A005169 -id:A005169 - cp's OEIS frontend

A230370 Voids left after packing 3 curves coins patterns (3c3s type) into fountain of coins base n.

0, 0, 3, 6, 13, 19, 39, 54, 66, 85, 100, 123, 141, 168, 189, 220, 244, 279, 306, 345, 375, 418, 451, 498, 534, 585, 624, 679, 721, 780, 825, 888, 936, 1003, 1054, 1125, 1179, 1254, 1311, 1390, 1450, 1533, 1596, 1683

Offset: 1

Kival Ngaokrajang, Oct 17 2013

nonn

Refer to arrangement same as A005169: "A fountain is formed by starting with a row of coins, then stacking additional coins on top so that each new coin touches two in the previous row". The 3 curves coins patterns consist of a part of each coin circumference and forms a continuous area. There are total 4 distinct patterns. For selected pattern, I would like to call "3c3s" type as it cover 3 coins and symmetry. When packing 3c3s into fountain of coins base n, the total number of 3c3s is A008805, the coins left is A008795 and voids left is a(n). See illustration in links.

Kival Ngaokrajang, Illustration of initial terms (V)

A001399, A230267, A230276 (5-curves coins patterns); A074148, A229093, A220154 (4-curves coins patterns); A008795 (3-curves coins patterns).

G.f.: x^3*(11*x^8 - 5*x^7 - 21*x^6 + 6*x^5 + 9*x^4 + x^2 + 3*x + 3)/((1-x)*(1-x^2)^2) (conjectured). Ralf Stephan, Oct 19 2013

A291148 Expansion of 1 - x/(1 - x^2/(1 - x^3/(1 - x^4/(1 - x^5/(1 - x^6/(1 - ... - x^n/(1 - ...))))))), a continued fraction.

Original entry on oeis.org

1, -1, 0, -1, 0, -1, -1, -1, -2, -2, -4, -4, -7, -9, -13, -19, -25, -38, -51, -75, -104, -149, -211, -298, -426, -600, -857, -1211, -1724, -2444, -3471, -4930, -6995, -9940, -14104, -20038, -28444, -40397, -57362, -81453, -115675, -164250, -233262, -331227

Offset: 0

Seiichi Manyama, Aug 18 2017

sign

The sequence b(n>=1) = 1, 0, 1, 0, 1, 1, 1, 2, 2, 4, 4, 7, ... of absolute values counts fountains of n coins that cannot be separated into two or more fountains by cutting vertically through the fountain without splitting a coin. (This separation requires that the fountain is a left-right sequence of more elementary fountains counted by b(n).) A005169(n) = Sum_{compositions n=n1+n2+n3+...} Product b(n1)*b(n2)*.... - R. J. Mathar, Aug 22 2018

			G.f. = 1 - x - x^3 - x^5 - x^6 - x^7 - 2*x^8 - 2*x^9 - ...

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A005169, A290975, A291147.

Convolution inverse of A005169.

a(n) ~ c * d^n, where d = 1.42009048763893649946106129818306075366296460727614... and c = -0.093433697175825717154301151812109730023054876584907211486145769... - Vaclav Kotesovec, Oct 16 2017

A138158 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and path length k; 0 <= k <= n(n+1)/2.

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 2, 1, 1, 0, 0, 0, 0, 1, 3, 3, 3, 2, 1, 1, 0, 0, 0, 0, 0, 1, 4, 6, 7, 7, 5, 5, 3, 2, 1, 1, 0, 0, 0, 0, 0, 0, 1, 5, 10, 14, 17, 16, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 6, 15, 25, 35, 40, 43, 44, 40, 37, 32, 28, 22, 18, 13, 11, 7, 5, 3, 2, 1, 1

Offset: 0

Emeric Deutsch, Mar 21 2008

T(n,k) is the number of Dyck paths of semilength n for which the sum of the heights of the vertices that terminate an upstep (i.e. peaks and doublerises) is k. Example: T(4,7)=3 because we have UUDUDUDD, UDUUUDDD and UUUDDDUD.
See related triangle A227543.
Row n contains 1+n(n+1)/2 terms.
The maximum in each row of the triangle is A274291. - Torsten Muetze, Nov 28 2018
It appears that for j = 0,1,...,n-1 the first j terms of the rows in reversed order are given by A000041(j), the partition numbers. - Geoffrey Critzer, Jul 14 2020

