A036289 -id:A036289 - cp's OEIS frontend

A162007 Third left hand column of the EG1 triangle A162005.

1, 270, 36096, 4766048, 704357760, 120536980224, 24060789342208, 5590122715250688, 1503080384197754880, 464520829174515630080, 163839204411117787938816, 65500849343294249018327040

Offset: 3

Johannes W. Meijer, Jun 27 2009

Third left hand column of the EG1 triangle A162005.
Other left hand columns are A000182 and A162006.
Related to A094665, A083061 and A156919.
A000079, A036289 and A100381 appear in the a(n, 3) formula.
A001789, A003472, A054849, A002409, A054851, A140325 and A140354 (scaled by 2^(m-1)) appear one by one in the a(n, m) formulas for m= 4 and higher .

nmax := 14; mmax := nmax: imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1)*(x+1)*T1(i-1, x+1) - 2*x^2*T1(i-1, x)): dx := degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1 = 0..dx): od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do A094665(n+1, k+1) := A083061(n, k) od: od: A094665(0, 0) := 1: for n from 1 to nmax do A094665(n, 0) := 0 od: for m from 1 to mmax do A156919(0, m) := 0 end do: for n from 0 to nmax do A156919(n, 0) := 2^n end do: for n from 1 to nmax do for m from 1 to mmax do A156919(n, m) := (2*m+2)*A156919(n-1, m) + (2*n-2*m+1) * A156919(n-1, m-1) end do end do: m:=3; for n from m to nmax do a(n, m) := sum((-1)^(m-p1-1)*sum(2^(n-q-1)*binomial(n-q-1, m-p1-1) * A094665(n-1, q) * A156919(q, p1), q=1..n-m+p1), p1=0..m-1) od: seq(a(n, m), n = m..nmax);
# Maple program edited by Johannes W. Meijer, Sep 25 2012

a(n) = sum((-1)^(m-p-1)*sum(2^(n-q-1)*binomial(n-q-1,m-p-1)*A094665(n-1,q)* A156919(q,p),q=1..n-m+p), p=0..m-1) with m = 3.

A093968 Inverse binomial transform of n*Pell(n).

Original entry on oeis.org

0, 1, 2, 6, 8, 20, 24, 56, 64, 144, 160, 352, 384, 832, 896, 1920, 2048, 4352, 4608, 9728, 10240, 21504, 22528, 47104, 49152, 102400, 106496, 221184, 229376, 475136, 491520, 1015808, 1048576, 2162688, 2228224, 4587520, 4718592, 9699328, 9961472, 20447232, 20971520

Offset: 0

Paul Barry, Apr 21 2004

Binomial transform is A093967.
Binomial transform of (-1)^(n+1)(n*Pell(n-2)) (see A093969).
S-D transform of A001477 (cf. A051159). - Philippe Deléham, Aug 01 2006
a(n) is also the number of projective permutations of vertices of regular n-gons. A permutation of n vertices (AFB...CD) is considered 'projective' if there exists a line so that all the vertices can be projected onto it and the resulted points can be read in the same order: A'F'B'...C'D'. - Anton Zakharov, Jul 25 2016

			a(3) = 6, as there are only 6 projective permutations of vertices in a triangle ABC: ABC,CBA,ACB,BCA,CAB,BAC and it is equal to the number of simple permutations of three elements.
a(4) = 8, as there are only 8 permutations of vertices in a square, satisfying the projective criterion: ADBC,DACB,DCAB,CDBA,CBDA,BCAD,BACD,ABDC. ADCB is not allowed, cause there is no way to draw a line so that the projections A'B'C'D' of the original points form a line segment B'C' lying inside A'D' on this line. - _Anton Zakharov_, Jul 25 2016

Index entries for linear recurrences with constant coefficients, signature (0,4,0,-4).

