A099837 -id:A099837 - cp's OEIS frontend

A094717 a(n) = n! * Sum_{i+2j+3k=n} 1/(i!(2j)!(3k)!).

1, 1, 2, 5, 12, 36, 113, 351, 1080, 3281, 9882, 29646, 88817, 266085, 797526, 2391485, 7173360, 21520080, 64563521, 193700403, 581120892, 1743392201, 5230206126, 15690618378, 47071766561, 141215033961, 423644570442, 1270932914165, 3812797945332, 11438393835996

Offset: 0

Benoit Cloitre, May 23 2004

nonn

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,-12,10,-6,12,-9).

Cf. A024493, A049347, A057083, A057682, A094715, A099837.

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1-5*x+8*x^2-5*x^3+2*x^4-2*x^5)/((1-x)*(1-3*x)*(1+x+x^2)*(1-3*x+3*x^2)) )); // G. C. Greubel, Jul 14 2023

A094717_list := proc(n) local i; exp(z)*cosh(z)*(exp(z)+2*exp(-z/2)* cos(z*sqrt(3/4)))/3; series(%,z,n+2); seq(simplify(i!*coeff(%,z,i)),i=0..n) end: A094717_list(27); # Peter Luschny, Jul 11 2012

a[n_]:= n! Sum[Boole[i +2j +3k ==n]/(i! (2j)! (3k)!), {i,0,n}, {j,0,n}, {k,0,n}]; Table[a[n], {n,0,27}] (* Jean-François Alcover, Jul 06 2019 *)
LinearRecurrence[{6,-12,10,-6,12,-9}, {1,1,2,5,12,36}, 40] (* G. C. Greubel, Jul 14 2023 *)

a(n)=sum(i=0,n,sum(j=0,n,sum(k=0,n,if(n-i-2*j-3*k,0,n!/(i)!/(2*j)!/(3*k)!))))

def A094717_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( (1-5*x+8*x^2-5*x^3+2*x^4-2*x^5)/((1-x)*(1-3*x)*(1+x+x^2)*(1-3*x+3*x^2)) ).list()
A094717_list(40) # G. C. Greubel, Jul 14 2023

Limit_{n->oo} a(n)/3^n = 1/6.
E.g.f.: exp(z)*cosh(z)*(exp(z) + 2*exp(-z/2)*cos(z*sqrt(3/4)))/3. - Peter Luschny, Jul 11 2012
G.f.: (1-5*x+8*x^2-5*x^3+2*x^4-2*x^5)/((1-x)*(1-3*x)*(1+x+x^2)*(1-3*x+3*x^2)). - Colin Barker, Dec 24 2012
From G. C. Greubel, Jul 14 2023: (Start)
a(n) = (1/6)*(1 + 3^n + 2*A049347(n) + A049347(n-1) + 2*A057083(n) - 3*A057083(n-1)).
a(n) = (1/6)*(1 + 3^n + A099837(n+3) + A057682(n+3)). (End)

A117185 Riordan array ((1-x^2)/(1+x+2x^2+x^3+x^4),x/(1+x^2)).

Original entry on oeis.org

1, -1, 1, -2, -1, 1, 3, -3, -1, 1, 1, 4, -4, -1, 1, -4, 4, 5, -5, -1, 1, 1, -8, 8, 6, -6, -1, 1, 3, -3, -13, 13, 7, -7, -1, 1, -2, 11, -11, -19, 19, 8, -8, -1, 1, -1, 1, 24, -24, -26, 26, 9, -9, -1, 1, 1, -12, 12, 43, -43, -34, 34, 10, -10, -1, 1

Offset: 0

Paul Barry, Mar 01 2006

Row sums are A117188. Diagonal sums are A099837. Inverse of A117184.

			Triangle begins
1,
-1, 1,
-2, -1, 1,
3, -3, -1, 1,
1, 4, -4, -1, 1,
-4, 4, 5, -5, -1, 1,
1, -8, 8, 6, -6, -1, 1,
3, -3, -13, 13, 7, -7, -1, 1

A131132 a(n) = a(n-1) + a(n-2) + 1 if n is a multiple of 6, otherwise a(n) = a(n-1) + a(n-2).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 14, 22, 36, 58, 94, 152, 247, 399, 646, 1045, 1691, 2736, 4428, 7164, 11592, 18756, 30348, 49104, 79453, 128557, 208010, 336567, 544577, 881144, 1425722, 2306866, 3732588, 6039454, 9772042, 15811496, 25583539, 41395035, 66978574, 108373609

