A110331 -id:A110331 - 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

A007054 Super ballot numbers: 6(2n)!/(n!(n+2)!).

Original entry on oeis.org

3, 2, 3, 6, 14, 36, 99, 286, 858, 2652, 8398, 27132, 89148, 297160, 1002915, 3421710, 11785890, 40940460, 143291610, 504932340, 1790214660, 6382504440, 22870640910, 82334307276, 297670187844, 1080432533656, 3935861372604, 14386251913656, 52749590350072

Offset: 0

N. J. A. Sloane, Mira Bernstein, Ira M. Gessel

Hankel transform is 2n+3. The Hankel transform of a(n+1) is n+2. The sequence a(n)-2*0^n has Hankel transform A110331(n). - Paul Barry, Jul 20 2008

Number of pairs of Dyck paths of total length 2*n with heights differing by at most 1 (Gessel/Xin, p. 2). - Joerg Arndt, Sep 01 2012

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..200
Emily Allen and Irina Gheorghiciuc, A Weighted Interpretation for the Super Catalan Numbers, J. Int. Seq. 17 (2014) # 14.10.7.
David Callan, A combinatorial interpretation for a super-Catalan recurrence, arXiv:math/0408117 [math.CO], 2004.
David Callan, A Combinatorial Interpretation for a Super-Catalan Recurrence, Journal of Integer Sequences, Vol. 8 (2005), Article 05.1.8.
David Callan, A variant of Touchard's Catalan number identity, arXiv preprint arXiv:1204.5704 [math.CO], 2012. - From _N. J. A. Sloane_, Oct 10 2012
Ira M. Gessel, Letter to N. J. A. Sloane, Jul. 1992
Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194
Ira M. Gessel, Rational Functions With Nonnegative Integer Coefficients, 50th Séminaire Lotharingien de Combinatoire, 2003.
Ira M. Gessel and Guoce Xin, A Combinatorial Interpretation of the Numbers 6(2n)!/n!(n+2)!, Journal of Integer Sequences, Vol. 8 (2005), Article 05.2.3.
Ira M. Gessel and Guoce Xin, A Combinatorial Interpretation of The Numbers 6(2n)! /n! (n+2)!, arXiv:math/0401300v2 [math.CO], 2004.
Nicholas Pippenger and Kristin Schleich, Topological characteristics of random triangulated surfaces (section 7), Random Structures Algorithms 28 (2006) 247-288; arXiv:gr-qc/0306049, 2003.
G. Schaeffer, A combinatorial interpretation of super-Catalan numbers of order two, (2001).
Paveł Szabłowski, Beta distributions whose moment sequences are related to integer sequences listed in the OEIS, Contrib. Disc. Math. (2024) Vol. 19, No. 4, 85-109. See p. 96.

Cf. A002421, A007272, A091712, A000257, A001622.

[6*Factorial(2*n)/(Factorial(n)*Factorial(n+2)): n in [0..30]]; // Vincenzo Librandi, Aug 20 2011

seq(3*(2*n)!/(n!)^2/binomial(n+2,n), n=0..22); # Zerinvary Lajos, Jun 28 2007
A007054 := n -> 6*4^n*GAMMA(1/2+n)/(sqrt(Pi)*GAMMA(3+n)):
seq(A007054(n),n=0..28); # Peter Luschny, Dec 14 2015

Table[6(2n)!/(n!(n+2)!),{n,0,30}] (* or *) CoefficientList[Series[ (-1+Sqrt[1-4*x]+(6-4*Sqrt[1-4*x])*x)/(2*x^2),{x,0,30}],x] (* Harvey P. Dale, Oct 05 2011 *)

a(n)=6*(2*n)!/(n!*(n+2)!); /* Joerg Arndt, Sep 01 2012 */

def A007054(n): return (-4)^(2 + n)*binomial(3/2, 2 + n)/2
print([A007054(n) for n in range(29)])  # Peter Luschny, Nov 04 2021

