(11+2*sqrt(30))^(2*n))/(12*(11+2*sqrt(30))^n). - Gerry Martens, Jun 07 2015
a(n) = ChebyshevU(n-1, 11) - ChebyshevU(n-2, 11). - G. C. Greubel, Jan 14 2020

Edited by Alois P. Heinz, Sep 09 2011

A165253 Triangle T(n,k), read by rows given by [1,0,1,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 6, 5, 1, 0, 1, 10, 15, 7, 1, 0, 1, 15, 35, 28, 9, 1, 0, 1, 21, 70, 84, 45, 11, 1, 0, 1, 28, 126, 210, 165, 66, 13, 1, 0, 1, 36, 210, 462, 495, 286, 91, 15, 1, 0, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 0, 1, 55, 495, 1716, 3003, 3003, 1820, 680

Offset: 0

Philippe Deléham, Sep 10 2009

Mirror image of triangle in A121314.

			Triangle begins:
  1;
  1,    0;
  1,    1,    0;
  1,    3,    1,    0;
  1,    6,    5,    1,    0;
  1,   10,   15,    7,    1,    0;
  1,   15,   35,   28,    9,    1,    0;
  1,   21,   70,   84,   45,   11,    1,    0;
  1,   28,  126,  210,  165,   66,   13,    1,    0;
  1,   36,  210,  462,  495,  286,   91,   15,    1,    0,
  1,   45,  330,  924, 1287, 1001,  455,  120,   17,    1,    0;

Indranil Ghosh, Rows 0..125 of triangle, flattened

Cf. A054142, A085478, A121314.

m = 13;
(* DELTA is defined in A084938 *)
DELTA[Join[{1, 0, 1}, Table[0, {m}]], Join[{0, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)

T(0,0)=1, T(n,k) = binomial(n-1+k,2k) for n >= 1.
Sum {k=0..n} T(n,k)*x^k = A000012(n), A001519(n), A001835(n), A004253(n), A001653(n), A049685(n-1), A070997(n-1), A070998(n-1), A072256(n), A078922(n), A077417(n-1), A085260(n), A001570(n) for x = 0,1,2,3,4,5,6,7,8,9,10,11,12 respectively.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000007(n), A001519(n), A047849(n), A165310(n), A165311(n), A165312(n), A165314(n), A165322(n), A165323(n), A165324(n) for x= 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 26 2009
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k), T(0,0)=T(1,0)=1, T(1,1)=0. - Philippe Deléham, Feb 18 2012
G.f.: (1-x-y*x)/((1-x)^2-y*x). - Philippe Deléham, Feb 19 2012

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

Original entry on oeis.org

1, 14, 209, 3121, 46606, 695969, 10392929, 155197966, 2317576561, 34608450449, 516809180174, 7717529252161, 115246129602241, 1720974414781454, 25699370092119569, 383769576967012081, 5730844284413061646, 85578894689228912609, 1277952576054020627489

