A000466 -id:A000466 - cp's OEIS frontend

A066830 a(n) = lcm(n+1, n-1).

0, 3, 4, 15, 12, 35, 24, 63, 40, 99, 60, 143, 84, 195, 112, 255, 144, 323, 180, 399, 220, 483, 264, 575, 312, 675, 364, 783, 420, 899, 480, 1023, 544, 1155, 612, 1295, 684, 1443, 760, 1599, 840, 1763, 924, 1935, 1012, 2115, 1104, 2303, 1200, 2499, 1300

Offset: 1

Benoit Cloitre, Jan 20 2002

a(n+2) is the order of rowmotion on a poset obtained by adjoining a unique minimal and maximal element to a disjoint union of at least two chains of n elements. - Nick Mayers, Jun 01 2018

Harry J. Smith, Table of n, a(n) for n = 1..1000
J. Striker and N. Williams, Promotion and Rowmotion, arXiv preprint arXiv:1108.1172 [math.CO], 2011-2012.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A000466, A046092.

A066830:=n->lcm(n+1,n-1): seq(A066830(n), n=1..100); # Wesley Ivan Hurt, Jul 09 2017

Table[LCM[n-1,n+1],{n,100}] (* Zak Seidov, Oct 23 2009 *)
a[n_] := If[EvenQ[n], n(n+2)/2, n(n+2)]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Jun 13 2017 *)

a(n) = { lcm(n+1, n-1) } \\ Harry J. Smith, Mar 30 2010

concat(0, Vec(x^2*(x^4-6*x^2-4*x-3)/((x-1)^3*(x+1)^3) + O(x^100))) \\ Colin Barker, Nov 05 2014

a(2n) = A000466(n); a(2n+1) = A046092(n).
From Colin Barker, Nov 05 2014: (Start)
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n > 6.
a(n) = (3+(-1)^n)*(-1+n^2)/4.
G.f.: x^2*(x^4 - 6*x^2 - 4*x - 3) / ((x-1)^3*(x+1)^3). (End)
From Amiram Eldar, Aug 09 2022: (Start)
a(n) = numerator((n^2 - 1)/2).
Sum_{n>=2} 1/a(n) = 1. (End)
E.g.f.: (2 - (2 - x - 2*x^2)*cosh(x) - (1 - 2*x - x^2)*sinh(x))/2. - Stefano Spezia, Aug 04 2025

A214297 a(0)=-1, a(1)=0, a(2)=-3; thereafter a(n+2) - 2*a(n+1) + a(n) has period 4: repeat -4, 8, -4, 2.

Original entry on oeis.org

-1, 0, -3, 2, 3, 6, 5, 12, 15, 20, 21, 30, 35, 42, 45, 56, 63, 72, 77, 90, 99, 110, 117, 132, 143, 156, 165, 182, 195, 210, 221, 240, 255, 272, 285, 306, 323, 342, 357, 380, 399, 420, 437, 462, 483, 506, 525, 552, 575, 600, 621, 650, 675, 702, 725, 756, 783, 812, 837, 870, 899, 930, 957, 992, 1023, 1056, 1085, 1122, 1155, 1190

Offset: 0

Paul Curtz, Jul 11 2012

Let a(n)/A000290(n) = [-1/0, 0/1, -3/4, 2/9, 3/16, 6/25, 5/36, 12/49, 15/64, 20/81, 21/100, 30/121, ...] = a(n)/b(n) (say).
Then b(n)-4*a(n)=4, 1, 16, 1 (period of length 4).
Permutation from a(n) to A061037(n): 1, 3, 2, 7, 5, 11, 4, 15, 9, 19, 6, ... = shifted A145979 + 1.
A061037(n) - a(n) = 0, 3, -3, -3, 0, -15, 3, -33, 0 -57, 15, -87, 0, -123, ...
First 3 rows:
-1 0 -3 2  3  6  5 12 15 20 21 30 35
1 -3  5 1  3 -1  7  3  5  1  9  5  7
-4 8 -4 2 -4  8 -4  2 -4  8 -4  2 -4.
Note that the terms of a(n) increase from 12. Compare to increasing terms permutation of A061037(n): -3,-1,0,2,3,5,6,12,15, .... and A129647.
c(n) =  0, -1, 0, -1, 2, 1, 2, 1, 4, 3, 4, 3, 6, 5, 6, 5, ... (cf. A134967)
d(n) = -1,  1, 1,  3, 1, 3, 3, 5, 3, 5, 5, 7, 5, 7, 7, 9, ..., hence:
a(n) = c(n+1) * d(n+1).

