A061347 -id:A061347 - cp's OEIS frontend

A130777 Coefficients of first difference of Chebyshev S polynomials.

1, -1, 1, -1, -1, 1, 1, -2, -1, 1, 1, 2, -3, -1, 1, -1, 3, 3, -4, -1, 1, -1, -3, 6, 4, -5, -1, 1, 1, -4, -6, 10, 5, -6, -1, 1, 1, 4, -10, -10, 15, 6, -7, -1, 1, -1, 5, 10, -20, -15, 21, 7, -8, -1, 1, -1, -5, 15, 20, -35, -21, 28, 8, -9, -1, 1, 1, -6, -15, 35, 35, -56, -28, 36, 9, -10, -1, 1

Offset: 0

Philippe Deléham, Jul 14 2007

Inverse of triangle in A061554.
Signed version of A046854.
From Paul Barry, May 21 2009: (Start)
Riordan array ((1-x)/(1+x^2),x/(1+x^2)).
This triangle is the coefficient triangle for the Hankel transforms of the family of generalized Catalan numbers that satisfy a(n;r)=r*a(n-1;r)+sum{k=1..n-2, a(k)*a(n-1-k;r)}, a(0;r)=a(1;r)=1. The Hankel transform of a(n;r) is h(n)=sum{k=0..n, T(n,k)*r^k} with g.f. (1-x)/(1-r*x+x^2). These sequences include A086246, A000108, A002212. (End)
From Wolfdieter Lang, Jun 11 2011: (Start)
The Riordan array ((1+x)/(1+x^2),x/(1+x^2)) with entries Phat(n,k)= ((-1)^(n-k))*T(n,k) and o.g.f. Phat(x,z)=(1+z)/(1-x*z+z^2) for the row polynomials Phat(n,x) is related to Chebyshev C and S polynomials as follows.
Phat(n,x) = (R(n+1,x)-R(n,x))/(x+2) = S(2*n,sqrt(2+x))
  with R(n,x)=C_n(x) in the Abramowitz and Stegun notation, p. 778, 22.5.11. See A049310 for the S polynomials. Proof from the o.g.f.s.
Recurrence for the row polynomials Phat(n,x):
  Phat(n,x) = x*Phat(n-1,x) - Phat(n-2,x) for n>=1; Phat(-1,x)=-1, Phat(0,x)=1.
The A-sequence for this Riordan array Phat (see the W. Lang link under A006232 for A- and Z-sequences for Riordan matrices) is given by 1, 0, -1, 0, -1, 0, -2, 0, -5,.., starting with 1 and interlacing the negated A000108 with zeros (o.g.f. 1/c(x^2) = 1-c(x^2)*x^2, with the o.g.f. c(x) of A000108).
The Z-sequence has o.g.f. sqrt((1-2*x)/(1+2*x)), and it is given by A063886(n)*(-1)^n.
The A-sequence of the Riordan array T(n,k) is identical with the one for the Riordan array Phat, and the Z-sequence is -A063886(n).
(End)
The row polynomials P(n,x) are the characteristic polynomials of the adjacency matrices of the graphs which look like P_n (n vertices (nodes), n-1 lines (edges)), but vertex no. 1 has a loop. - Wolfdieter Lang, Nov 17 2011
From Wolfdieter Lang, Dec 14 2013: (Start)
The zeros of P(n,x) are x(n,j) = -2*cos(2*Pi*j/(2*n+1)), j=1..n. From P(n,x) = (-1)^n*S(2*n,sqrt(2-x)) (see, e.g., the Lemma 6 of the W. Lang link).
The discriminants of the P-polynomials are given in A052750. (End)

