A186424 -id:A186424 - cp's OEIS frontend

A016957 a(n) = 6*n + 4.

4, 10, 16, 22, 28, 34, 40, 46, 52, 58, 64, 70, 76, 82, 88, 94, 100, 106, 112, 118, 124, 130, 136, 142, 148, 154, 160, 166, 172, 178, 184, 190, 196, 202, 208, 214, 220, 226, 232, 238, 244, 250, 256, 262, 268, 274, 280, 286, 292, 298, 304, 310, 316, 322, 328

Offset: 0

N. J. A. Sloane

Number of 2 X n binary matrices avoiding simultaneously the right-angled numbered polyomino patterns (ranpp) (00;1), (01,1) and (11;0). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i1 < i2, j1 < j2 and these elements are in the same relative order as those in the triple (x,y,z). In general, the number of m X n 0-1 matrices in question is given by (n+2)*2^(m-1) + 2*m*(n-1) - 2 for m > 1 and n > 1. - Sergey Kitaev, Nov 12 2004
If Y is a 4-subset of an n-set X then, for n >= 4, a(n-4) is the number of 3-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 08 2007
4th transversal numbers (or 4-transversal numbers): Numbers of the 4th column of positive numbers in the square array of nonnegative and polygonal numbers A139600. Also, numbers of the 4th column in the square array A057145. - Omar E. Pol, May 02 2008
a(n) is the maximum number such that there exists an edge coloring of the complete graph with a(n) vertices using n colors and every subgraph whose edges are of the same color (subgraph induced by edge color) is planar. - Srikanth K S, Dec 18 2010
Also numbers having two antecedents in the Collatz problem: 12*n+8 and 2*n+1 (respectively A017617(n) and A005408(n)). - Michel Lagneau, Dec 28 2012
a(n) = 6n+4 has three undirected edges e1 = (3n+2, 6n+4), e2 = (6n+4, 12n+8) and e3 = (2n+1, 6n+4) in the Collatz graph of A006370. - Heinz Ebert, Mar 16 2021
Conjecture: this sequence contains some but not all, even numbers with odd abundance A088827. They appear in this sequence at indices A186424(n) - 1. - John Tyler Rascoe, Jul 09 2022

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189. - From N. J. A. Sloane, Dec 01 2012

Heinz Ebert, A Graph Theoretical Approach to the Collatz Problem, arXiv:1905.07575 [math.GM], 2019-2020.
Tanya Khovanova, Recursive Sequences.
Sergey Kitaev, On multi-avoidance of right angled numbered polyomino patterns, Integers: Electronic Journal of Combinatorial Number Theory, Vol. 4 (2004), Article A21, 20pp.
Index to sequences related to polygonal numbers.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A000217, A005408, A006370, A008588, A008615, A016921, A016789, A016933, A016945, A016958, A016969, A017617, A017329, A057145, A067412, A088827, A139600, A139606, A157176, A186424.

a016957 = (+ 4) . (* 6)  -- Reinhard Zumkeller, Jul 05 2013

seq(6*n+4, n = 0 .. 50) # Matt C. Anderson, Jun 09 2017

Range[4, 1000, 6] (* Vladimir Joseph Stephan Orlovsky, May 27 2011 *)

makelist(6*n+4, n, 0, 30); /* Martin Ettl, Nov 12 2012 */

a(n)=6*n+4 \\ Charles R Greathouse IV, Jul 10 2016

A008615(a(n)) = n+1. - Reinhard Zumkeller, Feb 27 2008
a(n) = A016789(n)*2. - Omar E. Pol, May 02 2008
A157176(a(n)) = A067412(n+1). - Reinhard Zumkeller, Feb 24 2009
a(n) = sqrt(A016958(n)). - Zerinvary Lajos, Jun 30 2009
a(n) = 2*(6*n+1) - a(n-1) (with a(0)=4). - Vincenzo Librandi, Nov 20 2010
a(n) = floor((sqrt(36*n^2 - 36*n + 1) + 6*n + 1)/2). - Srikanth K S, Dec 18 2010
From Colin Barker, Jan 30 2012: (Start)
G.f.: 2*(2+x)/(1-2*x+x^2).
a(n) = 2*a(n-1) - a(n-2). (End)
A089911(2*a(n)) = 9. - Reinhard Zumkeller, Jul 05 2013
a(n) = 3 * A005408(n) + 1. - Fred Daniel Kline, Oct 24 2015
a(n) = A057145(n+2,4). - R. J. Mathar, Jul 28 2016
a(4*n+2) = 4 * a(n). - Zhandos Mambetaliyev, Sep 22 2018
Sum_{n>=0} (-1)^n/a(n) = sqrt(3)*Pi/18 - log(2)/6. - Amiram Eldar, Dec 10 2021
E.g.f.: 2*exp(x)*(2 + 3*x). - Stefano Spezia, May 29 2024

