A000337 -id:A000337 - cp's OEIS frontend

A163940 Triangle related to the divergent series 1^m1! - 2^m2! + 3^m3! - 4^m4! + ... for m >= -1.

1, 1, 0, 1, 2, 0, 1, 5, 3, 0, 1, 9, 17, 4, 0, 1, 14, 52, 49, 5, 0, 1, 20, 121, 246, 129, 6, 0, 1, 27, 240, 834, 1039, 321, 7, 0, 1, 35, 428, 2250, 5037, 4083, 769, 8, 0, 1, 44, 707, 5214, 18201, 27918, 15274, 1793, 9, 0, 1, 54, 1102, 10829, 54111, 133530, 145777, 55152, 4097, 10, 0

Offset: 0

Johannes W. Meijer, Aug 13 2009

The divergent series g(x,m) = Sum_{k >= 1} (-1)^(k+1)*k^m*k!*x^k, m >= -1, are related to the higher order exponential integrals E(x,m,n=1), see A163931.
Hardy, see the link below, describes how Euler came to the rather surprising conclusion that g(x,-1) = exp(1/x)*Ei(1,1/x) with Ei(1,x) = E(x,m=1,n=1). From this result it follows inmediately that g(x,0) = 1 - g(x,-1). Following in Euler's footsteps we discovered that g(x,m) = (-1)^(m) * (M(x,m)*x - ST(x,m)* Ei(1,1/x) * exp(1/x))/x^(m+1), m => -1.
So g(x=1,m) = (-1)^m*(A040027(m) - A000110 (m+1)*A073003), with A040027(m = -1) = 0. We observe that A073003 = - exp(1)*Ei(-1) is Gompertz's constant, A000110 are the Bell numbers and A040027 was published a few years ago by Gould.
The polynomial coefficients of the M(x,m) = Sum_{k = 0..m} a(m,k) * x^k, for m >= 0, lead to the triangle given above. We point out that M(x,m=-1) = 0.
The polynomial coefficients of the ST(x,m) = Sum_{k = 0..m+1} S2(m+1, k) * x^k, m >= -1, lead to the Stirling numbers of the second kind, see A106800.
The formulas that generate the coefficients of the left hand columns lead to the Minkowski numbers A053657. We have a closer look at them in A163972.
The right hand columns have simple generating functions, see the formulas. We used them in the first Maple program to generate the triangle coefficients (n >= 0 and 0 <= k <= n). The second Maple program calculates the values of g(x,m) for m >= -1, at x=1.

			The first few triangle rows are:
  [1]
  [1, 0]
  [1, 2, 0]
  [1, 5, 3, 0]
  [1, 9, 17, 4, 0]
  [1, 14, 52, 49, 5, 0]
The first few M(x,m) are:
  M(x,m=0) = 1
  M(x,m=1) = 1 + 0*x
  M(x,m=2) = 1 + 2*x + 0*x^2
  M(x,m=3) = 1 + 5*x + 3*x^2 + 0*x^3
The first few ST(x,m) are:
  ST(x,m=-1) = 1
  ST(x,m=0) = 1 + 0*x
  ST(x,m=1) = 1 + 1*x + 0*x^2
  ST(x,m=2) = 1 + 3*x + x^2 + 0*x^3
  ST(x,m=3) = 1 + 6*x + 7*x^2 + x^3 + 0*x^4
The first few g(x,m) are:
  g(x,-1) = (-1)*(- (1)*Ei(1,1/x)*exp(1/x))/x^0
  g(x,0) = (1)*((1)*x - (1)*Ei(1,1/x)*exp(1/x))/x^1
  g(x,1) = (-1)*((1)*x - (1+ x)*Ei(1,1/x)*exp(1/x))/x^2
  g(x,2) = (1)*((1+2*x)*x - (1+3*x+x^2)*Ei(1,1/x)*exp(1/x))/x^3
  g(x,3) = (-1)*((1+5*x+3*x^2)*x - (1+6*x+7*x^2+x^3)*Ei(1,1/x)*exp(1/x))/x^4

G. H. Hardy, Divergent Series, Oxford University Press, 1949. pp. 26-29 and pp. 7-8.
Maurice de Gosson, Branko Dragovich and Andrei Khrennikov, Some p-adic differential equations, (see Section 5), arxiv:math-ph/0010023, Oct 2000.