			The triangle T(n,k) begins:
n\k  0   1   1   3    4    5    6    7    8    9  10  11  12  13 14 15 ...
0:   1
1:  -1   1
2:  -1  -1   1
3:   1  -2  -1   1
4:   1   2  -3  -1    1
5:  -1   3   3  -4   -1    1
6:  -1  -3   6   4   -5   -1    1
7:   1  -4  -6  10    5   -6   -1    1
8:   1   4 -10 -10   15    6   -7   -1    1
9:  -1   5  10 -20  -15   21    7   -8   -1    1
10: -1  -5  15  20  -35  -21   28    8   -9   -1   1
11:  1  -6 -15  35   35  -56  -28   36    9  -10  -1   1
12:  1   6 -21 -35   70   56  -84  -36   45   10 -11  -1   1
13: -1   7  21 -56  -70  126   84 -120  -45   55  11 -12  -1   1
14: -1  -7  28  56 -126 -126  210  120 -165  -55  66  12 -13  -1  1
15:  1  -8 -28  84  126 -252 -210  330  165 -220 -66  78  13 -14 -1  1
...  reformatted and extended - _Wolfdieter Lang_, Jul 31 2014.
---------------------------------------------------------------------------
From _Paul Barry_, May 21 2009: (Start)
Production matrix is
-1, 1,
-2, 0, 1,
-2, -1, 0, 1,
-4, 0, -1, 0, 1,
-6, -1, 0, -1, 0, 1,
-12, 0, -1, 0, -1, 0, 1,
-20, -2, 0, -1, 0, -1, 0, 1,
-40, 0, -2, 0, -1, 0, -1, 0, 1,
-70, -5, 0, -2, 0, -1, 0, -1, 0, 1 (End)
Row polynomials as first difference of S polynomials:
P(3,x) = S(3,x) - S(2,x) = (x^3 - 2*x) - (x^2 -1) = 1 - 2*x - x^2 +x^3.
Alternative triangle recurrence (see a comment above): T(6,2) = T(5,2) + T(5,1) = 3 + 3 = 6. T(6,3) = -T(5,3) + 0*T(5,1) = -(-4) = 4. - _Wolfdieter Lang_, Jul 31 2014

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964. Tenth printing, Wiley, 2002 (also electronically available).

Hyeong-Kwan Ju, On the sequence generated by a certain type of matrices, Honam Math. J. 39, No. 4, 665-675 (2017), Theorem 2.16.
Wolfdieter Lang, The field Q(2cos(pi/n)), its Galois group and length ratios in the regular n-gon, arXiv:1210.1018 [math.GR], 2012-2017; see Definition 1, Lemma 6 and Remark 4.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), p. 22-31 (formula 5).
Index entries for sequences related to Chebyshev polynomials.

Cf. A066170, A046854, A057077 (first column).

Row sums: A010892(n+1); repeat(1,0,-1,-1,0,1). Alternating row sums: A061347(n+2); repeat(1,-2,1).

A130777 := proc(n,k): (-1)^binomial(n-k+1,2)*binomial(floor((n+k)/2),k) end: seq(seq(A130777(n,k), k=0..n), n=0..11); # Johannes W. Meijer, Aug 08 2011

T[n_, k_] := (-1)^Binomial[n - k + 1, 2]*Binomial[Floor[(n + k)/2], k];
Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 14 2017, from Maple *)

@CachedFunction
def A130777(n,k):
    if n< 0: return 0
    if n==0: return 1 if k == 0 else 0
    h = A130777(n-1,k) if n==1 else 0
    return A130777(n-1,k-1) - A130777(n-2,k) - h
for n in (0..9): [A130777(n,k) for k in (0..n)] # Peter Luschny, Nov 20 2012

Number triangle T(n,k) = (-1)^C(n-k+1,2)*C(floor((n+k)/2),k). - Paul Barry, May 21 2009
From Wolfdieter Lang, Jun 11 2011: (Start)
Row polynomials: P(n,x) = sum(k=0..n, T(n,k)*x^k) = R(2*n+1,sqrt(2+x)) / sqrt(2+x), with Chebyshev polynomials R with coefficients given in A127672 (scaled T-polynomials).
R(n,x) is called C_n(x) in Abramowitz and Stegun's handbook, p. 778, 22.5.11.
P(n,x) = S(n,x)-S(n-1,x), n>=0, S(-1,x)=0, with the Chebyshev S-polynomials (see the coefficient triangle A049310).
O.g.f. for row polynomials: P(x,z):= sum(n>=0, P(n,x)*z^n ) = (1-z)/(1-x*z+z^2).
  (from the o.g.f. for R(2*n+1,x), n>=0, computed from the o.g.f. for the R-polynomials (2-x*z)/(1-x*z+z^2) (see A127672))
Proof of the Chebyshev connection from the o.g.f. for Riordan array property of this triangle (see the P. Barry comment above).
For the A- and Z-sequences of this Riordan array see a comment above. (End)
abs(T(n,k)) = A046854(n,k) = abs(A066170(n,k)) T(n,n-k) = A108299(n,k); abs(T(n,n-k)) = A065941(n,k). - Johannes W. Meijer, Aug 08 2011
From Wolfdieter Lang, Jul 31 2014: (Start)
Similar to the triangles A157751, A244419 and A180070 one can give for the row polynomials P(n,x) besides the usual three term recurrence another one needing only one recurrence step. This uses also a negative argument, namely P(n,x) = (-1)^(n-1)*(-1 + x/2)*P(n-1,-x) + (x/2)*P(n-1,x), n >= 1, P(0,x) = 1. Proof by computing the o.g.f. and comparing with the known one. This entails the alternative triangle recurrence T(n,k) = (-1)^(n-k)*T(n-1,k) + (1/2)*(1 + (-1)^(n-k))*T(n-1,k-1), n >= m >= 1, T(n,k) = 0 if n < k and T(n,0) = (-1)^floor((n+1)/2) = A057077(n+1). [P(n,x) recurrence corrected Aug 03 2014]
(End)

