A034856 -id:A034856 - cp's OEIS frontend

A193737 Mirror of the triangle A193736.

1, 1, 1, 1, 2, 1, 2, 4, 3, 1, 3, 8, 8, 4, 1, 5, 15, 19, 13, 5, 1, 8, 28, 42, 36, 19, 6, 1, 13, 51, 89, 91, 60, 26, 7, 1, 21, 92, 182, 216, 170, 92, 34, 8, 1, 34, 164, 363, 489, 446, 288, 133, 43, 9, 1, 55, 290, 709, 1068, 1105, 826, 455, 184, 53, 10, 1, 89, 509, 1362, 2266, 2619, 2219, 1414, 682, 246, 64, 11, 1

Offset: 0

Clark Kimberling, Aug 04 2011

This triangle is obtained by reversing the rows of the triangle A193736.

			First six rows:
  1;
  1,  1;
  1,  2,  1;
  2,  4,  3,  1;
  3,  8,  8,  4,  1;
  5, 15, 19, 13,  5,  1;

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

Cf. A000007, A011782 (diagonal sums), A019590, A052542 (row sums).

Cf. A034856, A193736, A321620, A324969, A330793.

function T(n,k) // T = A193737
  if k lt 0 or n lt 0 then return 0;
  elif n lt 3 then return Binomial(n,k);
  else return T(n - 1, k) + T(n - 1, k - 1) + T(n - 2, k);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 24 2023

(* First program *)
z=20;
p[0, x_]:= 1;
p[n_, x_]:= Fibonacci[n+1, x] /; n > 0
q[n_, x_]:= (x + 1)^n;
t[n_, k_]:= Coefficient[p[n, x], x^(n-k)];
t[n_, n_]:= p[n, x] /. x -> 0;
w[n_, x_]:= Sum[t[n, k]*q[n-k+1, x], {k,0,n}]; w[-1, x_] := 1;
g[n_]:= CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n,-1,z}]] (* A193736 *)
TableForm[Table[g[n], {n,-1,z}]]
Flatten[Table[g[n], {n,-1,z}]]          (* A193737 *)
(* Additional programs *)
(* Function RiordanSquare defined in A321620. *)
RiordanSquare[1 + 1/(1 - x - x^2), 11]//Flatten  (* Peter Luschny, Feb 27 2021 *)
T[n_, k_]:= T[n, k]= If[n<3, Binomial[n, k], T[n-1,k] + T[n-1,k-1] + T[n-2,k]];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 24 2023 *)

def T(n,k): # T = A193737
    if (n<3): return binomial(n,k)
    else: return T(n-1,k) +T(n-1,k-1) +T(n-2,k)
flatten([[T(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 24 2023

Write w(n,k) for the triangle at A193736.  This is then given by w(n,n-k).
T(0,0) = T(1,0) = T(1,1) = T(2,0) = 1; T(n,k) = 0 if k<0 or k>n; T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k). - Philippe Deléham, Feb 13 2020
From G. C. Greubel, Oct 24 2023: (Start)
T(n, 0) = Fibonacci(n) + [n=0] = A324969(n+1).
T(n, n-1) = n, for n >= 1.
T(n, n-2) = A034856(n-1), for n >= 2.
T(2*n, n) = A330793(n).
Sum_{k=0..n} T(n,k) = A052542(n).
Sum_{k=0..n} (-1)^k * T(n,k) = A000007(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A011782(n).
Sum_{k=0..floor(n/2)} (-1)^k * T(n-k,k) = A019590(n). (End)

A248156 Inverse Riordan triangle of A106513: Riordan ((1 - 2*x^2 )/(1 + x), x/(1+x)).

Original entry on oeis.org

1, -2, 1, 1, -3, 1, 0, 4, -4, 1, -1, -4, 8, -5, 1, 2, 3, -12, 13, -6, 1, -3, -1, 15, -25, 19, -7, 1, 4, -2, -16, 40, -44, 26, -8, 1, -5, 6, 14, -56, 84, -70, 34, -9, 1, 6, -11, -8, 70, -140, 154, -104, 43, -10, 1, -7, 17, -3, -78, 210, -294, 258, -147, 53, -11, 1, 8, -24, 20, 75, -288, 504, -552, 405, -200, 64, -12, 1

Offset: 0

Wolfdieter Lang, Oct 05 2014

Row sums have o.g.f. (1 - 2*x)/(1 + x): [1, -1, repeat(-1, 1)].