A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

Original entry on oeis.org

1, 4, 15, 32, 65, 108, 175, 256, 369, 500, 671, 864, 1105, 1372, 1695, 2048, 2465, 2916, 3439, 4000, 4641, 5324, 6095, 6912, 7825, 8788, 9855, 10976, 12209, 13500, 14911, 16384, 17985, 19652, 21455, 23328, 25345, 27436, 29679, 32000, 34481, 37044, 39775, 42592

Offset: 1

Stefano Spezia, Aug 01 2018

Terms are obtained as partial sums in an algorithm for the generation of the sequence of the fourth powers (A000583). Starting with the sequence of the positive integers (A000027), it is necessary to delete every 4th term and to consider the partial sums of the obtained sequence, then to delete every 3rd term, and lastly to consider again the partial sums (see References).
a(n) is the trace of an n X n square matrix M(n) formed by writing the numbers 1, ..., n^2 successively forward and backward along the rows in zig-zag pattern as shown in the examples below. Specifically, M(n) is defined as M[i,j,n] = j + n*(i-1) if i is odd and M[i,j,n] = n*i - j + 1 if i is even, and it has det(M(n)) = 0 for n > 2 (proved).
From Saeed Barari, Oct 31 2021: (Start)
Also the sum of the entries in an n X n matrix whose elements start from 1 and increase as they approach the center. For instance, in case of n=5, the entries of the following matrix sum to 65:
  1 2 3 2 1
  2 3 4 3 2
  3 4 5 4 3
  2 3 4 3 2
  1 2 3 2 1. (End)
The n X n square matrix of the preceding comment is defined as: A[i,j,n] = n - abs((n + 1)/2 - j) - abs((n + 1)/2 - i). - Stefano Spezia, Nov 05 2021

			For n = 1 the matrix M(1) is
  1
with trace Tr(M(1)) = a(1) = 1.
For n = 2 the matrix M(2) is
  1, 2
  4, 3
with Tr(M(2)) = a(2) = 4.
For n = 3 the matrix M(3) is
  1, 2, 3
  6, 5, 4
  7, 8, 9
with Tr(M(3)) = a(3) = 15.

Edward A. Ashcroft, Anthony A. Faustini, Rangaswami Jagannathan, and William W. Wadge, Multidimensional Programming, Oxford University Press 1995, p. 12.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
G. Polya, Mathematics and Plausible Reasoning: Induction and analogy in mathematics, Princeton University Press 1990, p. 118.
Shailesh Shirali, A Primer on Number Sequences, Universities Press (India) 2004, p. 106.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Exercise 3.7.3 on pages 122-123.

Stefano Spezia, Table of n, a(n) for n = 0..10000
Alfred Moessner, Eine Bemerkung über die Potenzen der natürlichen Zahlen, München 1952. Sitzungsberichte: 1951,3.
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A000583, A000027, A186424 (first differences).
Cf. A000578, A000035, A193356, A210378.
Cf. A006003, A317297, A001489, A322844.
Cf. related to the M matrices: A074147 (antidiagonals), A130130 (rank), A241016 (row sums), A317617 (column sums), A322277 (permanent), A323723 (subdiagonal sums), A323724 (superdiagonal sums).

a_n:=List([1..nmax], n->(1/2)*(n^3 + n*RemInt(n, 2)));

List([1..50],n->(1/2)*(n^3+n*(n mod 2))); # Muniru A Asiru, Aug 24 2018

[IsEven(n) select n^3/2 else (n^3+n)/2: n in [1..50]]; // Vincenzo Librandi, Aug 07 2018

a:=n->(1/2)*(n^3+n*modp(n,2)): seq(a(n),n=1..50); # Muniru A Asiru, Aug 24 2018

CoefficientList[Series[1/4 E^-x (1 + 3 E^(2 x) + 6 E^(2 x) x + 2 E^(2 x) x^2), {x, 0, 45}], x]*Table[(k + 1)!, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)*(1 + x)^3), {x, 0, 45}], x]*Table[(k + 1)*(-1)^k, {k, 0, 45}]
CoefficientList[Series[-(1 + x^2)/((-1 + x)^3*(1 + x)), {x, 0, 45}], x]*Table[(k + 1), {k, 0, 45}]
From Robert G. Wilson v, Aug 01 2018: (Start)
a[i_, j_, n_] := If[OddQ@ i, j + n (i - 1), n*i - j + 1]; f[n_] := Tr[Table[a[i, j, n], {i, n}, {j, n}]]; Array[f, 45]
CoefficientList[Series[(x^4 + 2x^3 + 6x^2 + 2x + 1)/((x - 1)^4 (x + 1)^2), {x, 0, 45}], x]
LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 4, 15, 32, 65, 108}, 45]
(End)

a(n):=(1/2)*(n^3 + n*mod(n,2))$ makelist(a(n), n, 1, nmax);