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

[(2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8: n in [0..100]]; // G. C. Greubel, Sep 19 2018

A214297 := proc(n)
    option remember;
    if n <=5 then
        op(n+1,[-1,0,-3,2,3,6]) ;
    else
        2*procname(n-1)-procname(n-2)+procname(n-4)-2*procname(n-5)+procname(n-6) ;
    end if;
end proc: # R. J. Mathar, Jun 28 2013

Table[(2 n^2 - 11 - 9 (-1)^n + 6 ((-1)^((2 n + 1 - (-1)^n)/4) + (-1)^((2 n - 1 + (-1)^n)/4)))/8, {n, 0, 69}] (* or *)
CoefficientList[Series[-(1 - 2 x + 4 x^2 - 8 x^3 + 3 x^4)/((1 - x)^2*(1 - x^4)), {x, 0, 69}], x] (* Michael De Vlieger, Mar 24 2017 *)

vector(100, n, n--; (2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8) \\ G. C. Greubel, Sep 19 2018

a(k+4)  - a(k)    =  2*k + 4.
a(k+2)  - a(k-2)  =  2*k.
a(k+6)  - a(k-6)  =  6*k.
a(k+10) - a(k-10) = 10*k.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*k)   = -1, -3, followed by 3, 5, 15, 21, 35, 45, ... (A142717);
a(2*k+1) = k*(k+1) (see A002378).
A198442(n) = -1,0,0,2,3,6,8,12,  minus 3 at A198442(4*n+2).
G.f. -( 1-2*x+4*x^2-8*x^3+3*x^4 )/( (1-x)^2*(1-x^4) ). - R. J. Mathar, Jul 17 2012; edited by N. J. A. Sloane, Jul 22 2012
From R. J. Mathar, Jun 28 2013: (Start)
a(4*k)   = A000466(k);
a(4*k+1) = A002943(k);
a(4*k+2) = A078371(k-1) for k>0;
a(4*k+3) = A002939(k+1). (End)
a(n) = (2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8. - Luce ETIENNE, Oct 27 2016

Edited by N. J. A. Sloane, Jul 22 2012

A225948 a(0) = -1; for n>0, a(n) = numerator(1/4 - 4/n^2).

Original entry on oeis.org

-1, -15, -3, -7, 0, 9, 5, 33, 3, 65, 21, 105, 2, 153, 45, 209, 15, 273, 77, 345, 6, 425, 117, 513, 35, 609, 165, 713, 12, 825, 221, 945, 63, 1073, 285, 1209, 20, 1353, 357, 1505, 99, 1665, 437, 1833, 30, 2009, 525, 2193, 143

Offset: 0

Paul Curtz, May 21 2013

Denominators are in A226008.
Fractions in lowest terms for n>0: -15/4, -3/4, -7/36, 0/1, 9/100, 5/36, 33/196, 3/16, 65/324, 21/100, 105/484, 2/9, 153/676, 45/196, 209/900, 15/64,...
If t(n) is the sequence with period 8: 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, ... (see A226044), then A226008(n) = 4*a(n) + t(n).

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

Cf. A000466, A028566, A061037, A061041, A078371, A106609, A142705, A145923, A226008.

[-1] cat [Numerator(1/4-4/n^2): n in [1..50]]; // Bruno Berselli, May 22 2013

Join[{-1}, Table[Numerator[1/4 - 4/n^2], {n, 50}]] (* Bruno Berselli, May 24 2013 *)

concat([-1], vector(100, n, numerator(1/4 - 4/n^2))) \\ G. C. Greubel, Sep 19 2018