			The triangle T(n,k) begins:
  n\k  0   1   2   3    4    5    6    7    8    9
  0:   1
  1:  -2   1
  2:   1  -3   1
  3:   0   4  -4   1
  4:  -1  -4   8  -5    1
  5:   2   3 -12  13   -6    1
  6:  -3  -1  15 -25   19   -7    1
  7:   4  -2 -16  40  -44   26   -8    1
  8:  -5   6  14 -56   84  -70   34   -9    1
  9:   6 -11  -8  70 -140  154 -104   43  -10    1
  ...
For more rows see the link.
Recurrence from A-sequence: T(5,2) = T(4,1) - T(4,2) = -4 - 8 = -12.
Recurrence from the Z-sequence: T(5,0) = -(2*(-1) + 3*(-4) + 7*8 + 17*(-5) + 41*1) = 2.
Standard recurrence for T(n,0): T(3,0) = -2*T(2,0) - T(1,0) = -2*1 - (-2) = 0.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Wolfdieter Lang, First 13 rows of the triangle

Cf. A006232, A083318, A106513, A248157, A248158, A248159, A248160, A248161.
Columns: A248157 (k=0), A248158 (k=1), A248159 (k=2), A248160 (k=3).
Diagonals: A000012 (k=n), A022958(n+3) (k=n-1), -A034856(n-1) (k=n-2), A000297(n-4) (k=n-3), A014309(n-3) (k=n-4).
Sums: (-1)^n*A001611(n) (diagonal), (-1)^n*A083318(n) (alternating sign row).

function T(n,k) // T = A248156
  if k eq n then return 1;
  elif k eq 0 then return (-1)^n*(3-n);
  else return T(n-1,k-1) - T(n-1,k);
  end if;
end function;
[T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, May 27 2025

T[n_, k_] := SeriesCoefficient[x^k*(1 - 2*x^2)/(1 + x)^(k + 2), {x, 0, n}]; Table[T[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 09 2014 *)
T[n_, k_]:= T[n, k]= If[k==n,1, If[k==0,(-1)^n*(3-n), T[n-1,k-1]-T[n-1,k]]];
Table[T[n,k], {n,0,25}, {k,0,n}]//Flatten (* G. C. Greubel, May 27 2025 *)

def T(n,k): # T = A248156
    if (k==n): return 1
    elif (k==0): return (-1)^n*(3-n)
    else: return T(n-1,k-1) - T(n-1,k)
print(flatten([[T(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, May 27 2025

O.g.f. row polynomials R(n,x) = Sum_{k=0..n} T(n,k)*x^k = [(-z)^n] (1 - 2*z^2)/( (1 + z)*(1 + (1-x)*z)).
O.g.f. column m: x^m*(1 - 2*x^2)/(1 + x)^(m+2), m >= 0.
The A-sequence is [1, -1], implying the recurrence T(n,k) = T(n-1, k-1) - T(n-1, k), n >= k > = 1.
The Z-sequence is -[2, 3, 7, 17, 41, 99, 239, 577, 1393, ...] = A248161, implying the recurrence T(n, 0) = Sum_{k=0..n-1} T(n-1,k)*Z(k). See the W. Lang link under A006232 for Riordan A- and Z-sequences.
The standard recurrence for the sequence for column k=0 is T(0,0) = 1 and T(n,0) = -2*T(n-1,0) - T(n-2,0), n >= 3, with T(1,0) = -2 and T(2,0) = 1.
From G. C. Greubel, May 27 2025: (Start)
Sum_{k=0..n} T(n, k) = (-1)^(n+1) + 2*([n=0] - [n=1]).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = the repeated pattern of [1, -2, 0, 3, -4, 2]. (End)

A141539 Square array A(n,k) of numbers of length n binary words with at least k "0" between any two "1" digits (n,k >= 0), read by antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 3, 8, 1, 2, 3, 5, 16, 1, 2, 3, 4, 8, 32, 1, 2, 3, 4, 6, 13, 64, 1, 2, 3, 4, 5, 9, 21, 128, 1, 2, 3, 4, 5, 7, 13, 34, 256, 1, 2, 3, 4, 5, 6, 10, 19, 55, 512, 1, 2, 3, 4, 5, 6, 8, 14, 28, 89, 1024, 1, 2, 3, 4, 5, 6, 7, 11, 19, 41, 144, 2048, 1, 2, 3, 4, 5, 6, 7, 9, 15, 26, 60, 233, 4096