G.f.: c(x)*(4-c(x)), where c(x) = g.f. for Catalan numbers A000108; Convolution of Catalan numbers with negative Catalan numbers but -C(0)=-1 replaced by 3. - Wolfdieter Lang
E.g.f. in Maple notation: exp(2*x)*(4*x*(BesselI(0, 2*x)-BesselI(1, 2*x))-BesselI(1, 2*x))/x. Integral representation as n-th moment of a positive function on [0, 4], in Maple notation: a(n)=int(x^n*(4-x)^(3/2)/x^(1/2), x=0..4)/(2*Pi), n=0, 1, ... This representation is unique. - Karol A. Penson, Oct 10 2001
E.g.f.: Sum_{n>=0} a(n)*x^(2*n) = 3*BesselI(2, 2x).
a(n) = A000108(n)*6/(n+2). - Philippe Deléham, Oct 30 2007
a(n+1) = 2*(A000108(n+2) - A000108(n+1))/(n+1). - Paul Barry, Jul 20 2008
G.f.: ((6-4*sqrt(1-4*x))*x+sqrt(1-4*x)-1)/(2*x^2) - Harvey P. Dale, Oct 05 2011
a(n) = 4*A000108(n) - A000108(n+1) (Gessel/Xin, p. 2). - Joerg Arndt, Sep 01 2012
D-finite with recurrence (n+2)*a(n) +2*(-2*n+1)*a(n-1)=0. - R. J. Mathar, Dec 03 2012
G.f.: 1/(x^2*G(0)) + 3/x - (1/2)/x^2, where G(k) = 1 + 1/(1 - 2*x*(2*k+3)/(2*x*(2*k+3) + (k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 06 2013
G.f.: 3/x - 1/(2*x^2) + G(0)/(4*x^2), where G(k) = 1 + 1/(1 - 2*x*(2*k-3)/(2*x*(2*k-3) + (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 18 2013
0 = a(n)*(+16*a(n+1) - 14*a(n+2)) + a(n+1)*(+6*a(n+1) + a(n+2)) for all n in Z. - Michael Somos, Sep 18 2014
A002421(n+2) = 2*a(n) for all n in Z. - Michael Somos, Sep 18 2014
a(n) = 3*(2*n)!*[x^(2*n)]hypergeometric([],[3],x^2). - Peter Luschny, Feb 01 2015
a(n) = 6*4^n*Gamma(1/2+n)/(sqrt(Pi)*Gamma(3+n)). - Peter Luschny, Dec 14 2015
a(n) = (-4)^(2 + n)*binomial(3/2, 2 + n)/2. - Peter Luschny, Nov 04 2021
From Amiram Eldar, May 16 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + 20*Pi/(81*sqrt(3)).
Sum_{n>=0} (-1)^n/a(n) = 3/25 - 8*log(phi)/(25*sqrt(5)), where phi is the golden ratio (A001622). (End)
a(n-1) = 3*A000984(n)/((2*n-1)*(n+1)). - R. J. Mathar, Jul 12 2024

Corrected and extended by Vincenzo Librandi, Aug 20 2011

A165900 a(n) = n^2 - n - 1.

Original entry on oeis.org

-1, -1, 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

Offset: 0

Philippe Deléham, Sep 29 2009

Previous name was: Values of Fibonacci polynomial n^2 - n - 1.
Shifted version of the array denoted rB(0,2) in A132382, whose e.g.f. is exp(x)(1-x)^2. Taking the derivative gives the e.g.f. of this sequence. - Tom Copeland, Dec 02 2013
The Fibonacci numbers are generated by the series x/(1 - x - x^2). - T. D. Noe, Dec 04 2013
Absolute value of expression f(k)*f(k+1) - f(k-1)*f(k+2) where f(1)=1, f(2)=n. Sign is alternately +1 and -1. - Carmine Suriano, Jan 28 2014 [Can anybody clarify what is meant here? - Joerg Arndt, Nov 24 2014]
Carmine's formula is a special case related to 4 consecutive terms of a Fibonacci sequence. A generalization of this formula is |a(n)| = |f(k+i)*f(k+j) - f(k)*f(k+i+j)|/F(i)*F(j), where f denotes a Fibonacci sequence with the initial values 1 and n, and F denotes the original Fibonacci sequence A000045. The same results can be obtained with the simpler formula |a(n)| = |f(k+1)^2 - f(k)^2 - f(k+1)*f(k)|. Everything said so far is also valid for Fibonacci sequences f with the initial values f(1) = n - 2, f(2) = 2*n - 3. - Klaus Purath, Jun 27 2022
a(n) is the total number of dollars won when using the Martingale method (bet $1, if win then continue to bet $1, if lose then double next bet) for n trials of a wager with exactly one loss, n-1 wins. For the case with exactly one win, n-1 losses, see A070313. - Max Winnick, Jun 28 2022
Numbers m such that 4*m+5 is a square b^2, where b = 2*n -1, for m = a(n). - Klaus Purath, Jul 23 2022