Vec(x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2) + O(x^40)) \\ Colin Barker, Aug 02 2018

M(i, j, n) = if (i % 2, j + n*(i-1), n*i - j + 1);
a(n) = sum(k=1, n, M(k, k, n)); \\ Michel Marcus, Aug 07 2018

for (n in 1:nmax){
   a <- (n^3+n*n%%2)/2
   output <- c(n, a)
   cat(output, "\n")
}
(MATLAB and FreeMat)
for(n=1:nmax); a=(n^3+n*mod(n,2))/2; fprintf('%d\t%0.f\n',n,a); end

a(n) = (1/2)*(A000578(n) + n*A000035(n)).
a(n) = A006003(n) - (n/2)*(1 - (n mod 2)).
a(n) = Sum_{k=1..n} T(n,k), where T(n,k) = ((n + 1)*k - n)*(n mod 2) + ((n - 1)*k + 1)*(1 - (n mod 2)).
E.g.f.: E(x) = (1/4)*exp(-x)*x*(1 + 3*exp(2*x) + 6*exp(2*x)*x + 2*exp(2*x)*x^2).
L.g.f.: L(x) = -x*(1 + x^2)/((-1 + x)*(1 + x)^3).
H.l.g.f.: LH(x) = -x*(1 + x^2)/((-1 + x)^3*(1 + x)).
Dirichlet g.f.: (1/2)*(Zeta(-3 + s) + 2^(-s)*(-2 + 2^s)*Zeta(-1 + s)).
From Colin Barker, Aug 02 2018: (Start)
G.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4) / ((1 - x)^4*(1 + x)^2).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
a(n) = n^3/2 for n even.
a(n) = (n^3+n)/2 for n odd. (End)
a(2*n) = A317297(n+1) + A001489(n). - Stefano Spezia, Dec 28 2018
Sum_{n>0} 1/a(n) = (1/2)*(-2*polygamma(0, 1/2) + polygamma(0, (1-i)/2)+ polygamma(0, (1+i)/2)) + zeta(3)/4 approximately equal to 1.3959168891658447368440622669882813003351669... - Stefano Spezia, Feb 11 2019
a(n) = (A000578(n) + A193356(n))/2. - Stefano Spezia, Jun 27 2022
a(n) = A210378(n-1)/n. - Stefano Spezia, Jul 15 2024

A091999 Numbers that are congruent to {2, 10} mod 12.

Original entry on oeis.org

2, 10, 14, 22, 26, 34, 38, 46, 50, 58, 62, 70, 74, 82, 86, 94, 98, 106, 110, 118, 122, 130, 134, 142, 146, 154, 158, 166, 170, 178, 182, 190, 194, 202, 206, 214, 218, 226, 230, 238, 242, 250, 254, 262, 266, 274, 278, 286, 290, 298, 302, 310, 314, 322, 326, 334

Offset: 1

Ray Chandler, Feb 21 2004

Numbers divisible by 2 but not by 3 or 4. - Robert Israel, Apr 24 2015
For n > 1, a(n) is representable as a sum of four but no fewer consecutive nonnegative integers, i.e., 10 = 1 + 2 + 3 + 4, 14 = 2 + 3 + 4 + 5, 22 = 4 + 5 + 6 + 7, etc. (see A138591). - Martin Renner, Mar 14 2016
Essentially the same as A063221. - Omar E. Pol, Aug 16 2023

David Lovler, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Second row of A092260.
Cf. A109761 (subsequence).
Cf. A002193, A007310, A101263, A138591, A186424.
Cf. A017545, A017641, A063221.

a091999 n = a091999_list !! (n-1)
a091999_list = 2 : 10 : map (+ 12) a091999_list
-- Reinhard Zumkeller, Jan 21 2013

[6*n-3+(-1)^n : n in [1..100]]; // Wesley Ivan Hurt, Apr 23 2015

A091999:=n->6*n-3+(-1)^n: seq(A091999(n), n=1..100); # Wesley Ivan Hurt, Apr 23 2015