Offset: 0

N. J. A. Sloane, May 25 2008

nonn

Also: convolution of A000045 with the period-6 sequence (0,0,0,0,0,0, 1,...). - R. J. Mathar, May 30 2008
Sequences of the form s(0)=a, s(1)= b, s(n) = s(n-1) + s(n-2) + k if n mod m = p, else s(n) = s(n-1) + s(n-2) have a form s(n) = fibonacci(n-1)*a + fibonacci(n)*b + P(x)*k. a(n) is the P(x) sequence for m=6. s(n) = fib(n)*a + fib(n-1)*b + a(n-6-p)*k. - Gary Detlefs, Dec 05 2010
a(n) is the number of compositions of n where the order of the 2 and the 3 does not matter. - Gregory L. Simay, May 18 2017

			Since 5 is not a multiple of 6, a(5) = a(4) + a(3) = 5 + 3 = 8. Since 6 is a multiple of 6, a(6) = a(5) + a(4) + 1 = 8 + 5 + 1 = 14. - _Michael B. Porter_, Jul 26 2016

H. Matsui et al., Problem B-1035, Fibonacci Quarterly, Vol. 45, Number 2; 2007; p. 182.
Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,0,1,-1,-1).

Cf. A052952, A004695, A080239, A124502, A066983.

A131132:=proc(n) option remember; local t1; if n <= 2 then RETURN(1); fi: if n mod 6 = 1 then t1:=1 else t1:=0; fi: procname(n-1)+procname(n-2)+t1; end; [seq(A131132(n), n=1..100)]; # N. J. A. Sloane, May 25 2008; Typo corrected by R. J. Mathar, May 30 2008

Print[Table[Round[(1 + Sqrt[5])/8 Fibonacci[n + 3]], {n, 0, 50}]] ;
Print[RecurrenceTable[{c[n] == c[n - 1] + c[n - 2] + c[n - 6] - c[n - 7] - c[n - 8], c[0] == 1, c[1] == 1, c[2] == 2, c[3] == 3, c[4] == 5, c[5] == 8, c[6] == 14, c[7] == 22}, c, {n, 0, 50}]] ;  (* John M. Campbell, Jul 07 2016 *)
LinearRecurrence[{1, 1, 0, 0, 0, 1, -1, -1}, {1, 1, 2, 3, 5, 8, 14, 22}, 40] (* Vincenzo Librandi, Jul 07 2016 *)

O.g.f.: 1/((1-x^6)(1 - x - x^2)). - R. J. Mathar, May 30 2008
a(n) = ((-1)^n-1)/6 + A099837(n+3)/12 + A000045(n+4)/4 + A057079(n)/12. - R. J. Mathar, Dec 07 2010
a(n) = floor(A066983(n+4)/3). - Gary Detlefs, Dec 19 2010
a(n) = round((1 + sqrt(5))/8 A000045(n+3)). - John M. Campbell, Jul 06 2016
a(n) = (number of compositions of n consisting of only 1 or 2 or 6) - (number of compositions with only 7 or ((1 or 2) and 7)) - (number of compositions with only 8 or ((1 or 2) and 8)). The "or" is inclusive. - Gregory L. Simay, May 25 2017

More specific name from R. J. Mathar, Dec 09 2009

A143975 a(n) = floor(n*(n+3)/3).

Original entry on oeis.org

1, 3, 6, 9, 13, 18, 23, 29, 36, 43, 51, 60, 69, 79, 90, 101, 113, 126, 139, 153, 168, 183, 199, 216, 233, 251, 270, 289, 309, 330, 351, 373, 396, 419, 443, 468, 493, 519, 546, 573, 601, 630, 659, 689, 720, 751, 783, 816, 849, 883, 918, 953, 989, 1026, 1063, 1101

Offset: 1

Clark Kimberling, Sep 06 2008

Fourth diagonal of A143974, associated with counting unit squares in a lattice.

			Main diagonal of A143974: (0,1,3,5,8,12,...) = A000212;
2nd diagonal: (0,2,4,6,10,14,18,...) = A128422;
3rd diagonal: (1,2,5,8,11,16,21,...) = A032765;
4th diagonal: (1,3,6,9,13,18,23,...) = A143975.

Vincenzo Librandi, Table of n, a(n) for n = 1..3000
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A099837, A143974.