The row sums equal A040027 (Gould).
A000007, A000027, A000337, A163941 and A163942 are the first five right hand columns.
A000012, A000096, A163943 and A163944 are the first four left hand columns.
Cf. A163931, A163972, A106800 (Stirling2), A000110 (Bell), A073003 (Gompertz), A053657 (Minkowski), A014619.

nmax := 10; for p from 1 to nmax do Gf(p) := convert(series(1/((1-(p-1)*x)^2*product((1-k1*x), k1=1..p-2)), x, nmax+1-p), polynom); for q from 0 to nmax-p do a(p+q-1, q) := coeff(Gf(p), x, q) od: od: seq(seq(a(n, k), k=0..n), n=0..nmax-1);
# End program 1
nmax1:=nmax; A040027 := proc(n): if n = -1 then 0 elif n= 0 then 1 else add(binomial(n, k1-1)*A040027(n-k1), k1 = 1..n) fi: end: A000110 := proc(n) option remember; if n <= 1 then 1 else add( binomial(n-1, i) * A000110(n-1-i), i=0..n-1); fi; end: A073003 := - exp(1) * Ei(-1): for n from -1 to nmax1 do g(1, n) := (-1)^n * (A040027(n) - A000110(n+1) * A073003) od;
# End program 2

nmax = 11;
For[p = 1, p <= nmax, p++, gf = 1/((1-(p-1)*x)^2*Product[(1-k1*x), {k1, 1, p-2}]) + O[x]^(nmax-p+1) // Normal; For[q = 0, q <= nmax-p, q++, a[p+q-1, q] = Coefficient[gf, x, q]]];
Table[a[n, k], {n, 0, nmax-1}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 02 2019, from 1st Maple program *)

The generating functions of the right hand columns are Gf(p, x) = 1/((1 - (p-1)*x)^2 * Product_{k = 1..p-2} (1-k*x) ); Gf(1, x) = 1. For the first right hand column p = 1, for the second p = 2, etc..
From Peter Bala, Jul 23 2013: (Start)
Conjectural explicit formula: T(n,k) = Stirling2(n,n-k) + (n-k)*Sum_{j = 0..k-1} (-1)^j*Stirling2(n, n+1+j-k)*j! for 0 <= k <= n.
The n-th row polynomial R(n,x) appears to satisfy the recurrence equation R(n,x) = n*x^(n-1) + Sum_{k = 1..n-1} binomial(n,k+1)*x^(n-k-1)*R(k,x). The row polynomials appear to have only real zeros. (End)

Edited by Johannes W. Meijer, Sep 23 2012

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1

Offset: 0

Abdullahi Umar, Aug 25 2008

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)

Peter Bala, A 4-parameter family of embedded Riordan arrays
Peter Bala, A note on the diagonals of a proper Riordan Array
Harry Crane, Left-right arrangements, set partitions, and pattern avoidance, Australasian Journal of Combinatorics, 61(1) (2015), 57-72.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8.

Cf. A007051, A118801, A135233, columns: A000225, A000337, A055580, A027608.
Cf. also A038207, A250118, A250119.
Cf. A007318, A167374, A145661, A029635.

A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011

Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024

A206817 Sum_{0

Original entry on oeis.org

1, 10, 73, 520, 3967, 33334, 309661, 3166468, 35416555, 430546642, 5655609529, 79856902816, 1206424711303, 19419937594990, 331860183278677, 6000534640290364, 114462875817046051, 2297294297649673738, 48394006967070653425

Offset: 2

Clark Kimberling, Feb 12 2012

nonn

In the following guide to related sequences,

c(n) = Sum_{0

t(n) = Sum_{0

s(k).................c(n)........t(n)

k....................A000217.....A000292

k^2..................A016061.....A004320

k^3..................A206808.....A206809

k^4..................A206810.....A206811

k!...................A206816.....A206817

prime(k).............A152535.....A062020

prime(k+1)...........A185382.....A206803

2^(k-1)..............A000337.....A045618

k(k+1)/2.............A007290.....A034827

k-th quarter-square..A049774.....A206806

			a(3) = (2-1) + (6-1) + (6-2) = 10.