Offset: 0

Alois P. Heinz, Aug 15 2008

A(n,k+1) = A(n,k) - A143291(n,k).
From Gary W. Adamson, Dec 19 2009: (Start)
Alternative method generated from variants of an infinite lower triangle T(n) = A000012 = (1; 1,1; 1,1,1; ...) such that T(n) has the leftmost column shifted up n times. Then take lim_{k->infinity} T(n)^k, obtaining a left-shifted vector considered as rows of an array (deleting the first 1) as follows:
  1, 2, 4, 8, 16, 32, 64, 128, 256, ... = powers of 2
  1, 1, 2, 3,  5,  8, 13,  21,  34, ... = Fibonacci numbers
  1, 1, 1, 2,  3,  4,  6,   9,  13, ... = A000930
  1, 1, 1, 1,  2,  3,  4,   5,   7, ... = A003269
  ... with the next rows A003520, A005708, A005709, ... such that beginning with the Fibonacci row, the succession of rows are recursive sequences generated from a(n) = a(n-1) + a(n-2); a(n) = a(n-1) + a(n-3), ... a(n) = a(n-1) + a(n-k); k = 2,3,4,... Last, columns going up from the topmost 1 become rows of triangle A141539. (End)

			A(4,2) = 6, because 6 binary words of length 4 have at least 2 "0" between any two "1" digits: 0000, 0001, 0010, 0100, 1000, 1001.
Square array A(n,k) begins:
    1,  1,  1,  1,  1,  1,  1,  1, ...
    2,  2,  2,  2,  2,  2,  2,  2, ...
    4,  3,  3,  3,  3,  3,  3,  3, ...
    8,  5,  4,  4,  4,  4,  4,  4, ...
   16,  8,  6,  5,  5,  5,  5,  5, ...
   32, 13,  9,  7,  6,  6,  6,  6, ...
   64, 21, 13, 10,  8,  7,  7,  7, ...
  128, 34, 19, 14, 11,  9,  8,  8, ...

Alois P. Heinz, Table of n, a(n) for n = 0..140, flattened

Cf. column k=0: A000079, k=1: A000045(n+2), k=2: A000930(n+2), A068921, A078012(n+5), k=3: A003269(n+4), A017898(n+7), k=4: A003520(n+4), A017899(n+9), k=5: A005708(n+5), A017900(n+11), k=6: A005709(n+6), A017901(n+13), k=7: A005710(n+7), A017902(n+15), k=8: A005711(n+7), A017903(n+17), k=9: A017904(n+19), k=10: A017905(n+21), k=11: A017906(n+23), k=12: A017907(n+25), k=13: A017908(n+27), k=14: A017909(n+29).
Main diagonal gives A000027(n+1).
A(2n,n) gives A000217(n+1)
A(3n,n) gives A008778.
A(3n,2n) gives A034856(n+1).
A(2n,3n) gives A005408.
A(2^n-1,n) gives A376697.
See also A143291.

A:= proc(n, k) option remember;
      if k=0 then 2^n
    elif n<=k and n>=0 then n+1
    elif n>0 then A(n-1, k) +A(n-k-1, k)
    else          A(n+1+k, k) -A(n+k, k)
      fi
    end:
seq(seq(A(n, d-n), n=0..d), d=0..15);

a[n_, k_] := a[n, k] = Which[k == 0, 2^n, n <= k && n >= 0, n+1, n > 0, a[n-1, k] + a[n-k-1, k], True, a[n+1+k, k] - a[n+k, k]]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 15}] // Flatten (* Jean-François Alcover, Dec 17 2013, translated from Maple *)

G.f. of column k: x^(-k)/(1-x-x^(k+1)).

A(n,k) = 2^n if k=0, otherwise A(n,k) = n+1 if n<=k, otherwise A(n,k) = A(n-1,k) + A(n-k-1,k).

A176145 a(n) = n(n-3)(n^2-7*n+14)/8.

