A059100 -id:A059100 - cp's OEIS frontend

A189834 a(n) = n^2 + 9.

9, 10, 13, 18, 25, 34, 45, 58, 73, 90, 109, 130, 153, 178, 205, 234, 265, 298, 333, 370, 409, 450, 493, 538, 585, 634, 685, 738, 793, 850, 909, 970, 1033, 1098, 1165, 1234, 1305, 1378, 1453, 1530, 1609, 1690, 1773, 1858, 1945

Offset: 0

Vladimir Joseph Stephan Orlovsky, Apr 28 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A059100, A117950, A087475, A154533.

[n^2+9: n in [0..50]]; // Vincenzo Librandi, Aug 31 2011

Table[n^2+9,{n,0,100}]
LinearRecurrence[{3,-3,1},{9,10,13},50] (* Harvey P. Dale, Aug 21 2020 *)

a(n)=n^2+9 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = A154533(n+1). - R. J. Mathar, May 16 2011
G.f.: ( -9+17*x-10*x^2 ) / (x-1)^3 . - R. J. Mathar, Aug 31 2011
E.g.f.: (9 + x + x^2)*exp(x). - G. C. Greubel, Jan 13 2018
From Amiram Eldar, Nov 02 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + 3*Pi*coth(3*Pi))/18.
Sum_{n>=0} (-1)^n/a(n) = (1 + 3*Pi*cosech(3*Pi))/18. (End)
From Amiram Eldar, Feb 12 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = (2/3)*sqrt(2)*sinh(2*sqrt(2)*Pi)/sinh(3*Pi).
Product_{n>=0} (1 + 1/a(n)) = (sqrt(10)/3)*sinh(sqrt(10)*Pi)/sinh(3*Pi). (End)

A290743 Maximum number of distinct Lyndon factors that can appear in words of length n over an alphabet of size 2.

Original entry on oeis.org

2, 3, 4, 6, 8, 11, 14, 18, 22, 27, 32, 38, 44, 51, 58, 66, 74, 83, 92, 102, 112, 123, 134, 146, 158, 171, 184, 198, 212, 227, 242, 258, 274, 291, 308, 326, 344, 363, 382, 402, 422, 443, 464, 486, 508, 531, 554, 578, 602, 627, 652, 678, 704, 731, 758

Offset: 1

N. J. A. Sloane, Aug 11 2017

See theorem 1 of reference for formula.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Amy Glen, Jamie Simpson, and W. F. Smyth, Counting Lyndon Factors, Electronic Journal of Combinatorics 24(3) (2017), #P3.28.
Ryo Hirakawa, Yuto Nakashima, Shunsuke Inenaga, and Masayuki Takeda, Counting Lyndon Subsequences, arXiv:2106.01190 [math.CO], 2021. See MDF(n, s).
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A290744, A290745, A290746, A014206 (bisection), A059100 (bisection).

[Binomial(n+1,2)-(2-(n-2*Floor(n/2)))*Binomial(Floor(n/2)+1,2)-(n-2*Floor(n/2))*Binomial(Floor(n/2)+2,2)+2: n in [1..60]]; // Vincenzo Librandi, Oct 04 2017

Table[(Binomial[n+1,2] - (2-(n - 2 Floor[n/2])) Binomial[Floor[n/2]+1, 2] - (n-2 Floor[n/2]) Binomial[Floor[n/2]+2, 2] + 2), {n, 60}] (* Vincenzo Librandi, Oct 04 2017 *)

a(n)=(s->my(m=n\s,p=n%s); binomial(n+1,2)-(s-p)*binomial(m+1,2)-p*binomial(m+2,2)+s)(2); \\ Andrew Howroyd, Aug 14 2017