New name and Chebyshev comments by Wolfdieter Lang, Jun 11 2010

A174709 Partial sums of floor(n/6).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 16, 18, 21, 24, 27, 30, 33, 36, 40, 44, 48, 52, 56, 60, 65, 70, 75, 80, 85, 90, 96, 102, 108, 114, 120, 126, 133, 140, 147, 154, 161, 168, 176, 184, 192

Offset: 0

Mircea Merca, Nov 30 2010

Partial sums of A152467.

			a(7) = floor(0/6) + floor(1/6) + floor(2/6) + floor(3/6) + floor(4/6) + floor(5/6) + floor(6/6) + floor(7/6) = 0 + 0 + 0 + 0 + 0 + 0 + 1 + 1 = 2.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions, J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,1,-2,1).

Cf. A008724, A152467.

[Round(n*(n-4)/12): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

Maple
```
a(n):=round(n*(n-4)/12)
```

Table[Sum[Floor[k/6], {k,0,n}], {n,0,25}] (* G. C. Greubel, Dec 13 2016 *)

a(n)=(n-2)^2\12 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = round(n*(n-4)/12) = round((2*n^2 - 8*n - 1)/24).
a(n) = floor((n-2)^2/12).
a(n) = ceiling((n+1)*(n-5)/12).
a(n) = a(n-6) + n - 5, n > 5.
From R. J. Mathar, Nov 30 2010: (Start)
a(n) = 2*a(n-1) - a(n-2) + a(n-6) - 2*a(n-7) + a(n-8).
G.f.: -x^6 / ( (1+x)*(x^2-x+1)*(1+x+x^2)*(x-1)^3 ).
a(n) = -n/3 + 5/72 + n^2/12 + (-1)^n/24 + A057079(n+5)/6 + A061347(n)/18. (End)
a(6n) = A000567(n), a(6n+1) = 2*A000326(n), a(6n+2) = A033428(n), a(6n+3) = A049451(n), a(6n+4) = A045944(n), a(6n+5) = A028896(n). - Philippe Deléham, Mar 26 2013
a(n) = A008724(n-2). - R. J. Mathar, Jul 10 2015
Sum_{n>=6} 1/a(n) = Pi^2/18 - Pi/(2*sqrt(3)) + 49/12. - Amiram Eldar, Aug 13 2022

A131713 Period 3: repeat [1, -2, 1].

Original entry on oeis.org

1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1, 1, -2, 1

Offset: 0

Paul Curtz, Sep 14 2007

Second differences of A131534. Binomial transform of 1, -3, 6, -9, 9, 0, ..., A057083 signed.

Nonsimple continued fraction expansion of sqrt(2)-1 = 0.414213562... - R. J. Mathar, Mar 08 2012

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

Cf. A057083, A061347, A131534.

&cat [[1, -2, 1]^^30]; // Wesley Ivan Hurt, Jul 01 2016

seq(op([1, -2, 1]), n=0..50); # Wesley Ivan Hurt, Jul 01 2016

f[n_] := If[ Mod[n, 3] == 1, -2, 1]; Array[f, 105, 0]
CoefficientList[Series[(1 - x)/(1 + x + x^2), {x, 0, 104}], x]
PadRight[{}, 120, {1,-2,1}] (* Harvey P. Dale, Jan 25 2014 *)

a(n)=[1,-2,1][1+n%3] \\ Jaume Oliver Lafont, Mar 24 2009

a(n)=1-3*(n%3==1) \\ Jaume Oliver Lafont, Mar 24 2009

a(n) = b(2*n + 1) where b(n) is multiplicative with b(2^n) = 0^n, b(3^n) = 3 * 0^n - 2, b(p^n) = 1 if p > 3. - Michael Somos, Jan 02 2011
G.f.: (1-x)/(x^2+x+1). - R. J. Mathar, Nov 14 2007
a(n) = 2*cos((2n+1)*Pi/3). - Jaume Oliver Lafont, Nov 23 2008
a(n) = A117188(2*n). - R. J. Mathar, Jun 13 2011
a(n) + a(n-1) + a(n-2) = 0 for n>1, a(n) = a(n-3) for n>2. - Wesley Ivan Hurt, Jul 01 2016
a(n) = (1/4^n) * Sum_{k = 0..n} binomial(2*n+1,2*k)*(-3)^k. - Peter Bala, Feb 06 2019
E.g.f.: 2*exp(-x/2)*cos(sqrt(3)*x/2 + Pi/3). - Fabian Pereyra, Oct 17 2024

