A162997 -id:A162997 - cp's OEIS frontend

A028387 a(n) = n + (n+1)^2.

1, 5, 11, 19, 29, 41, 55, 71, 89, 109, 131, 155, 181, 209, 239, 271, 305, 341, 379, 419, 461, 505, 551, 599, 649, 701, 755, 811, 869, 929, 991, 1055, 1121, 1189, 1259, 1331, 1405, 1481, 1559, 1639, 1721, 1805, 1891, 1979, 2069, 2161, 2255, 2351, 2449, 2549, 2651

Offset: 0

Patrick De Geest

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005
Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006
A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007
Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007
Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008
a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009
sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009
When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010
a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010
The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011
Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012
Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012
Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013
a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013
a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014
A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015
An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015
Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015
Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017
The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017
From Klaus Purath, Mar 18 2019: (Start)
Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with
x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.
But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)
a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019
a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021
Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021
(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022
The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

			From _Ilya Gutkovskiy_, Apr 13 2016: (Start)
Illustration of initial terms:
                                        o               o
                        o           o   o o           o o
            o       o   o o       o o   o o o       o o o
    o   o   o o   o o   o o o   o o o   o o o o   o o o o
o   o o o   o o o o o   o o o o o o o   o o o o o o o o o
n=0  n=1       n=2           n=3               n=4
(End)
From _Klaus Purath_, Mar 18 2019: (Start)
Examples:
a(0) = 1: 1^1-0*1 = 1, 0+1 = 1 (Fibonacci A000045).
a(1) = 5: 3^2-1*4 = 5, 1+3 = 4 (Lucas A000032).
a(2) = 11: 4^2-1*5 = 11, 1+4 = 5 (A000285); 5^2-2*7 = 11, 2+5 = 7 (A001060).
a(3) = 19: 5^2-1*6 = 19, 1+5 = 6 (A022095); 7^2-3*10 = 19, 3+7 = 10 (A022120).
a(4) = 29: 6^2-1*7 = 29, 1+6 = 7 (A022096); 9^2-4*13 = 29, 4+9 = 13 (A022130).
a(11)/5 = 31: 7^2-2*9 = 31, 2+7 = 9 (A022113); 8^2-3*11 = 31, 3+8 = 11 (A022121).
a(24)/11 = 59: 9^2-2*11 = 59, 2+9 = 11 (A022114); 12^2-5*17 = 59, 5+12 = 17 (A022137).
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Patrick De Geest, World!Of Numbers, Palindromes of the form n+(n+1)^2.
Adalbert Kerber, A matrix of combinatorial numbers related to the symmetric groups, Discrete Math., 21 (1978), 319-321. [Annotated scanned copy]
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Nandini Nilakantan and Anurag Singh, Homotopy type of neighborhood complexes of Kneser graphs, KG_{2,k}, Proceeding-Mathematical Sciences, 128, Article number: 53(2018).
Yanni Pei and Jiang Zeng, Counting signed derangements with right-to-left minima and excedances, arXiv:2206.11236 [math.CO], 2022.
Popular Computing (Calabasas, CA), The CSR Function, Vol. 4 (No. 34, Jan 1976), pages PC34-10 to PC34-11. Annotated and scanned copy.
Zdzislaw Skupień and Andrzej Żak, Pair-sums packing and rainbow cliques, in Topics In Graph Theory, A tribute to A. A. and T. E. Zykovs on the occasion of A. A. Zykov's 90th birthday, ed. R. Tyshkevich, Univ. Illinois, 2013, pages 131-144, (in English and Russian).
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Complement of A028392. Third column of array A094954.
Cf. A000217, A002522, A062392, A062786, A127701, A135223, A143596, A052905, A162997, A062938 (squares of this sequence).
A110331 and A165900 are signed versions.
Cf. A002327 (primes), A094210.
Frobenius number for k successive numbers: this sequence (k=2), A079326 (k=3), A138984 (k=4), A138985 (k=5), A138986 (k=6), A138987 (k=7), A138988 (k=8).
Cf. also A135223, A176271, A214604, A105728, A005408, A002378, A084990, A253607, A260910, A089270.

