A034007 -id:A034007 - cp's OEIS frontend

A045891 First differences of A045623.

1, 1, 3, 7, 16, 36, 80, 176, 384, 832, 1792, 3840, 8192, 17408, 36864, 77824, 163840, 344064, 720896, 1507328, 3145728, 6553600, 13631488, 28311552, 58720256, 121634816, 251658240, 520093696, 1073741824, 2214592512, 4563402752

Offset: 0

Wolfdieter Lang

Let M_n be the n X n matrix m_(i,j) = 3 + abs(i-j), then det(M_n) =(-1)^(n+1)*a(n+1). - Benoit Cloitre, May 28 2002
If X_1, X_2, ..., X_n are 2-blocks of a (2n+3)-set X then, for n>=1, a(n+2) is the number of (n+1)-subsets of X intersecting each X_i, (i=1..n). - Milan Janjic, Nov 18 2007
Equals row sums of triangle A152194. - Gary W. Adamson, Nov 28 2008
An elephant sequence, see A175655. For the central square 16 A[5] vectors, with decimal values between 19 and 400, lead to this sequence (without the first leading 1). For the corner squares these vectors lead to the companion sequence A045623. - Johannes W. Meijer, Aug 15 2010
a(n) is the total number of runs of 1 in the compositions of n+1. For example, a(3) = A045623(3) - A045623(2) = 12 - 5 = 7 runs of only 1 in the compositions of 4, enumerated "()" as follows: 3,(1); (1),3; 2,(1,1);(1),2,(1); (1,1),2; (1,1,1,1). More generally, the total number of runs of only part k in the compositions of n+k is A045623(n) - A045623(n-k). - Gregory L. Simay, May 02 2017
This is essentially the p-INVERT of (1,1,1,1,1,...) for p(S) = 1 - S - S^2 + S^3; see A291000.  - Clark Kimberling, Aug 24 2017

			G.f. = 1 + x + 3*x^2 + 7*x^3 + 16*x^4 + 36*x^5 + 80*x^6 + ... - _Michael Somos_, Mar 26 2022

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Paul Barry, A Catalan Transform and Related Transformations on Integer Sequences, Journal of Integer Sequences, Vol. 8 (2005), Article 05.4.5.
Frank Ellermann, Illustration of binomial transforms
Milan Janjic, Two Enumerative Functions
Milan Janjic and B. Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
Thomas Selig and Haoyue Zhu, New combinatorial perspectives on MVP parking functions and their outcome map, arXiv:2309.11788 [math.CO], 2023. See p. 29.
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Cf. A027656, A034007, A045623, A152194, A175655.

[1,1] cat [(n+4)*2^(n-3): n in [2..40]]; // G. C. Greubel, Sep 27 2022

Join[{1,1,a=3,b=7},Table[c=4*b-4*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 15 2011 *)
Table[ If[n<2, 1, 2^(n-3)*(n+4)], {n, 0, 30}] (* Jean-François Alcover, Sep 12 2012 *)
LinearRecurrence[{4,-4},{1,1,3,7},40] (* Harvey P. Dale, May 03 2019 *)

v=[1,1,3,7];for(i=1,99,v=concat(v,4*(v[#v]-v[#v-1])));v \\ Charles R Greathouse IV, Jun 01 2011

[1,1]+[(n+4)*2^(n-3) for n in range(2,40)] # G. C. Greubel, Sep 27 2022

a(n) = Sum_{k=0..n-2} (k+3)*binomial(n-2,k) for n >= 2. - N. J. A. Sloane, Jan 30 2008
a(n) = (n+4)*2^(n-3), n >= 2, with a(0) = a(1) = 1.
G.f.: (1-x)^3/(1-2*x)^2.
Equals binomial transform of A027656.
Starting 1, 3, 7, 16, ... this is ((n+5)*2^n - 0^n)/4, the binomial transform of (1, 2, 2, 3, 3, ...). - Paul Barry, May 20 2003
From Paul Barry, Nov 29 2004: (Start)
a(n) = ((n+4)*2^(n-1) + 3*C(0, n) - C(1, n))/4;
a(n) = Sum_{k=0..floor(n/2)} C(n, 2*k)*(k+1). (End)
a(n) = A045623(n-1) + 2^(n-2) = A034007(n+1) - 2^(n-2) for n>=2. - Philippe Deléham, Apr 20 2009
G.f.: 1 + Q(0)*x/(1-x)^2, where Q(k)= 1 + (k+1)*x/(1 - x - x*(1-x)/(x + (k+1)*(1-x)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, Apr 25 2013
a(n) = Sum_{k=0..n} (k+1)*C(n-2,n-k). Peter Luschny, Apr 20 2015
From Amiram Eldar, Jan 13 2021: (Start)
Sum_{n>=0} 1/a(n) = 128*log(2) - 1292/15.
Sum_{n>=0} (-1)^n/a(n) = 782/15 - 128*log(3/2). (End)
E.g.f.: (2 - x + exp(2*x)*(2 + x))/4. - Stefano Spezia, Mar 26 2022

A062111 Upper-right triangle resulting from binomial transform calculation for nonnegative integers.