			T(2,2)=1 because /\ is the only ordered tree with 2 edges and path length 2.
Triangle starts
 1,
 0, 1,
 0, 0, 1, 1,
 0, 0, 0, 1, 2, 1, 1,
 0, 0, 0, 0, 1, 3, 3, 3, 2, 1, 1,
 0, 0, 0, 0, 0, 1, 4, 6, 7, 7, 5, 5, 3, 2, 1, 1,
 0, 0, 0, 0, 0, 0, 1, 5, 10, 14, 17, 16, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1,
 0, 0, 0, 0, 0, 0, 0, 1, 6, 15, 25, 35, 40, 43, 44, 40, 37, 32, 28, 22, 18, 13, 11, 7, 5, 3, 2, 1, 1,
... [_Joerg Arndt_, Feb 21 2014]

Seiichi Manyama, Rows n = 0..38, flattened
Ron M. Adin and Yuval Roichman, On maximal chains in the non-crossing partition lattice, arXiv:1201.4669 [math.CO], 2012-2013.
Luca Ferrari, Unimodality and Dyck paths, arXiv:1207.7295 [math.CO], 2012.
FindStat - Combinatorial Statistic Finder, The bounce statistic of a Dyck path, The dinv statistic of a Dyck path, The area of a Dyck path.
Philippe Flajolet and Robert Sedgewick, Analytic Combinatorics, Cambridge Univ. Press, 2009, page 185.

Cf. A227543, A000108, A005169, A000346, A047998, A274291.

P[0]:=1: for n to 7 do P[n]:=sort(expand(t*(sum(P[j]*P[n-j-1]*t^(n-j-1),j= 0.. n-1)))) end do: for n from 0 to 7 do seq(coeff(P[n], t, j),j=0..(1/2)*n*(n+1)) end do; # yields sequence in triangular form

nmax = 7;
P[0] = 1; P[n_] := P[n] = t*Sum[P[j]*P[n-j-1]*t^(n-j-1), {j, 0, n-1}];
row[n_] := row[n] = CoefficientList[P[n] + O[t]^(n(n+1)/2 + 1), t];
T[n_, k_] := row[n][[k+1]];
Table[T[n, k], {n, 0, nmax}, {k, 0, n(n+1)/2}] // Flatten (* Jean-François Alcover, Jul 11 2018, from Maple *)
nn = 10; f[z_, u_] := Sum[Sum[a[n, k] u^k z^n, {k, 0, Binomial[n, 2]}], {n, 1, nn}]; sol = SolveAlways[Series[0 == f[z, u] - z/(1 - f[u z, u]) , {z, 0, nn}], {z, u}];Level[Table[Table[a[n, k], {k, 0, Binomial[n, 2]}], {n, 1, nn}] /.
sol, {2}] // Grid (* Geoffrey Critzer, Jul 14 2020 *)

G.f. G(t,z) satisfies G(t,z) = 1+t*z*G(t,z)*G(t,t*z).
Row generating polynomials P[n]=P[n](t) are given by P[0]=1, P[n] = t * Sum( P[j]*P[n-j-1]*t^(n-1-j), j=0..n-1 ) (n>=1).
Row sums are the Catalan numbers (A000108).
Sum of entries in column n = A005169(n).
Sum_{k=0..n(n+1)/2} k*T(n,k) = A000346(n-1).
T(n,k) = A047998(k,n).
G.f.: 1/(1 - x*y/(1 - x*y^2/(1 - x*y^3/(1 - x*y^4/(1 - x*y^5)/(1 - ... ))))), a continued fraction. - Ilya Gutkovskiy, Apr 21 2017

A005166 a(0) = 1; a(n) = (1 + a(0)^3 + ... + a(n-1)^3)/n (not always integral!).

Original entry on oeis.org

1, 2, 5, 45, 22815, 2375152056927, 2233176271342403475345148513527359103

Offset: 0

N. J. A. Sloane

Terms are integers until n=A097398(2,2)=89.

Guy states that by computing the sequence modulo 89 it is easy to show that a(89) is not integral. - T. D. Noe, Sep 17 2007

R. K. Guy, Unsolved Problems in Number Theory, Springer, 1st edition, 1981. See section E15.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n=0..9
R. K. Guy, Letter to N. J. A. Sloane, Sep 25 1986.
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712.
R. K. Guy, The strong law of small numbers. Amer. Math. Monthly 95 (1988), no. 8, 697-712. [Annotated scanned copy]
N. Lygeros & M. Mizony, Study of primality of terms of a_k(n)=(1+(sum from 1 to n-1)(a_k(i)^k))/(n-1) [dead link]
Alex Stone, The Astonishing Behavior of Recursive Sequences, Quanta Magazine, Nov 16 2023, 13 pages.
Eric Weisstein's World of Mathematics, Goebel's Sequence.