[Floor(n*(n+3)/3): n in [1..60]]; // Vincenzo Librandi, May 08 2011

a[n_] := Floor[n*(n+3)/3]; Array[a, 60] (* Amiram Eldar, Oct 01 2022 *)

a(n) = floor(n*(n+3)/3).
From R. J. Mathar, Oct 05 2009: (Start)
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
G.f.: x*(-1 - x - x^2 + x^3)/( (1 + x + x^2) * (x-1)^3). (End)
9*a(n) = 3*n^2 + 9*n - 2 + A099837(n+3). - R. J. Mathar, Apr 26 2022
Sum_{n>=1} 1/a(n) = 4/3 + (tan((sqrt(13)+2)*Pi/6) - cot((sqrt(13)+1)*Pi/6)) * Pi/sqrt(13). - Amiram Eldar, Oct 01 2022
E.g.f.: (exp(x)*(3*x*(4 + x) - 2) + 2*exp(-x/2)*cos(sqrt(3)*x/2))/9. - Stefano Spezia, Oct 24 2022

A186575 Expansion of (1 + 2x + 6x^2)/(1 - x - x^2 - 2*x^3) in powers of x.

Original entry on oeis.org

1, 3, 10, 15, 31, 66, 127, 255, 514, 1023, 2047, 4098, 8191, 16383, 32770, 65535, 131071, 262146, 524287, 1048575, 2097154, 4194303, 8388607, 16777218, 33554431, 67108863, 134217730, 268435455, 536870911, 1073741826, 2147483647, 4294967295

Offset: 0

Vladimir Kruchinin, Feb 23 2011

From Kai Wang, May 23 2020: (Start)
Let f(t) = t^3 + u*t^2 + v*t + w and {x,y,z} be the simple roots of f(t).
For n >= 0, let p(n) = x^n/((x-y)*(x-z)) + y^n/((y-x)*(y-z)) + z^n/((z-x)*(z-y)) and q(n) = x^n + y^n + z^n.
Then for n >= 0, q(n) = 3*p(n+2) + 2*u*p(n+1) + v*p(n).
In this case, f(t) = t^3 - t^2 - t - 2. q(n) = 3*p(n+2) - 2*p(n+1) - p(n).
p(n) = {0, 0, 1, 1, 2, 5, 9,...}, q(n) = {3, 1, 3, 10, 15, 31,...}.
a(n) = q(n+1), A077939(n) = p(n+2). (End)

			G.f. = 1 + 3*x + 10*x^2 + 15*x^3 + 31*x^4 + 66*x^5 + 127*x^6 + 255*x^7 + ...

Colin Barker, Table of n, a(n) for n = 0..1000
Gamaliel Cerda-Morales, A note on Modified Third-order Jacobsthal numbers, arXiv:1905.00725 [math.CO], 2019. See pp. 3-4.
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.
Evren Eyican Polatlı and Yüksel Soykan, On generalized third-order Jacobsthal numbers, Asian Res. J. of Math. (2021) Vol. 17, No. 2, 1-19, Article No. ARJOM.66022.
Kai Wang, Closed Forms and Generating Functions For Power Sums, 2020.
Index entries for linear recurrences with constant coefficients, signature (1,1,2).

Cf. A099837.

R:=PowerSeriesRing(Integers(), 35); Coefficients(R!( (1 + 2*x + 6*x^2)/(1 - x - x^2 - 2*x^3))); // Marius A. Burtea, Jan 31 2020

CoefficientList[Series[(1+2x+6x^2)/(1-x-x^2-2x^3),{x,0,40}],x]  (* Harvey P. Dale, Mar 14 2011 *)

Vec((1 + 2*x + 6*x^2) / ((1 - 2*x)*(1 + x + x^2)) + O(x^40)) \\ Colin Barker, May 03 2019

polsym(polrecip(1 - x - x^2 - 2*x^3),44)[^1] \\ Joerg Arndt, Jun 23 2020

a(n+1) = n*Sum_{k=1..n} Sum_{j=n-3*k..k} 2^(k-j)*binomial(j,n-3*k+2*j)*binomial(k,j)/k.
G.f.: [log(1/(1 - x - x^2 - 2*x^3))]', (x + x^2 + 2*x^3)^k = Sum_{n>=k} Sum_{j=n-3*k..k} 2^(k-j)*binomial(j,n-3*k+2*j)*binomial(k,j)*x^n (see link).
a(n) = 2^(n+1) + A099837(n+1). - R. J. Mathar, Mar 18 2011
a(n) = a(n-1) + a(n-2) + 2*a(n-3) for n>2. - Colin Barker, May 03 2019
From Kai Wang, May 23 2020: (Start)
a(n) = 3*A077947(n+1) - 2*A077947(n) - A077947(n-1).
A077947(n) = (-8*a(n+3) + 27*a(n+2) - a(n+1))/147. (End)

More terms from Harvey P. Dale, Mar 14 2011

A192004 Alternating row sums of array A187360: minimal polynomial of 2*cos(Pi/n) evaluated for x=-1.