Offset: 1

Paul Weisenhorn, May 23 2009

This summarizes the case C=13 of common solutions to C*k+1=A^2, (C+4)*k+1=B^2.
The 2 equations are equivalent to the Pell equation x^2-C*(C+4)*y^2=1,
with x=(C*(C+4)*k+C+2)/2; y=A*B/2 and with smallest values x(1) = (C+2)/2, y(1)=1/2.
Generic recurrences are:
A(j+2)=(C+2)*A(j+1)-A(j) with A(1)=1; A(2)=C+1.
B(j+2)=(C+2)*B(j+1)-B(j) with B(1)=1; B(2)=C+3.
k(j+3)=(C+1)*(C+3)*( k(j+2)-k(j+1) )+k(j) with k(1)=0; k(2)=C+2; k(3)=(C+1)*(C+2)*(C+3).
x(j+2)=(C^2+4*C+2)*x(j+1)-x(j) with x(1)=(C+2)/2; x(2)=(C^2+4*C+1)*(C+2)/2;
Binet-type of solutions of these 2nd order recurrences are:
R=C^2+4*C; S=C*sqrt(R); T=(C+2); U=sqrt(R); V=(C+4)*sqrt(R);
A(j)=((R+S)*(T+U)^(j-1)+(R-S)*(T-U)^(j-1))/(R*2^j);
B(j)=((R+V)*(T+U)^(j-1)+(R-V)*(T-U)^(j-1))/(R*2^j);
x(j)+sqrt(R)*y(j)=((T+U)*(C^2*4*C+2+(C+2)*sqrt(R))^(j-1))/2^j;
k(j)=(((T+U)*(R+2+T*U)^(j-1)+(T-U)*(R+2-T*U)^(j-1))/2^j-T)/R. [Paul Weisenhorn, May 24 2009]
.C -A----- -B----- -k-----
01 A001519 A002878 A058038
02 A001653 A002315 A045899/2
03 A004253 A030221 A160695
04 A001653 A002315 A078522/4
05 A049685 A033890 A161582
06 A070997 A057080 A159683/2
07 A070998 A057081 A161585
08 A072256 A054320 A045502/4
09 A078922 A097783 A161586
10 A077417 A077416 A159681/2
11 A085260 A126816 A161588
12 A001570 A028230 A059989/4
13 A160682 A161591 A161584
14 A157456 A159678 A159679/2
15 A161595 A161599 A161583
16 A007805 A049629 A157459/4
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(13)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. [John M. Campbell, Jul 08 2011]
Positive values of x (or y) satisfying x^2 - 15xy + y^2 + 13 = 0. - Colin Barker, Feb 11 2014

Vincenzo Librandi, Table of n, a(n) for n = 1..200
J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
Index entries for the Pell equation
Index entries for linear recurrences with constant coefficients, signature (15,-1).

Cf. similar sequences listed in A238379.

I:=[1,14]; [n le 2 select I[n] else 15*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 12 2014

LinearRecurrence[{15,-1},{1,14},20] (* Harvey P. Dale, Oct 08 2012 *)
CoefficientList[Series[(1 - x)/(1 - 15 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)

a(n) = round((2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221)) \\ Colin Barker, Jul 25 2016

a(n) = 15*a(n-1)-a(n-2).
G.f.: (1-x)*x/(1-15*x+x^2).
a(n) = (2^(-1-n)*((15-sqrt(221))^n*(13+sqrt(221))+(-13+sqrt(221))*(15+sqrt(221))^n))/sqrt(221). - Colin Barker, Jul 25 2016

Edited, extended by R. J. Mathar, Sep 02 2009

First formula corrected by Harvey P. Dale, Oct 08 2012

A350916 Positive integers k such that (k+1)^4 has a divisor congruent to -1 modulo k.

Original entry on oeis.org

1, 2, 3, 5, 9, 11, 14, 17, 29, 35, 41, 43, 59, 65, 69, 125, 134, 139, 174, 194, 339, 386, 449, 461, 681, 901, 937, 1169, 1322, 1325, 1715, 1971, 2211, 3054, 6395, 7989, 8857, 9077, 10849, 11483, 12545, 13082, 20909, 21506, 23861, 35233, 54734, 62210, 66923, 89045, 129494, 143289, 172899, 174725, 203321, 332315, 375129, 390051, 426389, 493697, 561513, 982094

Offset: 1

Max Alekseyev, Jan 21 2022

nonn

For (k+1)^3 similar sequence is finite {1, 2, 3, 5, 9, 11, 14}, while for (k+1)^2 it is just {1, 2, 3, 5}. Starting with power 4 (this sequence), the number of values of k is infinite. One series of values for power 6 is given by A001570.
Formed by the union of 10 linear recurrent sequences satisfying b(n) = q*b(n-1) - b(n-2) - 4: A350919 (q=3), A350920 (q=4), A350921 (q=6), A350922 (q=7), A350923 (q=10), A103974 (q=14), A350924 (q=16), A350925 (q=16), A350926 (q=23), A350917 (q=23). Each of them give identities (b(n)+1)^4 = (b(n)*b(n-1)-1) * (b(n)*b(n+1)-1).
Only terms 1, 2, 5, 9, 11, 14, 29 are shared between two or more sequences, all others come from exactly one sequence.