Cf. A000129, A132314.

a[n_] := n*2^Floor[(n - 1)/2]; Array[a, 40, 0] (* Amiram Eldar, Feb 13 2023 *)

G.f.: x(1+2x+2x^2)/(1-2x^2)^2;
a(n) = 2^((n-4)/2)n((1+sqrt(2)) + (1-sqrt(2))(-1)^n).
a(2n) = A036289(n). a(2n+1) = A014480(n). - R. J. Mathar, Jun 02 2011
G.f.: x*G(0)/(1-x) where G(k) = 1 + x/(k+1 - 2*x*(k+1)*(k+2)/(2*x*(k+2) + 1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 01 2013
a(n) = n*2^floor((n-1)/2). - Anton Zakharov, Jul 25 2016
E.g.f.: x*(sqrt(2)*sinh(sqrt(2)*x) + 2*cosh(sqrt(2)*x))/2. - Ilya Gutkovskiy, Jul 25 2016
Sum_{n>=1} 1/a(n) = log(2) + sqrt(2)*log(1+sqrt(2)). - Amiram Eldar, Feb 13 2023

A162006 Second left hand column of the EG1 triangle A162005.

Original entry on oeis.org

1, 28, 1032, 52736, 3646208, 330545664, 38188155904, 5488365862912, 961530104709120, 201865242068910080, 50052995352723193856, 14476381898608390176768, 4831399425299156001882112

Offset: 2

Johannes W. Meijer, Jun 27 2009

Second left hand column of the EG1 triangle A162005.
Other left hand columns are A000182 and A162007.
Related to A094665, A083061 and A156919.
A000079 and A036289 appear in the Maple program.

nmax := 14; mmax := nmax: imax := nmax: T1(0, x) := 1: T1(0, x+1) := 1: for i from 1 to imax do T1(i, x) := expand((2*x+1)*(x+1)*T1(i-1, x+1) - 2*x^2*T1(i-1, x)): dx := degree(T1(i, x)): for k from 0 to dx do c(k) := coeff(T1(i, x), x, k) od: T1(i, x+1) := sum(c(j1)*(x+1)^(j1), j1 = 0..dx): od: for i from 0 to imax do for j from 0 to i do A083061(i, j) := coeff(T1(i, x), x, j) od: od: for n from 0 to nmax do for k from 0 to n do A094665(n+1, k+1) := A083061(n, k) od: od: A094665(0, 0) := 1: for n from 1 to nmax do A094665(n, 0) := 0 od: for m from 1 to mmax do A156919(0, m) := 0 end do: for n from 0 to nmax do A156919(n, 0) := 2^n end do: for n from 1 to nmax do for m from 1 to mmax do A156919(n, m) := (2*m+2)*A156919(n-1, m) + (2*n-2*m+1) * A156919(n-1, m-1) end do end do: m:=2; for n from m to nmax do a(n, m) := sum((-1)^(m-p1-1)*sum(2^(n-q-1)*binomial(n-q-1, m-p1-1) * A094665(n-1, q) * A156919(q, p1), q=1..n-m+p1), p1=0..m-1) od: seq(a(n, m), n = m..nmax);
# Maple program edited by Johannes W. Meijer, Sep 25 2012

a(n) = sum((-1)^(m-p-1)*sum(2^(n-q-1)*binomial(n-q-1,m-p-1)*A094665(n-1,q)*A156919(q,p),q=1..n-m+p), p=0..m-1) with m = 2.

A103904 a(n) = n(n-1)/2 2^(n*(n-1)/2).

