A005803 -id:A005803 - cp's OEIS frontend

A356185 The difference between number of even and number of odd Grassmannian permutations of size n.

1, 1, 0, 1, 0, 3, 2, 9, 8, 23, 22, 53, 52, 115, 114, 241, 240, 495, 494, 1005, 1004, 2027, 2026, 4073, 4072, 8167, 8166, 16357, 16356, 32739, 32738, 65505, 65504, 131039, 131038, 262109, 262108, 524251, 524250, 1048537, 1048536, 2097111, 2097110, 4194261, 4194260

Offset: 0

Per W. Alexandersson, Jul 28 2022

A permutation is Grassmann if it has at most one descent. A closed-form formula was proved by J. B. Gil and J. A. Tomasko.

			For n=3, 123, 231, 312 are even Grassmann permutations, and 132, 213 are the odd ones. Hence a(3) = 1.

Juan B. Gil and Jessica A. Tomasko, Restricted Grassmannian permutations, arXiv:2112.03338 [math.CO], 2021.
Juan B. Gil and Jessica A. Tomasko, Restricted Grassmannian permutations, Enum. Combin. Appl. 2 (2022), no. 4, Article #S4PP6.
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,2).

Cf. A000325, A001710, A032085, A060546, A122746, A233411.

Bisections give: A005803 (even part), A183155 (odd part).

Table[2^Floor[1 + (n - 1)/2] - n, {n, 1, 80}]

a(n) = 2^(1+floor((n-1)/2))-n.
From Alois P. Heinz, Jul 28 2022: (Start)
G.f.: -(4*x^3-3*x^2-x+1)/((2*x^2-1)*(x-1)^2).
a(n) = A000325(n) - A233411(n) = A060546(n) - n = 2^ceiling(n/2) - n.
a(n) = A000325(n) - 2*A032085(n) = A000325(n) - 2*A122746(n-2) for n>=2. (End)

A077866 Expansion of (1-x)^(-1)/(1 - x - 2x^2 + 2x^3).

Original entry on oeis.org

1, 2, 5, 8, 15, 22, 37, 52, 83, 114, 177, 240, 367, 494, 749, 1004, 1515, 2026, 3049, 4072, 6119, 8166, 12261, 16356, 24547, 32738, 49121, 65504, 98271, 131038, 196573, 262108, 393179, 524250, 786393, 1048536, 1572823, 2097110, 3145685, 4194260, 6291411, 8388562

Offset: 0

N. J. A. Sloane, Nov 17 2002

Equals triangle A122196 * [1,2,4,8,16,...]. - Gary W. Adamson, Nov 29 2008

Conjecture: let b(n) be the number of subsets S of {1,2,...,n} having more than one element such that (sum of least two elements of S) = max(S). Then b(0) = b(1) = b(2) = 0 and b(n+3) = a(n) for n >= 0. - Clark Kimberling Sep 27 2022

			G.f. = 1 + 2*x + 5*x^2 + 8*x^3 + 15*x^4 + 22*x^5 + 37*x^6 + ... - _Michael Somos_, Aug 11 2021

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,1,-4,2).

Bisections are A005803 and A050488.

Cf. A052551 (first differences), A122196.

CoefficientList[Series[(1-x)^(-1)/(1-x-2x^2+2x^3),{x,0,50}],x] (* or *) LinearRecurrence[{2,1,-4,2},{1,2,5,8},50] (* Harvey P. Dale, Feb 16 2013 *)

Vec((1-x)^(-1)/(1-x-2*x^2+2*x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = 2^(n/2)*(3 + 2*sqrt(2) + (3 - 2*sqrt(2))*(-1)^n) - n - 5. - Paul Barry, Apr 23 2004
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + 2*a(n-4); a(0)=1, a(1)=2, a(2)=5, a(3)=8. - Harvey P. Dale, Feb 16 2013
a(2n) = 3*2^(n+1) - 2(n+1) - 3 = A050488(n+1) and a(2n+1) = 2^(n+3) - 2(n+3) = A005803(n+3). Also, a(2n+1) - a(2n) = 2^(n+1) - 1 = a(2n) - a(2n - 1). - Gregory L. Simay, Feb 07 2021
E.g.f.: 6*cosh(sqrt(2)*x) + 4*sqrt(2)*sinh(sqrt(2)*x) - exp(x)*(5 + x). - Stefano Spezia, Feb 08 2021
G.f.: 1/((1 - x)^2 * (1 - 2*x^2)). - Michael Somos, Aug 11 2021

A130102 E.g.f.: (e^x - x)^2.