Original entry on oeis.org

0, 1, 5, 18, 49, 110, 216, 385, 638, 999, 1495, 2156, 3015, 4108, 5474, 7155, 9196, 11645, 14553, 17974, 21965, 26586, 31900, 37973, 44874, 52675, 61451, 71280, 82243, 94424, 107910, 122791, 139160, 157113, 176749, 198170, 221481, 246790, 274208, 303849

Offset: 3

Michel Lagneau, Apr 10 2010

Number of points of intersection of diagonals of a general convex n-polygon. (both inside and outside the polygon).

n(n-3)/2 (n >= 3) is the number of diagonals of an n-gon (A080956). The number of points (inside or outside), distinct of tops, where these diagonals intersect is : (1/2)( n(n-3)/2)(n(n-3)/2 - 2(n-4) -1) = n(n-3)(n^2 - 7n + 14) / 8.

			For n=3, a(3) = 0 (no diagonals in triangle),
For n=4, a(4) = 1 (2 diagonals => 1 point of intersection),
For n=5, a(5) = 5 (5 diagonals => 5 points of intersection),
For n=6, a(6) = 18 (9 diagonals => 18 points of intersection).

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 797.

Vincenzo Librandi, Table of n, a(n) for n = 3..10000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A080956, A055504.

Cf. A000217, A034856, A000124, A005581-A005584.

[n*(n-3)*(n^2 - 7*n + 14) / 8: n in [3..60]]; // Vincenzo Librandi, May 21 2011

for n from 3 to 50 do: x:=n*(n-3)*(n^2 - 7*n +14)/8 : print(x):od:

Table[n*(n - 3)*(n^2 - 7*n + 14)/8, {n, 3, 42}] (* Bobby Milazzo, Jun 24 2013 *)
Drop[CoefficientList[Series[x^4(1+3x^2-x^3)/(1-x)^5,{x,0,50}],x],3] (* or *) LinearRecurrence[{5,-10,10,-5,1},{0,1,5,18,49},50] (* Harvey P. Dale, Mar 14 2022 *)

vector(100,n,(n+2)*(n-1)*(n^2-3*n+4)/8) \\ Derek Orr, Jan 21 2015

G.f.: x^4*(1+3*x^2-x^3)/(1-x)^5. [Colin Barker, Jan 31 2012]
a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) + a(n-5), with a(3)= 0, a(4)= 1, a(5)=5, a(6)= 18, a(7) = 49. [Bobby Milazzo, Jun 24 2013]
a(n) = Sum_{k=(n-3)..(n-2)*(n-3)/2} k. - J. M. Bergot, Jan 21 2015

Edited by N. J. A. Sloane, Apr 19 2010

A323224 A(n, k) = [x^k] C^nx/(1 - x) where C = 2/(1 + sqrt(1 - 4x)), square array read by ascending antidiagonals with n >= 0 and k >= 0.

Original entry on oeis.org

0, 0, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 8, 9, 1, 0, 1, 5, 13, 22, 23, 1, 0, 1, 6, 19, 41, 64, 65, 1, 0, 1, 7, 26, 67, 131, 196, 197, 1, 0, 1, 8, 34, 101, 232, 428, 625, 626, 1, 0, 1, 9, 43, 144, 376, 804, 1429, 2055, 2056, 1

Offset: 0

Peter Luschny, Jan 24 2019

Equals A096465 when the leading column (k = 0) is removed. - Georg Fischer, Jul 26 2023