Max Alekseyev, Table of n, a(n) for n = 1..5000

Union of A350917, A350919, A350920, A350921, A350922, A350923, A103974, A350924, A350925, A350926.

Cf. A001570.

{ for(k=1,10^6, fordiv((k+1)^4,d, if(Mod(d,k)==-1, print1(k,", "); break)) ); }

A133607 Triangle read by rows: T(n, k) = qStirling2(n, k, q) for q = -1, with 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, -1, 0, 1, -1, -1, 0, 1, -1, -2, 1, 0, 1, -1, -3, 2, 1, 0, 1, -1, -4, 3, 3, -1, 0, 1, -1, -5, 4, 6, -3, -1, 0, 1, -1, -6, 5, 10, -6, -4, 1, 0, 1, -1, -7, 6, 15, -10, -10, 4, 1, 0, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1, 0, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1

Offset: 0

Philippe Deléham, Dec 27 2007

Previous name: Triangle T(n,k), 0<=k<=n, read by rows given by [0, 1, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, -2, 1, 0, 0, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938.

			Triangle begins:
  1;
  0, 1;
  0, 1, -1;
  0, 1, -1, -1;
  0, 1, -1, -2, 1;
  0, 1, -1, -3, 2, 1;
  0, 1, -1, -4, 3, 3, -1;
  0, 1, -1, -5, 4, 6, -3, -1;
  0, 1, -1, -6, 5, 10, -6, -4, 1;
  0, 1, -1, -7, 6, 15, -10, -10, 4, 1;
  0, 1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
  0, 1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
  0, 1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
  ...
Triangle A103631 begins:
  1;
  0, 1;
  0, 1, 1;
  0, 1, 1, 1;
  0, 1, 1, 2, 1;
  0, 1, 1, 3, 2, 1;
  0, 1, 1, 4, 3, 3, 1;
  0, 1, 1, 5, 4, 6, 3, 1;
  0, 1, 1, 6, 5, 10, 6, 4, 1;
  0, 1, 1, 7, 6, 15, 10, 10, 4, 1;
  0, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1;
  0, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1;
  0, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1;
  ...
Triangle A108299 begins:
  1;
  1, -1;
  1, -1, -1;
  1, -1, -2, 1;
  1, -1, -3, 2, 1;
  1, -1, -4, 3, 3, -1;
  1, -1, -5, 4, 6, -3, -1;
  1, -1, -6, 5, 10, -6, -4, 1;
  1, -1, -7, 6, 15, -10, -10, 4, 1;
  1, -1, -8, 7, 21, -15, -20, 10, 5, -1;
  1, -1, -9, 8, 28, -21, -35, 20, 15, -5, -1;
  1, -1, -10, 9, 36, -28, -56, 35, 35, -15, -6, 1;
  ...

Another version is A108299.

Unsigned version is A103631 (T(n,k) = A103631(n,k)*A057077(k)).

m = 13
(* DELTA is defined in A084938 *)
DELTA[Join[{0, 1}, Table[0, {m}]], Join[{1, -2, 1}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)
qStirling2[n_, k_, q_] /; 1 <= k <= n := q^(k-1) qStirling2[n-1, k-1, q] + Sum[q^j, {j, 0, k-1}] qStirling2[n-1, k, q];
qStirling2[n_, 0, _] := KroneckerDelta[n, 0];
qStirling2[0, k_, _] := KroneckerDelta[0, k];
qStirling2[, , _] = 0;
Table[qStirling2[n, k, -1], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Mar 10 2020 *)

from sage.combinat.q_analogues import q_stirling_number2
for n in (0..9):
    print([q_stirling_number2(n,k).substitute(q=-1) for k in [0..n]])
# Peter Luschny, Mar 09 2020