Original entry on oeis.org

0, 2, 24, 384, 10240, 491520, 44040192, 7516192768, 2473901162496, 1583296743997440, 1981583836043018240, 4869940435459321626624, 23574053482485268906770432, 225305087149939210031640608768

Offset: 1

Ralf Stephan, Feb 21 2005

nonn

a(n) is the number of birooted graphs on n labeled nodes. - Andrew Howroyd, Nov 23 2020

Old (incorrect) name was: "Number of perfect matchings of an n X (n+1) Aztec rectangle with the third vertex in the topmost row removed". See Mathematics Stack Exchange for the discussion. - Andrey Zabolotskiy, Jun 05 2022

M. Ciucu, Enumeration of perfect matchings in graphs with reflective symmetry, J. Combin. Theory Ser. A 77 (1997), no. 1, 67-97, doi:10.1006/jcta.1996.2725.
N. Elkies, G. Kuperberg, M. Larsen and J. Propp, Alternating sign matrices and domino tilings, Journal of Algebraic Combinatorics 1 (1992), 111-132 (Part I), 219-234 (Part II); arXiv:math/9201305 [math.CO], 1992.
H. Helfgott and I. M. Gessel, Enumeration of tilings of diamonds and hexagons with defects, arXiv:math/9810143 [math.CO], 1998.
C. Krattenthaler, Schur function identities and the number of perfect matchings of Aztec holey rectangles, arXiv:math/9712204 [math.CO], 1997.
Mathematics Stack Exchange, Mistake in OEIS A103904?, 2021.

Cf. A000217, A006125, A038094, A095340, A095351, A134401, A036289, A303831.

a(n)={binomial(n,2)*2^binomial(n,2)} \\ Andrew Howroyd, Nov 23 2020

a(n) = A000217(n-1) * A006125(n).
a(n) = 2*A095351(n). - Andrew Howroyd, Nov 23 2020
a(n) = A036289(n*(n-1)/2). - Michael Somos, Feb 28 2021

Name replaced by a formula, a(1) changed from 1 to 0, and entry edited by Andrey Zabolotskiy, Jun 05 2022

A129178 Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} such that invc(p)=k (n >= 0; 0 <= k <= (n-1)(n-2)/2) (see comment for invc definition).

Original entry on oeis.org

1, 1, 2, 4, 2, 8, 8, 6, 2, 16, 24, 28, 26, 16, 8, 2, 32, 64, 96, 120, 126, 110, 82, 52, 26, 10, 2, 64, 160, 288, 432, 564, 658, 680, 638, 542, 416, 284, 172, 90, 38, 12, 2, 128, 384, 800, 1376, 2072, 2824, 3526, 4058, 4344, 4346, 4066, 3562, 2912, 2218, 1566, 1016, 598

Offset: 0

Emeric Deutsch, Apr 11 2007

invc(p) is defined (by Carlitz) in the following way: express p in standard cycle form (i.e., cycles ordered by increasing smallest elements with each cycle written with its smallest element in the first position), then remove the parentheses and count the inversions in the obtained word.
Row n has 1+(n-1)*(n-2)/2 - delta_{0,n} terms. Row sums are the factorials (A000142). T(n,0) = 2^(n-1) = A011782(n) = A000079(n-1). T(n,1) = (n-2)*2^(n-2) = A036289(n-2) for n>=2. T(n,k) = A121552(n,n+k).
It appears that Sum_{k>=0} k*T(n,k) = A126673(n).

			T(3,0)=4, T(3,1)=2 because we have 123=(1)(2)(3), 132=(1)(23), 213=(12)(3), 231=(123) with the resulting word (namely 123) having 0 inversions and 312=(132) and (321)=(13)(2) with the resulting word (namely 132) having 1 inversion.
Triangle starts:
   1;
   1;
   2;
   4,   2;
   8,   8,   6,   2;
  16,  24,  28,  26,  16,   8,   2;
  32,  64,  96, 120, 126, 110,  82,  52,  26,  10,  2;
  ...

L. Carlitz, Generalized Stirling numbers, Combinatorial Analysis Notes, Duke University, 1968, 1-7.

Alois P. Heinz, Rows n = 0..50, flattened
Toufik Mansour, Mark Shattuck, A q-analog of the hyperharmonic numbers, Afrika Matematika 25.1 (2014): 147-160.
M. Shattuck, Parity theorems for statistics on permutations and Catalan words, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 5, Paper A07, 2005.