Original entry on oeis.org

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

Offset: 1

Wolfdieter Lang, Jul 14 2011

It seems that after a(1) = 1, -1's occur only at the positions 2^k (with k >= 1) and -2's only at positions 3*2^k (with k >= 0, A007283), with everything else being 1. It would be nice to know whether this is true. - Antti Karttunen, May 27 2017
From Wolfdieter Lang, May 29 2017: (Start)
The preceding conjecture can be checked by using for even n Theorem 1A, eq. (41), and for odd n Theorem 2A, eq. (50) of the W. Lang arXiv link given in A187360 putting x = -1.
One uses for the polynomials that (A127672) and q (A130777) appearing there the result that(n, -1) = A099837(n+3), i.e., = 2 if n == 0 (mod 3), = -1 if n == 1 or 2 (mod 3), and q(n, -1) = A061347(n+2), i.e., = 1 if n == 0 or 2 (mod 3) and = -2 if n == 1 (mod 3).
E.g., n = 2^k, k >= 1: C(2^k, -1) = that(2^(k-1), -1) = -1 because 2^(k-1) == 1 or 2 (mod 3).
n = 3*2^k, k >= 1: C(2^k*3) = that(2^(k-1)*3, -1) / that(2^(k-1), -1) = 2/(-1) = -2 because 2^(k-1)*3 == 0 (mod 3), and the previous congruence. C(3, -1) = -2 also, by theorem 2A, see the next example.
n = 3^k, k >= 1: C(3^k, -1) = q((3^k-1)/2, -1) / q((3^(k-1)-1)/2, -1) = (-2)/1 = -2 if k = 1, and = (-2)/(-2) = +1 if k >= 2. (End)

R. J. Mathar, Table of n, a(n) for n = 1..220

Cf. A055034, A192003.

a(n) = Sum_{m=0..A055034(n)} (-1)^m*A187360(n,m), n >= 1.

a(n) = C(n,x=-1), with the minimal (monic and integer) polynomial C(n,x) of 2*cos(Pi/n).

A213172 Floor of the Euclidean distance of a point on the (1, 2, 3; 4, 5, 6) 3D walk.

Original entry on oeis.org

0, 1, 2, 3, 6, 9, 12, 16, 21, 26, 32, 38, 45, 52, 61, 69, 78, 88, 99, 110, 121, 133, 146, 159, 173, 188, 203, 218, 234, 251, 268, 286, 305, 324, 343, 364, 384, 406, 428, 450, 473, 497, 521, 546, 571, 597, 624, 651, 679, 707, 736, 765, 795, 826, 857

Offset: 0

Jon Perry, Apr 14 2013

nonn

Consider a standard 3-dimensional Euclidean lattice. We take 1 step along the positive x-axis, 2 along the positive y-axis, 3 along the positive z-axis, 4 along the positive x-axis, and so on. This sequence gives the floor of the Euclidean distance to the origin after n steps.

The (x,y,z) coordinates are (1,0,0), (1,2,0), (1,2,3), (5,2,3), (5,7,3), (5,7,9), (12,7,9) etc, where the x values run through A000326, the y-values through A005449, and the z-values through A045943. The squared Euclidean distances are s(n) = 1, 5, 14, 38, 83, 155, 274, 450,..., which obey the recurrence s(n) = 3*s(n-1) -3*s(n-2) +3*s(n-3) -6*s(n-4) +6*s(n-5) -3*s(n-6) +3*s(n-7) -3*s(n-8) +s(n-9), s(n) = (3*n^2+9*n+10)^2/108 +4*A099837(n+3)/27 -2*(-1)^n*A165202(n)/9, with a = floor(sqrt(s(n))). - R. J. Mathar, May 02 2013

			For a(4) we are at [5,2,3], so a(n) = floor(sqrt(25+4+9)) = 6.

Cf. A054925, A224985, A225215.

p=new Array(0,0,0);
for (a=0;a<100;a++) {
p[a%3]+=a;
document.write(Math.floor(Math.sqrt(p[0]*p[0]+p[1]*p[1]+p[2]*p[2]))+", ");
}

a(n) ~ n^2 sqrt(3)/6. - Charles R Greathouse IV, May 02 2013

A264938 a(n) = n(2n-1) + floor(n/3).