Danny Rorabaugh, Table of n, a(n) for n = 2..400

Cf. A000142, A206816.

s[k_] := k!; t[1] = 0;
p[n_] := Sum[s[k], {k, 1, n}];
c[n_] := n*s[n] - p[n];
t[n_] := t[n - 1] + (n - 1) s[n] - p[n - 1];
Table[c[n], {n, 2, 32}]          (* A206816 *)
Flatten[Table[t[n], {n, 2, 20}]] (* A206817 *)

a(n)=sum(j=1,n,j!*(2*j-n-1)) \\ Charles R Greathouse IV, Oct 11 2015

a(n)=my(t=1); sum(j=1,n,t*=j; t*(2*j-n-1)) \\ Charles R Greathouse IV, Oct 11 2015

[sum([sum([factorial(k)-factorial(j) for j in range(1,k)]) for k in range(2,n+1)]) for n in range(2,21)] # Danny Rorabaugh, Apr 18 2015

a(n) = a(n-1)+(n-1)s(n)-p(n-1), where s(n) = n! and p(k) = 1!+2!+...+k!.

a(n) = Sum_{k=2..n} A206816(k).

A336725 A(n,k) is the n-th number that is a sum of k positive k-th powers; square array A(n,k), n>=1, k>=1, read by antidiagonals.

Original entry on oeis.org

1, 2, 2, 3, 5, 3, 4, 10, 8, 4, 5, 19, 17, 10, 5, 6, 36, 34, 24, 13, 6, 7, 69, 67, 49, 29, 17, 7, 8, 134, 132, 98, 64, 36, 18, 8, 9, 263, 261, 195, 129, 84, 43, 20, 9, 10, 520, 518, 388, 258, 160, 99, 55, 25, 10, 11, 1033, 1031, 773, 515, 321, 247, 114, 62, 26, 11, 12, 2058, 2056, 1542, 1028, 642, 384, 278, 129, 66, 29, 12

Offset: 1

Alois P. Heinz, Aug 01 2020

			Square array A(n,k) begins:
   1,  2,  3,   4,   5,   6,    7,    8,    9,   10, ...
   2,  5, 10,  19,  36,  69,  134,  263,  520, 1033, ...
   3,  8, 17,  34,  67, 132,  261,  518, 1031, 2056, ...
   4, 10, 24,  49,  98, 195,  388,  773, 1542, 3079, ...
   5, 13, 29,  64, 129, 258,  515, 1028, 2053, 4102, ...
   6, 17, 36,  84, 160, 321,  642, 1283, 2564, 5125, ...
   7, 18, 43,  99, 247, 384,  769, 1538, 3075, 6148, ...
   8, 20, 55, 114, 278, 734,  896, 1793, 3586, 7171, ...
   9, 25, 62, 129, 309, 797, 2193, 2048, 4097, 8194, ...
  10, 26, 66, 164, 340, 860, 2320, 6568, 4608, 9217, ...

Alois P. Heinz, Antidiagonals n = 1..141, flattened

Columns k=1-11 give: A000027, A000404, A003072, A003338, A003350, A003362, A003374, A003386, A003398, A004810, A004822.
Rows n=1-3 give: A000027, A052944, A145071.
Main diagonal gives A000337.
Cf. A336820.

A:= proc() local l, w, A; l, w, A:= proc() [] end, proc() [] end,
      proc(n, k) option remember; local b; b:=
        proc(x, y) option remember; `if`(x=0, {0}, `if`(y<1, {},
          {b(x, y-1)[], map(t-> t+l(k)[y], b(x-1, y))[]}))
        end;
        while nops(w(k)) < n do forget(b);
          l(k):= [l(k)[], (nops(l(k))+1)^k];
          w(k):= sort([select(h-> h

nmax = 12;
pow[n_, k_] := IntegerPartitions[n, {k}, Range[n^(1/k) // Ceiling]^k];
col[k_] := col[k] = Reap[Module[{j = k, n = 1, p}, While[n <= nmax, p = pow[j, k]; If[p =!= {}, Sow[j]; n++]; j++]]][[2, 1]];
A[n_, k_] := col[k][[n]];
Table[A[n-k+1, k], {n, 1, nmax}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Dec 03 2020 *)

A036826 a(n) = A036800(n)/2.

Original entry on oeis.org

0, 1, 9, 45, 173, 573, 1725, 4861, 13053, 33789, 84989, 208893, 503805, 1196029, 2801661, 6488061, 14876669, 33816573, 76283901, 170917885, 380633085, 843055101, 1858076669, 4076863485, 8908701693, 19394461693, 42077257725, 90999619581, 196226318333

Offset: 0

N. J. A. Sloane

Binomial transform of A054569 (with leading 0). Partial sums of A014477 (with leading 0). - Paul Barry, Jun 11 2003

This sequence is related to A000337 by a(n) = n*A000337(n) - Sum_{i=0..n-1} A000337(i). - Bruno Berselli, Mar 06 2012

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-18,20,-8).