			G.f. = -1 - x + x^2 + 5*x^3 + 11*x^4 + 19*x^5 + 29*x^6 + 41*x^7 + ... - _Michael Somos_, Mar 23 2023

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Joseph J. Heed and Lucille Kelly, An interesting sequence of Fibonacci sequence generators, Fibonacci Quarterly, 13 (1975), pp. 29-30.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

A028387 and A110331 are very similar sequences.
Cf. A000045, A005843, A015614, A070313, A132382, A152015, A188652, A214803.
Other quadratics: A002061, A002378, A005563, A008865, A014206, A014209, A028552, A034856.

a165900 n = n * (n - 1) - 1  -- Reinhard Zumkeller, Jul 29 2012

Table[n^2 - n - 1, {n, 0, 50}] (* Ron Knott, Oct 27 2010 *)
LinearRecurrence[{3,-3,1},{-1,-1,1},60] (* Harvey P. Dale, Jul 05 2021 *)

a(n)=n^2-n-1 \\ Charles R Greathouse IV, Jan 12 2012

a(n+2) = (n+1)*a(n+1) - (n+2)*a(n).
G.f.: (x^2+2*x-1)/(1-x)^3.
E.g.f.: exp(x)*(x^2-1).
a(n) = - A188652(2*n) for n > 0. - Reinhard Zumkeller, Apr 13 2011
a(n) = A214803(A015614(n+1)) for n > 0. - Reinhard Zumkeller, Jul 29 2012
a(n+1) = a(n) + A005843(n) = A002378(n) - 1. - Ivan N. Ianakiev, Feb 18 2013
a(n+2) = A028387(n). - Michael B. Porter, Sep 26 2018
From Klaus Purath, Aug 25 2022: (Start)
a(2*n) = n*(a(n+1) - a(n-1)) -1.
a(2*n+1)  = (2*n+1)*(a(n+1) - a(n)) - 1.
a(n+2) = a(n) + 4*n + 2.
a(n) = A014206(n-1) - 3 = A002061(n-1) - 2.
a(n) = A028552(n-2) + 1 = A014209(n-2) + 2 = 2* A034856(n-2) + 3.
a(n) = A008865(n-1) + n = A005563(n-1) - n.
a(n) = A014209(n-3) + 2*n = A028387(n-1) - 2*n.
a(n) = A152015(n)/n, n != 0.
(a(n+k) - a(n-k))/(2*k) = 2*n-1, for any k.
(End)
For n > 1, 1/a(n) = Sum_{k>=1} F(k)/n^(k+1), where F(n) = A000045(n). - Diego Rattaggi, Nov 01 2022
a(n) = a(1-n) for all n in Z. - Michael Somos, Mar 23 2023
For n > 1, 1/a(n) = Sum_{k>=1} F(2k)/((n+1)^(k+1)), where F(2n) = A001906(n). - Diego Rattaggi, Jan 20 2025
From Amiram Eldar, May 11 2025: (Start)
Sum_{n>=1} 1/a(n) = tan(sqrt(5)*Pi/2)*Pi/sqrt(5).
Product_{n>=3} 1 - 1/a(n) = -sec(sqrt(5)*Pi/2)*Pi/6.
Product_{n>=2} 1 + 1/a(n) = -sec(sqrt(5)*Pi/2)*Pi. (End)