Original entry on oeis.org

1, 0, 2, 2, 8, 22, 52, 114, 240, 494, 1004, 2026, 4072, 8166, 16356, 32738, 65504, 131038, 262108, 524250, 1048536, 2097110, 4194260, 8388562, 16777168, 33554382, 67108812, 134217674, 268435400, 536870854, 1073741764, 2147483586

Offset: 0

Paul Barry, May 07 2007

a(n) is the number of length n binary sequences in which no symbol occurs exactly once. (The Rosenthal formula takes 2^n for the total number of binary sequences and subtracts n for each sequence of length n with a single 0 or 1.) - Geoffrey Critzer, Dec 03 2011
From Ambrosio Valencia-Romero, Mar 08 2022: (Start)
a(n), for n > 1, is the number of pure Nash equilibria in the symmetric n-player two-strategy normal-form unanimity game. Let i be a player in set N = {1, 2, 3, ..., n} and s(i) in set S = {0, 1} be i's strategy. Then i's payoff, u(i), in this game is given by:
u(i) = 1 if s(1) = s(2) = ... = s(n-1) = s(n); otherwise, u(i) = 0.
Only two of the a(n) pure equilibria in this unanimity game are strict: s = <0, 0, ..., 0, 0> and s = <1, 1, ..., 1, 1>; these are the diagonal collective strategies where all actors obtain the payoff u(i) = 1.
The other a(n)-2 pure equilibria are weak and produce an individual payoff of u(i) = 0; these correspond to the collective strategy outcomes where more than one and fewer than n-1 individual strategies differ. For instance, for n = 4, the a(4)-2 = 6 weak pure equilibria are <0, 0, 1, 1>, <0, 1, 0, 1>, <0, 1, 1, 0>, <1, 0, 0, 1>, <1, 0, 1, 0>, and <1, 1, 0, 0>. (End)

			a(4) = 8 because there are 8 sequences on {0,1} such that neither 0 nor 1 occurs exactly once: {0,0,0,0}, {0,0,1,1}, {0,1,0,1}, {0,1,1,0}, {1,0,0,1}, {1,0,1,0}, {1,1,0,0}, {1,1,1,1}. - _Geoffrey Critzer_, Dec 03 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A005803, A060576.

I:=[1, 0, 2, 2, 8, 22]; [n le 6 select I[n] else 4*Self(n-1)-5*Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 28 2012

a=Exp[x]-x; Range[0,20]! CoefficientList[Series[a^2, {x,0,20}], x] (* Geoffrey Critzer, Dec 03 2011 *)
CoefficientList[Series[1+2*x^2-2*x^3/((2*x-1)*(x-1)^2),{x,0,40}],x] (* Vincenzo Librandi, Jun 28 2012 *)

a(n) = 2^n - 2*n for n <> 2 (cf. A005803). - Rainer Rosenthal, Feb 14 2010.
E.g.f.: e^(2*x) - 2*x*e^x + x^2.
a(n) = Sum_{k=0..n} binomial(n,k)*A060576(k)*A060576(n-k).
G.f. 1 + 2*x^2 - 2*x^3/((2*x - 1)*(x - 1)^2). - R. J. Mathar, Dec 04 2011

A156901 Triangle formed by coefficients of the expansion of p(x, n), where p(x,n) = (1 + 2x - x^2)^(n + 1)Sum_{j >= 0} (j+1)^n(-2x + x^2)^j.