Sum_{k, 0<=k<=n}T(n,k)*x^(n-k)= A057077(n), A010892(n), A000012(n), A001519(n), A001835(n), A004253(n), A001653(n), A049685(n-1), A070997(n-1), A070998(n-1), A072256(n), A078922(n), A077417(n-1), A085260(n), A001570(n-1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14 respectively .
Sum_{k, 0<=k<=n}T(n,k)*x^k = A000007(n), A010892(n), A133631(n), A133665(n), A133666(n), A133667(n), A133668(n), A133669(n), A133671(n), A133672(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively .
G.f.: (1-x+y*x)/(1-x+y^2*x^2). - Philippe Deléham, Mar 14 2012
T(n,k) = T(n-1,k) - T(n-2,k-2), T(0,0) = T(1,1) = T(2,1) = 1, T(1,0) = T(2,0) = 0, T(2,2) = -1 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 14 2012

New name from Peter Luschny, Mar 09 2020

A103200 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) 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

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

nonn

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

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.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Index entries for linear recurrences with constant coefficients, signature (1,8,-8,-1,1).

Cf. A103201, A177187 (first differences).

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

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

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 *)

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

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

A105038 Nonnegative n such that 6n^2 + 6n + 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

Gerald McGarvey, Apr 03 2005

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (I).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (II).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (III).
Editors, L'Intermédiaire des Mathématiciens, Query 4500: The equation x(x+1)/2 = y*(y+1)/3, L'Intermédiaire des Mathématiciens, 22 (1915), 255-260 (IV).
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A001652, A001570, A049629, A105040, A104240, A077288, A105036, A103200, A105037.

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 *)

for(n=0,427284,if(issquare(6*n*(n+1)+1),print1(n,",")))

Vec(4*x/(1-11*x+11*x^2-x^3)+O(x^99)) \\ Charles R Greathouse IV, Nov 13 2012

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.

More terms from Vladeta Jovovic, Apr 05 2005

Incorrect conjectures deleted by N. J. A. Sloane, Nov 24 2010

A188646 Array of a(n)=a(n-1)*k-((k-1)/(k^n)) where a(0)=1 and k=(sqrt(x^2-1)+x)^2 for integers x>=1.

Original entry on oeis.org

1, 1, 1, 1, 13, 1, 1, 181, 33, 1, 1, 2521, 1121, 61, 1, 1, 35113, 38081, 3781, 97, 1, 1, 489061, 1293633, 234361, 9505, 141, 1, 1, 6811741, 43945441, 14526601, 931393, 20021, 193, 1, 1, 94875313, 1492851361, 900414901, 91267009, 2842841, 37441, 253, 1

Offset: 0

Charles L. Hohn, Apr 06 2011

Conjecture: Given function f(x, y)=(sqrt(x^2+y)+x)^2; constant k=f(x, y); and initial term a(0)=1; then for all integers x>=1 and y=[+-]1, k may be irrational, but sequence a(n)=a(n-1)*k-((k-1)/(k^n)) always produces integer sequences; y=-1 results shown here; y=1 results are A188647.

Also square array A(n,k), n >= 1, k >= 0, read by antidiagonals, where A(n,k) is (1/n) * T_{2*k+1}(n), with the Chebyshev polynomials of the first kind (type T). - Seiichi Manyama, Jan 01 2019

			Square array begins:
     | 0    1       2          3             4
-----+---------------------------------------------
   1 | 1,   1,      1,         1,            1, ...
   2 | 1,  13,    181,      2521,        35113, ...
   3 | 1,  33,   1121,     38081,      1293633, ...
   4 | 1,  61,   3781,    234361,     14526601, ...
   5 | 1,  97,   9505,    931393,     91267009, ...
   6 | 1, 141,  20021,   2842841,    403663401, ...
   7 | 1, 193,  37441,   7263361,   1409054593, ...
   8 | 1, 253,  64261,  16322041,   4145734153, ...
   9 | 1, 321, 103361,  33281921,  10716675201, ...
  10 | 1, 397, 158005,  62885593,  25028308009, ...
  11 | 1, 481, 231841, 111746881,  53861764801, ...
  12 | 1, 573, 328901, 188788601, 108364328073, ...
  13 | 1, 673, 453601, 305726401, 206059140673, ...
  14 | 1, 781, 610741, 477598681, 373481557801, ...
  15 | 1, 897, 805505, 723342593, 649560843009, ...
  ...