a(n) = 3*a(n-8) -3*a(n-16) +a(n-24).
a(2n) = A061037(n), a(2n+1) = A145923(n-2) for A145923(-2)=-15, A145923(-1)=-7.
a(4n)   = A142705(n) for A142705(0)=-1, a(8n) = A000466(n);
a(4n+1) = A028566(4n-3) for A028566(-3)=-15;
a(4n+2) = A078371(n-1) for A078371(-1)=-3;
a(4n+3) = A028566(4n-1) for A028566(-1)=-7.
a(n+4)  = A106609(n) * A106609(n+8).
G.f.: -(1 +15*x +3*x^2 +7*x^3 -9*x^5 -5*x^6 -33*x^7 -6*x^8 -110*x^9 -30*x^10 -126*x^11 -2*x^12 -126*x^13 -30*x^14 -110*x^15 -3*x^16 -33*x^17 -5*x^18 -9*x^19 +7*x^21 +3*x^22 +15*x^23)/(1-x^8)^3. - Bruno Berselli, May 22 2013
a(n) = (n^2-16)*(6*cos(Pi*n/4)-54*cos(Pi*n/2)+6*cos(3*Pi*n/4)-219*(-1)^n+293)/512. - Bruno Berselli, May 22 2013
a(n+10) = a(n+2)*(n+14)/(n-2) for n=0,1 and n>2. - Bruno Berselli, May 22 2013

Edited by Bruno Berselli, May 22 2013

A242850 32n^5 - 32n^3 + 6*n.

Original entry on oeis.org

0, 6, 780, 6930, 30744, 96030, 241956, 526890, 1032240, 1866294, 3168060, 5111106, 7907400, 11811150, 17122644, 24192090, 33423456, 45278310, 60279660, 79015794, 102144120, 130395006, 164575620, 205573770, 254361744, 312000150, 379641756, 458535330, 550029480, 655576494

Offset: 0

Vincenzo Librandi, May 29 2014

Chebyshev polynomial of the second kind U(5,n).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A000466, A144138, A144139, A242851.

Magma
```
[32*n^5-32*n^3+6*n: n in [0..40]];
```

Table[32 n^5 - 32 n^3 + 6 n, {n, 0, 40}] (* or *) Table[ChebyshevU[5, n], {n, 0, 40}]

G.f.: x*(6 + 744*x + 2340*x^2 + 744*x^3 + 6*x^4)/(1 - x)^6.

a(n) = 2*n*(2*n-1)*(2*n+1)*(4*n^2-3).

Edited by Bruno Berselli, May 29 2014

A242851 64n^6 - 80n^4 + 24*n^2 - 1.

Original entry on oeis.org

-1, 7, 2911, 40391, 242047, 950599, 2883167, 7338631, 16451071, 33489287, 63202399, 112211527, 189447551, 306634951, 478821727, 724955399, 1068505087, 1538129671, 2168392031, 3000519367, 4083209599, 5473483847, 7237584991, 9451922311, 12204062207, 15593764999

Offset: 0

Vincenzo Librandi, May 29 2014

Chebyshev polynomial of the second kind U(6,n).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A000466, A144138, A144139, A242850.

Magma
```
[64*n^6-80*n^4+24*n^2-1: n in [0..40]];
```

Table[64 n^6 - 80 n^4 + 24 n^2 - 1, {n, 0, 40}] (* or *) Table[ChebyshevU[6, n], {n, 0, 40}]

G.f.: (-1 + 14*x + 2841*x^2 + 20196*x^3 + 20161*x^4 + 2862*x^5 + 7*x^6)/(1 - x)^7.

a(n) = (8*n^3-4*n^2-4*n+1)*(8*n^3+4*n^2-4*n-1).

Edited by Bruno Berselli, May 29 2014

A278310 Numbers m such that T(m) + 3*T(m+1) is a square, where T = A000217.

Original entry on oeis.org

3, 143, 4899, 166463, 5654883, 192099599, 6525731523, 221682772223, 7530688524099, 255821727047183, 8690408031080163, 295218051329678399, 10028723337177985443, 340681375412721826703, 11573138040695364122499, 393146012008229658338303, 13355391270239113019379843

Offset: 1

Bruno Berselli, Nov 17 2016

Equivalently, both m+1 and 2*m+3 are squares for nonnegative m.
Corresponding triangular numbers T(m): 6, 10296, 12002550, 13855048416, 15988853699286, 18451128064030200, 21292585958400815526, ...
Square roots of T(m) + 3*T(m+1) are listed by A082405 (after 0).
Negative values of m for which T(m) + 3*T(m+1) is a square: -1, -2, -26, -842, -28562, -970226, -32959082, ...

