A014301 -id:A014301 - cp's OEIS frontend

A172063 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^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

Richard Choulet, Jan 24 2010

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

			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.

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

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).

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

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;

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 *)

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

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

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

A172064 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=7.

Original entry on oeis.org

1, 8, 46, 230, 1068, 4744, 20476, 86662, 361711, 1494384, 6126818, 24972326, 101320712, 409609664, 1651162688, 6640469816, 26655382802, 106830738224, 427612715516, 1709790470780, 6830461107736, 27266848437608

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 7th 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+7 that are 5 steps higher than the preceding peak. - David Scambler, Apr 22 2013

			a(4) = C(15,4) - C(14,3) + C(13,2) - C(12,1) + C(11,0) = 7*13*15 - 14*13*2 + 78 - 12 + 1 = 1068.

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

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), A172063 (k=6), A172065 (k=8), A172066 (k=9), A172067 (k=10).

k:=7; 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

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;

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^7, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=7; 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

k=7; ((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j, n-j), with k=7.
a(n) ~ 2^(2*n+8)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+7)*(3*n+11)*a(n) -(21*n^3+212*n^2+719*n+840)*a(n-1) -2*(2*n+5)*(n+3)*(3*n+14)*a(n-2)=0. - R. J. Mathar, Feb 19 2016

A172065 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=8.

Original entry on oeis.org

1, 9, 56, 297, 1444, 6656, 29618, 128603, 548591, 2309467, 9624964, 39799813, 163556776, 668796712, 2723729944, 11055878188, 44753742226, 180746332690, 728571706240, 2932018571370, 11783070278816, 47297147250204

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 8th 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.

			a(4) = C(16,4) - C(15,3) + C(14,2) - C(13,1) + C(12,0) = 20*91 - 35*13 + 91 - 13 + 1 = 1820 - 455 + 79 = 1444.

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

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), A172063 (k=6), A172064 (k=7), A172066 (k=9), A172067 (k=10)

k:=8; 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

a:= n-> add((-1)^(p)*binomial(2*n-p+8, n-p), p=0..n):
seq(a(n), n=0..40);
# 2nd program
a:= n-> coeff(series((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))
        /(2*z))^8, z, n+20), z, n):
seq(a(n), n=0..40);

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^8, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=8; 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

k=8; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j *binomial(2*n+k-j, n-j), with k=8.
a(n) ~ 2^(2*n+9)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+8)*(3*n+13)*a(n) -(21*n^3 + 247*n^2 + 980*n + 1344)*a(n-1) - 2*(n+3)*(3*n+16)*(2*n+7)*a(n-2) = 0. - R. J. Mathar, Feb 29 2016

A172066 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=9.

Original entry on oeis.org

1, 10, 67, 376, 1912, 9142, 41941, 186880, 815083, 3498146, 14827487, 62236064, 259187048, 1072567256, 4415408372, 18098359424, 73915594466, 300958990724, 1222228100590, 4952609171080, 20030298812596, 80876902778482

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 9th 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.

			a(4) = C(17,4) - C(16,3) + C(15,2) - C(14,1) + C(13,0) = 17*4*5*7 - 16*5*7 + 105 - 14 + 1 = 5*7*(68-16) + 92 = 1912.

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

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), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172067 (k=10).

k:=9; 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

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;

CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^9, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=9; 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

k=9; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 17 2019

a(n) = Sum_{j=0..n} (-1)^j * binomial(2*n+k-j,n-j), with k=9.
a(n) ~ 2^(2*n+10)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+9)*(n+5)*a(n) -(7*n^3+94*n^2+427*n+672)*a(n-1) -2*(2*n+7)*(n+6)*(n+4)*a(n-2)=0. - R. J. Mathar, Feb 19 2016

A172067 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=10.

Original entry on oeis.org

1, 11, 79, 468, 2486, 12323, 58277, 266492, 1188679, 5202523, 22436251, 95630272, 403770544, 1691678428, 7042481236, 29161852240, 120212658034, 493656394350, 2020590599710, 8247228533780, 33579755528278, 136434358356201

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 10th 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.

			a(4) = C(18,4) - C(17,3) + C(16,2) - C(15,1) + C(14,0) = 60*51 - 680 + 120 - 15 + 1 = 2486.

G. C. Greubel, Table of n, a(n) for n = 0..1000

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), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), this sequence (k=10).

k:=10; 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 27 2019

a:= n-> add((-1)^(p)*binomial(2*n-p+10, n-p), p=0..n):
seq(a(n), n=0..40);
# 2nd program
a:= n-> coeff(series((2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))
        /(2*z))^10, z, n+20), z, n):
seq(a(n), n=0..40);

With[{k=10}, CoefficientList[Series[(2/(3*Sqrt[1-4*x]-1+4*x))*((1-Sqrt[1-4*x])/(2*x))^k, {x, 0, 30}], x]] (* G. C. Greubel, Feb 27 2019 *)

k=10; 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 27 2019

k=10; m=30; a=((2/(3*sqrt(1-4*x)-1+4*x))*((1-sqrt(1-4*x))/(2*x))^k ).series(x, m+2).coefficients(x, sparse=False); a[0:m] # G. C. Greubel, Feb 27 2019

a(n) = Sum_{j=0..n} (-1)^j*binomial(2*n+k-j,n-j), with k=10.