Corrected and extended by Michael Somos, Jan 02 2011

A319014 a(n) = 123 + 456 + 789 + 101112 + 131415 + 161718 + ... + (up to n).

Original entry on oeis.org

1, 2, 6, 10, 26, 126, 133, 182, 630, 640, 740, 1950, 1963, 2132, 4680, 4696, 4952, 9576, 9595, 9956, 17556, 17578, 18062, 29700, 29725, 30350, 47250, 47278, 48062, 71610, 71641, 72602, 104346, 104380, 105536, 147186, 147223, 148592, 202020, 202060, 203660

Offset: 1

Wesley Ivan Hurt, Sep 07 2018

In general, for sequences that multiply the first k natural numbers, and then add the product of the next k natural numbers (preserving the order of operations up to n), we have a(n) = Sum_{i=1..floor(n/k)} (k*i)!/(k*i-k)! + Sum_{j=1..k-1} (1-sign((n-j) mod k)) * (Product_{i=1..j} n-i+1). Here, k=3.

			a(1)  = 1;
a(2)  = 1*2 = 2;
a(3)  = 1*2*3 = 6;
a(4)  = 1*2*3 + 4 = 10;
a(5)  = 1*2*3 + 4*5 = 26;
a(6)  = 1*2*3 + 4*5*6 = 126;
a(7)  = 1*2*3 + 4*5*6 + 7 = 133;
a(8)  = 1*2*3 + 4*5*6 + 7*8 = 182;
a(9)  = 1*2*3 + 4*5*6 + 7*8*9 = 630;
a(10) = 1*2*3 + 4*5*6 + 7*8*9 + 10 = 640;
a(11) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11 = 740;
a(12) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 = 1950;
a(13) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13 = 1963;
a(14) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14 = 2132;
a(15) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 = 4680;
a(16) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16 = 4696;
a(17) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16*17 = 4952;
a(18) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16*17*18 = 9576;
a(19) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16*17*18 + 19 = 9595;
etc.

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

Cf. A093361, (k=1) A000217, (k=2) A228958, (k=3) this sequence, (k=4) A319205, (k=5) A319206, (k=6) A319207, (k=7) A319208, (k=8) A319209, (k=9) A319211, (k=10) A319212.

Cf. A049347, A061347, A268685 (trisection).

CoefficientList[Series[(1 + x + 4*x^2 + 12*x^4 + 84*x^5 - 3*x^6 - 9*x^7 + 72*x^8 + 2*x^9 - 4*x^10 + 2*x^11)/((1 - x)^5*(1 + x + x^2)^4), {x, 0, 50}], x] (* after Colin Barker *)

Vec(x*(1 + x + 4*x^2 + 12*x^4 + 84*x^5 - 3*x^6 - 9*x^7 + 72*x^8 + 2*x^9 - 4*x^10 + 2*x^11) / ((1 - x)^5*(1 + x + x^2)^4) + O(x^50)) \\ Colin Barker, Sep 08 2018

a(n) = Sum_{i=1..floor(n/3)} (3*i)!/(3*i-3)! + Sum_{j=1..2} (1-sign((n-j) mod 3)) * (Product_{i=1..j} n-i+1).
From Colin Barker, Sep 08 2018: (Start)
G.f.: x*(1 + x + 4*x^2 + 12*x^4 + 84*x^5 - 3*x^6 - 9*x^7 + 72*x^8 + 2*x^9 - 4*x^10 + 2*x^11) / ((1 - x)^5*(1 + x + x^2)^4).
a(n) = a(n-1) + 4*a(n-3) - 4*a(n-4) - 6*a(n-6) + 6*a(n-7) + 4*a(n-9) - 4*a(n-10) - a(n-12) + a(n-13) for n>13.
(End)
a(3*k) = 3*k*(k+1)*(3*k-2)*(3*k+1)/4, a(3*k+1) = a(3*k) + 3*k + 1, a(3*k+2) = a(3*k) + (3*k+2)*(3*k+1). - Giovanni Resta, Sep 08 2018
a(n) = (3*n^4 - 6*n^3 + 9*n^2 + 6*n - 8 - 2*(3*n^3 - 6*n^2 - 6*n + 2)*A061347(n-3) + 6*(n^3 - 6*n^2 + 6*n + 2)*A049347(n-2))/36. - Stefano Spezia, Apr 23 2023

A002621 Expansion of 1 / ((1-x)^2(1-x^2)(1-x^3)*(1-x^4)).