a(22) corrected by Reinhard Zumkeller, Apr 13 2011

Better name from Joerg Arndt, Oct 26 2024

A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, -1, -1, 1, -2, 1, 1, 1, -3, 5, -1, 0, 1, -4, 11, -13, 1, -1, 1, -5, 19, -41, 34, -1, 1, 1, -6, 29, -91, 153, -89, 1, 0, 1, -7, 41, -169, 436, -571, 233, -1, -1, 1, -8, 55, -281, 985, -2089, 2131, -610, 1, 1, 1, -9, 71, -433, 1926, -5741, 10009, -7953, 1597, -1, 0

Offset: 1

L. Edson Jeffery, Bob Selcoe and Andrew Hone, Feb 01 2018

This array is used to compute A269252: A269252(n) = least k such that |A(n,k)| is a prime, or -1 if no such k exists.
For detailed theory, see [Hone].
The array can be extended to k<0 with A(n, k) = -A(n, -k-1) for all k in Z. - Michael Somos, Jun 19 2023

			Array begins:
1   0  -1     1     0      -1       1         0        -1           1
1  -1   1    -1     1      -1       1        -1         1          -1
1  -2   5   -13    34     -89     233      -610      1597       -4181
1  -3  11   -41   153    -571    2131     -7953     29681     -110771
1  -4  19   -91   436   -2089   10009    -47956    229771    -1100899
1  -5  29  -169   985   -5741   33461   -195025   1136689    -6625109
1  -6  41  -281  1926  -13201   90481   -620166   4250681   -29134601
1  -7  55  -433  3409  -26839  211303  -1663585  13097377  -103115431
1  -8  71  -631  5608  -49841  442961  -3936808  34988311  -310957991
1  -9  89  -881  8721  -86329  854569  -8459361  83739041  -828931049

Robert Price, Table of n, a(n) for n = 1..5050
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.

Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A269251, A269252, A269253, A269254, A294099, A298675, A298677, A298878, A299045, A299071.
Cf. A094954 (unsigned version of this array, but missing the first row).
Cf. Rows: A057078, A033999, A099496, A079935 (or A001835), A004253, A001653, A049685, A070997, A070998, A138288 (or A072256), ...
Cf. Columns: A000012, A001477 (A000027), A110331 (A165900), A123972, A192398, ...

(* Array: *)
Grid[Table[LinearRecurrence[{-n, -1}, {1, 1 - n}, 10], {n, 10}]]
(*Array antidiagonals flattened (gives this sequence):*)
A299045[n_, k_] := Sum[(-1)^(Floor[j/2]) Binomial[k - Floor[(j + 1)/2], Floor[j/2]] (-n)^(k - j), {j, 0, k}]; Flatten[Table[A299045[n - k, k], {n, 11}, {k, 0, n - 1}]]

{A(n, k) = sum(j=0, k, (-1)^(j\2)*binomial(k-(j+1)\2, j\2)*(-n)^(k-j))}; /* Michael Somos, Jun 19 2023 */

G.f. for row n: (1 + x)/(1 + n*x + x^2), n >= 1.

A(n, k) = B(-n, k) where B = A294099. - Michael Somos, Jun 19 2023

A110330 Inverse of a number triangle related to the Pell numbers.

Original entry on oeis.org

1, -2, 1, -2, -4, 1, 0, -6, -6, 1, 0, 0, -12, -8, 1, 0, 0, 0, -20, -10, 1, 0, 0, 0, 0, -30, -12, 1, 0, 0, 0, 0, 0, -42, -14, 1, 0, 0, 0, 0, 0, 0, -56, -16, 1, 0, 0, 0, 0, 0, 0, 0, -72, -18, 1, 0, 0, 0, 0, 0, 0, 0, 0, -90, -20, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -110, -22, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -132, -24, 1

Offset: 0

Paul Barry, Jul 20 2005

This is the matrix inverse of A110327.