Cf. A000142, A011782, A000079, A036289, A121552, A126673.

s:=j->2+sum(t^i, i=1..j): for n from 0 to 9 do P[n]:=sort(expand(simplify(product(s(j), j=0..n-2)))) od: for n from 0 to 9 do seq(coeff(P[n], t, j), j=0..degree(P[n])) od;  # yields sequence in triangular form

nMax = 9; s[j_] := 2 + Sum[t^i, {i, 1, j}]; P[0] = P[1] = 1; P[2] = 2; For[ n = 3, n <= nMax, n++, P[n] = Sort[Expand[Simplify[Product[s[j], {j, 0, n-2}]]]]]; Table[Coefficient[P[n], t, j], {n, 0, nMax}, {j, 0, Exponent[ P[n], t]}] // Flatten (* Jean-François Alcover, Jan 24 2017, adapted from Maple *)

Generating polynomial of row n is P[n](t) = 2*(2+t)*(2+t+t^2)*...*(2 + t + t^2 + ... + t^(n-2)) for n >= 3, P[1](t)=1, P[2](t)=2.

One term for row n=0 prepended by Alois P. Heinz, Dec 16 2016

A068566 Numerator of Sum_{k=1..n} 1/(k * 2^k).

Original entry on oeis.org

1, 5, 2, 131, 661, 1327, 1163, 148969, 447047, 44711, 983705, 7869871, 102309709, 204620705, 31972079, 32739453941, 556571077357, 556571247527, 10574855234543, 42299423848079, 42299425233749, 84598851790183

Offset: 1

Benoit Cloitre, Mar 25 2002

Sum_{k>=1} 1/(k * 2^k) = log(2).
From Paul Curtz, Jun 11 2019: (Start)
(Link) page 9:
   T0 =           1/2   =    1/2
   T1 =   1/2   + 1/8   =    5/8
   T2 =   5/8   + 1/24  =    2/3
   T3 =   2/3   + 1/64  =  131/192
   T4 = 131/192 + 1/160 =  661/960
  (T5 = 661/960 + 1/384 = 1327/1920)
... .
a(n)/A068565(n) is the first and the third column.
The denominators of the second column are essentially A036289, A097064 and A134401. (End)

Robert Israel, Table of n, a(n) for n = 1..1228
Paul Curtz, Accélération de la convergence de certaines séries alternées à l'aide des fonctions de sommation, Thèse de 3ème Cycle d'Analyse Numérique, Faculté des Sciences de l'Université de Paris, 4 mai 1965.

Cf. A068565.

Cf. A036289, A097064, A134401.

List([1..30], n-> NumeratorRat( Sum([1..n], k-> 1/(2^k*k)) ) ) # G. C. Greubel, Jun 30 2019

[Numerator( (&+[1/(2^k*k): k in [1..n]]) ): n in [1..30]]; // G. C. Greubel, Jun 30 2019

map(numer, ListTools:-PartialSums([seq(1/k/2^k,k=1..100)])); # Robert Israel, Jul 10 2015

Numerator[Accumulate[Table[1/(k 2^k),{k,30}]]] (* Harvey P. Dale, May 11 2013 *)
a[n_]:=Log[2]-Hypergeometric2F1[1+n,1+n,2+n,-1]/(1+n);
Numerator[Table[Simplify[a[n]],{n,1,30}]] (* Gerry Martens, Aug 06 2015 *)

vector(30, n, numerator(sum(k=1,n, 1/(k * 2^k)))) \\ Michel Marcus, Aug 07 2015

[numerator( sum(1/(2^k*k) for k in (1..n)) ) for n in (1..30)] # G. C. Greubel, Jun 30 2019

From Peter Bala, Feb 05 2024: (Start)
Integral_{x = 0..1} x^n/(1 + x)^(n+1) dx = log(2) - Sum_{k = 1..n} 1/(k * 2^k).
Hence a(n) = the numerator of Integral_{x = 0..1} ((1 + x)^n - x^n)/(1 + x)^(n+1) dx.
Integral_{x = 0..1/2} x^n/(1 - x) dx = Integral_{x >= 2} 1/(x^(n+2) - x^(n+1)) dx = log(2) - a(n)/A068565(n). (End)

A134400 M * A007318, where M = triangle with (1, 1, 2, 3, ...) in the main diagonal and the rest zeros.