			The square array starts:
   [n\k]  0  1   2   3    4     5     6      7       8       9
    ---------------------------------------------------------------
    [0]   0, 1,  1,  1,   1,    1,    1,     1,      1,      1, ... A057427
    [1]   0, 1,  2,  4,   9,   23,   65,   197,    626,   2056, ... A014137
    [2]   0, 1,  3,  8,  22,   64,  196,   625,   2055,   6917, ... A014138
    [3]   0, 1,  4, 13,  41,  131,  428,  1429,   4861,  16795, ... A001453
    [4]   0, 1,  5, 19,  67,  232,  804,  2806,   9878,  35072, ... A114277
    [5]   0, 1,  6, 26, 101,  376, 1377,  5017,  18277,  66727, ... A143955
    [6]   0, 1,  7, 34, 144,  573, 2211,  8399,  31655, 118865, ...
    [7]   0, 1,  8, 43, 197,  834, 3382, 13378,  52138, 201364, ...
    [8]   0, 1,  9, 53, 261, 1171, 4979, 20483,  82499, 327656, ...
    [9]   0, 1, 10, 64, 337, 1597, 7105, 30361, 126292, 515659, ...
.
Triangle given by ascending antidiagonals:
    0;
    0, 1;
    0, 1, 1;
    0, 1, 2,  1;
    0, 1, 3,  4,   1;
    0, 1, 4,  8,   9,   1;
    0, 1, 5, 13,  22,  23,   1;
    0, 1, 6, 19,  41,  64,  65,   1;
    0, 1, 7, 26,  67, 131, 196, 197,   1;
    0, 1, 8, 34, 101, 232, 428, 625, 626, 1;
.
The difference table of a column successively gives the preceding columns, here starting with column 6.
col(6) = 1, 65, 196, 428, 804, 1377, 2211, 3382, 4979, 7105, ...
col(5) =    64, 131, 232, 376,  573,  834, 1171, 1597, 2126, ...
col(4) =         67, 101, 144,  197,  261,  337,  426,  529, ...
col(3) =              34,  43,   53,   64,   76,   89,  103, ...
col(2) =                    9,   10,   11,   12,   13,   14, ...
col(1) =                          1,    1,    1,    1,    1, ...
col(0) =                                0,    0,    0,    0, ...
.
Example for the sum formula: C(0) = 1, C(1) = 1, C(2) = 2 and C(3) = 5.
X(3, 4) = {{0,0,0}, {0,0,1}, {0,1,0}, {1,0,0}, {0,0,2}, {0,1,1}, {0,2,0}, {1,0,1},
{1,1,0}, {2,0,0}, {0,0,3}, {0,1,2}, {0,2,1}, {0,3,0}, {1,0,2}, {1,1,1}, {1,2,0},
{2,0,1}, {2,1,0}, {3,0,0}}. T(3,4) = 1+1+1+1+2+1+2+1+1+2+5+2+2+5+2+1+2+2+2+5 = 41.

The coefficients of the polynomials generating the columns are in A323233.
Sums of antidiagonals and row 1 are A014137. Main diagonal is A242798.
Rows: A057427 (n=0), A014137 (n=1), A014138 (n=2), A001453 (n=3), A114277 (n=4), A143955 (n=5).
Columns: A000027 (k=2), A034856 (k=3), A323221 (k=4), A323220 (k=5).
Similar array based on central binomials is A323222.
Cf. A096465.

Row := proc(n, len) local C, ogf, ser; C := (1-sqrt(1-4*x))/(2*x);
ogf := C^n*x/(1-x); ser := series(ogf, x, (n+1)*len+1);
seq(coeff(ser, x, j), j=0..len) end:
for n from 0 to 9 do Row(n, 9) od;
# Alternatively by recurrence:
B := proc(n, k) option remember; if n <= 0 or k < 0 then 0
elif n = k then 1 else B(n-1, k) + B(n, k-1) fi end:
A := (n, k) -> B(n + k, k): seq(lprint(seq(A(n, k), k=0..9)), n=0..9);

(* Illustrating the sum formula, not efficient. *) T[0, K_] := Boole[K != 0];
T[N_, K_] := Module[{}, r[n_, k_] := FrobeniusSolve[ConstantArray[1, n], k];
X[n_] := Flatten[Table[r[N, j], {j, 0, n - 1}], 1];
Sum[Product[CatalanNumber[m[[i]]], {i, 1, N}], {m , X[K]}]];
Trow[n_] := Table[T[n, k], {k, 0, 9}]; Table[Trow[n], {n, 0, 9}]

For n>0 and k>0 let X(n, k) denote the set of all tuples of length n with elements from {0, ..., k-1} with sum < k. Let C(m) denote the m-th Catalan number. Then: A(n, k) = Sum_{(j1,...,jn) in X(n, k)} C(j1)*C(j2)*...*C(jn).

A(n, k) = T(n + k, k) with T(n, k) = T(n-1, k) + T(n, k-1) with T(n, k) = 0 if n <= 0 or k < 0 and T(n, n) = 1.

A323254 The determinant of an n X n Toeplitz matrix M(n) whose first row consists of successive positive integer numbers 2n - 1, n - 1, ..., 1 and whose first column consists of 2n - 1, 2*n - 2, ..., n.

