A034299 -id:A034299 - cp's OEIS frontend

A160156 Partial sums of A007583.

1, 4, 15, 58, 229, 912, 3643, 14566, 58257, 233020, 932071, 3728274, 14913085, 59652328, 238609299, 954437182, 3817748713, 15270994836, 61083979327, 244335917290, 977343669141, 3909374676544, 15637498706155, 62549994824598

Offset: 0

Omar E. Pol, May 27 2009

This sequence is one of 104 sequences mentioned in the Lang's paper; see page 4. - Omar E. Pol, Jun 13 2012
Also 1 plus the total number of toothpicks of the first n toothpick structures of A139250 in which the number of exposed toothpicks that are orthogonals to the initial toothpick is equal to 4. - Omar E. Pol, Jun 16 2012
This is the sequence A(1,4;5,-4;-1,n) of the family of sequences [a,b:c,d:k] considered by Gary Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Nov 16 2013

			G.f. = 1 + 4*x + 15*x^2 + 58*x^3 + 229*x^4 + 912*x^5 + 3643*x^6 + ... - _Michael Somos_, Oct 16 2020

Hacène Belbachir and El-Mehdi Mehiri, Enumerating moves in the optimal solution of the Tower of Hanoi, arXiv:2210.08657 [math.CO], 2022.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS
Index entries for sequences related to toothpick sequences
Index entries for linear recurrences with constant coefficients, signature (6,-9,4).

Cf. A002450, A007583, A034299, A139250.

a := proc (n) options operator, arrow: (1/3)*n+1/9+(1/9)*2^(2*n+3) end proc: seq(a(n), n = 0 .. 25); # Emeric Deutsch, Jun 20 2009

LinearRecurrence[{6,-9,4},{1,4,15},30] (* Harvey P. Dale, Oct 04 2018 *)

{a(n) = (2^(2*n + 3) + 3*n + 1)/9}; /* Michael Somos, Oct 16 2020 */

a(n) = (3n + 1 + 2^(2n+3))/9. - Emeric Deutsch, Jun 20 2009
G.f.: ( -1+2*x ) / ( (-1+4*x)*(x-1)^2 ). - R. J. Mathar, Jun 28 2012
From Wolfdieter Lang, Nov 16 2013: (Start)
a(n) = 6*a(n-1) - 9*a(n-2) + 4*a(n-3), n >= 2, a(-1)=0, a(0)=1, a(1)=4.
a(n) = 5*a(n-1) - 4*a(n-2) -1, n >= 2, a(0)=1, a(1)=4. (End)
a(n) = A034299(2*n). - Michael Somos, Oct 16 2020

More terms from Emeric Deutsch, Jun 20 2009

A103196 a(n) = (1/9)(2^(n+3)-(-1)^n(3n-1)).

Original entry on oeis.org

1, 2, 3, 8, 13, 30, 55, 116, 225, 458, 907, 1824, 3637, 7286, 14559, 29132, 58249, 116514, 233011, 466040, 932061, 1864142, 3728263, 7456548, 14913073, 29826170, 59652315, 119304656, 238609285, 477218598, 954437167

Offset: 0

Creighton Dement, Mar 18 2005

A floretion-generated sequence relating to the Jacobsthal sequence A001045 as well as to A095342 (Number of elements in n-th string generated by a Kolakoski(5,1) rule starting with a(1)=1). (a(n)) may be seen as the result of a certain transform of the natural numbers (see program code).

Floretion Algebra Multiplication Program, FAMP Code: 4jesleftforseq[A*B] with A = + 'i + 'j + i' + j' + 'ii' + 'jj' + 'ij' + 'ji' + e and B = - .25'i + .25'j + .25'k + .25i' - .25j' + .25k' - .25'ii' + .25'jj' + .25'kk' + .25'ij' + .25'ik' + .25'ji' + .25'jk' - .25'ki' - .25'kj' - .25e; 1vesforseq[A*B](n) = n, ForType: 1A.

Index entries for linear recurrences with constant coefficients, signature (0,3,2).

Cf. A001045, A095342, A053088, A034299, A052953, A026644.

Table[(2^(n+3)-(-1)^n (3n-1))/9,{n,0,30}] (* or *) LinearRecurrence[ {0,3,2},{1,2,3},40] (* Harvey P. Dale, Jul 09 2018 *)

G.f. (2x+1)/((1-2x)(x+1)^2); Superseeker results: a(n) + a(n+1) = A001045(n+3); a(n+1) - a(n) = A095342(n+1); a(n+2) - a(n+1) - a(n) = A053088(n+1) = A034299(n+1) - A034299(n); a(n) + 2a(n+1) + a(n+2) = 2^(n+3); a(n+2) - 2a(n+1) + a(n) = A053088(n+1) - A053088(n); a(n+2) - a(n) = A001045(n+4) - A001045(n+3) = A052953(n+3) - A052953(n+2) = A026644(n+2) - A026644(n+1);

a(n)=sum{k=0..n+2, (-1)^(n-k)*C(n+2, k)phi(phi(3^k))}; a(n)=sum{k=0..n+2, (-1)^(n-k)*C(n+2, k)(2*3^k/9+C(1, k)/3+4*C(0, k)/9)}; a(n)=sum{k=0..n+2, J(n-k+3)((-1)^(k+1)-2C(1, k)+4C(0, k))} where J(n)=A001045(n); a(n)=A113954(n+2). - Paul Barry, Nov 09 2005

A116914 Number of UUDD's, where U=(1,1) and D=(1,-1), in all hill-free Dyck paths of semilength n (a hill in a Dyck path is a peak at level 1).