Cf. A000337, A014477, A036800, A054569, A209359.

m:=28; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!((1+2*x)/((1-x)*(1-2*x)^3))); // Bruno Berselli, Mar 06 2012

A036826:= n-> 2^n*(3-2*n+n^2) -3; seq(A036826(n), n=0..30); # G. C. Greubel, Mar 31 2021

LinearRecurrence[{7,-18,20,-8}, {0,1,9,45}, 29] (* Bruno Berselli, Mar 06 2012 *)

for(n=0, 28, print1(2^n*(n^2-2*n+3)-3", ")); \\ Bruno Berselli, Mar 06 2012

[2^n*(3-2*n+n^2) -3 for n in (0..30)] # G. C. Greubel, Mar 31 2021

From Paul Barry, Jun 11 2003: (Start)
G.f.: x*(1+2*x)/((1-x)*(1-2*x)^3).
a(n) = 2^n*(n^2-2*n+3) - 3.
a(n) = Sum_{k=0..n} k^2*2^(k-1). (End)
a(n) = 7*a(n-1) -18*a(n-2) +20*a(n-3) -8*a(n-4). - Harvey P. Dale, Mar 04 2015
E.g.f.: -3*exp(x) + (3 -2*x +4*x^2)*exp(2*x). - G. C. Greubel, Mar 31 2021

A058922 a(n) = n2^n - 2^n = 2^n(n-1).

Original entry on oeis.org

0, 4, 16, 48, 128, 320, 768, 1792, 4096, 9216, 20480, 45056, 98304, 212992, 458752, 983040, 2097152, 4456448, 9437184, 19922944, 41943040, 88080384, 184549376, 385875968, 805306368, 1677721600, 3489660928, 7247757312, 15032385536, 31138512896, 64424509440, 133143986176

Offset: 1

N. J. A. Sloane, Jan 12 2001

A hierarchical sequence (S(W'2{2}*c) - see A059126).

Harry J. Smith, Table of n, a(n) for n = 1..200
Jonas Wallgren, Hierarchical sequences, 2001.
Index entries for linear recurrences with constant coefficients, signature (4,-4).

A001787(n) = a(n+1)/4. A073346(n, n-2) = a(n-2).

Cf. A000337. - Omar E. Pol, Feb 22 2010

a058922 n = (n - 1) * 2 ^ n
a058922_list = zipWith (*) [0..] $ tail a000079_list
-- Reinhard Zumkeller, Jul 11 2014

Table[n*2^n-2^n,{n,100}] (* Vladimir Joseph Stephan Orlovsky, Jan 15 2011 *)

a(n) = { n*2^n - 2^n } \\ Harry J. Smith, Jun 24 2009

a(n) = -det(M(n+1)) where M(n) is the n X n matrix with m(i,i)=1, m(i,j)=-i/j for i != j. - Benoit Cloitre, Feb 01 2003
With offset 0, this is 4n*2^(n-1), the binomial transform of 4n. - Paul Barry, May 20 2003
a(1)=0, a(n) = 2*a(n-1) + 2^n for n>1. - Philippe Deléham, Apr 20 2009
a(n) = A000337(n) - 1. - Omar E. Pol, Feb 22 2010
From R. J. Mathar, Mar 01 2010: (Start)
a(n)= 4*a(n-1) - 4*a(n-2).
G.f.: 4*x^2/(2*x-1)^2. (End)
From Amiram Eldar, Jan 12 2021: (Start)
Sum_{n>=2} 1/a(n) = log(2)/2.
Sum_{n>=2} (-1)^n/a(n) = log(3/2)/2. (End)

A162797 a(n) = difference between the number of toothpicks of A139250 that are orthogonal to the initial toothpick and the number of toothpicks that are parallel to the initial toothpick, after n even rounds.