Flatten[#+{2,10}&/@(12*Range[0,30])] (* or *) LinearRecurrence[{1,1,-1},{2,10,14},60] (* Harvey P. Dale, Jun 24 2013 *)

a(n) = 6*n - 3 + (-1)^n \\ David Lovler, Jul 16 2022

a(n) = 2*A007310(n).
a(n) = A186424(n) - A186424(n-2), for n > 1.
a(n) = 12*(n-1) - a(n-1), with a(1)=2. - Vincenzo Librandi, Nov 16 2010
G.f.: 2*x*(1+4*x+x^2) / ( (1+x)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
a(n) = a(n-1) + a(n-2) - a(n-3); a(1)=2, a(2)=10, a(3)=14. - Harvey P. Dale, Jun 24 2013
a(n) = 6*n - 3 + (-1)^n. - Wesley Ivan Hurt, Apr 23 2015
E.g.f.: 2 + (6*x - 2)*cosh(x) + 2*(3*x - 2)*sinh(x). - Stefano Spezia, May 09 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(4*sqrt(3)). - Amiram Eldar, Dec 13 2021
E.g.f.: 2 + (6*x - 3)*exp(x) + exp(-x). - David Lovler, Aug 08 2022
a(n) = A063221(n), n > 1. - Omar E. Pol, Aug 15 2023
From Amiram Eldar, Nov 24 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(2) (A002193).
Product_{n>=1} (1 + (-1)^n/a(n)) = 2*sin(Pi/12) (A101263). (End)

A080859 a(n) = 6n^2 + 4n + 1.

Original entry on oeis.org

1, 11, 33, 67, 113, 171, 241, 323, 417, 523, 641, 771, 913, 1067, 1233, 1411, 1601, 1803, 2017, 2243, 2481, 2731, 2993, 3267, 3553, 3851, 4161, 4483, 4817, 5163, 5521, 5891, 6273, 6667, 7073, 7491, 7921, 8363, 8817, 9283, 9761, 10251, 10753, 11267

Offset: 0

Paul Barry, Feb 23 2003

The old definition of this sequence was "Generalized polygonal numbers".
Column T(n,4) of A080853.
Sequence found by reading the line from 1, in the direction 1, 11, ..., in the square spiral whose vertices are the generalized pentagonal numbers A001318. - Omar E. Pol, Sep 08 2011

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

Subsequence of A186424.
Cf. A000384, A001318, A033579, A033581.
Cf. A220083 for a list of numbers of the form n*P(s,n)-(n-1)*P(s,n-1), where P(s,n) is the n-th polygonal number with s sides.

[6*n^2+4*n+1: n in [0..50]]; // Vincenzo Librandi, Apr 29 2016

Table[6 n^2 + 4 n + 1, {n, 0, 50}] (* Vincenzo Librandi, Apr 29 2016 *)

a(n)=6*n^2+4*n+1 \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (C(3,0) + (C(5,2) - 2)*x + C(3,2)*x^2)/(1-x)^3 = (1 + 8*x + 3*x^2)/(1-x)^3.
E.g.f.: (1 + 10*x + 6*x^2)*exp(x). - Vincenzo Librandi, Apr 29 2016
a(n) = C(4,0) + C(4,1)n + C(4,2)n^2.
a(n) = A186424(2*n).
a(n) = 12*n + a(n-1) - 2 with n > 0, a(0)=1. - Vincenzo Librandi, Aug 08 2010
a(n) = (n+1)*A000384(n+1) - n*A000384(n). - Bruno Berselli, Dec 10 2012
a(n) = (n+1)^4 mod n^3 for n >= 7. - J. M. Bergot, Aug 14 2017
a(n) = (2*n+1)^2 + 2*n^2. - Robert FERREOL, Jan 13 2024

Definition replaced with the closed form by Bruno Berselli, Dec 10 2012

A126587 a(n) is the number of integer lattice points inside the right triangle with legs 3n and 4n (and hypotenuse 5n).

Original entry on oeis.org

3, 17, 43, 81, 131, 193, 267, 353, 451, 561, 683, 817, 963, 1121, 1291, 1473, 1667, 1873, 2091, 2321, 2563, 2817, 3083, 3361, 3651, 3953, 4267, 4593, 4931, 5281, 5643, 6017, 6403, 6801, 7211, 7633, 8067, 8513, 8971, 9441, 9923, 10417, 10923, 11441

Offset: 1

Zak Seidov, Jan 05 2007

Row sums of triangle A193832. - Omar E. Pol, Aug 22 2011

			At n=1, three lattice points (1,1), (1,2) and (2,1) are inside the triangle with vertices at the points (0,0), (3n,0) and (0,4n); hence a(1)=3.

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

Cf. A000326, A186424, A193832.

