A172063 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=6.

1, 7, 37, 174, 771, 3300, 13820, 57044, 233108, 945793, 3817351, 15347362, 61520899, 246052888, 982365976, 3916739872, 15599504614, 62076995998, 246866382826, 981218764540, 3898442536366, 15483778158792, 61482966826992

Offset: 0

Author

Richard Choulet, Jan 24 2010

Keywords

• easy
• nonn

Comments

This sequence is the 6th diagonal below the main diagonal (which itself is A026641) in the array which grows with "Pascal rule" given here by rows: 1,0,1,0,1,0,1,0,1,0,1,0,1,0, 1,1,1,1,1,1,1,1,1,1,1,1,1,1, 1,1,2,2,3,3,4,4,5,5,6,6,7,7, 1,2,4,6,9,12,16,20,25,30, 1,3,7,13,22,34,50,70,95. The Maple programs give the first diagonals of this array.

Apparently the number of peaks in all Dyck paths of semilength n+6 that are 4 steps higher than the preceding peak. - David Scambler, Apr 22 2013

Examples

			a(4) = C(14,4) - C(13,3) + C(12,2) - C(11,1) + C(10,0) = 7*13*11 - 26*11 + 66 - 11 + 1 = 771.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200

Crossrefs

Cf. A091526 (k=-2), A072547 (k=-1), A026641 (k=0), A014300 (k=1), A014301 (k=2), A172025 (k=3), A172061 (k=4), A172062 (k=5), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

Programs

Magma

k:=6; m:=30; R:=PowerSeriesRing(Rationals(), m); Coefficients(R!( (2/(3*Sqrt(1-4*x)-1+4*x))*((1-Sqrt(1-4*x))/(2*x))^k )); // G. C. Greubel, Feb 17 2019

Maple

for k from 0 to 20 do for n from 0 to 40 do a(n):=sum('(-1)^(p)*binomial(2*n-p+k, n-p)', p=0..n): od:seq(a(n), n=0..40):od;
# 2nd program
for k from 0 to 40 do taylor((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k, z=0, 40+k):od;

Mathematica

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^6, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

PARI

k=6; my(x='x+O('x^30)); Vec((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k) \\ G. C. Greubel, Feb 17 2019

Sage

k=6; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k).series(x, 20).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

Formula

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j, n-j), with k=6.

a(n) ~ 2^(2*n+7)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014

Conjecture: 2*n*(n+6)*(n+3)*a(n) -(7*n^3+59*n^2+166*n+160)*a(n-1) -2*(2*n+5)*(n+4)*(n+2)*a(n-2)=0. - R. J. Mathar, Feb 19 2016