A072547 -id:A072547 - cp's OEIS frontend

A172062 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=5.

1, 6, 29, 128, 541, 2232, 9076, 36568, 146446, 584082, 2322967, 9220544, 36548573, 144732176, 572756312, 2265577184, 8959034798, 35421613196, 140035644602, 553606049024, 2188652065586, 8653317051056, 34216118389384

Offset: 0

Richard Choulet, Jan 24 2010

This sequence is the 5th 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+5 that are 3 steps higher than the preceding peak. - David Scambler, Apr 22 2013
Apparently half the sum of all height differences between adjacent peaks in all Dyck paths of semilength n+3. - David Scambler, Apr 22 2013

			a(4) = C(13,4) - C(12,3) + C(11,2) - C(10,1) + C(9,0) = 13*11*5 - 20*11 + 55 - 10 + 1 = 541.

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), A172063 (k=6), A172064 (k=7), A172065 (k=8), A172066 (k=9), A172067 (k=10).

k:=5; 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))^5, {x, 0, 20}], x] (* Vaclav Kotesovec, Apr 19 2014 *)

k=5; 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=5; ((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=5.
a(n) ~ 2^(2*n+6)/(3*sqrt(Pi*n)). - Vaclav Kotesovec, Apr 19 2014
Conjecture: 2*n*(n+5)*(3*n+7)*a(n) - (n+3)*(21*n^2+79*n+80)*a(n-1) - 2*(3*n+10)*(2*n+3)*(n+2)*a(n-2) = 0. - R. J. Mathar, Feb 19 2016

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

Original entry on oeis.org

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

A112465 Riordan array (1/(1+x), x/(1-x)).

Original entry on oeis.org

1, -1, 1, 1, 0, 1, -1, 1, 1, 1, 1, 0, 2, 2, 1, -1, 1, 2, 4, 3, 1, 1, 0, 3, 6, 7, 4, 1, -1, 1, 3, 9, 13, 11, 5, 1, 1, 0, 4, 12, 22, 24, 16, 6, 1, -1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, -1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1

Offset: 0

Paul Barry, Sep 06 2005

Inverse is A112466. Note that C(n,k) = Sum_{j = 0..n-k} C(j+k-1, j).

			Triangle starts
   1;
  -1, 1;
   1, 0, 1;
  -1, 1, 1,  1;
   1, 0, 2,  2,  1;
  -1, 1, 2,  4,  3,  1;
   1, 0, 3,  6,  7,  4,  1;
  -1, 1, 3,  9, 13, 11,  5, 1;
   1, 0, 4, 12, 22, 24, 16, 6, 1;
Production matrix begins
  -1, 1;
   0, 1, 1;
   0, 0, 1, 1;
   0, 0, 0, 1, 1;
   0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 0, 1, 1; - _Paul Barry_, Apr 08 2011

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Roland Bacher, Chebyshev polynomials, quadratic surds and a variation of Pascal's triangle, arXiv:1509.09054 [math.CO], 2015.
E. Deutsch, L. Ferrari and S. Rinaldi, Production Matrices, Advances in Mathematics, 34 (2005) pp. 101-122.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A059260, A108561, A112466, A112468.
Columns: A033999(n) (k=0), A000035(n) (k=1), A004526(n) (k=2), A002620(n-1) (k=3), A002623(n-4) (k=4), A001752(n-5) (k=5), A001753(n-6) (k=6), A001769(n-7) (k=7), A001779(n-8) (k=8), A001780(n-9) (k=9), A001781(n-10) (k=10), A001786(n-11) (k=11), A001808(n-12) (k=12).
Diagonals: A000012(n) (k=n), A023443(n) (k=n-1), A152947(n-1) (k=n-2), A283551(n-3) (k=n-3).
Main diagonal: A072547.
Sums: A078008 (row), A078024 (diagonal), A092220 (signed diagonal), A280560 (signed row).

a112465 n k = a112465_tabl !! n !! k
a112465_row n = a112465_tabl !! n
a112465_tabl = iterate f [1] where
   f xs'@(x:xs) = zipWith (+) ([-x] ++ xs ++ [0]) ([0] ++ xs')
-- Reinhard Zumkeller, Jan 03 2014

A112465:= func< n,k | (-1)^(n+k)*(&+[(-1)^j*Binomial(j+k-1,j): j in [0..n-k]]) >;
[A112465(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Apr 18 2025

T[n_, k_]:= Sum[Binomial[j+k-1, j]*(-1)^(n-k-j), {j, 0, n-k}];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jul 23 2018 *)

def A112465(n,k): return (-1)^(n+k)*sum((-1)^j*binomial(j+k-1,j) for j in range(n-k+1))
print(flatten([[A112465(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 18 2025

Number triangle T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*C(j+k-1, j).
T(2*n, n) = A072547(n) (main diagonal). - Paul Barry, Apr 08 2011
From Reinhard Zumkeller, Jan 03 2014: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k), 0 < k < n, with T(n, 0) = (-1)^n and T(n, n) = 1.
T(n, k) = A108561(n, n-k). (End)
T(n, k) = T(n-1, k-1) + T(n-2, k) + T(n-2, k-1), T(0, 0) = 1, T(1, 0) = -1, T(1, 1) = 1, T(n, k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 11 2014
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(-1 + x + x^2/2! + x^3/3!) = -1 + 2*x^2/2! + 6*x^3/3! + 13*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014

A112556 Sums of squared terms in rows of triangle A112555.