[6*n^2 - 4*n + 1: n in [1..50] ]; // Vincenzo Librandi, May 23 2011

nip[a_,b_]:=Sum[Floor[b-b*i/a-10^-6],{i,a-1}] Table[nip[3k,4k],{k,100}]
Table[6*n^2-4*n+1, {n,1,50}] (* G. C. Greubel, Mar 06 2018 *)

a(n)=6*n^2-4*n+1 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = A186424(2*n-1).
By Pick's theorem, a(n) = 6*n^2 - 4*n + 1. - Nick Hobson, Mar 13 2007
O.g.f.: x*(3+8*x+x^2)/(1-x)^3 = -1 - 12/(-1+x)^3 - 11/(-1+x) - 22/(-1+x)^2. - R. J. Mathar, Dec 10 2007
E.g.f.: exp(x)*(1 + 2*x + 6*x^2) - 1. - Stefano Spezia, May 09 2021
a(n) = (A000326(2n-1) + A000326(2n))/2. - Charlie Marion, Apr 17 2024

A186423 Partial sums of A186421.

Original entry on oeis.org

0, 1, 3, 4, 8, 11, 17, 20, 28, 33, 43, 48, 60, 67, 81, 88, 104, 113, 131, 140, 160, 171, 193, 204, 228, 241, 267, 280, 308, 323, 353, 368, 400, 417, 451, 468, 504, 523, 561, 580, 620, 641, 683, 704, 748, 771, 817, 840, 888, 913, 963, 988, 1040, 1067, 1121, 1148, 1204, 1233, 1291, 1320

Offset: 0

Reinhard Zumkeller, Feb 21 2011

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1,1,-1,-1,1).

A062717 is the subsequence of even terms.

A186424 is the subsequence of odd terms.

List([0..65], n-> (6*n^2 +6*n +3 +(-1)^n*(2*n+1) -4*(-1)^Binomial(n+1, 2))/16); # G. C. Greubel, Oct 09 2019

a186423 n = a186423_list !! n
a186423_list = scanl1 (+) a186421_list

[(6*n^2 +6*n +3 +(-1)^n*(2*n+1) -4*(-1)^Binomial(n+1, 2))/16: n in [0..65]]; // G. C. Greubel, Oct 09 2019

A087960 := proc(n) op((n mod 4)+1,[1,-1,-1,1]) ; end proc:
A186423 := proc(n) 3*n*(n+1)/8 +3/16 +(-1)^n*(2*n+1)/16 -A087960(n)/4 ; end proc: # R. J. Mathar, Feb 28 2011

CoefficientList[Series[x(1+2x+2x^3+x^4)/((1-x)^3(1+x)^2(1+x^2)),{x, 0, 65}],x]  (* Harvey P. Dale, Mar 13 2011 *)
Table[(6*n^2 +6*n +3 +(-1)^n*(2*n+1) -4*(-1)^Binomial[n+1, 2])/16, {n, 0, 65}] (* G. C. Greubel, Oct 09 2019 *)

vector(66, n, my(m=n-1); (6*m^2 +6*m +3 +(-1)^m*(2*m+1) -4*(-1)^binomial(m+1, 2))/16) \\ G. C. Greubel, Oct 09 2019

def A186423(n): return (6*n*(n+1)+3+(-2*n-1 if n&1 else 2*n+1)+(4 if n+1&2 else -4))>>4 # Chai Wah Wu, Jan 31 2023

[(6*n^2 +6*n +3 +(-1)^n*(2*n+1) -4*(-1)^binomial(n+1, 2))/16 for n in (0..65)] # G. C. Greubel, Oct 09 2019

From R. J. Mathar, Feb 28 2011: (Start)
G.f.: x*(1 + 2*x + 2*x^3 + x^4)/( (1+x^2)*(1+x)^2*(1-x)^3 ).
a(n) = (6*n*(n+1) + 3 + (-1)^n*(2*n+1) - 4*A087960(n))/16. (End)
E.g.f.: ((2 + 5*x + 3*x^2)*cosh(x) + (1 + 7*x + 3*x^2)*sinh(x) + 2*sin(x) - 2*cos(x))/8. - G. C. Greubel, Oct 09 2019

More terms added by G. C. Greubel, Oct 09 2019

A006010 Number of paraffins (see Losanitsch reference for precise definition).