			3 is in the sequence because T(3) + 3*T(4) = 6 + 3*10 = 6^2.
For n=5 is a(5) = 5654883, therefore floor(sqrt(5654883)) = 2377 = A182189(5) - 2 = 2379 - 2.

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Subsequence of A000466.
Cf. A000217, A001109, A029546, A046176, A082405, A156164, A182189.
Cf. A278438: numbers m such that T(m) + 2*T(m+1) is a square.
Cf. A078522: numbers m such that 3*T(m) + T(m+1) is a square.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: A084703 (k=-1), A076218 (k=3), this sequence (k=-5).

Iv:=[3,143]; [n le 2 select Iv[n] else 34*Self(n-1)-Self(n-2)+40: n in [1..20]];

P:=proc(q) local n; for n from 3 to q do if type(sqrt(2*n^2+5*n+3),integer) then print(n); fi; od; end: P(10^9); # Paolo P. Lava, Nov 18 2016

Table[((1 + Sqrt[2])^(4 n) + (1 - Sqrt[2])^(4 n))/8 - 5/4, {n, 1, 20}]
RecurrenceTable[{a[1] == 3, a[2] == 143, a[n] == 34 a[n - 1] - a[n - 2] + 40}, a, {n, 1, 20}]
LinearRecurrence[{35, -35, 1}, {3, 143, 4899}, 50] (* G. C. Greubel, Nov 20 2016 *)

Vec(x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)) + O(x^50)) \\ G. C. Greubel, Nov 20 2016

def A278310():
    a, b = 3, 143
    yield a
    while True:
        yield b
        a, b = b, 34*b - a + 40
a = A278310(); print([next(a) for  in range(18)]) # _Peter Luschny, Nov 18 2016

O.g.f.: x*(3 + 38*x - x^2)/((1 - x)*(1 - 34*x + x^2)).
E.g.f.: (exp((1-sqrt(2))^4*x) + exp((1+sqrt(2))^4*x) - 10*exp(x))/8 + 1.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n>3.
a(n) = 34*a(n-1) - a(n-2) + 40 for n>2.
a(n) = a(-n) = ((1 + sqrt(2))^(4*n) + (1 - sqrt(2))^(4*n))/8 - 5/4.
a(n) = 4*A001109(n)^2 - 1.
a(n) = -A029546(n) + 38*A029546(n-1) + 3*A029546(n-2) for n>1.
Lim_{n -> infinity} a(n)/a(n-1) = A156164.
Floor(sqrt(a(n))) = A182189(n) - 2.
a(n) - a(n-1) = 4*A046176(n) for n>1.

A069074 a(n) = (2n+2)(2n+3)(2n+4) = 24A000330(n+1).

Original entry on oeis.org

24, 120, 336, 720, 1320, 2184, 3360, 4896, 6840, 9240, 12144, 15600, 19656, 24360, 29760, 35904, 42840, 50616, 59280, 68880, 79464, 91080, 103776, 117600, 132600, 148824, 166320, 185136, 205320, 226920, 249984, 274560, 300696, 328440, 357840

Offset: 0

Benoit Cloitre, Apr 05 2002

sqrt((Sum_{k=0..n} 2*a(k)) + 1) = A056220(n+2). - Doug Bell, Mar 09 2009
Second leg of Pythagorean triangles with hypotenuse a square: A057769(n)^2 + a(n-1)^2 = A007204(n)^2. - Martin Renner, Nov 12 2011
Numbers which are both the sum of 2*n + 4 consecutive odd integers and the sum of the 2*n + 2 immediately higher consecutive odd integers.  In general, let f(k,n) = 3*k^3*A000330(n).  Then f(k,n) is both the sum of k*n + k consecutive terms from the arithmetic progression with first term A000217(k) and constant difference k and the immediately higher k*n terms from the same progression.  When k = 1, f(k,n) = A059270(n). - Charlie Marion, Aug 23 2021

Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 53.
T. J. I'a. Bromwich, Introduction to the Theory of Infinite Series, Macmillan, 2nd. ed. 1949, p. 190.
Jolley, Summation of Series, Dover (1961).
Konrad Knopp, Theory and application of infinite series, Dover, p. 269