Original entry on oeis.org

1, 2, 4, 7, 12, 18, 27, 38, 53, 71, 94, 121, 155, 194, 241, 295, 359, 431, 515, 609, 717, 837, 973, 1123, 1292, 1477, 1683, 1908, 2157, 2427, 2724, 3045, 3396, 3774, 4185, 4626, 5104, 5615, 6166, 6754, 7386, 8058, 8778, 9542, 10358, 11222, 12142, 13114

Offset: 0

N. J. A. Sloane

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
E. Fix and J. L. Hodges, Jr., Significance probabilities of the Wilcoxon test, Annals Math. Stat., 26 (1955), 301-312.
E. Fix and J. L. Hodges, Significance probabilities of the Wilcoxon test, Annals Math. Stat., 26 (1955), 301-312. [Annotated scanned copy]
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 199
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Thomas Wieder, The number of certain k-combinations of an n-set, Applied Mathematics Electronic Notes, vol. 8 (2008).
Index entries for linear recurrences with constant coefficients, signature (2, 0, -1, 0, -2, 2, 0, 1, 0, -2, 1).

Partial sums of A001400. Column 4 of A092905.

A002621:=-1/(z**2+1)/(z**2+z+1)/(z+1)**2/(z-1)**5; # Simon Plouffe in his 1992 dissertation
with(combstruct):ZL:=[st, {st=Prod(left, right), left=Set(U, card=r+2), right=Set(U, card=1)}, unlabeled]: subs(r=2, stack): seq(count(subs(r=2, ZL), size=m), m=4..51) ; # Zerinvary Lajos, Feb 07 2008
A057077 := proc(n) (-1)^floor(n/2) ; end proc:
A061347 := proc(n) op(1+(n mod 3),[1,1,-2]) ; end proc:
A002621 := proc(n) 83/288*n^2+55/64*n+2815/3456+11/288*n^3+1/576*n^4+11/128*(-1)^n+1/64*(-1)^n*n; %+ A057077(n)/16 +A061347(n)/27; end proc:
seq(A002621(n),n=0..10) ; # R. J. Mathar, Mar 15 2011

CoefficientList[Series[1/((1-x)^2*(1-x^2)*(1-x^3)*(1-x^4)),{x,0,60}],x] (* Stefan Steinerberger, Jun 10 2007 *)
LinearRecurrence[{2, 0, -1, 0, -2, 2, 0, 1, 0, -2, 1}, {1, 2, 4, 7, 12, 18, 27, 38, 53, 71, 94}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 23 2012 *)

a(n)=(n+1)*(9*(-1)^n+n^3+21*n^2+145*n+350)\/576 \\ Charles R Greathouse IV, May 23 2013

a(n) = +2*a(n-1) - a(n-3) - 2*a(n-5) + 2*a(n-6) + a(n-8) - 2*a(n-10) + a(n-11).
a(n) = 83*n^2/288 +55*n/64 +2815/3456 +11*n^3/288 +n^4/576 +11*(-1)^n/128 +(-1)^n*n/64 + A057077(n)/16 +A061347(n)/27. - R. J. Mathar, Mar 15 2011
a(n)=floor((n+1)*(9*(-1)^n + n^3 + 21*n^2 + 145*n + 350)/576 + 1/2). - Tani Akinari, Nov 10 2012

A004482 Tersum n + 1 (answer recorded in base 10).

Original entry on oeis.org

1, 2, 0, 4, 5, 3, 7, 8, 6, 10, 11, 9, 13, 14, 12, 16, 17, 15, 19, 20, 18, 22, 23, 21, 25, 26, 24, 28, 29, 27, 31, 32, 30, 34, 35, 33, 37, 38, 36, 40, 41, 39, 43, 44, 42, 46, 47, 45, 49, 50, 48, 52, 53, 51, 55, 56, 54, 58, 59, 57, 61, 62

Offset: 0

N. J. A. Sloane

Tersum m + n: write m and n in base 3 and add mod 3 with no carries; e.g., 5 + 8 = "21" + "22" = "10" = 1.