a(n) = binomial(n+1,2) - (s-p)*binomial(m+1,2) - p*binomial(m+2,2) + s where s=2, m=floor(n/s), p=n-m*s. - Andrew Howroyd, Aug 14 2017
From Colin Barker, Oct 03 2017: (Start)
G.f.: x*(2 - x - 2*x^2 + 2*x^3) / ((1 - x)^3*(1 + x)).
a(n) = (2*n^2 + 16) / 8 for n even.
a(n) = (2*n^2 + 14) / 8 for n odd.
a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 4. (End)
E.g.f.: ((8 + x + x^2)*cosh(x) + (7 + x + x^2)*sinh(x) - 8)/4. - Stefano Spezia, Jul 06 2021
Sum_{n>=1} 1/a(n) = coth(sqrt(2)*Pi)*Pi/(2*sqrt(2)) + tanh(sqrt(7)*Pi/2)*Pi/sqrt(7) - 1/4. - Amiram Eldar, Sep 16 2022

a(11)-a(55) from Andrew Howroyd, Aug 14 2017

A102305 a(n) = n^2 + 2*n + 3.

Original entry on oeis.org

6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027, 2118, 2211, 2306, 2403

Offset: 1

Ralf Stephan, Jan 03 2005

Essentially a duplicate of A059100.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A010000, A027578, A059100, A129388.

[(n+1)^2+2: n in [1..60]]; // G. C. Greubel, Feb 03 2024

A102305:=n->n^2+2*n+3: seq(A102305(n), n=1..100); # Wesley Ivan Hurt, Jan 22 2017

Table[n^2+2n+3,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{6,11,18},50] (* Harvey P. Dale, Aug 05 2015 *)

a(n)=n^2+2*n+3 \\ Charles R Greathouse IV, Oct 16 2015

[n^2+2*n+3 for n in range(1,61)] # G. C. Greubel, Feb 03 2024

a(n) = (1/5) * A027578(n-1).
a(n) = 2*n + a(n-1) + 1 (with a(1)=6). - Vincenzo Librandi, Nov 16 2010
a(n) = A059100(n+1). - Reinhard Zumkeller, Mar 21 2008
a(n) = A010000(n+1) for n >= 1. - Georg Fischer, Nov 02 2018
From Amiram Eldar, Sep 14 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi * coth(sqrt(2)*Pi)/(2*sqrt(2)) - 7/12.
Sum_{n>=1} (-1)^(n+1)/a(n) = cosech(sqrt(2)*Pi)*Pi/(2*sqrt(2)) + 1/12. (End)
From G. C. Greubel, Feb 03 2024: (Start)
G.f.: (3 - 3*x + 2*x^2)/(1-x)^3.
E.g.f.: (3 + 3*x + x^2)*exp(x). (End)

A114964 a(n) = n^2 + 30.

Original entry on oeis.org

30, 31, 34, 39, 46, 55, 66, 79, 94, 111, 130, 151, 174, 199, 226, 255, 286, 319, 354, 391, 430, 471, 514, 559, 606, 655, 706, 759, 814, 871, 930, 991, 1054, 1119, 1186, 1255, 1326, 1399, 1474, 1551, 1630, 1711, 1794, 1879, 1966, 2055, 2146, 2239, 2334, 2431, 2530

Offset: 0

Cino Hilliard, Feb 21 2006

x^2 + 30 != y^n for all x,y and n > 1, so this is a subsequence of A007916.
From Bruno Berselli, May 12 2014: (Start)
This is the case k=5 of the identity n^2 + k*(k+1) = (Sum_{i=-k..k} (n+i)^3)/((2*k+1)*n).
Similar sequences: A059100 (k=1), A114949 (k=2), A241748 (k=3), A241850 (k=4). (End)
The old name of this sequence was: Numbers of the form x^2 + 30. Also numbers that are not a perfect power.

			11*4*a(4) = (-1)^3 + 0^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3 + 8^3 + 9^3 = 2024. - _Bruno Berselli_, May 12 2014

Shawn A. Broyles, Table of n, a(n) for n = 0..1000
J. H. E. Cohn, The diophantine equation x^2 + C = y^n, Acta Arithmetica LXV.4 (1993).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A007916, A059100, A114949, A241748, A241850.

