A230584 -id:A230584 - cp's OEIS frontend

A298375 Partial sums of A230584.

2, 5, 11, 18, 29, 43, 61, 84, 111, 145, 183, 230, 281, 343, 409, 488, 571, 669, 771, 890, 1013, 1155, 1301, 1468, 1639, 1833, 2031, 2254, 2481, 2735, 2993, 3280, 3571, 3893, 4219, 4578, 4941, 5339, 5741, 6180, 6623, 7105, 7591, 8118, 8649, 9223, 9801, 10424

Offset: 1

Gerald Hillier, Jan 18 2018

			For n = 5 then a(5) = 2+3+6+7+11 = 29.

Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Cf. A230584.

CoefficientList[ Series[(2 + x - x^2 - x^3 + 2x^5 - x^6)/((x -1)^4 (x + 1)^2), {x, 0, 50}], x] (* Robert G. Wilson v, Jan 18 2018 *)

Vec(x*(2 + x - x^2 - x^3 + 2*x^5 - x^6) / ((1 - x)^4*(1 + x)^2) + O(x^50)) \\ Colin Barker, Jan 18 2018

Let g = n + ((n + 1) mod 2), then for n > 1,
a(n) = (g^3 + 6*g^2 + 11*g + 18) / 12 - If(n mod 2 = 1, 0, ((n + 2) / 2)^2 + 2).
From Colin Barker, Jan 18 2018: (Start)
G.f.: x*(2 + x - x^2 - x^3 + 2*x^5 - x^6) / ((1 - x)^4*(1 + x)^2).
a(n) = (n^3 + 6*n^2 + 14*n) / 12 for n>1 and even.
a(n) = (n^3 + 6*n^2 + 11*n + 18) / 12 for n>1 and odd.
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.
(End)

A109613 Odd numbers repeated.

Original entry on oeis.org

1, 1, 3, 3, 5, 5, 7, 7, 9, 9, 11, 11, 13, 13, 15, 15, 17, 17, 19, 19, 21, 21, 23, 23, 25, 25, 27, 27, 29, 29, 31, 31, 33, 33, 35, 35, 37, 37, 39, 39, 41, 41, 43, 43, 45, 45, 47, 47, 49, 49, 51, 51, 53, 53, 55, 55, 57, 57, 59, 59, 61, 61, 63, 63, 65, 65, 67, 67, 69, 69, 71, 71, 73

Offset: 0

Reinhard Zumkeller, Aug 01 2005

The number of rounds in a round-robin tournament with n competitors. - A. Timothy Royappa, Aug 13 2011
Diagonal sums of number triangle A113126. - Paul Barry, Oct 14 2005
When partitioning a convex n-gon by all the diagonals, the maximum number of sides in resulting polygons is 2*floor(n/2)+1 = a(n-1) (from Moscow Olympiad problem 1950). - Tanya Khovanova, Apr 06 2008
The inverse values of the coefficients in the series expansion of f(x) = (1/2)*(1+x)*log((1+x)/(1-x)) lead to this sequence; cf. A098557. - Johannes W. Meijer, Nov 12 2009
From Reinhard Zumkeller, Dec 05 2009: (Start)
First differences: A010673; partial sums: A000982;
A059329(n) = Sum_{k = 0..n} a(k)*a(n-k);
A167875(n) = Sum_{k = 0..n} a(k)*A005408(n-k);
A171218(n) = Sum_{k = 0..n} a(k)*A005843(n-k);
A008794(n+2) = Sum_{k = 0..n} a(k)*A059841(n-k). (End)
Dimension of the space of weight 2n+4 cusp forms for Gamma_0(5). - Michael Somos, May 29 2013
For n > 4: a(n) = A230584(n) - A230584(n-2). - Reinhard Zumkeller, Feb 10 2015
The arithmetic function v+-(n,2) as defined in A290988. - Robert Price, Aug 22 2017
For n > 0, also the chromatic number of the (n+1)-triangular (Johnson) graph. - Eric W. Weisstein, Nov 17 2017
a(n-1), for n >= 1, is also the upper bound a_{up}(b), where b = 2*n + 1, in the first (top) row of the complete coach system Sigma(b) of Hilton and Pedersen [H-P]. All odd numbers <= a_{up}(b) of the smallest positive restricted residue system of b appear once in the first rows of the c(2*n+1) = A135303(n) coaches. If b is an odd prime a_{up}(b) is the maximum. See a comment in the proof of the quasi-order theorem of H-P, on page 263 ["Furthermore, every possible a_i < b/2 ..."]. For an example see below. - Wolfdieter Lang, Feb 19 2020
Satisfies the nested recurrence a(n) = a(a(n-2)) + 2*a(n-a(n-1)) with a(0) = a(1) = 1. Cf. A004001. - Peter Bala, Aug 30 2022
The binomial transform is 1, 2, 6, 16, 40, 96, 224, 512, 1152, 2560,.. (see A057711). - R. J. Mathar, Feb 25 2023

			G.f. = 1 + x + 3*x^2 + 3*x^3 + 5*x^4 + 5*x^5 + 7*x^6 + 7*x^7 + 9*x^8 + 9*x^9 + ...
Complete coach system for (a composite) b = 2*n + 1 = 33: Sigma(33) ={[1; 5], [5, 7, 13; 2, 1, 2]} (the first two rows are here 1 and 5, 7, 13), a_{up}(33) = a(15) = 15. But 15 is not in the reduced residue system modulo 33, so the maximal (odd) a number is 13. For the prime b = 31, a_{up}(31) = a(14) = 15 appears as maximum of the first rows. - _Wolfdieter Lang_, Feb 19 2020

Peter Hilton and Jean Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge University Press, 2010, 3rd printing 2012, pp. (260-281).

