A079414 -id:A079414 - cp's OEIS frontend

A144139 Chebyshev polynomial of the second kind U(4,n).

1, 5, 209, 1189, 3905, 9701, 20305, 37829, 64769, 104005, 158801, 232805, 330049, 454949, 612305, 807301, 1045505, 1332869, 1675729, 2080805, 2555201, 3106405, 3742289, 4471109, 5301505, 6242501, 7303505, 8494309, 9825089, 11306405

Offset: 0

Vladimir Joseph Stephan Orlovsky, Sep 11 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A057722, A079414, A165900, A193250.

[16*n^4-12*n^2+1: n in [0..40]]; // Vincenzo Librandi, May 29 2014

lst={}; Do[AppendTo[lst, ChebyshevU[4, n]], {n, 0, 9^2}]; lst
CoefficientList[Series[(1 + 194 x^2 + 184 x^3 + 5 x^4)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, May 29 2014 *)

G.f.: (1 + 194*x^2 + 184*x^3 + 5*x^4)/(1 - x)^5. - Vincenzo Librandi, May 29 2014
a(n) = 16*n^4-12*n^2+1 = (4*n^2-2*n-1)*(4*n^2+2*n-1). - Vincenzo Librandi, May 29 2014
From Klaus Purath, Sep 08 2022: (Start)
a(n) = A165900(2*n)*A165900(2*n+1).
a(n) = A057722(2*n).
a(n) = 4*(Sum_{i=1..n} A193250(i)) + 1 = 4*A079414(n) + 1.
(End)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Wesley Ivan Hurt, Aug 04 2025

Changed offset from 1 to 0 by Vincenzo Librandi, May 29 2014

A130284 Integers j > 0 such that (2j+1)^2(m^2-1) + 1 is a square for some integer m > 1.

Original entry on oeis.org

7, 17, 31, 49, 71, 97, 104, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 594, 647, 721, 799, 881, 967, 1057, 1151, 1249, 1351, 1455, 1457, 1567, 1681, 1799, 1921, 1952, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 3041, 3199, 3361, 3527, 3697, 3871, 4049

Offset: 1

M. F. Hasler, May 24 2007, May 29 2007

nonn

All terms > 4 in A130283 are odd squares, but not all odd squares are in that sequence: This sequence here gives the exceptions as (2a(n)+1)^2. The sequence consists mainly of the subsequences: (1) A056220(k) = 2k^2-1 with k>1: {7,17,31,49,...}, for which m=k gives (1+2*A056220(k))^2(k^2-1)+1 = k^2(4k^2-3)^2; (2) 2*A079414(k) = 2k^2(4k^2-3) with k>1: {104,594,1952,4850,...}, for which m=k gives (1+4*A079414(k))^2(k^2-1)+1 = k^2(16k^4-20k^2+5)^2. A third subsequence starts {1455,20195,...}; up to 20195, all terms are in one of these subsequences.

			Up to k=17, a(k)=P[1](k+1) with P[1] = 2x^2 - 1, A130280(a(k)) = k+1.
a(18) = P[2](2) < P[1](19) with P[2] = 2x^2*(4x^2 - 3), A130280(a(18)) = 2.
a(106) = P[1](100) < a(107) = P[3](3) < a(108) = P[4](2) < a(109) = P[1](101).

Cf. A084702, A130280, A130284, A130288.

Cf. A130280, A130283, A130281.

r[n_] := Reduce[m>1 && k>1 && (2n+1)^2*(m^2-1)+1 == k^2, {m, k}, Integers];
Reap[For[n=1, n <= 5000, n++, If[r[n] =!= False, Print[n]; Sow[n]]]][[2,1]] (* Jean-François Alcover, May 12 2017 *)

A130284( LIM=9999, START=1 )={ local(N); for( n=START, LIM, N=(2*n+1)^2; for( m=2, sqrtint(n>>1+1), if(!issquare( N*(m^2-1)+1 ), next); print1(n", "); next(2))) }

{Q(k,x=x)=if(m>0,(4*x^2-2)*Q(k-1,x)-Q(k-2,x),1)} {P(k,x=x)=if(type(x=(x^2*Q(k,x)^2-1)/(x^2-1))!="t_POL",sqrtint(x)\2,((-1)^k*Pol(sqrt(x))-1)/2)}

A130284 = { P[k](m) ; k=1,2,3,..., m=2,3,4,... } where P[k] = (sqrt((X^2 Q[k]^2 - 1)/(X^2 - 1))-1)/2 and Q[0] = Q[-1] = 1, Q[k+1] = (4X^2 -2)*Q[k] - Q[k-1]. Furthermore, (2P[k](m)+1)^2 (m^2 - 1)+1 = m^2 Q[k](m)^2, thus A130280(P[k](m)) <= m. So far, no case is known where we have strict inequality.

A193250 Small rhombicuboctahedron with faces of centered polygons.

Original entry on oeis.org

1, 51, 245, 679, 1449, 2651, 4381, 6735, 9809, 13699, 18501, 24311, 31225, 39339, 48749, 59551, 71841, 85715, 101269, 118599, 137801, 158971, 182205, 207599, 235249, 265251, 297701, 332695, 370329, 410699, 453901, 500031, 549185, 601459, 656949, 715751

Offset: 1

Craig Ferguson, Jul 19 2011

The sequence starts with a central dot and expands outward with (n-1) centered polygonal pyramids producing a small rhombicuboctahedron. Each iteration requires the addition of (n-2) edges and (n-1) vertices to complete the centered polygon of each face. [centered triangles (A005448) and centered squares (A001844)]

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
OEIS, (Centered_polygons) pyramidal numbers
Wikipedia, Tetrahedral number
Wikipedia, Triangular number
Wikipedia, Centered polygonal number
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A079414 (partial sums).

Cf. A005448, A001844.

copy and paste the following formula   =16 *ROW()^3-24 *ROW()^2+10*ROW()-1 fill down to desired size.

[16*n^3-24*n^2+10*n-1: n in [1..50]]; // Vincenzo Librandi, Aug 30 2011

Table[16n^3-24n^2+10n-1,{n,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{1,51,245,679},40] (* Harvey P. Dale, Aug 27 2011 *)

vector(40, n, 16*n^3-24*n^2+10*n-1) \\ G. C. Greubel, Nov 10 2018

a(n) = 16*n^3 - 24*n^2 + 10*n - 1.
G.f.: x*(1+x)*(x^2 + 46*x + 1) / (x-1)^4. - R. J. Mathar, Aug 26 2011
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4); a(0)=1, a(1)=51, a(2)=245, a(3)=679. - Harvey P. Dale, Aug 27 2011
E.g.f.: 1 + (-1 + 2*x + 24*x^2 + 16*x^3)*exp(x). - G. C. Greubel, Nov 10 2018

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A144139 Chebyshev polynomial of the second kind U(4,n).

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A130284 Integers j > 0 such that (2j+1)^2(m^2-1) + 1 is a square for some integer m > 1.

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A193250 Small rhombicuboctahedron with faces of centered polygons.

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