Sprague-Grundy values for game of Wyt Queens.

E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982, see p. 76.

F. Michel Dekking, Jeffrey Shallit, and N. J. A. Sloane, Queens in exile: non-attacking queens on infinite chess boards, Electronic J. Combin., 27:1 (2020), Article P1.52.
Andreas Dress, Achim Flammenkamp, and Norbert Pink, Additive periodicity of the Sprague-Grundy function of certain Nim games, Adv. Appl. Math., 22, p. 249-270 (1999).
Gabriel Nivasch, More on the Sprague-Grundy function for Wythoff's game, pages 377-410 in "Games of No Chance 3", MSRI Publications Volume 56, 2009. See Table 1.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

This sequence is row 1 of table A004481.
a(n) = A061347(n+1) + n.
Third column of triangle A296339.

LinearRecurrence[{1,0,1,-1},{1,2,0,4},70] (* or *) Table[3*Floor[n/3]+ Mod[ n+1,3],{n,0,70}] (* Harvey P. Dale, Nov 29 2014 *)

Periodic with period 3 and saltus 3: a(n) = 3*floor(n/3) + ((n+1) mod 3).
a(n) = n - 2*cos(2*(n+1)*Pi/3). - Wesley Ivan Hurt, Sep 29 2017
Sum_{n>=3} (-1)^(n+1)/a(n) = 1/2 - log(2)/3. - Amiram Eldar, Aug 21 2023

More terms from Erich Friedman

A173173 a(n) = ceiling(Fibonacci(n)/2).

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 4, 7, 11, 17, 28, 45, 72, 117, 189, 305, 494, 799, 1292, 2091, 3383, 5473, 8856, 14329, 23184, 37513, 60697, 98209, 158906, 257115, 416020, 673135, 1089155, 1762289, 2851444, 4613733, 7465176, 12078909, 19544085, 31622993, 51167078, 82790071

Offset: 0

Roger L. Bagula, Nov 22 2010

Also the independence number of the n-Fibonacci cube graph. - Eric W. Weisstein, Sep 06 2017
Also the edge cover number of the (n-2)-Fibonacci cube graph. - Eric W. Weisstein, Dec 26 2017
Also the calque covering number of the (n-2)-Fibonacci cube graph. - Eric W. Weisstein, Apr 20 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..280
Eric Weisstein's World of Mathematics, Clique Covering Number
Eric Weisstein's World of Mathematics, Edge Cover Number
Eric Weisstein's World of Mathematics, Fibonacci Cube Graph
Eric Weisstein's World of Mathematics, Independence Number
Index entries for linear recurrences with constant coefficients, signature (1,1,1,-1,-1).

Column m=3 of A185646.

[Fibonacci(n) - Floor(Fibonacci(n)/2): n in [0..50]]; // Vincenzo Librandi, Apr 24 2011

with(combinat,fibonacci): seq(ceil(fibonacci(n)/2),n=0..33) # Mircea Merca, Jan 04 2010

Table[Fibonacci[n] - Floor[Fibonacci[n]/2], {n, 0, 40}] (* Harvey P. Dale, Jun 09 2013 *)
(* Start from Eric W. Weisstein, Sep 06 2017 *)
Table[Ceiling[Fibonacci[n]/2], {n, 0, 20}]
Ceiling[Fibonacci[Range[0, 20]]/2]
LinearRecurrence[{1, 1, 1, -1, -1}, {1, 2, 3, 4, 7}, 20]
CoefficientList[Series[(1 + x - 2 x^3 - x^4)/(1 - x - x^2 - x^3 + x^4 + x^5), {x, 0, 20}], x]
(* End *)

/* Continued Fraction: */
{a(n)=my(CF); CF=1+x; for(k=0, n, CF=1/(1 - x^(n-k+1)*(1 - x^(n-k+4)) *CF +x*O(x^n) )); polcoeff(x*CF, n)} \\ Paul D. Hanna, Jul 08 2013