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Chromatic Number
Eric Weisstein's World of Mathematics, Johnson Graph
Eric Weisstein's World of Mathematics, Triangular Graph
Wikipedia, Round-robin tournament
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000217, A063196, A110660, A186421, A186422, A211955, A230584, A290988.
Complement of A052928 with respect to the universe A004526. - Guenther Schrack, Aug 21 2018
First differences of A000982, A061925, A074148, A105343, A116940, and A179207. - Guenther Schrack, Aug 21 2018

a109613 = (+ 1) . (* 2) . (`div` 2)
a109613_list = 1 : 1 : map (+ 2) a109613_list
-- Reinhard Zumkeller, Oct 27 2012, Feb 21 2011

A109613:=n->2*floor(n/2)+1; seq(A109613(k), k=0..100); # Wesley Ivan Hurt, Oct 22 2013

Flatten@ Array[{2# - 1, 2# - 1} &, 37] (* Robert G. Wilson v, Jul 07 2012 *)
(# - Boole[EvenQ[#]] &) /@ Range[80] (* Alonso del Arte, Sep 11 2019 *)
With[{c=2*Range[0,40]+1},Riffle[c,c]] (* Harvey P. Dale, Jan 02 2020 *)

A109613(n)=n>>1<<1+1 \\ Charles R Greathouse IV, Feb 24 2011

def a(n) : return( len( CuspForms( Gamma0( 5), 2*n + 4, prec=1). basis())); # Michael Somos, May 29 2013

((1 to 49) by 2) flatMap { List.fill(2)() } // _Alonso del Arte, Sep 11 2019

a(n) = 2*floor(n/2) + 1.
a(n) = A052928(n) + 1 = 2*A004526(n) + 1.
a(n) = A028242(n) + A110654(n).
a(n) = A052938(n-2) + A084964(n-2) for n > 1. - Reinhard Zumkeller, Aug 27 2005
G.f.: (1 + x + x^2 + x^3)/(1 - x^2)^2. - Paul Barry, Oct 14 2005
a(n) = 2*a(n-2) - a(n-4), a(0) = 1, a(1) = 1, a(2) = 3, a(3) = 3. - Philippe Deléham, Nov 03 2008
a(n) = A001477(n) + A059841(n). - Philippe Deléham, Mar 31 2009
a(n) = 2*n - a(n-1), with a(0) = 1. - Vincenzo Librandi, Nov 13 2010
a(n) = R(n, -2), where R(n, x) is the n-th row polynomial of A211955. a(n) = (-1)^n + 2*Sum_{k = 1..n} (-1)^(n - k - 2)*4^(k-1)*binomial(n+k, 2*k). Cf. A084159. - Peter Bala, May 01 2012
a(n) = A182579(n+1, n). - Reinhard Zumkeller, May 06 2012
G.f.: ( 1 + x^2 ) / ( (1 + x)*(x - 1)^2 ). - R. J. Mathar, Jul 12 2016
E.g.f.: x*exp(x) + cosh(x). - Ilya Gutkovskiy, Jul 12 2016
From Guenther Schrack, Sep 10 2018: (Start)
a(-n) = -a(n-1).
a(n) = A047270(n+1) - (2*n + 2).
a(n) = A005408(A004526(n)). (End)
a(n) = A000217(n) / A004526(n+1), n > 0. - Torlach Rush, Nov 10 2023

A230586 a(n) = n^5 - 5n^3 + 5n.

Original entry on oeis.org

1, 2, 123, 724, 2525, 6726, 15127, 30248, 55449, 95050, 154451, 240252, 360373, 524174, 742575, 1028176, 1395377, 1860498, 2441899, 3160100, 4037901, 5100502, 6375623, 7893624, 9687625, 11793626, 14250627, 17100748, 20389349, 24165150, 28480351, 33390752

Offset: 1

Ralf Stephan, Oct 24 2013

Numbers k such that the polynomial x^10 - k*x^5 + 1 is reducible (see paper in A231123).

Second row of A231123.

Kelvin Voskuijl, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (6, -15, 20, -15, 6, -1).

Cf. A230584, A231123.

[n^5 - 5*n^3 + 5*n: n in [1..40]]; // Vincenzo Librandi, Jan 11 2015

Table[n^5-5n^3+5n,{n,30}] (* or *) LinearRecurrence[{6,-15,20,-15,6,-1},{1,2,123,724,2525,6726},30] (* Harvey P. Dale, Jan 10 2015 *)
CoefficientList[Series[(1 - 4 x + 126 x^2 - 4 x^3 + x^4) / (1 - x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Jan 11 2015 *)

for(n=1,10^10,if(!polisirreducible(x^10-n*x^5+1),print1(n,", ")));

G.f.: x*(1 - 4*x + 126*x^2 - 4*x^3 + x^4)/(1 - x)^6. - Vincenzo Librandi, Jan 11 2015

More terms from Vincenzo Librandi, Jan 11 2015

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A298375 Partial sums of A230584.

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A109613 Odd numbers repeated.

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A230586 a(n) = n^5 - 5*n^3 + 5*n.

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A230586 a(n) = n^5 - 5n^3 + 5n.