Original entry on oeis.org

1, 1, 1, -2, 1, 1, -8, 8, -4, 1, 1, -22, 55, -52, 23, -6, 1, 1, -52, 290, -472, 394, -188, 50, -8, 1, 1, -114, 1265, -3624, 4838, -3668, 1750, -536, 97, -10, 1, 1, -240, 4884, -24092, 49239, -56448, 40664, -19320, 6231, -1360, 180, -12, 1, 1, -494, 17419, -142124, 441625, -730898, 749723, -515944, 247067, -83122, 19673, -3244, 331, -14, 1

Offset: 0

Roger L. Bagula, Feb 17 2009

			Irregular triangle begins as:
  1;
  1;
  1,   -2,    1;
  1,   -8,    8,     -4,     1;
  1,  -22,   55,    -52,    23,     -6,     1;
  1,  -52,  290,   -472,   394,   -188,    50,     -8,    1;
  1, -114, 1265,  -3624,  4838,  -3668,  1750,   -536,   97,   -10,   1;
  1, -240, 4884, -24092, 49239, -56448, 40664, -19320, 6231, -1360, 180, -12, 1;

G. C. Greubel, Rows n = 0..50 of the irregular triangle, flattened

Cf. A005803, A156890, A156896, A156918.

p[x_, n_]= (1+2*x-x^2)^(n+1)*Sum[(k+1)^n*(-2*x+x^2)^k, {k,0,Infinity}];
Table[CoefficientList[p[x, n], x], {n,0,10}]//Flatten

def T(n, k): return ( (1+2*x-x^2)^(n+1)*sum((j+1)^n*(x^2-2*x)^j for j in (0..2*n+1)) ).series(x, 2*n+2).list()[k]
flatten([1]+[[T(n, k) for k in (0..2*n-2)] for n in (1..12)]) # G. C. Greubel, Jan 07 2022

T(n, k) = coefficients of the expansion of p(x, n), where p(x,n) = (1 + 2*x - x^2)^(n + 1)*Sum_{j >= 0} (j+1)^n*(-2*x + x^2)^j.

T(n, 1) = (-1)*A005803(n) for n >= 2.

Edited by G. C. Greubel, Jan 07 2022

A222403 Triangle read by rows: left and right edges are A000217, interior entries are filled in using the Pascal triangle rule.

Original entry on oeis.org

0, 1, 1, 3, 2, 3, 6, 5, 5, 6, 10, 11, 10, 11, 10, 15, 21, 21, 21, 21, 15, 21, 36, 42, 42, 42, 36, 21, 28, 57, 78, 84, 84, 78, 57, 28, 36, 85, 135, 162, 168, 162, 135, 85, 36, 45, 121, 220, 297, 330, 330, 297, 220, 121, 45, 55, 166, 341, 517, 627, 660, 627, 517, 341, 166, 55

Offset: 0

N. J. A. Sloane, Feb 18 2013

In general, if the sequence defining the left and right edges is [a_0, a_1, ...], the row sums [s_0, s_1, ...] are given by s_0=a_0 and, for n>0,
s_n = 2a_n + Sum_{i=1..n-1} 2^(n-i) a_i.
Conversely, given the rows sums [s_0, s_1, ...], the edge sequence is [a_0, a_1, ...] where a_0=s_0 and, for n>0, a_n = (s_n - Sum_{i=1..n-1} s_i)/2.

			Triangle begins:
0
1, 1
3, 2, 3
6, 5, 5, 6
10, 11, 10, 11, 10
15, 21, 21, 21, 21, 15
21, 36, 42, 42, 42, 36, 21
28, 57, 78, 84, 84, 78, 57, 28
...

Robert Israel, Table of n, a(n) for n = 0..10010

Other triangles of this type: A007318, A051666, A134634, A222404, A222405.
Cf. A000217.
Row sums are A005803.

d:=[seq(n*(n+1)/2,n=0..14)];
f:=proc(d) local T,M,n,i;
M:=nops(d);
T:=Array(0..M-1,0..M-1);
for n from 0 to M-1 do T[n,0]:=d[n+1]; T[n,n]:=d[n+1]; od:
for n from 2 to M-1 do
for i from 1 to n-1 do T[n,i]:=T[n-1,i-1]+T[n-1,i]; od: od:
lprint("triangle:");
for n from 0 to M-1 do lprint(seq(T[n,i],i=0..n)); od:
lprint("row sums:");
lprint([seq( add(T[i,j],j=0..i), i=0..M-1)]);
end;
f(d);

t[n_, n_] := n*(n+1)/2; t[n_, 0] := n*(n+1)/2; t[n_, k_] := t[n, k] = t[n-1, k-1] + t[n-1, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)

G.f. as triangle: (1+x-4*x*y+x*y^2+x^2*y^2)*y/((1-y)^2*(-x*y+1)^2*(-x*y-y+1)). - Robert Israel, Apr 04 2018

A287416 Number T(n,k) of set partitions of [n] such that the maximal value of all absolute differences between least elements of consecutive blocks and between consecutive elements within the blocks equals k; triangle T(n,k), n>=0, 0<=k<=max(n-1,0), read by rows.