Cf. A003504, A005167.

Cf. A108394.

a[0]=1; a[n_]:=(1 + Sum[a[k]^3, {k,0,n-1}])/n; Array[a,7,0] (* Stefano Spezia, Oct 13 2024 *)

A088357 G.f. = continued fraction: A(x)=1/(1-x/(1-2x^2/(1-3x^3/(1-4*x^4/(...))))).

Original entry on oeis.org

1, 1, 1, 3, 5, 11, 27, 55, 127, 285, 647, 1457, 3297, 7489, 16945, 38523, 87293, 198179, 449907, 1021135, 2318527, 5263581, 11950967, 27133985, 61609953, 139888777, 317629465, 721215027, 1637598485, 3718378619, 8443065363, 19171129327

Offset: 0

Paul D. Hanna, Sep 26 2003

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..100

Cf. A005169, A285381.

nmax = 50; CoefficientList[Series[1/Fold[(1 - #2/#1) &, 1, Reverse[Range[nmax + 1]*x^Range[nmax + 1]]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 25 2017 *)

S=1; L=30; for(k=1,L,m=L-k+1; S=1/(1-m*x^m*S)+x*O(x^L)); A(x)=S; a(n)=polcoeff(A(x),n,x)

G.f.: T(0), where T(k) = 1 - x^(k+1)*(k+1)/( x^(k+1)*(k+1) - 1/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 26 2013

a(n) ~ c * d^n, where d = 2.2706470084004562621321821916243432273516... and c = 0.1745837410025587240288929391139119506... - Vaclav Kotesovec, Aug 25 2017

A192729 G.f. satisfies: A(x) = 1/(1 - xA(x)^2/(1 - x^2A(x)^2/(1 - x^3A(x)^2/(1 - x^4A(x)^2/(1 - ...))))), a recursive continued fraction.

Original entry on oeis.org

1, 1, 3, 13, 63, 329, 1808, 10299, 60271, 360198, 2189111, 13488379, 84066176, 529037390, 3357014851, 21455604032, 137993279809, 892448240335, 5800266701499, 37864046563210, 248158092634265, 1632254493141021, 10771183395497445

Offset: 0

Paul D. Hanna, Jul 08 2011

nonn

			G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 63*x^4 + 329*x^5 + 1808*x^6 +...
which satisfies A(x) = P(x)/Q(x) where
P(x) = 1 - x^2*A(x)^2/(1-x) + x^6*A(x)^4/((1-x)*(1-x^2)) - x^12*A(x)^6/((1-x)*(1-x^2)*(1-x^3)) + x^20*A(x)^8/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) -+...
Q(x) = 1 - x*A(x)^2/(1-x) + x^4*A(x)^4/((1-x)*(1-x^2)) - x^9*A(x)^6/((1-x)*(1-x^2)*(1-x^3)) + x^16*A(x)^8/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) -+...
Explicitly, the above series begin:
P(x) = 1 - x^2 - 3*x^3 - 10*x^4 - 42*x^5 - 202*x^6 - 1060*x^7 - 5862*x^8 - 33592*x^9 - 197585*x^10 - 1185867*x^11 - 7233049*x^12 +...
Q(x) = 1 - x - 3*x^2 - 10*x^3 - 41*x^4 - 198*x^5 - 1041*x^6 - 5766*x^7 - 33074*x^8 - 194674*x^9 - 1168988*x^10 - 7132869*x^11 - 44097821*x^12 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..300
Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction.

Cf. A005169, A192728, A192730.

/* As a recursive continued fraction: */
{a(n)=local(A=1+x,CF);for(i=1,n,CF=1+x;for(k=0,n,CF=1/(1-x^(n-k+1)*A^2*CF+x*O(x^n)));A=CF);polcoeff(A,n)}

/* By Ramanujan's continued fraction identity: */
{a(n)=local(A=1+x,P,Q);for(i=1,n,
P=sum(m=0,sqrtint(n),x^(m*(m+1))/prod(k=1,m,1-x^k)*(-A^2+x*O(x^n))^m);
Q=sum(m=0,sqrtint(n),x^(m^2)/prod(k=1,m,1-x^k)*(-A^2+x*O(x^n))^m);A=P/Q);polcoeff(A,n)}