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Konrad Knopp, Theorie und Anwendung der unendlichen Reihen, Berlin, J. Springer, 1922. (Original german edition of "Theory and Application of Infinite Series")
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001844. A001844(n+1)^2 - a(n) and A001844(n+1)^2 + a(n) are both square numbers. - Doug Bell, Mar 08 2009

Cf. A000466. a(n) = Sum_{k=0..2n+3} (A000466(n+1) + 2k) which is the sum of 2n+4 consecutive odd integers starting at A000466(n+1). - Doug Bell, Mar 08 2009

[(2*n+2)*(2*n+3)*(2*n+4): n in [0..40]]; // Vincenzo Librandi, Oct 04 2011

LinearRecurrence[{4,-6,4,-1},{24,120,336,720},40] (* Harvey P. Dale, Apr 10 2017 *)

a(n)=6*binomial(2*n+4,3) \\ Charles R Greathouse IV, Mar 21 2015

Sum_{n>=0} (-1)^n/a(n) = (Pi-3)/4 = 0.03539816339... [Jolley, eq. 244]
Sum_{n>=0} 1/a(n) = 3/4 - log(2) = 0.05685281... [Jolley, eq. 249]
G.f.: ( 24+24*x ) / (x-1)^4. - R. J. Mathar, Oct 03 2011

A081350 First column in maze array of natural numbers A081349.

Original entry on oeis.org

1, 2, 3, 4, 15, 16, 35, 36, 63, 64, 99, 100, 143, 144, 195, 196, 255, 256, 323, 324, 399, 400, 483, 484, 575, 576, 675, 676, 783, 784, 899, 900, 1023, 1024, 1155, 1156, 1295, 1296, 1443, 1444, 1599, 1600, 1763, 1764, 1935, 1936, 2115, 2116, 2303, 2304, 2499

Offset: 0

Paul Barry, Mar 19 2003

Interleaves A000466 with even squares A016742.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A081351, A081352.

a(0)=1, a(1)=2, a(n)=(n-1)*(n+(-1)^n), n>1.

G.f.: -(2*x^6-10*x^4+x^3+x^2-x-1)/((1-x)^3*(1+x)^2). [Colin Barker, Sep 03 2012]

A160466 Row sums of the Eta triangle A160464.

Original entry on oeis.org

-1, -9, -87, -2925, -75870, -2811375, -141027075, -18407924325, -1516052821500, -153801543183750, -18845978136851250, -2744283682352086875, -468435979952504313750, -92643070481933918821875

Offset: 2

Johannes W. Meijer, May 24 2009

It is conjectured that the row sums of the Eta triangle depend on five different sequences.

Two Maple algorithms are given. The first one gives the row sums according to the Eta triangle A160464 and the second one gives the row sums according to our conjecture.

A160464 is the Eta triangle.

Row sum factors A119951, A000466, A043529, A045896 and A160467.

nmax:=15; c(2) := -1/3: for n from 3 to nmax do c(n):=(2*n-2)*c(n-1)/(2*n-1)-1/ ((n-1)*(2*n-1)) end do: for n from 2 to nmax do GCS(n-1) := ln(1/(2^(-(2*(n-1)-1-floor(ln(n-1)/ ln(2))))))/ln(2); p(n):=2^(-GCS(n-1))*(2*n-1)!; ETA(n, 1) := p(n)*c(n) end do: mmax:=nmax: for m from 2 to mmax do ETA(2, m) := 0 end do: for n from 3 to nmax do for m from 2 to mmax do q(n) := (1+(-1)^(n-3)*(floor(ln(n-1)/ln(2)) - floor(ln(n-2)/ln(2)))): ETA(n, m) := q(n)*(-ETA(n-1, m-1)+(n-1)^2*ETA(n-1, m)) end do end do: for n from 2 to nmax do s1(n):=0: for m from 1 to n-1 do s1(n) := s1(n) + ETA(n, m) end do end do: seq(s1(n), n=2..nmax);
# End first program.
nmax:=nmax; A160467 := proc(n): denom(4*(4^n-1)*bernoulli(2*n)/n) end: A043529 := proc(n): ceil(frac(log[2](n+1))+1) end proc: A000466 := proc(n): 4*n^2-1 end proc: A045896 := proc(n): denom((n)/((n+1)*(n+2))) end proc: A119951 := proc(n) : numer(sum(((2*k1)!/(k1!*(k1+1)!))/2^(2*(k1-1)), k1=1..n)) end proc: for n from 1 to nmax do SF(2*n+1):= A000466(n)/A043529(n-1); SF(2*n+2) := A045896(n-1)/A160467(n+1) end do: FF(2):=1: for n from 3 to nmax do FF(n) := SF(n) * FF(n-1) end do: for n from 2 to nmax do s2(n):= (-1)*A119951(n-1)*FF(n) end do: seq(s2(n), n=2..nmax);
# End second program.