Conjecture: 2*n*(n+10)*(3*n+17)*a(n) - (21*n^3 + 317*n^2 + 1622*n + 2880)*a(n-1) - 2*(3*n+20)*(n+4)*(2*n+9)*a(n-2) = 0. - R. J. Mathar, Feb 21 2016

A113187 Inverse of twin-prime related triangle A111125.

Original entry on oeis.org

1, -3, 1, 10, -5, 1, -35, 21, -7, 1, 126, -84, 36, -9, 1, -462, 330, -165, 55, -11, 1, 1716, -1287, 715, -286, 78, -13, 1, -6435, 5005, -3003, 1365, -455, 105, -15, 1, 24310, -19448, 12376, -6188, 2380, -680, 136, -17, 1, -92378, 75582, -50388, 27132, -11628, 3876, -969, 171, -19, 1, 352716, -293930, 203490

Offset: 0

Paul Barry, Oct 17 2005

Row sums are (-1)^n*A000984. Diagonal sums are (-1)^n*A014301(n+1). An interesting factorization is (1/sqrt(1+4x)),(sqrt(1+4x)-1)/2)(1/(1+x),x/(1+x)).
The Z-sequence for this Riordan array is [-3,1], and the A-sequence is [1,-2,1]. For the Z- and A-sequence of Riordan arrays see the W. Lang link, with references, under A006232.  - Wolfdieter Lang, Oct 18 2012
This triangle appears in the formula (x-1/x)^(2*n+1) = sum(T(n,k)*(x^(2*k+1) - 1/x^(2*k+1)),k=0..n), n >= 0. Proof from the inversion of the formula given in an Oct 18 2012 comment on A111125, due to the Riordan property. - Wolfdieter Lang, Nov 14 2012

			Triangle T(n,k) begins:
n\k     0      1      2     3      4    5    6   7   8  9 ...
0:      1
1:     -3      1
2:     10     -5      1
3:    -35     21     -7     1
4:    126    -84     36    -9      1
5:   -462    330   -165    55    -11    1
6:   1716  -1287    715  -286     78  -13    1
7:  -6435   5005  -3003  1365   -455  105  -15   1
8:  24310 -19448  12376 -6188   2380 -680  136 -17   1
9: -92378  75582 -50388 27132 -11628 3876 -969 171 -19  1
... Reformatted by Wolfdieter Lang, Oct 17 2012
From Wolfdieter Lang, Oct 18 2012: (Start)
Recurrence from the Z-sequence [-3,1] (see a comment above):  T(3,0) = -3*T(2,0) + 1*T(2,1) = -3*10 + (-5) = -35.
Recurrence from the A-sequence [1,-2,1]: T(5,1) = 1*T(4,0) -2*T(4,1) + 1*T(4,2) = 126 -2*(-84) +36 = 330. (End)

Indranil Ghosh, Rows 0..125 of triangle, flattened

Riordan array ((sqrt(1+4x)-1)/(2x*sqrt(1+4x)), (1+2x-sqrt(1+4x))/(2x)).
T(n, k)=(-1)^(n-k)*C(2n+1, n+k+1); T(n, k)=sum{j=0..n, (-1)^(n-k)*C(2n-j, n-j)C(j, k)}.
O.g.f. column k: ((2-c(-x))/(1+4*x))*(1-c(-x))^k, with the o.g.f. c(x) of A000108  (Catalan), k>=0. From the Riordan property given above. - Wolfdieter Lang, Oct 17 2012
O.g.f. of the row polynomials R(n,x) = sum(T(n,k)*x^k,k=0..n): ((2-c(-z))/(1+4*z))/(1-x*(1-c(-z))) =  1/((1+4*z)*(x-(1-x)^2*z))*(x+2*x*z-2*z + (1+x)*z*c(-z)), with the o.g.f. c(x) of A000108. - Wolfdieter Lang, Oct 18 2012

A109244 A tree-node counting triangle.

Original entry on oeis.org

1, 1, 1, 4, 2, 1, 13, 7, 3, 1, 46, 24, 11, 4, 1, 166, 86, 40, 16, 5, 1, 610, 314, 148, 62, 22, 6, 1, 2269, 1163, 553, 239, 91, 29, 7, 1, 8518, 4352, 2083, 920, 367, 128, 37, 8, 1, 32206, 16414, 7896, 3544, 1461, 541, 174, 46, 9, 1, 122464, 62292, 30086, 13672, 5776, 2232

Offset: 0

Paul Barry, Jun 23 2005

Columns include A026641,A014300,A014301. Inverse matrix is A109246. Row sums are A014300. Diagonal sums are A109245.