G.f. satisfies: A(x) = P(x)/Q(x) where
  P(x) = Sum_{n>=0} x^(n*(n+1)) * (-A(x)^2)^n / Product_{k=1..n} (1-x^k),
  Q(x) = Sum_{n>=0} x^(n^2) * (-A(x)^2)^n / Product_{k=1..n} (1-x^k),
due to Ramanujan's continued fraction identity.
a(n) ~ c * d^n / n^(3/2), where d = 7.0656326355634513691927118582399... and c = 0.2386935555822482686868972746... - Vaclav Kotesovec, Aug 25 2017

A192730 G.f. satisfies: A(x) = 1/(1 - xA(x)^3/(1 - x^2A(x)^3/(1 - x^3A(x)^3/(1 - x^4A(x)^3/(1 - ...))))), a recursive continued fraction.

Original entry on oeis.org

1, 1, 4, 23, 151, 1075, 8075, 62996, 505501, 4145684, 34594540, 292794156, 2507383158, 21686318745, 189162110341, 1662142617881, 14698913545378, 130723572694407, 1168419986539867, 10490326933563842, 94564400499455397, 855552893388047193

Offset: 0

Paul D. Hanna, Jul 08 2011

nonn

			G.f.: A(x) = 1 + x + 4*x^2 + 23*x^3 + 151*x^4 + 1075*x^5 + 8075*x^6 +...
which satisfies A(x) = P(x)/Q(x) where
P(x) = 1 - x^2*A(x)^3/(1-x) + x^6*A(x)^6/((1-x)*(1-x^2)) - x^12*A(x)^9/((1-x)*(1-x^2)*(1-x^3)) + x^20*A(x)^12/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) -+...
Q(x) = 1 - x*A(x)^3/(1-x) + x^4*A(x)^6/((1-x)*(1-x^2)) - x^9*A(x)^9/((1-x)*(1-x^2)*(1-x^3)) + x^16*A(x)^12/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) -+...
Explicitly, the above series begin:
P(x) = 1 - x^2 - 4*x^3 - 19*x^4 - 113*x^5 - 763*x^6 - 5557*x^7 - 42472*x^8 - 335804*x^9 - 2723164*x^10 - 22523476*x^11 - 189267247*x^12 +...
Q(x) = 1 - x - 4*x^2 - 19*x^3 - 112*x^4 - 757*x^5 - 5517*x^6 - 42188*x^7 - 333673*x^8 - 2706555*x^9 - 22390279*x^10 - 188175369*x^11 - 1602132261*x^12 +...

Vaclav Kotesovec, Table of n, a(n) for n = 0..280
Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction.

Cf. A005169, A192728, A192729.

/* As a recursive continued fraction: */
{a(n)=local(A=1+x,CF);for(i=1,n,CF=1+x;for(k=0,n,CF=1/(1-x^(n-k+1)*A^3*CF+x*O(x^n)));A=CF);polcoeff(A,n)}

/* By Ramanujan's continued fraction identity: */
{a(n)=local(A=1+x,P,Q);for(i=1,n,
P=sum(m=0,sqrtint(n),x^(m*(m+1))/prod(k=1,m,1-x^k)*(-A^3+x*O(x^n))^m);
Q=sum(m=0,sqrtint(n),x^(m^2)/prod(k=1,m,1-x^k)*(-A^3+x*O(x^n))^m);A=P/Q);polcoeff(A,n)}

G.f. satisfies: A(x) = P(x)/Q(x) where
  P(x) = Sum_{n>=0} x^(n*(n+1)) * (-A(x)^3)^n / Product_{k=1..n} (1-x^k),
  Q(x) = Sum_{n>=0} x^(n^2) * (-A(x)^3)^n / Product_{k=1..n} (1-x^k),
due to Ramanujan's continued fraction identity.
a(n) ~ c * d^n / n^(3/2), where d = 9.72359087408044730447308019524191930733163... and c = 0.151620024312256318854728680725808488795... - Vaclav Kotesovec, Nov 18 2017

A192737 G.f. satisfies: A(x) = 1/(1 - xA(x)/(1 - xA(x)^2/(1 - xA(x)^3/(1 - xA(x)^4/(1 - ...))))), a recursive continued fraction.