Original entry on oeis.org

1, 1, 1, 2, 4, 2, 3, 9, 9, 3, 4, 16, 24, 16, 4, 5, 25, 50, 50, 25, 5, 6, 36, 90, 120, 90, 36, 6, 7, 49, 147, 245, 245, 147, 49, 7, 8, 64, 224, 448, 560, 448, 224, 64, 8, 9, 81, 324, 756, 1134, 1134, 756, 324, 81, 9, 10, 100, 450, 1200, 2100, 2520, 2100, 1200, 450, 100, 10

Offset: 0

Gary W. Adamson, Oct 23 2007

Row sums = A134401: (1, 2, 8, 24, 64, 160, 384, ...).
Triangle T(n,k), read by rows, given by [1,1,-1,1,0,0,0,0,0,...] DELTA [1,1,-1,1,0,0,0,0,0,...] where DELTA is the operator defined in A084938. A134402*A007318 as infinite lower triangular matrices. - Philippe Deléham, Oct 26 2007
For n > 0, from n athletes, select a team of k players and then choose a coach who is allowed to be on the team or not. - Geoffrey Critzer, Mar 13 2010
Row sums are A036289 if first term changed to zero. Diagonal sums are A023610, starting with the 2nd diagonal. Partial sums of diagonals are A002940 if first term changed to zero. - John Molokach, Jul 06 2013
For n > 0, T(n,k) is the number of states in Sokoban puzzle with n non-obstacles cells and k boxes (see Russell and Norvig at page 157). - Stefano Spezia, Dec 03 2023

			First few rows of the triangle:
  1;
  1,  1;
  2,  4,   2;
  3,  9,   9,   3;
  4, 16,  24,  16,   4;
  5, 25,  50,  50,  25,   5;
  6, 36,  90, 120,  90,  36,  6;
  7, 49, 147, 245, 245, 147, 49, 7;
  ...

Stuart Russell and Peter Norvig, Artificial Intelligence: A Modern Approach, Fourth Edition, Hoboken: Pearson, 2021.

Wikipedia, Sokoban

Cf. A002940, A003506, A007318, A036289, A134401, A134402.

T(2n,n) give A002011(n-1) for n>=1.

with(combstruct): for n from 0 to 10 do seq(`if`(n=0, 1, n)* count( Combination(n), size=m), m=0..n) od; # Zerinvary Lajos, Apr 09 2008

Join[{1},Table[Table[n*Binomial[n, k], {k,0, n}], {n, 10}]] //Flatten (* Geoffrey Critzer, Mar 13 2010 adapted by Stefano Spezia, Dec 03 2023 *)

From Geoffrey Critzer, Mar 13 2010: (Start)
T(0,0) = 1 and T(n,k) = n * binomial(n,k) for n > 0.
E.g.f. for column k is: (x^k/k!)*exp(x)*(x+k). (End)
T(n,k) = A003506(n,k) + A003506(n,k-1). - Geoffrey Critzer, Mar 13 2010
G.f.: (1-x-x*y+x^2+x^2*y+x^2*y^2)/(1-2*x-2*x*y+x^2+2*x^2*y+x^2*y^2). - Philippe Deléham, Nov 14 2013
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) - 2*T(n-2,k-1) - T(n-2,k-2), T(0,0)=T(1,0)=T(1,1)=1, T(2,0)=T(2,2)=2, T(2,1)=4, T(n,k)=0 if k < 0 or if k > n. - Philippe Deléham, Nov 14 2013
E.g.f.: 1 + exp(y*x)*exp(x)*(y*x + x). - Geoffrey Critzer, Mar 15 2015

a(55)-a(65) from Stefano Spezia, Dec 03 2023

A171607 Expressible as A*B^A in a nontrivial way.