Range[0,60]^2+30 (* Harvey P. Dale, Oct 17 2022 *)

g(n,p) = for(x=0,n,y=x^2+p;print1(y","));

a(n) = n^2 + 30; \\ Altug Alkan, Apr 30 2018

From Amiram Eldar, Nov 04 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(30)*Pi*coth(sqrt(30)*Pi))/60.
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(30)*Pi*cosech(sqrt(30)*Pi))/60. (End)
From Elmo R. Oliveira, Dec 30 2024: (Start)
G.f.: (30 - 59*x + 31*x^2)/(1 - x)^3.
E.g.f.: (30 + x + x^2)*exp(x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

New name from Shawn A. Broyles and Altug Alkan, Apr 30 2018

A189833 a(n) = n^2 + 8.

Original entry on oeis.org

8, 9, 12, 17, 24, 33, 44, 57, 72, 89, 108, 129, 152, 177, 204, 233, 264, 297, 332, 369, 408, 449, 492, 537, 584, 633, 684, 737, 792, 849, 908, 969, 1032, 1097, 1164, 1233, 1304, 1377, 1452, 1529, 1608, 1689, 1772, 1857, 1944, 2033

Offset: 0

Vladimir Joseph Stephan Orlovsky, Apr 28 2011

From César Eliud Lozada, Mar 29 2021: (Start)
Numbers a(n) such that sqrt( a(n) + 4*n*sqrt(2) ) = n + 2*sqrt(2). Examples:
For n=1:  sqrt( 9 +  4*sqrt(2)) = 1 + 2*sqrt(2),
For n=2:  sqrt(12 +  8*sqrt(2)) = 2 + 2*sqrt(2),
For n=3:  sqrt(17 + 12*sqrt(2)) = 3 + 2*sqrt(2). (End)

G. C. Greubel, Table of n, a(n) for n = 0..10000 (terms 0..955 from Vincenzo Librandi)
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002522, A059100, A117950, A087475.

[n^2+8: n in [0..50]]; // Vincenzo Librandi, Apr 29 2011

Table[n^2+8,{n,0,100}]
LinearRecurrence[{3,-3,1},{8,9,12},50] (* Harvey P. Dale, Jun 21 2022 *)

a(n)=n^2+8 \\ Charles R Greathouse IV, Jun 17 2017

From G. C. Greubel, Jan 13 2018: (Start)
G.f.: (8 - 15*x + 9*x^2)/(1 - x)^3.
E.g.f.: (8 + x + x^2)*exp(x). (End)
From Amiram Eldar, Jul 04 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + 2*sqrt(2)*Pi*coth(2*sqrt(2)*Pi))/16.
Sum_{n>=0} (-1)^n/a(n) = (1 + 2*sqrt(2)*Pi*cosech(2*sqrt(2)*Pi))/16. (End)
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = (sqrt(7/2)/2)*sinh(sqrt(7)*Pi)/sinh(2*sqrt(2)*Pi).
Product_{n>=0} (1 + 1/a(n)) = (3/(2*sqrt(2)))*sinh(3*Pi)/sinh(2*sqrt(2)*Pi). (End)

Offset changed from 1 to 0 by Vincenzo Librandi, Apr 29 2011

A246453 Lucas numbers (A000204) of the form n^2 + 2.

Original entry on oeis.org

3, 11, 18, 123, 843, 5778, 39603, 271443, 1860498, 12752043, 87403803, 599074578, 4106118243, 28143753123, 192900153618, 1322157322203, 9062201101803, 62113250390418, 425730551631123, 2918000611027443, 20000273725560978, 137083915467899403, 939587134549734843

Offset: 1

Michel Lagneau, Aug 26 2014

a(n) = {11} union {A000204(2+4*n)} for n=0,1,...

Intersection of A000204 and A059100. - Michel Marcus, Aug 26 2014

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (7,-1).

Cf. A000204 (Lucas), A059100 (n^2+2).

Cf. quadrisection of A000032: A056854 (first), A056914 (second), this sequence (third, without 11), A288913 (fourth).