Original entry on oeis.org

1, 1, 5, 1, 5, 5, 17, 1, 5, 5, 17, 5, 17, 21, 49, 1, 5, 5, 17, 5, 17, 21, 49, 5, 17, 21, 49, 21, 53, 81, 129, 1, 5, 5, 17, 5, 17, 21, 49, 5, 17, 21, 49, 21, 53, 81, 129, 5, 17, 21, 49, 21, 53, 81, 129

Offset: 1

Omar E. Pol, Jul 14 2009

nonn

It appears that a(2^k) = 1, for k >= 0. [From Omar E. Pol, Feb 22 2010]

			Contribution from _Omar E. Pol_, Feb 22 2010: (Start)
If written as a triangle:
1;
1,5;
1,5,5,17;
1,5,5,17,5,17,21,49;
1,5,5,17,5,17,21,49,5,17,21,49,21,53,81,129;
1,5,5,17,5,17,21,49,5,17,21,49,21,53,81,129,5,17,21...
Rows converge to A173464.
(End)
Contribution from Omar E. Pol, Apr 01 2011 (Start):
It appears that the final terms of rows give A000337.
It appears that row sums give A006516.
(End)

Nathaniel Johnston, Table of n, a(n) for n = 1..94
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.]
N. J. A. Sloane, Catalog of Toothpick and Cellular Automata Sequences in the OEIS

Cf. A139250, A139251, A159791, A159792, A162793, A162794, A162795, A162796.

Cf. A000337, A058922, A173464. [From Omar E. Pol, Feb 22 2010]

a(n) = A162796(n) - A162795(n).

Edited by Omar E. Pol, Jul 18 2009
More terms from Omar E. Pol, Feb 22 2010
More terms (a(51)-a(55)) from Nathaniel Johnston, Mar 30 2011

A057728 A triangular table of decreasing powers of two (with first column all ones).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 8, 4, 2, 1, 1, 16, 8, 4, 2, 1, 1, 32, 16, 8, 4, 2, 1, 1, 64, 32, 16, 8, 4, 2, 1, 1, 128, 64, 32, 16, 8, 4, 2, 1, 1, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1, 1, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1

Offset: 1

Alford Arnold, Oct 29 2000

First differences of sequence A023758.
A023758 is the sequence of partial sums of a(n) with row sums A000337.
2^A004736(n) is a sequence closely related to a(n).
T(n,k) is the number of length n binary words having an odd number of 0's with exactly k 1's following the last 0, n >= 1, 0 <= k <= n - 1. - Geoffrey Critzer, Jan 28 2014

			Triangle starts:
  1,
  1,    1,
  1,    2,    1,
  1,    4,    2,   1,
  1,    8,    4,   2,   1,
  1,   16,    8,   4,   2,   1,
  1,   32,   16,   8,   4,   2,  1,
  1,   64,   32,  16,   8,   4,  2,  1,
  1,  128,   64,  32,  16,   8,  4,  2,  1,
  1,  256,  128,  64,  32,  16,  8,  4,  2, 1,
  1,  512,  256, 128,  64,  32, 16,  8,  4, 2, 1,
  1, 1024,  512, 256, 128,  64, 32, 16,  8, 4, 2, 1,
  1, 2048, 1024, 512, 256, 128, 64, 32, 16, 8, 4, 2, 1,
  ... - _Joerg Arndt_, May 04 2014
When viewed as a triangular array, row 8 of A023758 is 128 192 224 240 248 252 254 255 so row 8 here is 1 64 32 16 8 4 2 1
From _Mats Granvik_, Jan 19 2009: (Start)
Except for the first term the table can also be formatted as:
   1,
   1, 1,
   2, 1, 1,
   4, 2, 1, 1,
   8, 4, 2, 1, 1,
  16, 8, 4, 2, 1, 1,
  ...
(End)

Reinhard Zumkeller, Rows n = 1..100 of table, flattened

Cf. A000079, A004736, A023758 and A000337.

Cf. A155038 (essentially the same as this sequence). [Mats Granvik, Jan 19 2009]

a057728 n k = a057728_tabl !! (n-1) !! (k-1)
a057728_row n = a057728_tabl !! (n-1)
a057728_tabl = iterate
   (\row -> zipWith (+) (row ++ [0]) ([0] ++ tail row ++ [1])) [1]
-- Reinhard Zumkeller, Aug 08 2013

nn=10;Map[Select[#,#>0&]&,CoefficientList[Series[(x-x^2)/(1-2x)/(1-y x),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Jan 28 2014 *)
Module[{nn=12,ts},ts=2^Range[0,nn];Table[Join[{1},Reverse[Take[ts,n]]],{n,0,nn}]]//Flatten (* Harvey P. Dale, Jan 15 2022 *)