Original entry on oeis.org

8, 18, 24, 32, 50, 64, 72, 81, 98, 128, 160, 162, 192, 200, 242, 288, 324, 338, 375, 384, 392, 450, 512, 578, 648, 722, 800, 882, 896, 968, 1024, 1029, 1058, 1152, 1215, 1250, 1352, 1458, 1536, 1568, 1682, 1800, 1922, 2048, 2178, 2187, 2312, 2450, 2500, 2592

Offset: 1

Robert Munafo, Dec 12 2009

nonn

			8=2*2^2. 24=3*2^3. 375=3*5^3.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
R. Munafo, Generalized Cullen and Woodall Numbers

Cf. A171606. Union of the "KN^K" sequences A001105, A117642, A141046, ... or of the "NK^N" sequences A036289, A036290, A018215, A036291, ... but omitting the trivial initial terms.

is(n)=if(n<8, return(0)); for(a=2,logint(n\2,2), if(n%a==0 && ispower(n/a,a), return(1))); 0 \\ Charles R Greathouse IV, Feb 19 2017

list(lim)=my(v=List()); if(lim<8,return([])); for(a=2,logint(lim\2,2), for(b=2,sqrtnint(lim\a,a), listput(v,a*b^a))); Set(v) \\ Charles R Greathouse IV, Feb 19 2017

a(n) = 2n^2 - O(n^(5/3)). - Charles R Greathouse IV, Feb 19 2017

A099429 A Jacobsthal-Lucas convolution.

Original entry on oeis.org

0, 0, 2, 3, 12, 25, 66, 147, 344, 765, 1710, 3751, 8196, 17745, 38234, 81915, 174768, 371365, 786438, 1660239, 3495260, 7340025, 15379122, 32156323, 67108872, 139810125, 290805086, 603979767, 1252698804, 2594876065, 5368709130, 11095332171, 22906492256

Offset: 0

Paul Barry, Oct 15 2004

Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 02 2014

If we concatenate the lexicographically ordered bit strings of length n, then a(n) is the number of times 11 appears as a substring, if overlapping substrings are not considered as being separate. - John M. Campbell, Jan 18 2019

			G.f. = 2*x^2 + 3*x^3 + 12*x^4 + 25*x^5 + 66*x^6 + 147*x^7 + 344*x^8 + ...
If we concatenate the lexicographically ordered bit strings of length 4, we obtain the expression 0000000100100011010001010110011110001001101010111100110111101111, and we see that the substring 11 appears a total of a(4) = 12 times, with overlapping substrings not being considered as being separate. - _John M. Campbell_, Jan 18 2019

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3,-4,-4).

Cf. A001045, A014551, A193449.

m:=40; R:=PowerSeriesRing(Integers(), m); [0,0] cat Coefficients(R!( x^2*(2-x)/(1-x-2*x^2)^2 )); // G. C. Greubel, Feb 25 2019

CoefficientList[Series[x^2*(2-x)/(1-x-2x^2)^2, {x, 0, 32}], x] (* Michael De Vlieger, Jan 18 2019 *)

{a(n) = if( n>=0, polcoeff( x^2*(2-x)/((1+x)*(1-2*x))^2 + x*O(x^n), n), polcoeff( x*(1-2*x)/((1+x)*(2-x))^2 + x*O(x^-n), -n) )}; /* Michael Somos, Jun 02 2014 */

(x^2*(2-x)/(1-x-2*x^2)^2).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019