I:=[3,11,18,123]; [n le 4 select I[n] else 7*Self(n-1)-Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 21 2017

with(combinat,fibonacci):lst:={}:lst1:={}:nn:=5000:
  for n from 1 to nn do:
    lst:=lst union {2*fibonacci(n-1)+fibonacci(n)}:
  od:
   for m from 1 to nn do:
    if {m^2+2} intersect lst = {m^2+2}
    then
    lst1:=lst1 union {m^2+2}:
    else
    fi:
   od:
   print(lst1):

CoefficientList[Series[x*(3-10*x-56*x^2+8*x^3)/(1-7*x+x^2), {x,0,50}], x] (* or *) LinearRecurrence[{7,-1}, {3, 11, 18, 123}, 30] (* G. C. Greubel, Dec 21 2017 *)
Select[LucasL[Range[100]],IntegerQ[Sqrt[#-2]]&] (* Harvey P. Dale, Dec 31 2018 *)

lista(nn) = for (n=0, nn, luc = fibonacci(n+1) + fibonacci(n-1); if (issquare(luc-2), print1(luc, ", "))); \\ Michel Marcus, Mar 29 2016

Vec(x*(3 - 10*x - 56*x^2 + 8*x^3) / (1 - 7*x + x^2) + O(x^30)) \\ Colin Barker, Jun 20 2017

From Colin Barker, Jun 20 2017: (Start)
G.f.: x*(3 - 10*x - 56*x^2 + 8*x^3) / (1 - 7*x + x^2).
a(n) = (2^(-n)*((7+3*sqrt(5))^n*(-20+9*sqrt(5)) + (7-3*sqrt(5))^n*(20+9*sqrt(5)))) / sqrt(5) for n>2.
a(n) = 7*a(n-1) - a(n-2) for n>4. (End)
E.g.f.: 2*exp(7*x/2)*(9*cosh(3*sqrt(5)*x/2) - 4*sqrt(5)*sinh(3*sqrt(5)*x/2)) + 4*x^2 - 18. - Stefano Spezia, Apr 14 2025

Corrected by Michel Marcus, Mar 29 2016

A085554 Greater of twin primes of the form x^2+2, x^2+4.

Original entry on oeis.org

5, 13, 229, 1093, 2029, 3253, 13693, 21613, 59053, 65029, 91813, 140629, 178933, 199813, 205213, 227533, 328333, 567013, 700573, 804613, 815413, 1071229, 2241013, 3629029, 4223029, 4347229, 4809253, 5212093, 5919493, 6185173

Offset: 1

Cino Hilliard, Jul 04 2003

Except for the first term, all a(n)=13 (mod 72) with x=3 (mod 6). The lesser of the twin prime pair is given by A253639, the x-values in A086381. - M. F. Hasler, Jan 18 2015

M. F. Hasler, Table of n, a(n) for n = 1..3044 (all terms below 10^12).

Cf. A085553, A086381, A087475, A059100.

Transpose[Select[Table[x^2+{2,4},{x,5000}],AllTrue[#,PrimeQ]&]][[2]] (* The program uses the AllTrue function from Mathematica version 10 *) (* Harvey P. Dale, Jan 15 2015 *)

is_A086381(x)=ispseudoprime(x^2+2)&&ispseudoprime(x^2+4) \\ or is_A067201(x)&&is_A007591(x)
A085554 = apply(A087475,select(is_A086381,vector(9999,n,n))) \\ A087475=x->x^2+4.
write(f="b085554.txt",c=1," 5"); forstep(x=3,1e6,6,is_A086381(x)&&write(f,c++" "x^2+4))
\\ M. F. Hasler, Jan 18 2015

A085554 = A087475 o A086381 = A020725^2 o A253639, i.e., a(n) = A087475(A086381(n)) = A253639(n)+2. - M. F. Hasler, Jan 18 2015

Edited by Don Reble, May 03 2006

Definition corrected by Harvey P. Dale and Franklin T. Adams-Watters, Jan 15 2015