a028387 n = n + (n + 1) ^ 2  -- Reinhard Zumkeller, Jul 17 2014

[n + (n+1)^2: n in [0..60]]; // Vincenzo Librandi, Apr 26 2011

FoldList[## + 2 &, 1, 2 Range@ 45] (* Robert G. Wilson v, Feb 02 2011 *)
Table[n + (n + 1)^2, {n, 0, 100}] (* Vincenzo Librandi, Oct 17 2012 *)
Table[ FrobeniusNumber[{n, n + 1}], {n, 2, 30}] (* Zak Seidov, Jan 14 2015 *)

a(n)=n^2+3*n+1 \\ Charles R Greathouse IV, Jun 10 2011

def a(n): return (n**2+3*n+1) # Torlach Rush, May 07 2024

[n+(n+1)^2 for n in range(0,48)] # Zerinvary Lajos, Jul 03 2008

a(n) = sqrt(A062938(n)). - Floor van Lamoen, Oct 08 2001
a(0) = 1, a(1) = 5, a(n) = (n+1)*a(n-1) - (n+2)*a(n-2) for n > 1. - Gerald McGarvey, Sep 24 2004
a(n) = A105728(n+2, n+1). - Reinhard Zumkeller, Apr 18 2005
a(n) = A109128(n+2, 2). - Reinhard Zumkeller, Jun 20 2005
a(n) = 2*T(n+1) - 1, where T(n) = A000217(n). - Gary W. Adamson, Aug 15 2007
a(n) = A005408(n) + A002378(n); A084990(n+1) = Sum_{k=0..n} a(k). - Reinhard Zumkeller, Aug 20 2007
Binomial transform of [1, 4, 2, 0, 0, 0, ...] = (1, 5, 11, 19, ...). - Gary W. Adamson, Sep 20 2007
G.f.: (1+2*x-x^2)/(1-x)^3. a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - R. J. Mathar, Jul 11 2009
a(n) = (n + 2 + 1/phi) * (n + 2 - phi); where phi = 1.618033989... Example: a(3) = 19 = (5 + .6180339...) * (3.381966...). Cf. next to leftmost column in A162997 array. - Gary W. Adamson, Jul 23 2009
a(n) = a(n-1) + 2*(n+1), with n > 0, a(0) = 1. - Vincenzo Librandi, Nov 18 2010
For k < n, a(n) = (k+1)*a(n-k) - k*a(n-k-1) + k*(k+1); e.g., a(5) = 41 = 4*11 - 3*5 + 3*4. - Charlie Marion, Jan 13 2011
a(n) = lower right term in M^2, M = the 2 X 2 matrix [1, n; 1, (n+1)]. - Gary W. Adamson, Jun 29 2011
G.f.: (x^2-2*x-1)/(x-1)^3 = G(0) where G(k) = 1 + x*(k+1)*(k+4)/(1 - 1/(1 + (k+1)*(k+4)/G(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Oct 16 2012
Sum_{n>0} 1/a(n) = 1 + Pi*tan(sqrt(5)*Pi/2)/sqrt(5). - Enrique Pérez Herrero, Oct 11 2013
E.g.f.: exp(x) (1+4*x+x^2). - Tom Copeland, Dec 02 2013
a(n) = A005408(A000217(n)). - Tony Foster III, May 31 2016
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2).
Product_{n>=1} (1 - 1/a(n)) = -Pi*sec(sqrt(5)*Pi/2)/6. (End)
a(5*n+1)/5 = A062786(n+1). - Torlach Rush, Jun 05 2024

Minor edits by N. J. A. Sloane, Jul 04 2010, following suggestions from the Sequence Fans Mailing List

A123972 a(n) = n^3 - n^2 - 2*n + 1.

Original entry on oeis.org

1, -1, 1, 13, 41, 91, 169, 281, 433, 631, 881, 1189, 1561, 2003, 2521, 3121, 3809, 4591, 5473, 6461, 7561, 8779, 10121, 11593, 13201, 14951, 16849, 18901, 21113, 23491, 26041, 28769, 31681, 34783, 38081, 41581, 45289, 49211, 53353, 57721, 62321, 67159, 72241

Offset: 0

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

a(n) is the determinant of the 3 X 3 matrix {{n,-1,0 },{-1,n,-1},{0,-1,n-1}}.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A162997.