G.f.: x^2*(2-x)/(1-x-2*x^2)^2. [Typo corrected by Colin Barker, Jun 16 2012]
a(n) = Sum_{k=0..n} J(n-k)*(2^(k-1) -(-1)^k +0^k/2).
a(n) = Sum_{k=0..n+1} J(n-k)*binomial(n-k+1, k)*binomial(1, (k+1)/2)*(1-(-1)^k)/2.
a(n) = A036289(n)/6 +(-1)^n*n/3. - R. J. Mathar, Sep 21 2012
a(-n) = (-2)^(-n-1) * A193449(n) for all n in Z. - Michael Somos, Jun 02 2014

A108891 Triangle read by rows: T(n,k) = number of Schroeder (or royal) n-paths (A006318) containing k returns to the diagonal y=x. (A northeast step lying on y=x contributes a return.)

Original entry on oeis.org

2, 2, 4, 6, 8, 8, 22, 28, 24, 16, 90, 112, 96, 64, 32, 394, 484, 416, 288, 160, 64, 1806, 2200, 1896, 1344, 800, 384, 128, 8558, 10364, 8952, 6448, 4000, 2112, 896, 256, 41586, 50144, 43392, 31616, 20160, 11264, 5376, 2048, 512

Offset: 1

David Callan, Jul 25 2005

			Table begins
  n\k  1    2    3    4    5    6
  -------------------------------
  1 |  2
  2 |  2    4
  3 |  6    8    8
  4 | 22   28   24   16
  5 | 90  112   96   64   32
  6 |394  484  416  288  160   64
The paths DD, END, DEN, ENEN each have 2 returns (E=east, N=north, D=northeast); so T(2,2)=4.
From _Philippe Deléham_, Nov 02 2013: (Start)
Triangle (0, 1, 2, 1, 2, 1, 2, ...) DELTA (1, 0, 0, 0, ...) begins:
  1;
  0,   2;
  0,   2,   4;
  0,   6,   8,   8;
  0,  22,  28,  24,  16;
  0,  90, 112,  96,  64,  32;
  0, 394, 484, 416, 288, 160,  64; (End)

Michael De Vlieger, Table of n, a(n) for n = 1..11325 (rows n = 1..150, flattened)
Scott Balchin, Ethan MacBrough, and Kyle Ormsby, The combinatorics of N_oo operads for C_qp^n and D_p^n, arXiv:2209.06992 [math.AT], 2022.

Row sums are the large Schroeder numbers A006318. Column k=1 is twice the little Schroeder numbers A001003. The main diagonal consists of powers of 2, A000079. The first subdiagonal is A036289. The analogous Catalan triangle is A009766 (with rows reversed).

T[n_, k_] := (-1)^(n - k) Binomial[n, k] Hypergeometric2F1[k - n, n + 1, k + 2, 2]; Table[T[n - 1, k - 1]*2^k, {n, 9}, {k, n}] // Flatten (* Michael De Vlieger, Sep 21 2022, after Peter Luschny at A104219 *)

Column k is the k-fold convolution of column 1.
T(n, k) = A104219(n-1, k-1)*2^k. - Philippe Deléham, Jul 31 2005
Triangle T(n,k), 1 <= k <= n, read by rows given by (0, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Nov 02 2013

cp's OEIS Frontend

A162007 Third left hand column of the EG1 triangle A162005.

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A093968 Inverse binomial transform of n*Pell(n).

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A162006 Second left hand column of the EG1 triangle A162005.

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A103904 a(n) = n*(n-1)/2 * 2^(n*(n-1)/2).

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A129178 Triangle read by rows: T(n,k) is the number of permutations p of {1,2,...,n} such that invc(p)=k (n >= 0; 0 <= k <= (n-1)(n-2)/2) (see comment for invc definition).

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A068566 Numerator of Sum_{k=1..n} 1/(k * 2^k).

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A134400 M * A007318, where M = triangle with (1, 1, 2, 3, ...) in the main diagonal and the rest zeros.

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A103904 a(n) = n(n-1)/2 2^(n*(n-1)/2).