Original entry on oeis.org

1, 1, 3, 13, 68, 399, 2531, 16994, 119199, 865849, 6474177, 49616016, 388484212, 3100311228, 25172981053, 207665895001, 1738775327319, 14764815028481, 127076945426555, 1108103873824072, 9787004793441886, 87539719110388691

Offset: 0

Paul D. Hanna, Jul 08 2011

nonn

			G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 68*x^4 + 399*x^5 + 2531*x^6 +...
The g.f. A = A(x) satisfies A = P(x)/Q(x) where
P(x) = 1 - x*A^2/(1-A) + x^2*A^6/((1-A)*(1-A^2)) - x^3*A^12/((1-A)*(1-A^2)*(1-A^3)) + x^4*A^20/((1-A)*(1-A^2)*(1-A^3)*(1-A^4)) -+...
Q(x) = 1 - x*A/(1-A) + x^2*A^4/((1-A)*(1-A^2)) - x^3*A^9/((1-A)*(1-A^2)*(1-A^3)) + x^4*A^16/((1-A)*(1-A^2)*(1-A^3)*(1-A^4)) -+...
Explicitly, the above series begin:
P(x) = exp(1)*(1 - 1/4*x - 283/288*x^2 - 6223/1152*x^3 - 140734037/4147200*x^4 - 3826874463/16588800*x^5 - 290665690804549/175575859200*x^6 +...);
Q(x) = exp(1)*(1 - 5/4*x - 787/288*x^2 - 13731/1152*x^3 - 271921637/4147200*x^4 - 6765586315/16588800*x^5 - 481505562390493/175575859200*x^6 +...).

Vaclav Kotesovec, Table of n, a(n) for n = 0..250
Eric Weisstein's World of Mathematics, Rogers-Ramanujan Continued Fraction.

Cf. A005169, A192728, A192729, A192730.

/* As a recursive continued fraction: */
{a(n)=local(A=1+x, CF); for(i=1, n, CF=1+x; for(k=0, n, CF=1/(1-x*A^(n-k+1)*CF+x*O(x^n))); A=CF); polcoeff(A, n)}

/* By Ramanujan's continued fraction identity: */
{a(n)=local(A=1+x, P, Q); for(i=1, n,
P=sum(m=0, 2*n, (-x)^m*A^(m*(m+1))/prod(k=1, m, 1-A^k)/exp(1)+x*O(x^(2*n)));
Q=sum(m=0, 2*n, (-x)^m*A^(m^2)/prod(k=1, m, 1-A^k)/exp(1)+x*O(x^(2*n))); A=round(P/Q)); polcoeff(A, n)}

G.f. satisfies: A(x) = P(x)/Q(x) where
  P(x) = Sum_{n>=0} (-x)^n * A(x)^(n*(n+1)) / Product_{k=1..n} (1-A(x)^k),
  Q(x) = Sum_{n>=0} (-x)^n * A(x)^(n^2) / Product_{k=1..n} (1-A(x)^k),
due to Ramanujan's continued fraction identity.

A197870 Expansion of false theta series variation of Ramanujan theta function psi(x).

Original entry on oeis.org

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

Offset: 0

Michael Somos, Oct 18 2011

sign

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

			G.f. = 1 - x + x^3 - x^6 + x^10 - x^15 + x^21 - x^28 + x^36 - x^45 + x^55 + ...
G.f. = q - q^9 + q^25 - q^49 + q^81 - q^121 + q^169 - q^225 + q^289 - q^361 + ...

G. C. Greubel, Table of n, a(n) for n = 0..5000
Michael Somos, Introduction to Ramanujan theta functions
Eric Weisstein's World of Mathematics, Ramanujan Theta Functions

Cf. A010054.

Cf. A005169 (g.f.: G(0), where G(k) = 1/( 1 - q^(k+1)*G(k+1) ) ).

a[ n_] := If[ n < 0, 0, SeriesCoefficient[ Sum[ (-1)^k x^(k (k + 1)/2), {k, 0, (Sqrt[8 n + 1] - 1)/2}], {x, 0, n}]]; (* Michael Somos, Jul 21 2014 *)
a[n_] := Module[{r, k}, r = Reduce[k >= 0 && 2n == k(k+1), k, Integers]; If[r === False, 0, (-1)^r[[2]] ] ]; Table[a[n], {n, 0, 104}] (* Jean-François Alcover, Nov 27 2016 *)

{a(n) = local(x); if( issquare( 8*n + 1, &x), (-1)^(x\2), 0)};

G.f.: Sum_{k>=0} (-1)^k * x^(k*(k+1)/2). |a(n)| = A010054(n).
G.f.: G(0) where G(k) = 1 - q^(k+1)*G(k+1)  = 1 - q*(1 - q^2*(1 - q^3*(1 - q^4*(1 - ...)))). - Joerg Arndt, Jun 29 2013
G.f.: Sum_{k>=0} a(k) * x^(3*k) = 1 / (1+x) + x*(1-x) / ((1+x)*(1+x^2)*(1+x^3)) + x^2*(1-x)*(1-x^3) / ((1+x)*(1+x^2)*...*(1+x^5)) + ... - Michael Somos, Jul 21 2014

A224898 G.f.: Sum_{n>=0} (-1)^n* x^(n*(n+1)) / Product_{k=1..n} (1-x^k).