Original entry on oeis.org

1, 7, 58, 614, 8032, 125757, 2298208, 48075148, 1133554432, 29756555315, 860884417024, 27218972906226, 933850899349504, 34556209025624041, 1371957513591119872, 58174957356247084568, 2624017129323317493760, 125454378698728779884895, 6337442836338834419089408

Offset: 1

Stefano Spezia, Jan 09 2019

nonn

The trace of the matrix M(n) is A000384(n). [Corrected by Stefano Spezia, Dec 08 2019]
The sum of the first row of the matrix M(n) is A034856(n).
The sum of the first column of the matrix M(n) is A000326(n).
For n > 1, the sum of the superdiagonal of the matrix M(n) is A000290(n-1).
For n > 1, the sum of the subdiagonal of the matrix M(n) is A001105(n-1).

			For n = 1 the matrix M(1) is
   1
with determinant Det(M(1)) = 1.
For n = 2 the matrix M(2) is
   3, 1
   2, 3
with Det(M(2)) = 7.
For n = 3 the matrix M(3) is
   5, 2, 1
   4, 5, 2
   3, 4, 5
with Det(M(3)) = 58.

Vaclav Kotesovec, Table of n, a(n) for n = 1..300
Wikipedia, Toeplitz matrix

Cf. A000290, A000326, A000384, A001105, A001792.
Cf. A034856, A204235, A318173, A322908, A322909.
Cf. A323255 (permanent of matrix M(n)).

b[i_]:=i; a[n_]:=Det[ToeplitzMatrix[Join[{b[2*n-1]}, Array[b, n-1, {2*n-2,n}]], Join[{b[2*n-1]},Array[b, n-1, {n-1,1}]]]]; Array[a,20]

tm(n) = {my(m = matrix(n, n, i, j, if (j==1, 2*n-i, n-j+1))); for (i=2, n, for (j=2, n, m[i, j] = m[i-1, j-1]; ); ); m;}
a(n) = matdet(tm(n)); \\ Stefano Spezia, Dec 11 2019

a(n) ~ (5*exp(1) + exp(-1)) * n^n / 4. - Vaclav Kotesovec, Jan 10 2019

A099239 Square array read by antidiagonals associated with sections of 1/(1-x-x^k).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 3, 1, 1, 8, 8, 4, 1, 1, 16, 21, 13, 5, 1, 1, 32, 55, 41, 19, 6, 1, 1, 64, 144, 129, 69, 26, 7, 1, 1, 128, 377, 406, 250, 106, 34, 8, 1, 1, 256, 987, 1278, 907, 431, 153, 43, 9, 1, 1, 512, 2584, 4023, 3292, 1757, 686, 211, 53, 10, 1, 1, 1024, 6765, 12664, 11949, 7168, 3088, 1030, 281, 64, 11, 1

Offset: 0

Paul Barry, Oct 08 2004

Rows include A099242, A099253. Columns include A034856. Main diagonal is A099240. Sums of antidiagonals are A099241.

			Rows begin
  1, 1,  1,   1,   1, ...                               A000012;
  1, 2,  4,   8,  16, ...      1-section of 1/(1-x-x)   A000079;
  1, 3,  8,  21,  55, ....     bisection of 1/(1-x-x^2) A001906;
  1, 4, 13,  41, 129, ...     trisection of 1/(1-x-x^3) A052529; (essentially)
  1, 5, 19,  69, 250, ...  quadrisection of 1/(1-x-x^4) A055991;
  1, 6, 26, 106, 431, ...  quintisection of 1/(1-x-x^5) A079675; (essentially)

G. C. Greubel, Antidiagonal rows n = 0..50, flattened

Cf. A034856, A099240, A099241, A099242, A099253.

Cf. A000079, A001906, A052529, A055991, A079675.

A099239:= func< n,k | (&+[Binomial(k*(n-k) -(k-1)*(j-1), j): j in [0..n-k]]) >;
[A099239(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Mar 09 2021

T[n_, k_]:= Sum[Binomial[k*(n-k) - (k-1)*(j-1), j], {j,0,n-k}];
Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Mar 09 2021 *)

def A099239(n,k): return sum( binomial(k*(n-k) -(k-1)*(j-1), j) for j in (0..n-k) )
flatten([[A099239(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 09 2021

T(n, k) = Sum_{j=0..n} binomial(k*n -(k-1)*(j-1), j), n, k>=0. (square array)
T(n, k) = Sum_{j=0..n} binomial(k + (n-1)*(j+1), n*(j+1) -1), n>0. (square array)
T(n, k) = Sum_{j=0..n-k} binomial(k*(n-k) - (k-1)*(j-1), j). (number triangle)
Rows of the square array are generated by 1/((1-x)^k-x).
Rows satisfy a(n) = a(n-1) - Sum_{k=1..n} (-1)^(k^binomial(n, k)) * a(n-k).