Original entry on oeis.org

1, 5, 20, 52, 117, 225, 400, 656, 1025, 1525, 2196, 3060, 4165, 5537, 7232, 9280, 11745, 14661, 18100, 22100, 26741, 32065, 38160, 45072, 52897, 61685, 71540, 82516, 94725, 108225, 123136, 139520, 157505, 177157, 198612, 221940, 247285, 274721, 304400

Offset: 1

N. J. A. Sloane

This is also the square of the sum of the odd numbers plus the square of the sum of the even numbers, up to n. E.g., a(4) = (1+3)^2 + (2+4)^2 = 52. - Scott R. Shannon, Feb 20 2019

The area of a square whose side is a segment connecting the ends of a broken line (snake), the adjacent links of which are perpendicular and equal to the numbers 1, 2, 3, 4, ..., n. For example, a(5) = 9^2 + 6^2 = 117. - Nicolay Avilov, Aug 02 2022

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 = 1..1000
Nicolay Avilov, Illustration of a(1)-a(6)
Nicolay Avilov, The problem of a broken line in a square (in Russian).
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926.
S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. (Annotated scanned copy)
Index entries for linear recurrences with constant coefficients, signature (3,-1,-5,5,1,-3,1).

Cf. A005994, A186424 (2nd differences), A317614 (1st differences), A335648 (partial sums).

CoefficientList[Series[-(x^4 + 2 x^3 + 6 x^2 + 2 x + 1)/((x - 1)^5 (x + 1)^2), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 14 2013 *)
LinearRecurrence[{3,-1,-5,5,1,-3,1},{1,5,20,52,117,225,400},40] (* Harvey P. Dale, Dec 13 2018 *)

Vec(-x*(x^4+2*x^3+6*x^2+2*x+1)/((x-1)^5*(x+1)^2) + O(x^100)) \\ Colin Barker, Oct 05 2015

Sum of [ 1, 3, 9, ... ](A005994) + [ 0, 0, 1, 3, 9, ... ] + 2*[ 0, 1, 5, 15, 35, ... ](binomial(n, 4)).
If n is odd then a(n) = (1/8) * (n^4 + 2*n^3 + 2*n^2 + 2*n + 1) = Det(Transpose[M]*M) where M is the 2 X 3 matrix whose rows are [(n-1)/2, (n-1)/2], [(n-1)/2 + 1, 0] and [(n-1)/2 + 1, (n-1)/2 + 1]. If n is even then a(n) = (1/8) * (n^4 + 2*n^3 + 2*n^2) = Det(Transpose[M]*M) where M is the 2 X 3 matrix whose rows are [n/2, 0], [n/2, n/2] and [n/2 + 1, 0]. - Gerald McGarvey, Oct 30 2007
G.f.: -x*(x^4+2*x^3+6*x^2+2*x+1) / ((x-1)^5*(x+1)^2). - Colin Barker, Mar 20 2013
E.g.f.: (x*(7 + 15*x + 8*x^2 + x^3)*cosh(x) + (1 + 5*x + 15*x^2 + 8*x^3 + x^4)*sinh(x))/8. - Stefano Spezia, Jul 08 2020

More terms from David W. Wilson

A203568 a(n) = A026837(n) - A026838(n).

Original entry on oeis.org

0, 1, -1, 0, 0, 1, 0, -1, 0, 0, 0, 0, 1, 0, 0, -1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, -1, 0, 0, 0, 0

Offset: 0

Michael Somos, Jan 03 2012

sign

			G.f. = x - x^2 + x^5 - x^7 + x^12 - x^15 + x^22 - x^26 + x^35 - x^40 + x^51 - ...
G.f. = q^25 - q^49 + q^121 - q^169 + q^289 - q^361 + q^529 - q^625 + ..
From _Peter Bala_, Feb 13 2020: (Start)
G.f.s for the tails of A(x):
Sum_{n >= 1} (-1)^(n+1) * x^(2*n+3)*Product_{k = 2..n} 1 + x^k = x^5 - x^7 + x^12 - x^15 + x^22 - ....
Sum_{n >= 2} (-1)^n * x^(3*n+6)*Product_{k = 3..n} 1 + x^k = x^12 - x^15 + x^22 - x^26 + x^35 - ....
Sum_{n >= 3} (-1)^(n+1) * x^(4*n+10)*Product_{k = 4..n} 1 + x^k =
x^22 - x^26 + x^35 - x^40 + x^51 - .... (End)

Robert Israel, Table of n, a(n) for n = 0..10000
George E. Andrews, Euler's Pentagonal Number Theorem, Mathematics Magazine, Vol. 56, No. 5 (Nov., 1983), pp. 279-284.

Cf. A003406, A010815, A026837, A026838, A143062.