Original entry on oeis.org

1, 0, -1, -1, -1, -1, 0, 0, 1, 1, 2, 2, 2, 2, 2, 1, 1, 0, -1, -2, -2, -4, -4, -5, -5, -6, -5, -6, -4, -4, -3, -2, 1, 1, 4, 5, 8, 9, 12, 12, 15, 15, 17, 16, 18, 15, 16, 13, 13, 8, 7, 1, 0, -7, -9, -17, -19, -27, -29, -37, -38, -46, -46, -53, -51, -57, -53, -57, -51, -53, -45, -45, -32, -31

Offset: 0

Paul D. Hanna, Jul 24 2013

Conjecture: a(n+1) = A286744(n) - A286745(n). - George Beck May 13 2017

			G.f.: A(x) = 1 - x^2 - x^3 - x^4 - x^5 + x^8 + x^9 + 2*x^10 + 2*x^11 + 2*x^12 + 2*x^13 + 2*x^14 + x^15 + x^16 - x^18 +...
where
A(x) = 1 - x^2/(1-x) + x^6/((1-x)*(1-x^2)) - x^12/((1-x)*(1-x^2)*(1-x^3)) + x^20/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) - x^30/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)*(1-x^5)) +...

Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from Paul D. Hanna)

Cf. A227620, A227543, A039924, A005169, A286744, A286745, A003106.

a(n)=polcoeff(sum(m=0, sqrtint(n), (-1)^m*x^(m*(m+1))/prod(k=1, m, 1-x^k,1+x*O(x^n))),n)
for(n=0, 80, print1(a(n), ", "))

cp's OEIS Frontend

A230370 Voids left after packing 3 curves coins patterns (3c3s type) into fountain of coins base n.

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A291148 Expansion of 1 - x/(1 - x^2/(1 - x^3/(1 - x^4/(1 - x^5/(1 - x^6/(1 - ... - x^n/(1 - ...))))))), a continued fraction.

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A138158 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and path length k; 0 <= k <= n(n+1)/2.

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Maple

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A005166 a(0) = 1; a(n) = (1 + a(0)^3 + ... + a(n-1)^3)/n (not always integral!).

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A088357 G.f. = continued fraction: A(x)=1/(1-x/(1-2*x^2/(1-3*x^3/(1-4*x^4/(...))))).

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PARI

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A192729 G.f. satisfies: A(x) = 1/(1 - x*A(x)^2/(1 - x^2*A(x)^2/(1 - x^3*A(x)^2/(1 - x^4*A(x)^2/(1 - ...))))), a recursive continued fraction.

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PARI

PARI

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A192730 G.f. satisfies: A(x) = 1/(1 - x*A(x)^3/(1 - x^2*A(x)^3/(1 - x^3*A(x)^3/(1 - x^4*A(x)^3/(1 - ...))))), a recursive continued fraction.

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PARI

PARI

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A192737 G.f. satisfies: A(x) = 1/(1 - x*A(x)/(1 - x*A(x)^2/(1 - x*A(x)^3/(1 - x*A(x)^4/(1 - ...))))), a recursive continued fraction.

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A088357 G.f. = continued fraction: A(x)=1/(1-x/(1-2x^2/(1-3x^3/(1-4*x^4/(...))))).

A192729 G.f. satisfies: A(x) = 1/(1 - xA(x)^2/(1 - x^2A(x)^2/(1 - x^3A(x)^2/(1 - x^4A(x)^2/(1 - ...))))), a recursive continued fraction.

A192730 G.f. satisfies: A(x) = 1/(1 - xA(x)^3/(1 - x^2A(x)^3/(1 - x^3A(x)^3/(1 - x^4A(x)^3/(1 - ...))))), a recursive continued fraction.

A192737 G.f. satisfies: A(x) = 1/(1 - xA(x)/(1 - xA(x)^2/(1 - xA(x)^3/(1 - xA(x)^4/(1 - ...))))), a recursive continued fraction.