I:=[1, -1, 1, 13]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jun 27 2012

with(linalg): M:=n->matrix(3,3,[n,-1,0,-1,n,-1,0,-1,n-1]): seq(det(M(n)),n=0..42);

CoefficientList[Series[(1-5*x+11*x^2-x^3)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 27 2012 *)

a(n)=n^3-n^2-2*n+1 \\ Charles R Greathouse IV, Jun 30 2011

a(n) = (n + 2*cos(2*Pi/7)) * (n + 2*cos(4*Pi/7)) * (n + 2*cos(6*Pi/7)). Cf. 3rd column from the left in the array of A162997. - Gary W. Adamson, Jul 23 2009
a(n) equals the lower right term in M^3, M is the 2 X 2 matrix {{1, n-2}, {1, n-1}}. - Gary W. Adamson, Jun 29 2011
Starting (1, 13, 41, ...) = the binomial transform of (1, 12, 16, 6). - Gary W. Adamson, Jun 29 2011
G.f.: (1 - 5*x + 11*x^2 - x^3)/(1-x)^4. - Colin Barker, Jan 29 2012
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 27 2012

Edited by N. J. A. Sloane, Nov 01 2006 and Nov 24 2006

A192398 a(n) = n^4 + 3n^3 - 3n.

Original entry on oeis.org

1, 34, 153, 436, 985, 1926, 3409, 5608, 8721, 12970, 18601, 25884, 35113, 46606, 60705, 77776, 98209, 122418, 150841, 183940, 222201, 266134, 316273, 373176, 437425, 509626, 590409, 680428, 780361, 890910, 1012801, 1146784, 1293633, 1454146, 1629145, 1819476

Offset: 1

Gary W. Adamson, Jun 30 2011

Related to the 9-gon (nonagon). Following Steinbach's strategy re: "Diagonal Product Formulas" and applied to the 9-gon (nonagon), we extract the constants (a, b, c, d) as e-vals of the 4 X 4 tridiagaonal matrix with (1's in the super and subdiagonals), (1,2,2,2), and the rest zeros. The charpoly of this matrix is row 4 of A054142, a Morgan-Voyce polynomial: x^4 - 7*x^3 + 15*x^2 + 10*x - 1 = 0. Following Steinbach's procedure, let the matrix = M; then find the first four rows of M^n * [1,0,0,0,...] getting (1; 1,1; 2,3,1; 5,9,5,1). Using the SIMULT operation, we equate each of these rows to successive powers of the constant c (largest e-val of matrix M), 3.5320888...as follows: SIMULT: [1,0,0,0] = 1; [1,1,0,0] = c; [2,3,1,0] = c^2; [5,9,5,1] = c^3. Solving, we obtain the four distinct diagonals of the 9-gon (nonagon) with edge = 1: (1, 2.5320888,..., 2.879385,..., and 1.879385,...).
The sequence is column 3 in the array of A162997.
Analogous sequences using the matrix M^k generator -M^2 generates A028387: (1, 5, 11, 19, 29, 41,...); M^3 generates A123972: (1, 13, 41, 91, 169,...).

			a(5) = 5^4 + 3*5^3 - 3*5 = (625 + 375 - 15) = 985.
a(4) = 436 = (1, 3, 3, 1) dot (1, 33, 86, 78) = (1 + 99 + 258 + 78) = 436.
a(7) = 3409 = lower right term in M^4, M = {{1,6}{1,7}}.
a(4) = 436 = (3 + a) * (3 + b) * (3 + c) * (3 + d), = (5.347296...) * (3.120614...) * (4) * (6.532088...) = 436.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Peter Steinbach, Golden Fields: A Case for the Heptagon, Mathematics Magazine, Vol. 70, No. 1, Feb. 1997.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A028387, A054142, A123972, A162997.

[n^4 +3*n^3 -3*n: n in [1..45]]; // Vincenzo Librandi, Nov 25 2011

A192398:=n->n^4+3*n^3-3*n: seq(A192398(n), n=1..40); # Wesley Ivan Hurt, Sep 12 2014

LinearRecurrence[{5,-10,10,-5,1},{1,34,153,436,985},50] (* Vincenzo Librandi, Nov 25 2011 *)
Table[n^4+3n^3-3n,{n,40}] (* Harvey P. Dale, Feb 21 2023 *)

a(n) = n^4 +3*n^3 -3*n \\ Charles R Greathouse IV, Jun 30 2011

[n*(n^3+3*n^2-3) for n in range(1,51)] # G. C. Greubel, Jul 11 2023

G.f.: (1 +29*x -7*x^2 +x^3) / (1-x)^5. - R. J. Mathar, Jul 08 2011
a(n) = binomial transform of [1, 33, 86, 78, 24, 0, 0, 0,...].
a(n) = lower right term in the 2 X 2 matrix M^4, M = {{1,n-1}, {1,n}}.
a(n) = ((n-1) + a) * ((n-1) + b) * ((n-1) + c) * ((n-1) + d), where a, b, c, d, = {k=1,2,3,4} 4*cos^2 (2*Pi*k)/9.
E.g.f.: x*(1 + 16*x + 9*x^2 + x^3)*exp(x). - G. C. Greubel, Jul 11 2023

A121872 Triangle T(n, k) = (kChebyshevU(n, (k+2)/2) + 2ChebyshevT(n+1, (k+2)/2))/2.