Original entry on oeis.org

1, 2, 2, 4, 10, 32, 112, 408, 1514, 5680, 21472, 81644, 311896, 1196132, 4602236, 17757184, 68680170, 266200112, 1033703056, 4020716124, 15662273840, 61092127492, 238582873476, 932758045124, 3650336341240, 14298633670932, 56055986383412, 219931273282348, 863504076182884, 3392593262288780, 13337336618626132

Offset: 0

Paul D. Hanna, Sep 21 2005

nonn

First differences form A072547 and equals the unsigned central terms of triangle A112555.

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

Cf. A072547, A112555.

[(1/3)*(4 - (-1/2)^n) + (n+1)*Catalan(n) - (&+[(j+1)*Catalan(j)*(-1/2)^(n-j): j in [0..n]]): n in [0..30]]; // G. C. Greubel, Jan 13 2022

CoefficientList[Series[(2(1+x)/(1-x)+x/(1-4x)^(1/2))/(2+x), {x,0,30}], x] (* Harvey P. Dale, May 26 2011 *)

{a(n)=local(x=X+X*O(X^n)); polcoeff((2*(1+x)/(1-x)+x/(1-4*x)^(1/2))/(2+x),n,X)}

[(1/3)*(4 - (-1/2)^n) - sum( binomial(2*j, j)*(-1/2)^(n-j) for j in (0..n-1)) for n in (0..30)] # G. C. Greubel, Jan 13 2022

G.f.: ( 2*(1+x)/(1-x) + x/(1-4*x)^(1/2) )/(2+x).
a(n) ~ 2^(2*n) / (9*sqrt(Pi*n)). - Vaclav Kotesovec, Mar 26 2016
a(n) = (1/3)*(4 - (-1/2)^n) - Sum_{j=0..n-1} binomial(2*j, j)*(-1/2)^(n-j). - G. C. Greubel, Jan 13 2022

A092785 a(n) = sum(sum(binomial(j-n-1,m),m=0..n),j=0..n).

Original entry on oeis.org

1, -1, 7, -21, 81, -295, 1107, -4165, 15793, -60171, 230253, -884235, 3406105, -13154947, 50922987, -197519941, 767502945, -2987013067, 11641557717, -45429853651, 177490745985, -694175171647, 2717578296117, -10648297329691, 41757352712481, -163875286898935

Offset: 0

Francois Jooste (pin(AT)myway.com), Apr 23 2004

sign

Alois P. Heinz, Table of n, a(n) for n = 0..500

Differs from A072547 by -1, +1, -1, +1, -1, ... - Ralf Stephan, Apr 19 2004
Equals sum(m=0, n, (-1)^m*binomial(n+m, m)). - Henry Gould, Apr 23 2004
Let f(n) = (-1)^n a(n). Then 2f(n) + f(n-1) = (3n+1)C(n) + (-1)^n, where C(n) = (2n+1)!/n!(n+1)! is a Catalan number (A000108). - Henry Gould, Apr 24 2004
Recurrence: 2*(n+1)*(15*n^2 - 31*n + 12)*a(n) = -(5*n-3)*(15*n^2 - 19*n - 4)*a(n-1) + (165*n^3 - 266*n^2 - 11*n + 60)*a(n-2) - 2*(2*n-3)*(15*n^2 - n - 4)*a(n-3). - Vaclav Kotesovec, Sep 05 2014

A371813 a(n) = Sum_{k=0..n} (-1)^k * binomial(3*n-k-1,n-k).

Original entry on oeis.org

1, 1, 7, 40, 239, 1461, 9076, 57044, 361711, 2309467, 14827487, 95630272, 619111172, 4021011580, 26187682024, 170960159100, 1118406332655, 7330011083079, 48119501497909, 316354663355384, 2082573599282359, 13726029056757029, 90565080767425744

Offset: 0

Seiichi Manyama, Apr 06 2024

nonn

Cf. A072547, A371814.

Cf. A005809, A066380, A165817, A385004, A386700.

a(n) = sum(k=0, n, (-1)^k*binomial(3*n-k-1, n-k));

a(n) = [x^n] 1/((1+x) * (1-x)^(2*n)).
a(n) = binomial(3*n-1, n)*hypergeom([1, -n], [1-3*n], -1). - Stefano Spezia, Apr 07 2024
From Vaclav Kotesovec, Apr 07 2024: (Start)
Recurrence: 8*n*(2*n - 1)*(28*n^2 - 87*n + 67)*a(n) = 2*(1456*n^4 - 6008*n^3 + 8593*n^2 - 4949*n + 960)*a(n-1) + 3*(3*n - 5)*(3*n - 4)*(28*n^2 - 31*n + 8)*a(n-2).
a(n) ~ 3^(3*n + 1/2) / (sqrt(Pi*n) * 2^(2*n+2)). (End)
a(n) = Sum_{k=0..n} (-2)^(n-k) * binomial(3*n,k). - Seiichi Manyama, Jul 30 2025
G.f.: g/((-1+2*g) * (3-2*g)) where g = 1+x*g^3 is the g.f. of A001764. - Seiichi Manyama, Aug 13 2025
a(n) = Sum_{k=0..n} (-1)^k * 2^(n-k) * binomial(3*n,k) * binomial(3*n-k-1,n-k). - Seiichi Manyama, Aug 15 2025
G.f.: 1/(1 - x*g^2*(-3+4*g)) where g = 1+x*g^3 is the g.f. of A001764. - Seiichi Manyama, Aug 17 2025

cp's OEIS Frontend

A172062 Expansion of (2/(3*sqrt(1-4*z)-1+4*z))*((1-sqrt(1-4*z))/(2*z))^k with k=5.

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

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

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

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

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

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A172062 Expansion of (2/(3sqrt(1-4z)-1+4z))((1-sqrt(1-4z))/(2z))^k with k=5.

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.