			Rows begin:
  1;
  1,1;
  4,2,1;
  13,7,3,1;
  46,24,11,4,1;
  166,86,40,16,5,1;

G. C. Greubel, Rows n=0..100 of triangle, flattened

Flat(List([0..12], n-> List([0..n], k-> Sum([0..n], j-> (-1)^(n-j)*Binomial(n+j-k, j-k) )))); # G. C. Greubel, Feb 19 2019

[[(&+[(-1)^(n-j)*Binomial(n+j-k, j-k): j in [0..n]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, Feb 19 2019

Table[Sum[(-1)^(n-j)*Binomial[n+j-k, j-k], {j,0,n}], {n,0,12}, {k,0,n}] //Flatten  (* G. C. Greubel, Feb 19 2019 *)

{T(n,k) = sum(j=0,n, (-1)^(n-j)*binomial(n+j-k, j-k))};
for(n=0,12, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 19 2019

[[sum((-1)^(n-j)*binomial(n+j-k, j-k) for j in (0..n)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Feb 19 2019

Number triangle T(n, k) = Sum_{i=0..n} (-1)^(n-i)*binomial(n+i-k, i-k).
Riordan array (1/(1-x*c(x)-2*x^2*c(x)^2), x*c(x)) where c(x)=g.f. of A000108.
The production matrix M (discarding the zeros) is:
  1, 1;
  3, 1, 1;
  3, 1, 1, 1;
  3, 1, 1, 1, 1;
... such that the n-th row of the triangle is the top row of M^n. - Gary W. Adamson, Feb 16 2012

A143362 Triangle read by rows: T(n,k) is the number of ordered trees with n edges and k protected vertices (0<=k<=n-1). A protected vertex in an ordered tree is a vertex at least 2 edges away from its leaf descendants.

Original entry on oeis.org

1, 1, 1, 3, 1, 1, 6, 6, 1, 1, 17, 13, 10, 1, 1, 43, 50, 22, 15, 1, 1, 123, 141, 109, 33, 21, 1, 1, 343, 481, 325, 205, 46, 28, 1, 1, 1004, 1491, 1286, 631, 351, 61, 36, 1, 1, 2938, 4929, 4280, 2861, 1101, 562, 78, 45, 1, 1, 8791, 15840, 15662, 10025, 5676, 1783, 855, 97

Offset: 1

Emeric Deutsch, Aug 20 2008

Row sums are the Catalan numbers (A000108).
Sum(k*T(n,k),k>=0) = A014301(n).
T(n,0) = A143363(n)

			T(3,2)=1 because among the five ordered trees with 3 edges only the path tree has 2 vertices at least two edges away from the leaf.
Triangle starts:
1;
1,1;
3,1,1;
6,6,1,1;
17,13,10,1,1;
43,50,22,15,1,1;

Gi-Sang Cheon and Louis W. Shapiro, Protected points in ordered trees, Appl. Math. Letters, 21, 2008, 516-520.

Cf. A000108, A014301, A143363.

eq:=G-1/(1-z*G)-z*(t-1)*(G-1)/(1+z-z*G): G:=RootOf(eq,G): Gser:=simplify(series(G-1,z=0,13)): for n to 11 do P[n]:=sort(expand(coeff(Gser,z,n))) end do: for n to 11 do seq(coeff(P[n],t,j),j=0..n-1) end do; # yields sequence in triangular form

G.f.: G-1, where G=G(t,z) satisfies G = 1/(1-zG) + z(t-1)(G-1)/(1+z-zG).

A114596 Triangle read by rows: T(n,k) is the number of hill-free Dyck paths of semilength n and having abscissa of first return equal to 2k (2<=k<=n). A hill in a Dyck path is a peak at level 1.

Original entry on oeis.org

1, 0, 2, 1, 0, 5, 2, 2, 0, 14, 6, 4, 5, 0, 42, 18, 12, 10, 14, 0, 132, 57, 36, 30, 28, 42, 0, 429, 186, 114, 90, 84, 84, 132, 0, 1430, 622, 372, 285, 252, 252, 264, 429, 0, 4862, 2120, 1244, 930, 798, 756, 792, 858, 1430, 0, 16796, 7338, 4240, 3110, 2604, 2394, 2376, 2574, 2860, 4862, 0, 58786

Offset: 2

Emeric Deutsch, Dec 12 2005

Row sums are the Fine numbers (A000957). Column 2 yield the Fine numbers (A000957).