Original entry on oeis.org

0, 1, 6, 16, 29, 46, 68, 93, 122, 156, 193, 234, 280, 329, 382, 440, 501, 566, 636, 709, 786, 868, 953, 1042, 1136, 1233, 1334, 1440, 1549, 1662, 1780, 1901, 2026, 2156, 2289, 2426, 2568, 2713, 2862, 3016, 3173, 3334, 3500, 3669, 3842, 4020, 4201, 4386, 4576, 4769

Offset: 0

Paul Curtz, Nov 29 2015

Sequence extended to the left:
..., 133, 102, 76, 53, 34, 20, 9, 2, 0, 1, 6, 16, 29, 46, 68, 93, ...
Conjecture: after 0, a(n) provides the first bisection of A264041.
Conjecture: 2, 9, 20, 34, 53, 76, 102, 133, ... is A248121.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Second bisection of A260708.
Cf. A000217, A002264, A016777, A016969, A017197, A017641, A099837, A248121, A256320, A260699, A264041.
Cf. A171452: A000217(n-1) + A002264(n).

[n*(2*n-1)+Floor(n/3): n in [0..60]]; // Vincenzo Librandi, Dec 02 2015

seq(n*(2*n-1) + floor(n/3), n=0..100); # Robert Israel, Dec 02 2015

Table[n (2 n - 1) + Floor[n/3], {n, 0, 50}] (* Vincenzo Librandi, Dec 02 2015 *)
LinearRecurrence[{2,-1,1,-2,1},{0,1,6,16,29},60] (* Harvey P. Dale, Oct 13 2020 *)

concat(0, Vec(x*(1+x)^2*(1+2*x)/((1-x)^3*(1+x+x^2)) + O(x^100))) \\ Colin Barker, Dec 02 2015

a(n) = n*(2*n-1) + n\3; \\ Altug Alkan, Dec 01 2015

a(n) = a(n-3) + 12*n - 20 for n>2.
From Colin Barker, Dec 02 2015: (Start)
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5) for n>4.
G.f.: x*(1+x)^2*(1+2*x) / ((1-x)^3*(1+x+x^2)).
(End)
a(n) = A000217(2n-1) + A002264(n).
a(n) + a(-n) = 3*A256320(n).
a(n +8) - a(n -7) = 20*A016777(n).
a(n+16) - a(n-14) = 20*A016969(n).
a(n+23) - a(n-22) = 20*A017197(n).
a(n+31) - a(n-29) = 20*A017641(n).
Generalization of the previous four formulas:
a(n+30*k +8) - a(n-30*k -7) = 20*(4*k+1)*(3*n+1).
a(n+30*k+16) - a(n-30*k-14) = 20*(2*k+1)*(6*n+5).
a(n+30*k+24) - a(n-30*k-21) = 20*(4*k+3)*(3*n+4).
a(n+30*k+32) - a(n-30*k-28) = 20*(2*k+2)*(6*n+11).
E.g.f.: (6*x^2+4*x-1)*exp(x)/3 + (cos(sqrt(3)*x/2)/3 +sqrt(3)*sin(sqrt(3)*x/2)/9)*exp(-x/2). - Robert Israel, Dec 02 2015

Edited by Bruno Berselli, Dec 01 2015

A382682 Number of integer partitions of n with origin-to-boundary graph-distance equal to 3.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 4, 8, 15, 23, 32, 43, 54, 67, 82, 97, 114, 133, 152, 173, 196, 219, 244, 271, 298, 327, 358, 389, 422, 457, 492, 529, 568, 607, 648, 691, 734, 779, 826, 873, 922, 973, 1024, 1077, 1132, 1187, 1244, 1303, 1362, 1423, 1486, 1549, 1614, 1681, 1748, 1817, 1888, 1959

Offset: 0

N Guru Sharan, Jun 03 2025

Also the number of partitions of n with a fixed Durfee triangle of size 3.

Column k=3 of the triangle in A325188.

Michael De Vlieger, Table of n, a(n) for n = 0..10000
N. Guru Sharan and Armin Straub, Partitions with Durfee triangles of fixed size, arXiv:2507.19047 [math.CO], 2025. See p. 10.
Index entries for linear recurrences with constant coefficients, signature (2,-1,1,-2,1).