Original entry on oeis.org

1, 1, 5, 16, 58, 211, 781, 2920, 11006, 41746, 159154, 609324, 2341060, 9021559, 34855741, 134972368, 523689718, 2035462990, 7923732118, 30889008112, 120566373676, 471134916286, 1842964183570, 7216096752496, 28279240308268, 110913181145716, 435333520075796, 1709861650762900

Offset: 2

Emeric Deutsch, May 08 2006

nonn

Catalan transform of A034299. - R. J. Mathar, Jun 29 2009

			a(4)=5 because in the 6 (=A000957(5)) hill-free Dyck paths of semilength 4, namely UU(UUDD)DD, UUUDUDDD, UUD(UUDD)D, UUDUDUDD, U(UUDD)UDD and (UUDD)(UUDD) (U=(1,1), D=(1,-1)) we have altogether 5 UUDD's (shown between parentheses).

G. C. Greubel, Table of n, a(n) for n = 2..1000
David Anderson, E. S. Egge, M. Riehl, L. Ryan, R. Steinke, Y. Vaughan, Pattern Avoiding Linear Extensions of Rectangular Posets, arXiv:1605.06825 [math.CO], 2016.
Colin Defant, Proofs of Conjectures about Pattern-Avoiding Linear Extensions, arXiv:1905.02309 [math.CO], 2019.
E. Deutsch and L. Shapiro, A survey of the Fine numbers, Discrete Math., 241 (2001), 241-265.

Cf. A105640.

R:=PowerSeriesRing(Rationals(), 40); Coefficients(R!( x*(1 + 5*x-(1-x)*Sqrt(1-4*x))/(2*(2+x)^2*Sqrt(1-4*x)) )); // G. C. Greubel, May 08 2019

G:=z*(1+5*z-(1-z)*sqrt(1-4*z))/2/(2+z)^2/sqrt(1-4*z): Gser:=series(G,z=0,31): seq(coeff(Gser,z^n),n=2..28);

Rest[Rest[CoefficientList[Series[x*(1+5*x-(1-x)*Sqrt[1-4*x])/2/(2+x)^2/Sqrt[1-4*x], {x, 0, 40}], x]]] (* Vaclav Kotesovec, Mar 20 2014 *)

my(x='x+O('x^40)); Vec(x*(1+5*x-(1-x)*sqrt(1-4*x))/(2*(2+x)^2 *sqrt(1-4*x))) \\ G. C. Greubel, Feb 08 2017

a=(x*(1+5*x-(1-x)*sqrt(1-4*x))/(2*(2+x)^2*sqrt(1-4*x))).series(x, 40).coefficients(x, sparse=False); a[2:] # G. C. Greubel, May 08 2019

a(n) = Sum_{k=0..floor(n/2)} k*A105640(n,k).
G.f.: x*(1 + 5*x - (1-x)*sqrt(1-4*x))/(2*(2+x)^2*sqrt(1-4*x)).
a(n+2) = A126258(2*n,n). - Philippe Deléham, Mar 13 2007
a(n) ~ 2^(2*n-1)/(9*sqrt(Pi*n)). - Vaclav Kotesovec, Mar 20 2014
D-finite with recurrence +2*(-n+1)*a(n) +3*(-n+6)*a(n-1) +3*(13*n-44)*a(n-2) +10*(2*n-5)*a(n-3)=0. - R. J. Mathar, Jul 26 2022

A126258 Triangle generated by Pascal's rule with left diagonal = [1,0,2,0,3,0,4,...] (A000027 with interpolated zeros) and right diagonal = [1,1,1,1,1,...].

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 0, 3, 2, 1, 3, 3, 5, 3, 1, 0, 6, 8, 8, 4, 1, 4, 6, 14, 16, 12, 5, 1, 0, 10, 20, 30, 28, 17, 6, 1, 5, 10, 30, 50, 58, 45, 23, 7, 1, 0, 15, 40, 80, 108, 103, 68, 30, 8, 1, 6, 15, 55, 120, 188, 211, 171, 98, 38, 9, 1, 0, 21, 70, 175, 308, 399, 382, 269, 136, 47, 10, 1

Offset: 0

Philippe Deléham, Mar 08 2007

			Triangle begins:
  1;
  0, 1;
  2, 1, 1;
  0, 3, 2, 1;
  3, 3, 5, 3, 1;
  0, 6, 8, 8, 4, 1;
  4, 6, 14, 16, 12, 5, 1;
  0, 10, 20, 30, 28, 17, 6, 1;

Cf. A034299, A116914.

T(2*n,n) = A116914(n).

Sum_{k=0..n} T(n,k) = A034299(n).

More terms from Jason Yuen, Sep 19 2024

cp's OEIS Frontend

A160156 Partial sums of A007583.

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A103196 a(n) = (1/9)(2^(n+3)-(-1)^n(3n-1)).

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A116914 Number of UUDD's, where U=(1,1) and D=(1,-1), in all hill-free Dyck paths of semilength n (a hill in a Dyck path is a peak at level 1).

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A126258 Triangle generated by Pascal's rule with left diagonal = [1,0,2,0,3,0,4,...] (A000027 with interpolated zeros) and right diagonal = [1,1,1,1,1,...].

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