Original entry on oeis.org

1, 1, 0, 2, 0, 3, 2, 0, 4, 8, 3, 0, 5, 22, 19, 6, 0, 6, 52, 81, 48, 16, 0, 7, 114, 289, 267, 147, 53, 0, 8, 240, 941, 1250, 968, 529, 204, 0, 9, 494, 2894, 5310, 5469, 3919, 2174, 878, 0, 10, 1004, 8601, 21256, 28083, 25326, 17593, 9961, 4141

Offset: 0

Alois P. Heinz, May 24 2017

The maximal value is assumed to be zero if there are no consecutive blocks and no consecutive elements.

T(n,k) is defined for all n,k >= 0. The triangle contains only the terms for k <= max(n-1,0). T(n,k) = 0 if k>=n and k>0.

			T(4,1) = 4: 1234, 1|234, 1|2|34, 1|2|3|4.
T(4,2) = 8: 124|3, 12|34, 12|3|4, 134|2, 13|24, 13|2|4, 1|23|4, 1|24|3.
T(4,3) = 3: 123|4, 14|23, 14|2|3.
T(5,3) = 19: 1235|4, 123|45, 123|4|5, 125|34, 125|3|4, 134|25, 134|2|5, 13|24|5, 13|25|4, 145|23, 14|235, 14|23|5, 1|234|5, 145|2|3, 14|25|3, 14|2|35, 14|2|3|5, 1|25|34, 1|25|3|4.
Triangle T(n,k) begins:
  1;
  1;
  0, 2;
  0, 3,   2;
  0, 4,   8,   3;
  0, 5,  22,  19,    6;
  0, 6,  52,  81,   48,  16;
  0, 7, 114, 289,  267, 147,  53;
  0, 8, 240, 941, 1250, 968, 529, 204;
  ...

Alois P. Heinz, Rows n = 0..23, flattened
Wikipedia, Partition of a set

Columns k=1-2 give: A001477 (for n>1), A005803 (for n>0).
Row sums give A000110.
Cf. A287213, A287215, A287417, A287640.

b:= proc(n, k, l, t) option remember; `if`(n<1, 1, `if`(t-n>k, 0,
       b(n-1, k, map(x-> `if`(x-n>=k, [][], x), [l[], n]), n)) +add(
       b(n-1, k, sort(map(x-> `if`(x-n>=k, [][], x), subsop(j=n, l))),
       `if`(t-n>k, infinity, t)), j=1..nops(l)))
    end:
A:= (n, k)-> b(n, min(k, n-1), [], n):
T:= (n, k)-> A(n, k)-`if`(k=0, 0, A(n, k-1)):
seq(seq(T(n, k), k=0..max(n-1, 0)), n=0..12);

b[n_, k_, l_, t_] := b[n, k, l, t] = If[n < 1, 1, If[t - n > k, 0, b[n - 1, k, If[# - n >= k, Nothing, #]& /@ Append[l, n], n]] + Sum[b[n - 1, k, Sort[If[# - n >= k, Nothing, #]& /@ ReplacePart[l, j -> n]], If[t - n > k, Infinity, t]], {j, 1, Length[l]}]];
A[n_, k_] := b[n, Min[k, n - 1], {}, n];
T[n_, k_] := A[n, k] - If[k == 0, 0, A[n, k - 1]];
Table[T[n, k], {n, 0, 12}, { k, 0, Max[n - 1, 0]}] // Flatten (* Jean-François Alcover, May 24 2018, translated from Maple *)

T(n,k) = A287417(n,k) - A287417(n,k-1) for k>0, T(n,0) = 1.

T(n+2,n+1) = 1 + A000110(n).

A061761 a(n) = 2^n + 2*n - 1.