Cf. A130130, A325168, A325188.

CoefficientList[Series[(q^6 + 2 q^7 + q^8 + 2 q^9 - q^10 - q^12 - q^13 + q^14)/((1 - q)^3 (1 + q + q^2)), {q, 0, 200}],q]

G.f.: q^6*(1 + 2*q + q^2 + 2*q^3 - q^4 - q^6 - q^7 + q^8)/((1 - q)^3*(1 + q + q^2)).

9*a(n) = 2*A099837(n+3)+6*n^2+59-45*n for n>9. - R. J. Mathar, Jun 24 2025

A008749 Expansion of (1+x^6)/((1-x)(1-x^2)(1-x^3)).

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 8, 9, 12, 15, 18, 21, 26, 29, 34, 39, 44, 49, 56, 61, 68, 75, 82, 89, 98, 105, 114, 123, 132, 141, 152, 161, 172, 183, 194, 205, 218, 229, 242, 255, 268, 281, 296, 309, 324, 339, 354, 369

Offset: 0

N. J. A. Sloane

nonn

Conjecture: For n >= 1, A067628(a(n+2)) appears for the first time in A067628. Equivalently, A067628(a(n+2)) is the first T such that the minimal perimeter of polyiamonds of T triangles is a(n+2). - Winston C. Yang (winston(AT)cs.wisc.edu), Feb 05 2002

			Let n = 8. Then a(n+2) = a(10) = 18. Note A067628(18) = 12 and is the first appearance of 12 in A067628. Equivalently, 12 is the first T such that the min perimeter of polyiamonds of T triangles is 18.

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

Cf. A067628.

a:=[1,1,2,3,4,5];; for n in [7..60] do a[n]:=a[n-1]+a[n-2]-a[n-4] -a[n-5]+a[n-6]; od; a; # G. C. Greubel, Aug 03 2019

R:=PowerSeriesRing(Integers(), 60); Coefficients(R!( (1+x^6)/((1-x)*(1-x^2)*(1-x^3)) )); // G. C. Greubel, Aug 03 2019

CoefficientList[Series[(1+x^6)/((1-x)*(1-x^2)*(1-x^3)), {x,0,60}], x] (* G. C. Greubel, Aug 03 2019 *)

my(x='x+O('x^60)); Vec((1+x^6)/((1-x)*(1-x^2)*(1-x^3))) \\ G. C. Greubel, Aug 03 2019

((1+x^6)/((1-x)*(1-x^2)*(1-x^3))).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Aug 03 2019

Conjecture: Let b(n>=0) = (0, 1, 1, 1, 1, 3, 1, 3, 3, 3, 3, 5, 3, 5, 5, 5, 5, 7, 3, ...). Equivalently, let b(0) = 0, b(n>=1) = 2*floor((n-1)/6) + 1 + (2 if n+1=0 mod 6; 0 else). Then a(0) = 1, a(n>=1) = a(n-1) + b(n-1). - Winston C. Yang (winston(AT)cs.wisc.edu), Feb 05 2002

a(n) = (47 + 6*n^2 + 9*(-1)^n + 8*A099837(n+3))/36, n>0. - R. J. Mathar, Jun 24 2009

cp's OEIS Frontend

A094717 a(n) = n! * Sum_{i+2j+3k=n} 1/(i!*(2j)!*(3k)!).

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A117185 Riordan array ((1-x^2)/(1+x+2x^2+x^3+x^4),x/(1+x^2)).

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A131132 a(n) = a(n-1) + a(n-2) + 1 if n is a multiple of 6, otherwise a(n) = a(n-1) + a(n-2).

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A143975 a(n) = floor(n*(n+3)/3).

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A186575 Expansion of (1 + 2*x + 6*x^2)/(1 - x - x^2 - 2*x^3) in powers of x.

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A192004 Alternating row sums of array A187360: minimal polynomial of 2*cos(Pi/n) evaluated for x=-1.

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A213172 Floor of the Euclidean distance of a point on the (1, 2, 3; 4, 5, 6) 3D walk.

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A264938 a(n) = n*(2*n-1) + floor(n/3).

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A094717 a(n) = n! * Sum_{i+2j+3k=n} 1/(i!(2j)!(3k)!).

A186575 Expansion of (1 + 2x + 6x^2)/(1 - x - x^2 - 2*x^3) in powers of x.

A264938 a(n) = n(2n-1) + floor(n/3).

A008749 Expansion of (1+x^6)/((1-x)(1-x^2)(1-x^3)).