Row sums are A110331. Diagonal sums are A110322. Inverse of A110327. The result can be generalized as follows: The triangle whose columns have e.g.f. (x^k/k!)/(1-a*x-b*x^2) has inverse T(n,k)=if(n=k,1,if(n-k=1,-a*binomial(n,1),if(n-k=2,-2*b*binomial(n,2),0))).

			Rows begin
1;
-2,1;
-2,-4,1;
0,-6,-6,1;
0,0,-12,-8,1;
0,0,0,-20,-10,1;
0,0,0,0,-30,-12,1;

T[n_, k_] := Which[n == k, 1, n-k == 1, -2*Binomial[n, 1], n-k == 2, -2*Binomial[n, 2], True, 0]; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jul 09 2015 *)

{(T(n,k) = if(n==k, 1, if(n-k==1, -2*binomial(n, 1), if(n-k==2, -2*binomial(n, 2), 0)))); triangle(nMax) = for (n=0, nMax, for (k=0, n, print1(T(n,k), ", ")); print());} \\ Michel Marcus, Dec 02 2013

egfxy(n,k) = {x = xx + xx*O(xx^n); y = yy + yy*O(yy^k); n!*polcoeff(polcoeff(exp(x*y)*(1-2*x-x^2), n, xx), k, yy);} \\ Michel Marcus, Dec 02 2013

T(n,k) = if(n=k, 1, if(n-k=1, -2*binomial(n, 1), if(n-k=2, -2*binomial(n, 2), 0))).

E.g.f.: exp(x*y)(1-2x-x^2). This implies that the row polynomials form an Appell sequence. - Tom Copeland, Dec 02 2013

A110315 Diagonal sums of the Fibonacci related number triangle A110314.

Original entry on oeis.org

1, -1, -1, -2, -5, -3, -11, -4, -19, -5, -29, -6, -41, -7, -55, -8, -71, -9, -89, -10, -109, -11, -131, -12, -155, -13, -181, -14, -209, -15, -239, -16, -271, -17, -305, -18, -341, -19, -379, -20, -419, -21, -461, -22, -505, -23, -551, -24, -599, -25, -649

Offset: 0

Paul Barry, Jul 19 2005

Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

G.f.: (1-x-4x^2+x^3+x^4)/((1-x)^3*(1+x)^3).
a(n) = 3a(n-2)-3a(n-4)+a(n-6).
a(n) = (6-n^2)(-1)^n/8-(n^2+4n-2)/8.
a(2n) = A110331(n). - R. J. Mathar, Aug 22 2012

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

Original entry on oeis.org

-1, 0, 0, 2, 3, 6, 2, 12, 15, 20, 6, 30, 35, 42, 12, 56, 63, 72, 20, 90, 99, 110, 30, 132, 143, 156, 42, 182, 195, 210, 56, 240, 255, 272, 72, 306, 323, 342, 90, 380, 399, 420, 110, 462, 483, 506, 132, 552, 575, 600, 156

Offset: 0

Paul Curtz, Jul 23 2012

The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
Then c(n) = e(n)*f(n).
Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
For a(n), we divide by 4.
Note that A214297 is the reduced numerator of  1/4 - 1/A061038(n).
Note also that A168077(n) = A026741(n)^2.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000466, A002378, A131723.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* Harvey P. Dale, Jul 01 2019 *)

Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jan 22 2015

a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
a(n) = g(n) * h(n).
a(n) = k(n) * l(n).
G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jan 22 2015
From Luce ETIENNE, Apr 08 2017: (Start)
a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)

Edited by N. J. A. Sloane, Aug 04 2012

cp's OEIS Frontend

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A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)*binomial(k-floor((j+1)/2), floor(j/2))*(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.

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A110330 Inverse of a number triangle related to the Pell numbers.

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A110315 Diagonal sums of the Fibonacci related number triangle A110314.

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A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

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A299045 Rectangular array: A(n,k) = Sum_{j=0..k} (-1)^floor(j/2)binomial(k-floor((j+1)/2), floor(j/2))(-n)^(k-j), n >= 1, k >= 0, read by antidiagonals.