Original entry on oeis.org

0, 3, 7, 13, 23, 41, 75, 141, 271, 529, 1043, 2069, 4119, 8217, 16411, 32797, 65567, 131105, 262179, 524325, 1048615, 2097193, 4194347, 8388653, 16777263, 33554481, 67108915, 134217781, 268435511, 536870969, 1073741883, 2147483709

Offset: 0

Amarnath Murthy, May 20 2001

nonn

			a(5) = 2^5 + 2*5 - 1 = 32 + 10 - 1 = 41. - _Michael B. Porter_, Aug 18 2016

Harry J. Smith, Table of n, a(n) for n=0..200
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A000225, A005803.

Table[2^n+2*n-1,{n,0,60}] (* Vladimir Joseph Stephan Orlovsky, Feb 15 2011 *)

a(n) = { 2^n + 2*n - 1 } \\ Harry J. Smith, Jul 27 2009

G.f.: x(3-5x)/((1-x)^2*(1-2x)). Binomial transform of 0,3,1,1,... (1 continued). - R. J. Mathar, Sep 17 2008
a(n) = A000225(n+1) - A005803(n), for n>0. In other words, for n>0, a(n) is the sum of the elements on the perimeter of a Pascal's triangle of depth (n+1). - Ivan N. Ianakiev, Aug 18 2016
E.g.f.: exp(x)*(exp(x) + 2*x - 1). - Stefano Spezia, Dec 08 2024

A145654 Partial sums of A000918, starting from index 1.

Original entry on oeis.org

0, 2, 8, 22, 52, 114, 240, 494, 1004, 2026, 4072, 8166, 16356, 32738, 65504, 131038, 262108, 524250, 1048536, 2097110, 4194260, 8388562, 16777168, 33554382, 67108812, 134217674, 268435400, 536870854, 1073741764, 2147483586

Offset: 1

Vladimir Joseph Stephan Orlovsky, Oct 16 2008

			For n=7, a(7) = 6*2 + 5*2^2 + 4*2^3 + 3*2^4 + 2*2^5 + 1*2^6 = 240. - _Bruno Berselli_, Feb 10 2014
From _Bruno Berselli_, Jul 17 2018: (Start)
Row sums of the triangle:
   0   ......................................    0
   1,  1   ..................................    2
   3,  2,  3   ..............................    8
   6,  5,  5,  6   ..........................   22
  10, 11, 10, 11, 10   ......................   52
  15, 21, 21, 21, 21, 15   ..................  114
  21, 36, 42, 42, 42, 36, 21   ..............  240
  28, 57, 78, 84, 84, 78, 57, 28   ..........  494, etc.
(End)

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Cf. A000217, A000295, A000918, A005803, A121173.

a145654 n = a145654_list !! (n-1)
a145654_list = scanl1 (+) $ tail a000918_list
-- Reinhard Zumkeller, Nov 06 2013

Accumulate[2^Range[30] - 2] (* or *) LinearRecurrence[{4, -5, 2}, {0, 2, 8}, 30] (* Harvey P. Dale, Jul 15 2017 *)

a(n) = Sum_{i=1..n} A000918(i).
a(n+1) - a(n) = A000918(n+1).
a(n) = A005803(n+1). - R. J. Mathar, Oct 21 2008
From Colin Barker, Jan 11 2012: (Start)
a(n) = 2*(-1 + 2^n - n).
G.f.: 2*x^2/((1-x)^2*(1-2*x)). (End)
a(n+1) = A121173(2*n). - Reinhard Zumkeller, Nov 06 2013
a(n) = Sum_{i=1..n-1} (n-i)*2^i with a(1)=0. - Bruno Berselli, Feb 10 2014
a(n) = 2 * A000295(n). - Alois P. Heinz, May 28 2018

Edited by R. J. Mathar, Oct 21 2008

A321280 Number T(n,k) of permutations p of [n] with exactly k descents such that the up-down signature of p has nonnegative partial sums; triangle T(n,k), n>=0, 0<=k<=max(0,floor((n-1)/2)), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 8, 1, 22, 22, 1, 52, 172, 1, 114, 856, 604, 1, 240, 3488, 7296, 1, 494, 12746, 54746, 31238, 1, 1004, 43628, 330068, 518324, 1, 2026, 143244, 1756878, 5300418, 2620708, 1, 4072, 457536, 8641800, 43235304, 55717312, 1, 8166, 1434318, 40298572, 309074508, 728888188, 325024572