{a(n)=polcoeff( x*(1 - x^2 - x^3) / ((1-x^3)*(1 - x - x^2 +x*O(x^n))),n)} \\ Paul D. Hanna, Jul 18 2013

a(n)=(fibonacci(n)+1)\2 \\ Charles R Greathouse IV, Jun 11 2015

a(n) = ceiling(Fibonacci(n)/2). - Mircea Merca, Jan 04 2010
a(n) = a(n-1) +a(n-2) +a(n-3) -a(n-4) -a(n-5) - Joerg Arndt, Apr 24 2011
G.f.: x/(1 - x*(1-x^4)/(1 - x^2*(1-x^5)/(1 - x^3*(1-x^6)/(1 - x^4*(1-x^7)/(1 - x^5*(1-x^8)/(1 - x^6*(1-x^9)/(1 - x^7*(1-x^10)/(1 - x^8*(1-x^11)/(1 - ...))))))))), (continued fraction) - Paul D. Hanna, Jul 08 2013
G.f.: x*(1 - x^2 - x^3) / ((1-x^3)*(1 - x - x^2)). [Paul D. Hanna, Jul 18 2013, from Joerg Arndt's formula]
a(n) = A061347(n)/6 +1/3 +A000045(n)/2. - R. J. Mathar, Jul 19 2013
For n > 1, if n == 0 (mod 3) then a(n) = a(n-1) + a(n-2) - 1; otherwise a(n) = a(n-1) + a(n-2). - Franklin T. Adams-Watters, Jun 11 2018

Name simplified using Mircea Merca's formula by Eric W. Weisstein, Sep 06 2017

A096777 a(n) = a(n-1) + Sum_{k=1..n-1}(a(k) mod 2), a(1) = 1.

Original entry on oeis.org

1, 2, 3, 5, 8, 11, 15, 20, 25, 31, 38, 45, 53, 62, 71, 81, 92, 103, 115, 128, 141, 155, 170, 185, 201, 218, 235, 253, 272, 291, 311, 332, 353, 375, 398, 421, 445, 470, 495, 521, 548, 575, 603, 632, 661, 691, 722, 753, 785, 818, 851, 885, 920, 955, 991, 1028

Offset: 1

Reinhard Zumkeller, Jul 09 2004

a(n) = a(n-1) + (number of odd terms so far in the sequence). Example: 15 is 11 + 4 odd terms so far in the sequence (they are 1,3,5,11). See A007980 for the same construction with even integers. - Eric Angelini, Aug 05 2007

A016789 and A032766 give positions where even and odd terms occur; a(3*n)=A056106(n); a(3*n-1)=A077588(n); a(3*n-2)=A056108(n). - Reinhard Zumkeller, Dec 29 2007

			G.f. = x + 2*x^2 + 3*x^3 + 5*x^4 + 8*x^5 + 11*x^6 + 15*x^7 + 20*x^8 + ... - _Michael Somos_, Apr 18 2020

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J.-L. Baril, T. Mansour, A. Petrossian, Equivalence classes of permutations modulo excedances, 2014.
Eric Weisstein's World of Mathematics, Odd Number
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A004396, A007980, A016789, A032766, A056106, A056108, A061347, A077588, A097602, A131093.

a096777 n = a096777_list !! (n-1)
a096777_list = 1 : zipWith (+) a096777_list
                               (scanl1 (+) (map (`mod` 2) a096777_list))
-- Reinhard Zumkeller, Mar 11 2014

[Floor((n-2)^2/3)+n: n in [1..60]]; // Vincenzo Librandi, Dec 27 2015

A096777:=n->n + floor((n-2)^2/3); seq(A096777(n), n=1..100); # Wesley Ivan Hurt, Mar 06 2014

Table[n + Floor[(n-2)^2/3], {n, 100}] (* Wesley Ivan Hurt, Mar 06 2014 *)

a(n)=(n-2)^2\3+n \\ Charles R Greathouse IV, Mar 06 2014

a(n+1) - a(n) = A004396(n).
a(n) = floor(n/3) * (3*floor(n/3) + 2*(n mod 3) - 1) + n mod 3 + 0^(n mod 3). - Reinhard Zumkeller, Dec 29 2007
a(n) = floor((n-2)^2/3) + n. - Christopher Hunt Gribble, Mar 06 2014
G.f.: -x*(x^4+1) / ((x-1)^3*(x^2+x+1)). - Colin Barker, Mar 07 2014
Euler transform of finite sequence [2, 0, 1, 1, 0, 0, 0, -1]. - Michael Somos, Apr 18 2020
a(n) = (10 + 3*n*(n - 1) - A061347(n+1))/9. - Stefano Spezia, Sep 22 2022

A117444 Period 5: Repeat [0, 1, 2, -2, -1].

Original entry on oeis.org

0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0, 1, 2, -2, -1, 0

Offset: 0

Paul Barry, Mar 16 2006

See the comments in A203571 concerning formulas of the n-th term of periodic sequences. - M. F. Hasler, Jan 13 2013

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

Cf. A061347.