T(n, k) := if k = 0 then 1 else  2^(n - k - 1)$
create_list(T(n, k), n, 0, 12, k, 0, n - 1); /* Franck Maminirina Ramaharo, Jan 09 2019 */

G.f.: (x - x^2)/((1 - 2*x)*(1 - y*x)). - Geoffrey Critzer, Jan 28 2014 [This produces the triangle shown by Mats Granvik in example section. - Franck Maminirina Ramaharo, Jan 09 2019]
From Franck Maminirina Ramaharo, Jan 09 2019: (Start)
G.f.: x*(1 - 2*x + y*x^2)/((1 - x)*(1 - 2*x)*(1 - x*y)).
E.g.f.: (exp(2*x)*y - 2*exp(x*y))/(4 - 2*y) + exp(x) - 1/2. (End)

More terms from Larry Reeves (larryr(AT)acm.org), Oct 30 2000

A188553 T(n,k) = Number of n X k binary arrays without the pattern 0 1 diagonally, vertically, antidiagonally or horizontally.

Original entry on oeis.org

2, 3, 3, 4, 5, 4, 5, 8, 7, 5, 6, 12, 12, 9, 6, 7, 17, 20, 16, 11, 7, 8, 23, 32, 28, 20, 13, 8, 9, 30, 49, 48, 36, 24, 15, 9, 10, 38, 72, 80, 64, 44, 28, 17, 10, 11, 47, 102, 129, 112, 80, 52, 32, 19, 11, 12, 57, 140, 201, 192, 144, 96, 60, 36, 21, 12, 13, 68, 187, 303, 321, 256, 176

Offset: 1

R. H. Hardin, Apr 04 2011

From Miquel A. Fiol, Feb 06 2024: (Start)
Also, T(n,k) is the number of words of length k, x(1)x(2)...x(k), on the alphabet {0,1,...,n}, such that, for i=2,...,k, x(i)=either x(i-1) or x(i)=x(i-1)-1.
For the bijection between arrays and sequences, notice that the i-th column consists of 1's and then 0's, and there are x(i)=0 to n of 1's.
Such a bijection implies that all the empirical/conjectured formulas in A188554, A188555, A188556, A188557, A188558, and A188559 become correct.
(End)

			Table starts
..2..3..4..5...6...7...8...9...10...11...12....13....14....15....16.....17
..3..5..8.12..17..23..30..38...47...57...68....80....93...107...122....138
..4..7.12.20..32..49..72.102..140..187..244...312...392...485...592....714
..5..9.16.28..48..80.129.201..303..443..630...874..1186..1578..2063...2655
..6.11.20.36..64.112.192.321..522..825.1268..1898..2772..3958..5536...7599
..7.13.24.44..80.144.256.448..769.1291.2116..3384..5282..8054.12012..17548
..8.15.28.52..96.176.320.576.1024.1793.3084..5200..8584.13866.21920..33932
..9.17.32.60.112.208.384.704.1280.2304.4097..7181.12381.20965.34831..56751
.10.19.36.68.128.240.448.832.1536.2816.5120..9217.16398.28779.49744..84575
.11.21.40.76.144.272.512.960.1792.3328.6144.11264.20481.36879.65658.115402
Some solutions for 5 X 3:
  1 1 1   1 0 0   0 0 0   1 1 1   1 1 1   1 1 1   1 1 1
  1 1 1   0 0 0   0 0 0   1 1 1   1 1 1   1 1 1   1 1 1
  1 1 1   0 0 0   0 0 0   1 1 1   1 0 0   1 1 0   1 1 1
  1 1 1   0 0 0   0 0 0   1 1 0   0 0 0   1 0 0   1 1 1
  1 1 1   0 0 0   0 0 0   1 0 0   0 0 0   0 0 0   1 1 0
Some solutions for T(5,3): By taking the sums of the columns in the above arrays we get 555, 100, 000, 543, 322, 432, 554. - _Miquel A. Fiol_, Feb 04 2024

R. H. Hardin, Table of n, a(n) for n = 1..9934

Diagonal is A045623.
Column 4 is A086570.
Rows 2..8 are A022856(n+4), A188554, A188555, A188556, A188557, A188558, A188559.
Upper diagonals T(n,n+i) for i=1..8 give: A001792, A001787(n+1), A000337(n+1), A045618, A045889, A034009, A055250, A055251.
Lower diagonals T(n+i,n) for i=1..7 give: A045891(n+1), A034007(n+2), A111297(n+1), A159694(n-1), A159695(n-1), A159696(n-1), A159697(n-1).
Antidiagonal sums give A065220(n+5).