Row 1-8 give A000012, A001570, A077420, A302329, A302330, A302331, A302332, A253880.
Column 1 is A082109(n-1).
Cf. A188644, A188647 (f(x, y) as above with y=1).
Diagonal gives A322904.

A[n_, k_] := 1/n ChebyshevT[2k+1, n];
Table[A[n-k, k], {n, 1, 9}, {k, n-1, 0, -1}] // Flatten (* Jean-François Alcover, Jan 02 2019, after Seiichi Manyama *)

A(n,k) = 2 * A188644(n,k) - A(n,k-1).

A(n,k) = Sum_{j=0..k} binomial(2*k+1,2*j+1)*(n^2-1)^(k-j)*n^(2*j). - Seiichi Manyama, Jan 01 2019

Edited and extended by Seiichi Manyama, Jan 01 2019

A302329 a(0)=1, a(1)=61; for n>1, a(n) = 62*a(n-1) - a(n-2).

Original entry on oeis.org

1, 61, 3781, 234361, 14526601, 900414901, 55811197261, 3459393815281, 214426605350161, 13290990137894701, 823826961944121301, 51063980650397625961, 3165142973362708688281, 196187800367837541047461, 12160478479832564836254301, 753753477949251182306719201

Offset: 0

Bruno Berselli, Apr 05 2018

Centered hexagonal numbers (A003215) with index in A145607. Example: 35 is a member of A145607, therefore A003215(35) = 3781 is a term of this sequence.
Also, centered 10-gonal numbers (A062786) with index in A182432. Example: 28 is a member of A182432 and A062786(28) = 3781.
a(n) is a solution to the Pell equation (4*a(n))^2 - 15*b(n)^2 = 1. The corresponding b(n) are A258684(n). - Klaus Purath, Jul 19 2025

Colin Barker, Table of n, a(n) for n = 0..500
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (62,-1).

Cf. A003215, A145607; A062786, A182432.
Fourth row of the array A188646.
First bisection of A041449, A042859.
Similar sequences of the type cosh((2*n+1)*arccosh(k))/k: A000012 (k=1), A001570 (k=2), A077420 (k=3), this sequence (k=4), A302330 (k=5), A302331 (k=6), A302332 (k=7), A253880 (k=8).

Mathematica
```
LinearRecurrence[{62, -1}, {1, 61}, 20]
```

x='x+O('x^99); Vec((1-x)/(1-62*x+x^2)) \\ Altug Alkan, Apr 06 2018

G.f.: (1 - x)/(1 - 62*x + x^2).
a(n) = a(-1-n).
a(n) = cosh((2*n + 1)*arccosh(4))/4.
a(n) = ((4 + sqrt(15))^(2*n + 1) + 1/(4 + sqrt(15))^(2*n + 1))/8.
a(n) = (1/4)*T(2*n+1, 4), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Jul 08 2022
E.g.f.: exp(31*x)*(4*cosh(8*sqrt(15)*x) + sqrt(15)*sinh(8*sqrt(15)*x))/4. - Stefano Spezia, Jul 25 2025

A001922 Numbers k such that 3k^2 - 3k + 1 is both a square (A000290) and a centered hexagonal number (A003215).