Offset: 0

Alois P. Heinz, Nov 01 2018

			Triangle T(n,k) begins:
  1;
  1;
  1;
  1,     2;
  1,     8;
  1,    22,      22;
  1,    52,     172;
  1,   114,     856,       604;
  1,   240,    3488,      7296;
  1,   494,   12746,     54746,      31238;
  1,  1004,   43628,    330068,     518324;
  1,  2026,  143244,   1756878,    5300418,    2620708;
  1,  4072,  457536,   8641800,   43235304,   55717312;
  1,  8166, 1434318,  40298572,  309074508,  728888188,  325024572;
  1, 16356, 4438540, 180969752, 2026885824, 7589067592, 8460090160;
  ...

Alois P. Heinz, Rows n = 0..100, flattened
S. Spiro, Ballot Permutations, Odd Order Permutations, and a New Permutation Statistic, arXiv preprint arXiv:1810.00993 [math.CO], 2018.
David G. L. Wang, T. Zhao, The peak and descent statistics over ballot permutations, arXiv:2009.05973 [math.CO], 2020.

Columns k=0-3 give: A000012, A005803 (for n>0), A321268, A321269.
Row sums give A000246.
T(2n+1,n) gives A177042.
T(2n+2,n) gives A303285(n+1).
Cf. A262124, A262125.

b:= proc(u, o, c) option remember; `if`(c<0, 0, `if`(u+o=0, 1/x,
       add(expand(x*b(u-j, o-1+j, c-1)), j=1..u)+
       add(b(u+j-1, o-j, c+1), j=1..o)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(`if`(n=0, 1, b(n, 0, 1))):
seq(T(n), n=0..14);

b[u_, o_, c_] := b[u, o, c] = If[c < 0, 0, If[u + o == 0, 1/x, Sum[Expand[ x*b[u - j, o - 1 + j, c - 1]], {j, 1, u}] + Sum[b[u + j - 1, o - j, c + 1], {j, 1, o}]]];
T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][ If[n == 0, 1, b[n, 0, 1]]];
Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Dec 08 2018, after Alois P. Heinz *)

A046740 Triangle of number of permutations of [n] with 0 successions, by number of rises.

Original entry on oeis.org

1, 1, 1, 2, 1, 8, 2, 1, 22, 28, 2, 1, 52, 182, 72, 2, 1, 114, 864, 974, 164, 2, 1, 240, 3474, 8444, 4174, 352, 2, 1, 494, 12660, 57194, 61464, 15782, 732, 2, 1, 1004, 43358, 332528, 660842, 373940, 55286, 1496, 2, 1, 2026, 142552, 1747558, 5814124

Offset: 1

N. J. A. Sloane

The recurrence given by Roselle is wrong.

			Triangle begins:
  1;
  1;
  1,  2;
  1,  8,  2;
  1, 22, 28,  2;
  ...

D. P. Roselle, Permutations by number of rises and successions, Proc. Amer. Math. Soc., 19 (1968), 8-16.
D. P. Roselle, Permutations by number of rises and successions, Proc. Amer. Math. Soc., 19 (1968), 8-16. [Annotated scanned copy]

Cf. A046739, A000295. Row sums give A000255. Diagonals give A005803, A065340.

Row sums give A000255.

a[, 1] = 1; a[n, 2] := 2^n - 2*n; a[n_, r_] /; 1 <= r <= n-1 := a[n, r] = r*a[n-1, r] + (n-r)*a[n-1, r-1] + (n-2)*a[n-2, r-1]; a[, ] = 0;
row[1] = {{1}}; row[n_] := Table[a[n, r], {r, 1, n-1}];
Table[row[n], {n, 1, 11}] // Flatten (* Jean-François Alcover, Sep 07 2017 *)

a(n, 1) = 1; for r > 1, a(n, r) = r*a(n-1, r) + (n-r)*a(n-1, r-1) + (n-2)*a(n-2, r-1).

a(n, 2) = 2^n - 2*n = 2*A000295 = A005803, n >= 3.

More terms from Vladeta Jovovic, Jan 03 2003

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