N:= 1000: # to get a(0) to a(N)
V:= Array(0..N):
for k from 1 to floor((sqrt(1+24*N)-1)/6) do V[(3*k^2-k)/2]:= 1 od:
for k from 1 to floor((sqrt(1+24*N)+1)/6) do V[(3*k^2+k)/2]:= -1 od:
convert(V,list); # Robert Israel, Nov 24 2015

a[ n_] := Which[ n < 1, 0, SquaresR[ 1, 24 n + 1] == 2, -(-1)^Quotient[ Sqrt[24 n + 1], 3], True, 0]; (* Michael Somos, Jul 12 2015 *)

{a(n) = if( n<1, 0, if( issquare( 24*n + 1, &n), - kronecker( -12, n)))};

G.f.: Sum_{k in Z} sign(k) * x^(k * (3*k - 1) / 2).
G.f.: Sum_{k>0} x^(k * (3*k - 1) / 2) * (1 - x^k). - Michael Somos, Jul 12 2015
G.f.: x - x^2 * (1 + x) + x^3 * (1 + x) * (1 + x^2) - x^4 * (1 + x) * (1 + x^2) * (1 + x^3) + .... - Michael Somos, Jul 12 2015
G.f.: x / (1 + x) - x^3 / ((1 + x) * (1 + x^2)) + x^6 / ((1 + x) * (1 + x^2) * (1 + x^3)) - .... - Michael Somos, Jul 12 2015
G.f.: x / (1 + x^2) - x^2 / ((1 + x^2) * (1 + x^4)) + x^3 / ((1 + x^2 ) * (1 + x^4) * (1 + x^6)) - .... - Michael Somos, Jul 12 2015
a(n) = - A143062(n) unless n=0. - Michael Somos, Jul 12 2015
For k >= 1, a((3*k^2 - k)/2) = 1, a((3*k^2 + k)/2) = -1.  a(n) = 0 otherwise. - Robert Israel, Nov 24 2015
From Peter Bala, Feb 11 2021: (Start)
G.f.: A(x) = Sum_{n >= 1} x^(n*(2*n-1))/Product_{k = 1..2*n} 1 + x^k = x - x^2 + x^5 - x^7 + x^12 - x^15 + - ..., follows by adding terms in pairs in the above g.f. Sum_{n >= 1} (-1)^(n+1)*x^(n*(n+1)/2)/Product_{k = 1..n} 1 + x^k of Somos, dated Jul 12 2015.
G.f.: A(x) = 1/2 + (1/2)*Sum_{n >= 1} (-1)^n*x^(n*(n-1)/2)/Product_{k = 1..n} 1 + x^k.
A(x) = Sum_{n >= 0} (-1)^n * x^(n+1)*Product_{k = 1..n} 1 + x^k. (Set x = -1 in Andrews, equation 8. For similar results see the Examples below.)
Conjectural g.f: A(x) = Sum_{n >= 1} (-1)^(n+1) * x^(2*n-1)/Product_{k = 1..n} 1 + x^(2*k-1)  = x - x^2 + x^5 - x^7 + x^12 - x^15 + - ....
More generally, for positive integer N, we appear to have the identity
A(x) = Product_{j = 1..N-1} 1/(1 + x^(2*j)) * ( P(N,x) + Sum_{n >= 1} (-1)^(n+N) * x^(2*N*n-N)/Product_{k = 1..n} 1 + x^(2*k-1) ), where P(N,x) is a polynomial in x of degree N^2 - N - 1 for N > 1, with the first few values given empirically by P(1,x) = 0, P(2,x) = x, P(3,x) = x - x^2 + x^5 and P(4,x) = x - x^2 + x^3 + x^5 + x^7 - x^8 + x^11. Cf. A186424. (End)

A335648 Partial sums of A006010.

Original entry on oeis.org

0, 1, 6, 26, 78, 195, 420, 820, 1476, 2501, 4026, 6222, 9282, 13447, 18984, 26216, 35496, 47241, 61902, 80002, 102102, 128843, 160908, 199068, 244140, 297037, 358722, 430262, 512778, 607503, 715728, 838864, 978384, 1135889, 1313046, 1511658, 1733598, 1980883, 2255604

Offset: 0

Stefano Spezia, Jun 15 2020

Stefano Spezia, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-4,-4,10,-4,-4,4,-1).
Index entries for partial sums.

Cf. A000290, A001844, A005408, A007204, A053755, A058331, A059722.

Cf. A006010 (1st differences), A186424 (3rd differences), A317614 (2nd differences).