Original entry on oeis.org

1, 8, 105, 1456, 20273, 282360, 3932761, 54776288, 762935265, 10626317416, 148005508553, 2061450802320, 28712305723921, 399910829332568, 5570039304932025, 77580639439715776, 1080558912851088833, 15050244140475527880, 209622859053806301481

Offset: 0

N. J. A. Sloane

Also larger of two consecutive integers whose cubes differ by a square. Defined by a(n)^3 - (a(n) - 1)^3 = square.
Let m be the n-th ratio 2/1, 7/4, 26/15, 97/56, 362/209, ... Then a(n) = m*(2-m)/(m^2-3). The numerators 2, 7, 26, ... of m are A001075. The denominators 1, 4, 15, ... of m are A001353.
From Colin Barker, Jan 06 2015: (Start)
Also indices of centered triangular numbers (A005448) which are also centered square numbers (A001844).
Also indices of centered hexagonal numbers (A003215) which are also centered octagonal numbers (A016754).
Also positive integers x in the solutions to 3*x^2 - 4*y^2 - 3*x + 4*y = 0, the corresponding values of y being A156712.
(End)

			8 is in the sequence because 3*8^2 - 3*8 + 1 = 169 is a square and also a centered hexagonal number. - _Colin Barker_, Jan 07 2015

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..800
J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A001353, A001570, A001844, A001921, A003215.

Cf. A005448, A006051, A016754, A076139, A156712.

I:=[1, 8, 105]; [n le 3 select I[n] else 15*Self(n-1)-15*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Apr 16 2012

seq(simplify((1 +ChebyshevU(n,7) +ChebyshevU(n-1,7))/2), n=0..30); # G. C. Greubel, Oct 07 2022

With[{s1=3+2Sqrt[3],s2=3-2Sqrt[3],t1=7+4Sqrt[3],t2=7-4Sqrt[3]}, Simplify[ Table[(s1 t1^n+s2 t2^n+6)/12,{n,0,20}]]] (* or *) LinearRecurrence[ {15,-15,1},{1,8,105},21] (* Harvey P. Dale, Aug 14 2011 *)
CoefficientList[Series[(1-7*x)/(1-15*x+15*x^2-x^3),{x,0,30}],x] (* Vincenzo Librandi, Apr 16 2012 *)

Vec((1-7*x)/(1-15*x+15*x^2-x^3) + O(x^100)) \\ Colin Barker, Jan 06 2015

[(1+chebyshev_U(n,7) +chebyshev_U(n-1,7))/2 for n in range(30)] # G. C. Greubel, Oct 07 2022

a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3).
a(n) = (s1*t1^n + s2*t2^n + 6)/12 where s1 = 3 + 2*sqrt(3), s2 = 3 - 2*sqrt(3), t1 = 7 + 4*sqrt(3), t2 = 7 - 4*sqrt(3).
a(n) = A001075(n)*A001353(n+1).
G.f.: (1-7*x)/((1-x)*(1-14*x+x^2)). - Simon Plouffe (in his 1992 dissertation) and Colin Barker, Jan 01 2012
a(n) = A076139(n+1) - 7*A076139(n). - R. J. Mathar, Jul 14 2015
a(n) = (1/2)*(1 + ChebyshevU(n, 7) + ChebyshevU(n-1, 7)). G. C. Greubel, Oct 07 2022
a(n) = 1 - a(-1-n) = 1 + A001921(n) for all integers n. - Michael Somos, Jul 10 2025

Additional comments from James R. Buddenhagen, Mar 04 2001
Name improved by Colin Barker, Jan 07 2015
Edited by Robert Israel, Feb 20 2017

cp's OEIS Frontend

A157014 Expansion of x*(1-x)/(1 - 22*x + x^2).

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A165253 Triangle T(n,k), read by rows given by [1,0,1,0,0,0,0,0,0,...] DELTA [0,1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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A160682 The list of the A values in the common solutions to 13*k+1 = A^2 and 17*k+1 = B^2.

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A350916 Positive integers k such that (k+1)^4 has a divisor congruent to -1 modulo k.

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A133607 Triangle read by rows: T(n, k) = qStirling2(n, k, q) for q = -1, with 0 <= k <= n.

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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.

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

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A157014 Expansion of x(1-x)/(1 - 22x + x^2).

A160682 The list of the A values in the common solutions to 13k+1 = A^2 and 17k+1 = B^2.

A103200 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) for n >= 5.

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

A001922 Numbers k such that 3k^2 - 3k + 1 is both a square (A000290) and a centered hexagonal number (A003215).