Rowsums(n) = (-1) * A119951(n-1) * FF(n) for n >= 2.
FF(n) = SF(n) * FF(n-1) for n >= 3 with FF(2) =1.
SF(2*n) = A045896(n-2) / A160467(n) for n >= 2.
SF(2*n+1) = A000466(n) / A043529(n-1) for n >= 1.

A028874 Primes of form k^2 - 3.

Original entry on oeis.org

13, 61, 97, 193, 397, 673, 1021, 1153, 1597, 1933, 2113, 3361, 4093, 4621, 6397, 7393, 7741, 8461, 9601, 12097, 12541, 13921, 15373, 16381, 18493, 19597, 20161, 21313, 26893, 29581, 36097, 37633, 40801, 42433, 43261, 47521, 48397

Offset: 1

Patrick De Geest

nonn

Also primes equal to the product of two consecutive odd numbers (A000466) minus 2. - Giovanni Teofilatto, Feb 11 2010

All terms are of the form 6m + 1. - Zak Seidov, May 01 2014

			61 is prime and equal to 8^2 - 3, so it is in the sequence.
67 is prime but it's 8^2 + 3 = 9^2 - 14, so it is not in the sequence.
9^2 - 3 = 78 but it's composite, so it's not in the sequence either.

Nathaniel Johnston, Table of n, a(n) for n = 1..10000
Patrick De Geest, Palindromic Quasipronics of the form n(n+x).
R. J. Mathar, Solutions to the exponential Diophantine 1 + p_1^x + p_2^y + p_3^z = w^2 for distinct primes p_1, p_2, p_3, 2022.
Eric Weisstein's World of Mathematics, Near-Square Prime.

Cf. A002476 (Primes of form 6m + 1), A028871, A028872, A028873.

Primes terms in A082109. Subsequence of A068228. - Klaus Purath, Jan 09 2023

[a: n in [2..300] | IsPrime(a) where a is n^2-3 ]; // Vincenzo Librandi, Nov 08 2014

Select[Range[2, 250]^2 - 3, PrimeQ] (* Harvey P. Dale, Aug 07 2013 *)
Select[Table[n^2 - 3, {n, 2, 300}], PrimeQ] (* Vincenzo Librandi, Nov 08 2014 *)

select(isprime, vector(100,n,n^2-3)) \\ Charles R Greathouse IV, Nov 19 2014

A028872 INTERSECT A000040. - Klaus Purath, Dec 07 2020

a(n) = A028873(n)^2 - 3. - Amiram Eldar, Mar 01 2025

cp's OEIS Frontend

A066830 a(n) = lcm(n+1, n-1).

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A214297 a(0)=-1, a(1)=0, a(2)=-3; thereafter a(n+2) - 2*a(n+1) + a(n) has period 4: repeat -4, 8, -4, 2.

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A225948 a(0) = -1; for n>0, a(n) = numerator(1/4 - 4/n^2).

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A242850 32*n^5 - 32*n^3 + 6*n.

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A242851 64*n^6 - 80*n^4 + 24*n^2 - 1.

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A278310 Numbers m such that T(m) + 3*T(m+1) is a square, where T = A000217.

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A069074 a(n) = (2*n+2)*(2*n+3)*(2*n+4) = 24*A000330(n+1).

A242850 32n^5 - 32n^3 + 6*n.

A242851 64n^6 - 80n^4 + 24*n^2 - 1.

A069074 a(n) = (2n+2)(2n+3)(2n+4) = 24A000330(n+1).