A117938 Triangle, columns generated from Lucas Polynomials.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 4, 1, 4, 11, 14, 7, 1, 5, 18, 36, 34, 11, 1, 6, 27, 76, 119, 82, 18, 1, 7, 38, 140, 322, 393, 198, 29, 1, 8, 51, 234, 727, 1364, 1298, 478, 47, 1, 9, 66, 364, 1442, 3775, 5778, 4287, 1154, 76, 1, 10, 83, 536, 2599, 8886, 19602, 24476, 14159, 2786, 123

Offset: 1

Gary W. Adamson, Apr 03 2006

Companion triangle using Fibonacci polynomial generators = A073133. Inverse binomial transforms of the columns defines rows of A117937 (with some adjustments of offset).

A309220 is another version of the same triangle (except it omits the last diagonal), and perhaps has a clearer definition. - N. J. A. Sloane, Aug 13 2019

			First few rows of the triangle are:
  1;
  1, 1;
  1, 2,  3;
  1, 3,  6,   4;
  1, 4, 11,  14,   7;
  1, 5, 18,  36,  34,  11;
  1, 6, 27,  76, 119,  82,  18;
  1, 7, 38, 140, 322, 393, 198, 29;
  ...
For example, T(7,4) = 76 = f(4), x^3 + 3*x = 64 + 12 = 76.

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Cf. A104509, A114525, A117936, A117937, A118980, A118981, A309220.

Cf. A000204 (diagonal), A059100 (column 3), A061989 (column 4).

Lucas := proc(n,x) # see A114525
    option remember;
    if  n=0 then
        2;
    elif n =1 then
        x ;
    else
        x*procname(n-1,x)+procname(n-2,x) ;
    end if;
    expand(%) ;
end proc:
A117938 := proc(n::integer,k::integer)
    if k = 1 then
        1;
    else
        subs(x=n-k+1,Lucas(k-1,x)) ;
    end if;
end proc:
seq(seq(A117938(n,k),k=1..n),n=1..12) ; # R. J. Mathar, Aug 16 2019

T[n_, k_]:= LucasL[k-1, n-k+1] - Boole[k==1];
Table[T[n, k], {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Oct 28 2021 *)

def A117938(n,k): return 1 if (k==1) else round(2^(1-k)*( (n-k+1 + sqrt((n-k)*(n-k+2) + 5))^(k-1) + (n-k+1 - sqrt((n-k)*(n-k+2) + 5))^(k-1) ))
flatten([[A117938(n,k) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Oct 28 2021

Columns are f(x), x = 1, 2, 3, ..., of the Lucas Polynomials: (1, defined different from A034807 and A114525); (x); (x^2 + 2); (x^3 + 3*x); (x^4 + 4*x^2 + 2); (x^5 + 5*x^3 + 5*x); (x^6 + 6*x^4 + 9*x^2 + 2); (x^7 + 7*x^5 + 14*x^3 + 7*x); ...

Terms a(51) and a(52) corrected by G. C. Greubel, Oct 28 2021

A164897 a(n) = 4n(n+1) + 3.

Original entry on oeis.org

3, 11, 27, 51, 83, 123, 171, 227, 291, 363, 443, 531, 627, 731, 843, 963, 1091, 1227, 1371, 1523, 1683, 1851, 2027, 2211, 2403, 2603, 2811, 3027, 3251, 3483, 3723, 3971, 4227, 4491, 4763, 5043, 5331, 5627, 5931, 6243, 6563, 6891, 7227, 7571, 7923, 8283, 8651, 9027, 9411

Offset: 0

Paul Curtz, Aug 30 2009

One-fourth the sum of the three terms produced by the division of complex numbers (2*n-3+(2*n-1)*i)/(2*n+1+(2*n+3)*i). For (b+c*i)/(d+e*i) the three terms in parentheses are ((b*d+c*e)+(c*d-b*e)*i)/(d^2+e^2). By substituting b=2*n-3, c=2*n-1, d=2*n+1, and e=2*n+3 one gets a(n). - J. M. Bergot, Sep 10 2015

The continued fraction expansion of sqrt(a(n)) is [2n+1; {2n+1, 4n+2}]. - Magus K. Chu, Sep 08 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..900
R. M. Green and Tianyuan Xu, 2-roots for simply laced Weyl groups, arXiv:2204.09765 [math.RT], 2022.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000124, A008590, A010731, A016743, A069894.