I:=[0, 1, 6, 26, 78, 195, 420, 820]; [n le 8 select I[n] else 4*Self(n-1)-4*Self(n-2)-4*Self(n-3)+10*Self(n-4)-4*Self(n-5)-4*Self(n-6)+4*Self(n-7)-Self(n-8): n in [1..39]];

Table[(1+n)(5-5(-1)^n+8n+12n^2+8n^3+2n^4)/80,{n,0,38}]

a(n) = (1 + n)*(5 - 5*(-1)^n + 8*n + 12*n^2 + 8*n^3 + 2*n^4)/80;

(x*(1+2*x+6*x^2+2*x^3+x^4)/((1-x)^6*(1+x)^2)).series(x, 39).coefficients(x, False)

a(n) = (1 + n)*(5 - 5*(-1)^n + 8*n + 12*n^2 + 8*n^3 + 2*n^4)/80.
O.g.f.: x*(1 + 2*x + 6*x^2 + 2*x^3 + x^4)/((1 - x)^6*(1 + x)^2).
E.g.f.: (cosh(x) - sinh(x))*(-5 + 5*x + (5 + 65*x + 180*x^2 + 130*x^3 + 30*x^4 + 2*x^5)*(cosh(2*x) + sinh(2*x)))/80.
a(n) = 4*a(n-1) - 4*a(n-2) - 4*a(n-3) + 10*a(n-4) - 4*a(n-5) - 4*a(n-6) + 4*a(n-7) - a(n-8) for n > 7.
a(2*n-1) = n*A053755(n)/5 for n > 0.
a(2*n) = n*A005408(n)*A059722(n-1)/5.
a(2*n+1) - a(2*n-1) = A001844(n)^2 = A007204(n) for n > 0.
a(2*n) - a(2*n-2) = 2*A000290(n)*A058331(n) for n > 0.

A281445 Nonnegative k for which (2*k^2 + 1)/11 is an integer.

Original entry on oeis.org

4, 7, 15, 18, 26, 29, 37, 40, 48, 51, 59, 62, 70, 73, 81, 84, 92, 95, 103, 106, 114, 117, 125, 128, 136, 139, 147, 150, 158, 161, 169, 172, 180, 183, 191, 194, 202, 205, 213, 216, 224, 227, 235, 238, 246, 249, 257, 260, 268, 271, 279, 282, 290, 293, 301, 304, 312, 315

Offset: 1

Bruno Berselli, Apr 13 2017

For prime d < 11, (2*k^2 + 1)/d can provide integers when d = 3 (A186424).
Corresponding values of (2*k^2 + 1)/11 are listed in A179088.
All k == 4 or 7 (mod 11). - Robert Israel, Apr 25 2017

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

Cf. A179088.

Cf. A001651 (nonnegative k for which (2*k^2 + 1)/3 is an integer).

Magma
```
&cat [[11*n+4, 11*n+7]: n in [0..30]];
```

seq(seq(11*i+j,j=[4,7]),i=0..50); # Robert Israel, Apr 25 2017

Select[Range[400], IntegerQ[(2*#^2 + 1)/11] &]

[k for k in range(400) if ((2*k^2+1)/11).is_integer()]

O.g.f.: x*(4 + 3*x + 4*x^2)/((1 + x)*(1 - x)^2).
E.g.f.: 4 - 5*exp(-x)/4 - 11*(1 - 2*x)*exp(x)/4.
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = (22*n - 5*(-1)^n - 11)/4. Therefore: a(2*h) = 11*h - 4, a(2*h+1) = 11*h + 4.
If h>0,
h*a(n) + (6*h - 5*(-1)^h - 11)/4 = a(h*n) for odd n; otherwise:
h*a(n) + 4*(h - 1) = a(h*n). Some special cases:
h=2: 2*a(n) - 1 = a(2*n) for odd n, 2*a(n) +  4 = a(2*n) for even n;
h=3: 3*a(n) + 3 = a(3*n) for odd n, 3*a(n) +  8 = a(3*n) for even n;
h=4: 4*a(n) + 2 = a(4*n) for odd n, 4*a(n) + 12 = a(4*n) for even n;
h=5: 5*a(n) + 6 = a(5*n) for odd n, 5*a(n) + 16 = a(5*n) for even n, and so on.
Sum_{n>=1} (-1)^(n+1)/a(n) = tan(3*Pi/22)*Pi/11. - Amiram Eldar, Feb 27 2023
From Amiram Eldar, Nov 23 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = cos(Pi/22)*sec(3*Pi/22).
Product_{n>=1} (1 + (-1)^n/a(n)) = 2*sin(3*Pi/22). (End)

cp's OEIS Frontend

A016957 a(n) = 6*n + 4.

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A317614 a(n) = (1/2)*(n^3 + n*(n mod 2)).

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A091999 Numbers that are congruent to {2, 10} mod 12.

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A080859 a(n) = 6*n^2 + 4*n + 1.

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A126587 a(n) is the number of integer lattice points inside the right triangle with legs 3n and 4n (and hypotenuse 5n).

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A186423 Partial sums of A186421.

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A317614 a(n) = (1/2)(n^3 + n(n mod 2)).

A080859 a(n) = 6n^2 + 4n + 1.