Original entry on oeis.org

5, 13, 41, 34, 153, 436, 89, 571, 2089, 5741, 233, 2131, 10009, 33461, 90481, 610, 7953, 47956, 195025, 620166, 1663585, 1597, 29681, 229771, 1136689, 4250681, 13097377, 34988311, 4181, 110771, 1100899, 6625109, 29134601, 103115431, 310957991, 828931049

Offset: 1

Roger L. Bagula and Gary W. Adamson, Sep 09 2006

			Triangle begins as:
    5;
   13,   41;
   34,  153,   436;
   89,  571,  2089,  5741;
  233, 2131, 10009, 33461, 90481;

G. C. Greubel, Rows n = 1..100 of triangle, flattened

Cf. A094954, A162997.

Cf. A053117, A053120.

T:= func< n,k | ( k*Sinh((n+1)*Argcosh((k+2)/2))/Sinh(Argcosh((k+2)/2)) + 2*Cosh((n+1)*Argcosh((k+2)/2)) )/2 >;
[Round(T(n,k)): k in [1..n], n in [1..10]]; // G. C. Greubel, Oct 08 2019

seq(seq(simplify(( k*ChebyshevU(n, (k+2)/2) + 2*ChebyshevT(n+1, (k+2)/2) )/2), k=1..n), n=1..10); # G. C. Greubel, Oct 09 2019

f[k_]:= Sqrt[k*(k+4)]; T[n_, m_]:= T[n, m]= FullSimplify[((m+f[m])*(m+2 - f[m])^(n+2) - (m-f[m])*(m+2 + f[m])^(n+2))/(2^(n+3)*f[m])]; Table[T[n, m], {n,10}, {m,n}]//Flatten (* modified by G. C. Greubel, Oct 08 2019 *)
T[n_, k_]:= T[n, k]= (k*ChebyshevU[n, (k+2)/2] + 2*ChebyshevT[n+1, (k+ 2)/2])/2; Table[T[n, k], {n,10}, {k,n}]/Flatten (* G. C. Greubel, Oct 08 2019 *)

T(n,k)= ( k*sin((n+1)*acos((k+2)/2))/sin(acos((k+2)/2)) + 2*cos((n+1)*acos((k+2)/2)) )/2;
for(n=1,10, for(k=1,n, print1(round(T(n,k)), ", "))) \\ G. C. Greubel, Oct 08 2019

[[( k*chebyshev_U(n,(k+2)/2) + 2*chebyshev_T(n+1, (k+2)/2) )/2 for k in (1..n)] for n in (1..10)] # G. C. Greubel, Oct 08 2019

T(n, m) = ((m+f(m))*(m+2 - f(m))^(n+2) - (m-f(m))*(m+2 + f(m))^(n+2))/( 2^(n+3)*f(m)), where f(m) = sqrt(m*(m+4)).
From G. C. Greubel, Oct 08 2019: (Start)
T(n, k) = (k*ChebyshevU(n, (k+2)/2) + 2*ChebyshevT(n+1, (k+2)/2))/2;
T(n, k) = (k*Fibonacci(n+2, m+2, -1) + Lucas(n+2, m+2, -1))/2, where Fibonacci(n, x, y) and Lucas(n, x, y) are the bi-variate Fibonacci an Lucas polynomials, respectively. (End)

Major edit and new name, G. C. Greubel, Oct 08 2019

cp's OEIS Frontend

A162998 Row sums of triangle A162997.

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A028387 a(n) = n + (n+1)^2.

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A123972 a(n) = n^3 - n^2 - 2*n + 1.

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A192398 a(n) = n^4 + 3*n^3 - 3*n.

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A121872 Triangle T(n, k) = (k*ChebyshevU(n, (k+2)/2) + 2*ChebyshevT(n+1, (k+2)/2))/2.

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A218220 Array a(n,m) read by antidiagonals where a(0,m)=a(1,m)=1 and a(n,m) = m*a(n-1,m)-a(n-2,m) for n>=2.

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A192398 a(n) = n^4 + 3n^3 - 3n.

A121872 Triangle T(n, k) = (kChebyshevU(n, (k+2)/2) + 2ChebyshevT(n+1, (k+2)/2))/2.