[2-((2-n) mod 5) : n in [0..50]]; // Wesley Ivan Hurt, Jul 12 2014

A117444:=n->2 - ((2-n) mod 5): seq(A117444(n), n=0..50); # Wesley Ivan Hurt, Jul 12 2014

Table[2 - Mod[2 - n, 5], {n, 0, 50}] (* Wesley Ivan Hurt, Jul 12 2014 *)

A117444(n)=4315*5^(n%5)\1562%5-2  \\ M. F. Hasler, Jan 13 2013

A117444(n,p=[0,1,2,-2,-1])=p[n%#p+1] \\ M. F. Hasler, Jan 13 2013

G.f.: x*(1+3*x+x^2)/(1+x+x^2+x^3+x^4) = x*(1+2*x^2-2*x^3-x^4)/(1-x^5).
a(n) = (1/2)*Sum_{k=0..5} L(k*(k^2-n)/5), where L(j/p) is the Legendre symbol of j and p.
a(n) = sqrt(2+2/sqrt(5))*sin(2*Pi*n/5)-sqrt(2-2/sqrt(5))*sin(4*Pi*n/5).
a(n) = -2 + floor(23401/99999*10^(n+1)) mod 10. - Hieronymus Fischer, Jan 03 2013
a(n) = -2 + floor(863/1562*5^(n+1)) mod 5. - Hieronymus Fischer, Jan 04 2013
a(n) = 2 - ((2-n) mod 5). - Wesley Ivan Hurt, Jul 12 2014
a(n) = - a(n-1) - a(n-2) - a(n-3) - a(n-4). - Wesley Ivan Hurt, Sep 05 2022

More terms from Wesley Ivan Hurt, Jul 12 2014

A146535 Numerator of (2*n-1)/3.

Original entry on oeis.org

1, 1, 5, 7, 3, 11, 13, 5, 17, 19, 7, 23, 25, 9, 29, 31, 11, 35, 37, 13, 41, 43, 15, 47, 49, 17, 53, 55, 19, 59, 61, 21, 65, 67, 23, 71, 73, 25, 77, 79, 27, 83, 85, 29, 89, 91, 31, 95, 97, 33, 101, 103, 35, 107, 109, 37, 113, 115, 39, 119, 121, 41, 125, 127, 43, 131, 133, 45

Offset: 1

Artur Jasinski, Oct 31 2008

From Jaroslav Krizek, May 28 2010: (Start)
a(n+1) = numerators of antiharmonic mean of the first n positive integers for n >= 1.
See A169609(n-1) - denominators of antiharmonic mean of the first n positive integers for n >= 1. (End)

			Fractions begin with 1/6, 1/2, 5/6, 7/6, 3/2, 11/6, 13/6, 5/2, 17/6, 19/6, 7/2, 23/6, ...

Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Cf. A146306, A146307, A146308, A146309, A146310, A146311, A146312, A146313.
Cf. A061347, A102283.
Trisections: A016921, A005408, A016969. [R. J. Mathar, Nov 21 2008]

Table[Numerator[(2 n - 1)/6], {n, 1, 100}]
LinearRecurrence[{0,0,2,0,0,-1},{1,1,5,7,3,11},100] (* Harvey P. Dale, Feb 24 2015 *)

a(n) = numerator((2*n-1)/3); \\ Altug Alkan, Apr 13 2018

From R. J. Mathar, Nov 21 2008: (Start)
a(n) = 2*a(n-3) - a(n-6).
G.f.: x(1+x)(1+5x^2+x^4)/((1-x)^2*(1+x+x^2)^2). (End)
Sum_{k=1..n} a(k) ~ (7/9) * n^2. - Amiram Eldar, Apr 04 2024
a(n) = (2*n - 1)*(7 - A061347(n) +3*A102283(n))/9. - Stefano Spezia, Feb 14 2025

Name edited by Altug Alkan, Apr 13 2018

cp's OEIS Frontend

A130777 Coefficients of first difference of Chebyshev S polynomials.

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A174709 Partial sums of floor(n/6).

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A131713 Period 3: repeat [1, -2, 1].

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A319014 a(n) = 1*2*3 + 4*5*6 + 7*8*9 + 10*11*12 + 13*14*15 + 16*17*18 + ... + (up to n).

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A002621 Expansion of 1 / ((1-x)^2*(1-x^2)*(1-x^3)*(1-x^4)).

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A004482 Tersum n + 1 (answer recorded in base 10).

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A319014 a(n) = 123 + 456 + 789 + 101112 + 131415 + 161718 + ... + (up to n).

A002621 Expansion of 1 / ((1-x)^2(1-x^2)(1-x^3)*(1-x^4)).