T:= (n,k)-> `if`(k<=n+1, (2*n+3-k)*2^(k-2), (n+1-k)*binomial(k-1, n) * add(binomial(n, j-1)/(k-j)*T(n, j)*(-1)^(n-j), j=1..n+1)): seq(seq(T(n, 1+d-n), n=1..d), d=1..15); #Alois P. Heinz in the Sequence Fans Mailing List, Apr 04 2011 [We do not permit programs based on conjectures, but this program is now justified by Fiol's comment. - N. J. A. Sloane, Mar 09 2024]

Empirical: T(n,k) = (n+1)*2^(k-1) + (1-k)*2^(k-2) for k < n+3, and then the entire row n is a polynomial of degree n in k.
From Miquel A. Fiol, Feb 06 2024: (Start)
The above empirical formula is correct.
It can be proved that T(n,k) satisfies the recurrence
  T(n,k) = Sum_{r=1..n+1} (-1)^(r+1)*binomial(n+1,r)*T(n,k-r)
with initial values
  T(n,k) = Sum_{r=0..k-1} (n+1-r)*binomial(k-1,r) for k = 1..n+1. (End)

A097591 Triangle read by rows: T(n,k) is the number of permutations of [n] with exactly k increasing runs of odd length.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 5, 0, 1, 6, 0, 17, 0, 1, 0, 70, 0, 49, 0, 1, 90, 0, 500, 0, 129, 0, 1, 0, 1890, 0, 2828, 0, 321, 0, 1, 2520, 0, 23100, 0, 13930, 0, 769, 0, 1, 0, 83160, 0, 215292, 0, 62634, 0, 1793, 0, 1, 113400, 0, 1549800, 0, 1697430, 0, 264072, 0, 4097, 0, 1

Offset: 0

Emeric Deutsch, Aug 29 2004

			Triangle starts:
     1;
     0,    1;
     1,    0,     1;
     0,    5,     0,    1;
     6,    0,    17,    0,     1;
     0,   70,     0,   49,     0,   1;
    90,    0,   500,    0,   129,   0,   1;
     0, 1890,     0, 2828,     0, 321,   0, 1;
  2520,    0, 23100,    0, 13930,   0, 769, 0, 1;
  ...
Row n has n+1 entries.
Example: T(3,1) = 5 because we have (123), 13(2), (2)13, 23(1) and (3)12 (the runs of odd length are shown between parentheses).

Alois P. Heinz, Rows n = 0..140, flattened

Bisections of columns k=0-1 give: A000680, A302910.
Row sums give A000142.
T(n+1,n-1) gives A000337.
T(4n,2n) gives A308962.
Cf. A096654, A097592.

G:=t^2/(1-t*x-(1-t^2)*exp(-t*x)): Gser:=simplify(series(G,x=0,12)): P[0]:=1: for n from 1 to 11 do P[n]:=sort(expand(n!*coeff(Gser,x^n))) od: seq(seq(coeff(t*P[n],t^k),k=1..n+1),n=0..11);
# second Maple program:
b:= proc(u, o, t) option remember; `if`(u+o=0, x^t, expand(
      add(b(u+j-1, o-j, irem(t+1, 2)), j=1..o)+
      add(b(u-j, o+j-1, 1)*x^t, j=1..u)))
    end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..degree(p)))(b(n, 0, 1)):
seq(T(n), n=0..12);  # Alois P. Heinz, Nov 19 2013

b[u_, o_, t_] := b[u, o, t] = If[u+o == 0, x^t, Expand[Sum[b[u+j-1, o-j, Mod[t+1, 2]], {j, 1, o}] + Sum[b[u-j, o+j-1, 1]*x^t, {j, 1, u}]]]; T[n_] := Function[{p}, Table[Coefficient[p, x, i], {i, 1, Exponent[p, x]}]][b[n, 0, 1]]; Table[T[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Feb 19 2015, after Alois P. Heinz *)

E.g.f.: t^2/[1-tx-(1-t^2)exp(-tx)].

Sum_{k=1..n} k * T(n,k) = A096654(n-1) for n > 0. - Alois P. Heinz, Jul 03 2019

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A163940 Triangle related to the divergent series 1^m1! - 2^m2! + 3^m3! - 4^m4! + ... for m >= -1.

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