A113453 Triangle giving maximal permanent P(n,k) of an n X n lower Hessenberg (0,1)-matrix with exactly k 1's for 2 <= n <= k <= 2n, read by rows.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 4, 1, 1, 2, 2, 4, 4, 1, 1, 2, 2, 4, 4, 8, 1, 1, 2, 2, 4, 4, 8, 8, 1, 1, 2, 2, 4, 4, 8, 8, 16, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, 32, 32, 64

Offset: 0

Bryan Shader (bshader(AT)uwyo.edu), Jan 07 2006

G. C. Greubel, Table of n, a(n) for rows 2 to 50, flattened
D. D. Olesky, B. L. Shader and P. van den Driessche, Permanents of Hessenberg (0,1)-matrices, Electronic Journal of Combinatorics, 12 (2005) #R70.
B. Shader Table of known values of P(n,k) for n<=12.

Cf. A034856, A113452, A113453, A113454, A113455.

Table[2^(Floor[(k - n)/2]), {n, 2, 51}, {k, n, 2*n}] // Flatten (* G. C. Greubel, Mar 11 2017 *)

for(n=2,25, for(k=n,2*n, print1(2^(floor((k-n)/2)), ", "))) \\ G. C. Greubel, Mar 11 2017

P(n, k) = 2^(floor((k-n)/2)), if n <= k <= 2n.

A129936 a(n) = (n-2)(n+3)(n+2)/6.

Original entry on oeis.org

-2, -2, 0, 5, 14, 28, 48, 75, 110, 154, 208, 273, 350, 440, 544, 663, 798, 950, 1120, 1309, 1518, 1748, 2000, 2275, 2574, 2898, 3248, 3625, 4030, 4464, 4928, 5423, 5950, 6510, 7104, 7733, 8398, 9100, 9840, 10619, 11438, 12298, 13200, 14145, 15134, 16168, 17248

Offset: 0

Roger L. Bagula, Jun 09 2007

Essentially the same as A005586.

Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A000292, A002378, A005586, A023444, A034856, A214292.

seq(sum(i*(k-i+1), i=1..k+2), k=0..99); # Wesley Ivan Hurt, Sep 21 2013

f[n_] = Binomial[n + 3, 3] - (n + 3)*(n + 2)/2; Table[f[n], {n, 0, 30}]
LinearRecurrence[{4,-6,4,-1},{-2,-2,0,5},50] (* Harvey P. Dale, Jul 03 2020 *)

a(n)=(n-2)*(n+3)*(n+2)/6 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = binomial(n+3,3) - (n + 3)*(n + 2)/2.
a(n) = A214292(n+2,2). - Reinhard Zumkeller, Jul 12 2012
G.f.: (x^3-4*x^2+6*x-2)/(x-1)^4. - Colin Barker, Sep 05 2012
From Wesley Ivan Hurt, Sep 21 2013: (Start)
a(n) = Sum_{i=1..n+2} i*(n-i+1).
a(n+2) = A000292(n+1) + A034856(n), n>0. (End)
From Amiram Eldar, Sep 26 2022: (Start)
Sum_{n>=3} 1/a(n) = 77/200.
Sum_{n>=3} (-1)^(n+1)/a(n) = 363/200 - 12*log(2)/5. (End)
From Elmo R. Oliveira, Aug 04 2025: (Start)
E.g.f.: exp(x)*(x^3 + 6*x^2 - 12)/6.
a(n) = A023444(n)*A002378(n+2)/6.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). (End)

More terms from Wesley Ivan Hurt, Sep 21 2013

A165157 Zero followed by partial sums of A133622.