Odd-indexed terms of A059100.

[4*n*(n+1)+3: n in [0..50]]; // Vincenzo Librandi, Apr 24 2011

A164897:=n->4*n*(n+1)+3: seq(A164897(n), n=0..100); # Wesley Ivan Hurt, Sep 10 2015

Table[4 n (n + 1) + 3, {n, 0, 50}] (* Harvey P. Dale, Jan 23 2011 *)

a(n)=4*n*(n+1)+3 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = A000124(2*n) + A000124(2*n+1) = A069894(n)+1.
a(n+1) - a(n) = 8n+8 = A008590(n+1) (first differences).
a(n+1) - 2*a(n) + a(n-1) = 8 = A010731(n) (second differences).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n>2.
G.f.: (3+2*x+3*x^2) / (1-x)^3.
Sum_{k=n+1..2*n+1} a(k) - Sum_{k=0..n} a(k) = (2*n+2)^3. - Bruno Berselli, Jan 24 2011
E.g.f.: (4x^2 + 8x + 1)*exp(x). - G. C. Greubel, Sep 22 2015
a(n)^2 = A222465(n)*A222465(n+1) - 12. - Ezhilarasu Velayutham, Mar 18 2020
Sum_{n>=0} 1/a(n) = tanh(Pi/sqrt(2))*Pi/(4*sqrt(2)). - Amiram Eldar, Aug 21 2022
a(n) = A059100(2*n+1). - Dimitri Papadopoulos, Nov 21 2023

Definition simplified by R. J. Mathar, Sep 16 2009

A188892 Numbers n such that there is no triangular n-gonal number greater than 1.

Original entry on oeis.org

11, 18, 38, 102, 198, 326, 486, 678, 902, 1158, 1446, 1766, 2118, 2918, 3366, 3846, 4358, 4902, 5478, 6086, 6726, 7398, 8102, 8838, 9606, 10406, 11238, 12102, 12998, 13926, 14886, 15878, 16902, 17958, 19046, 20166, 21318, 22502, 24966, 26246

Offset: 1

T. D. Noe, Apr 13 2011

nonn

It is easy to find triangular numbers that are square, pentagonal, hexagonal, etc.  So it is somewhat surprising that there are no triangular 11-gonal numbers other than 0 and 1. For these n, the equation x^2 + x = (n-2)*y^2 - (n-4)*y has no integer solutions x>1 and y>1.
Chu shows how to transform the equation into a generalized Pell equation. When n has the form k^2+2 (A059100), then the Pell equation has only a finite number of solutions and it is simple to select the n that produce no integer solutions greater than 1.
The general case is in A188950.

Robert Israel, Table of n, a(n) for n = 1..10000
Wenchang Chu, Regular polygonal numbers and generalized pell equations, Int. Math. Forum 2 (2007), 781-802.

Cf. A051682 (11-gonal numbers), A051870 (18-gonal numbers), A188891, A188896.

filter:= n -> nops(select(t -> min(subs(t,[x,y]))>=2, [isolve(x^2 + x = (n-2)*y^2 - (n-4)*y)])) = 0:
select(filter, [seq(t^2+2,t=3..200)]); # Robert Israel, May 13 2018

cp's OEIS Frontend

A189834 a(n) = n^2 + 9.

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A290743 Maximum number of distinct Lyndon factors that can appear in words of length n over an alphabet of size 2.

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A102305 a(n) = n^2 + 2*n + 3.

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A114964 a(n) = n^2 + 30.

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A189833 a(n) = n^2 + 8.

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A246453 Lucas numbers (A000204) of the form n^2 + 2.

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A164897 a(n) = 4n(n+1) + 3.