			T(5,3)=2 because we have UUUDDD|UUDD and UUDUDD|UUDD, where U=(1,1), D=(1,-1) (first return is shown by a vertical bar).
Triangle begins:
   1;
   0,  2;
   1,  0,  5;
   2,  2,  0, 14;
   6,  4,  5,  0, 42;
  18, 12, 10, 14,  0, 132;

G. C. Greubel, Rows n = 2..100 of triangle, flattened
E. Deutsch and L. Shapiro, A survey of the Fine numbers, Discrete Math., 241 (2001), 241-265.

Cf. A000957, A000108, A014301.

c:=n->binomial(2*n,n)/(n+1): f:=n->3*sum(binomial(2*n-2*j,n),j=0..floor(n/2))-binomial(2*n+2,n+1): for n from 2 to 12 do seq(c(k-1)*f(n-k),k=2..n) od; # yields sequence in triangular form

f[n_]:= 3*Sum[Binomial[2*n-2*j, n], {j,0,Floor[n/2]}] - Binomial[2*n+2, n +1]; Table[CatalanNumber[k-1]*f[n-k], {n,2,12}, {k,2,n}] (* G. C. Greubel, Apr 06 2019 *)

{f(n) = 3*sum(j=0, floor(n/2), binomial(2*n-2*j, n)) - binomial(2*n+2, n+1)};
for(n=2,12, for(k=2,n, print1((binomial(2*(k-1),k)/(k-1))*f(n-k), ", "))) \\ G. C. Greubel, Apr 06 2019

@CachedFunction
def f(n):
  return 3*sum(binomial(2*n-2*j, n) for j in (0..floor(n/2))) - binomial(2*n+2, n+1)
def T(n,k): return catalan_number(k-1)*f(n-k)
[[T(n,k) for k in (2..n)] for n in (2..12)] # G. C. Greubel, Apr 06 2019

T(n,n) = Catalan(n-1) (A000108).
Sum_{k=2..n} k*T(n,k) = 2*A014301(n).
T(n, k) = Catalan(k-1)*f(n-k), for 2<=k<=n, where Catalan(n) are the Catalan numbers (A000108) and f(n) = 3*Sum_{j=0..floor(n/2)} ( binomial(2n-2j, n) ) - binomial(2n+2, n+1) (the Fine numbers, A000957).
G.f.: (2*(1+x-t*x) +sqrt(1-4*x) -sqrt(1-4*t*x))/(1 +2*x +sqrt(1-4*x)) -1.

Keyword tabf changed to tabl by Michel Marcus, Apr 09 2013

A143950 Triangle read by rows: T(n,k) is the number of Dyck n-paths containing k even-length ascents (0 <= k <= floor(n/2)).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 5, 7, 2, 12, 20, 10, 30, 61, 36, 5, 79, 182, 133, 35, 213, 547, 488, 168, 14, 584, 1668, 1728, 756, 126, 1628, 5116, 6020, 3240, 750, 42, 4600, 15752, 20812, 13200, 3960, 462, 13138, 48709, 71376, 52030, 19360, 3267, 132, 37871, 151164

Offset: 0

Emeric Deutsch, Oct 05 2008

Row n contains 1 + floor(n/2) entries.
Row sums are the Catalan numbers (A000108).
T(n,0) = A101785(n).
Sum_{k=0..floor(n/2)} k*T(n,k) = A014301(n).
For the Dyck path statistic "number of odd-length ascents" see A096793.

			T(4,1)=7 because we have UDUD(UU)DD, UD(UU)DDUD, UD(UU)DUDD, (UU)DDUDUD, (UU)DUDDUD, (UU)DUDUDD and (UUUU)DDDD (the even-length ascents are shown between parentheses).
Triangle starts:
   1;
   1;
   1,  1;
   2,  3;
   5,  7,  2;
  12, 20, 10;
  30, 61, 36,  5;

Cf. A000108, A014301, A096793, A101785.

eq:=G=1+(1+s*z*G)*z*G/(1-z^2*G^2): G:=RootOf(eq,G): Gser:=simplify(series(G,z =0,16)): for n from 0 to 13 do P[n]:=sort(expand(coeff(Gser,z,n))) end do: for n from 0 to 13 do seq(coeff(P[n],s,j),j=0..floor((1/2)*n)) end do; # yields sequence in triangular form

G.f. G=G(s,z) satisfies G = 1 + zG(1 + szG)/(1 - z^2*G^2).

The trivariate g.f. H=H(t,s,z), where t (s) marks odd-length (even-length) ascents satisfies H = 1 + zH(t+szH)/(1-z^2*H^2).

cp's OEIS Frontend

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A172064 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=7.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A172065 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=8.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A172066 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=9.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A172067 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=10.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A113187 Inverse of twin-prime related triangle A111125.

Views

Author

Keywords

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

A172064 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=7.

A172065 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=8.

A172066 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=9.

A172067 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=10.