Original entry on oeis.org

0, 1, 3, 4, 7, 8, 12, 13, 18, 19, 25, 26, 33, 34, 42, 43, 52, 53, 63, 64, 75, 76, 88, 89, 102, 103, 117, 118, 133, 134, 150, 151, 168, 169, 187, 188, 207, 208, 228, 229, 250, 251, 273, 274, 297, 298, 322, 323, 348, 349, 375, 376, 403, 404, 432, 433, 462, 463, 493, 494, 525

Offset: 0

Jaroslav Krizek, Sep 05 2009

A133622 is a toothed sequence.

Interleaving of A055998 and A034856.

			From _Stefano Spezia_, Jul 10 2020: (Start)
Illustration of the initial terms for n > 0:
o    o      o      o         o        o
     o o    o o    o o       o o      o o
            o      o         o        o
                   o o o     o o o    o o o
                             o        o
                                      o o o o
(1)  (3)   (4)    (7)       (8)      (12)
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Equals -1+A101881.
a(n) = A117142(n+2)-2 = A055802(n+6)-3 = A114220(n+5)-3 = A134519(n+3)-3.
Cf. A133622, A055998, A034856.

a165157 n = a165157_list !! n
a165157_list = scanl (+) 0 a133622_list
-- Reinhard Zumkeller, Feb 20 2015

m:=60; T:=[ 1+(1+(-1)^n)*n/4: n in [1..m] ]; [0] cat [ n eq 1 select T[1] else Self(n-1)+T[n]: n in [1..m] ]; // Klaus Brockhaus, Sep 06 2009

[ n le 2 select n-1 else n le 4 select n else 2*Self(n-2)-Self(n-4)+1: n in [1..61] ]; // Klaus Brockhaus, Sep 06 2009

a(0) = 0, a(2*n) = a(2*n-1) + n + 1, a(2*n+1) = a(2*n) + 1.
a(n) = (n^2+10*n)/8 if n is even, a(n) = (n^2+8*n-1)/8 if n is odd.
a(2*k) = A055998(k) = k*(k+5)/2; a(2*k+1) = A034856(k+1) = k*(k+5)/2+1.
a(n) = 2*a(n-2)-a(n-4)+1 for n > 3; a(0)=0, a(1)=1, a(2)=3, a(3)=4. - Klaus Brockhaus, Sep 06 2009
a(n) = (2*n*(n+9)-1+(2*n+1)*(-1)^n)/16. - Klaus Brockhaus, Sep 06 2009
a(n) = n+binomial(1+floor(n/2),2). - Mircea Merca, Feb 18 2012
G.f.: x*(1+2*x-x^2-x^3)/((1-x)^3*(1+x)^2). - Klaus Brockhaus, Sep 06 2009
From Stefano Spezia, Jul 10 2020: (Start)
E.g.f.: (x*(9 + x)*cosh(x) + (-1 + 11*x + x^2)*sinh(x))/8.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4. (End)

Edited and extended by Klaus Brockhaus, Sep 06 2009

cp's OEIS Frontend

A193737 Mirror of the triangle A193736.

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A248156 Inverse Riordan triangle of A106513: Riordan ((1 - 2*x^2 )/(1 + x), x/(1+x)).

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A141539 Square array A(n,k) of numbers of length n binary words with at least k "0" between any two "1" digits (n,k >= 0), read by antidiagonals.

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A176145 a(n) = n*(n-3)*(n^2-7*n+14)/8.

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A323224 A(n, k) = [x^k] C^n*x/(1 - x) where C = 2/(1 + sqrt(1 - 4*x)), square array read by ascending antidiagonals with n >= 0 and k >= 0.

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A323254 The determinant of an n X n Toeplitz matrix M(n) whose first row consists of successive positive integer numbers 2*n - 1, n - 1, ..., 1 and whose first column consists of 2*n - 1, 2*n - 2, ..., n.

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A099239 Square array read by antidiagonals associated with sections of 1/(1-x-x^k).

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A176145 a(n) = n(n-3)(n^2-7*n+14)/8.

A323224 A(n, k) = [x^k] C^nx/(1 - x) where C = 2/(1 + sqrt(1 - 4x)), square array read by ascending antidiagonals with n >= 0 and k >= 0.

A323254 The determinant of an n X n Toeplitz matrix M(n) whose first row consists of successive positive integer numbers 2n - 1, n - 1, ..., 1 and whose first column consists of 2n - 1, 2*n - 2, ..., n.

A129936 a(n) = (n-2)(n+3)(n+2)/6.