Original entry on oeis.org

0, 1, 1, 4, 3, 2, 12, 8, 5, 3, 32, 20, 12, 7, 4, 80, 48, 28, 16, 9, 5, 192, 112, 64, 36, 20, 11, 6, 448, 256, 144, 80, 44, 24, 13, 7, 1024, 576, 320, 176, 96, 52, 28, 15, 8, 2304, 1280, 704, 384, 208, 112, 60, 32, 17, 9, 5120, 2816, 1536, 832, 448, 240, 128, 68, 36, 19, 10

Offset: 0

Henry Bottomley, May 30 2001

From Philippe Deléham, Apr 15 2007: (Start)
This triangle can be found in the Laisant reference in the following form:
  .......................5...11..
  ...................4...9...20..
  ...............3...7..16...36..
  ...........2...5..12..28.......
  .......1...3...8..20..48.......
  ...0...1...4..12..32..80....... (End)
Triangle A152920 reversed. - Philippe Deléham, Apr 21 2009

			As a lower triangle (T(n, k)):
    0;
    1,   1;
    4,   3,   2;
   12,   8,   5,  3;
   32,  20,  12,  7,  4;
   80,  48,  28, 16,  9,  5;
  192, 112,  64, 36, 20, 11,  6;
  448, 256, 144, 80, 44, 24, 13, 7;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
F. Ellermann, Illustration of binomial transforms
C.-A. Laisant, Sur les tableaux de sommes - Nouvelles applications, Compt. Rendus de l'Association Francaise pour l'Avancement des Sciences, Aout 04 1893, pp. 206-216 (table given on p. 212).

Rows include (essentially) A001787, A001792, A034007, A045623, A045891.
Diagonals include (essentially) A001477, A005408, A008586, A008598, A017113.
Column sums are A058877.
Cf. A111297, A159694, A159695, A159696, A159697. - Philippe Deléham, Apr 21 2009
Cf. A058877, A073371, A130129, A152920.

[2^(n-k-1)*(n+k): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 28 2022

Table[2^(n-k-1)*(n+k), {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 28 2022 *)

def A062111(n,k): return 2^(n-k-1)*(n+k)
flatten([[A062111(n,k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Sep 28 2022

A(n, k) = A(n, k-1) + A(n+1, k) if k > n with A(n, n) = n.
A(n, k) = (k+n)*2^(k-n-1) if k >= n.
T(2*n, n) = 3*n*2^(n-1) = 3*A001787(n). - Philippe Deléham, Apr 21 2009
From G. C. Greubel, Sep 28 2022: (Start)
T(n, k) = 2^(n-k-1)*(n+k) for 0 <= k <= n, n >= 0.
T(m*n, n) = 2^((m-1)*n-1)*(m+1)*A001477(n), m >= 1.
T(2*n-1, n-1) = A130129(n-1).
T(2*n+1, n-1) = 12*A001787(n).
Sum_{k=0..n} T(n, k) = A058877(n+1).
Sum_{k=0..n} (-1)^k*T(n, k) = 3*A073371(n-2), n >= 2.
T(n, k) = A152920(n, n-k). (End)

A152920 Triangle read by rows: triangle A062111 reversed.

Original entry on oeis.org

0, 1, 1, 2, 3, 4, 3, 5, 8, 12, 4, 7, 12, 20, 32, 5, 9, 16, 28, 48, 80, 6, 11, 20, 36, 64, 112, 192, 7, 13, 24, 44, 80, 144, 256, 448, 8, 15, 28, 52, 96, 176, 320, 576, 1024, 9, 17, 32, 60, 112, 208, 384, 704, 1280, 2304, 10, 19, 36, 68, 128, 240, 448, 832, 1536, 2816, 5120

Offset: 0

Paul Curtz, Dec 15 2008

			Triangle starts:
  0;
  1,  1;
  2,  3,  4;
  3,  5,  8, 12;
  4,  7, 12, 20, 32;
  ...

Alois P. Heinz, Rows n = 0..150, flattened

Cf. A053220, A058877 (row sums), A193844, A212697.

Columns and diagonals: A001787, A001792, A034007, A045623, A045891, A111297, A159694, A159695, A159696, A159697.

[2^k*(n-k/2): k in [0..n], n in [0..12]]; // G. C. Greubel, Sep 27 2022

A062111 := proc(n,k) (k+n)*2^(k-n-1) ; end: A152920 := proc(n,k) A062111(n-k,n) ; end: for n from 0 to 15 do for k from 0 to n do printf("%d,",A152920(n,k)) ; od: od: # R. J. Mathar, Jan 22 2009
# second Maple program:
T:= proc(n, k) option remember;
     `if`(k=0, n, T(n, k-1)+T(n-1, k-1))
    end:
seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Sep 12 2022

t[0, k_]:= k; t[n_, k_]:= t[n, k]= t[n-1, k] + t[n-1, k+1];
Table[t[n-k, k], {n,0,10}, {k,n,0,-1}]//Flatten (* Jean-François Alcover, Sep 11 2016 *)

flatten([[2^(k-1)*(2*n-k) for k in range(n+1)] for n in range(12)]) # G. C. Greubel, Sep 27 2022

Row sums: (2^n-1)(n+1) = A058877(n). - R. J. Mathar, Jan 22 2009
T(2n, n) = 3*n*2^(n-1) = 3*A001787(n). - Philippe Deléham, Apr 20 2009
From Werner Schulte, Jul 31 2020: (Start)
T(n, k) = (2*n-k) * 2^(k-1) for 0 <= k <= n.
G.f.: Sum_{n>=0, k=0..n} T(n,k) * x^k * t^n = t*(1+x-3*x*t) / ((1-t)^2 * (1-2*x*t)^2).
Sum_{k=0..n} (-1)^k * binomial(n,k) * T(n,k) = 0 for n >= 0.
Sum_{k=0..n} binomial(n,k) * T(n,k) = 2*n * 3^(n-1) for n >= 0.
Define the array B(n,p) = (Sum_{k=0..n} binomial(p+k,p) * T(n,k))/(n+p+1) for n >= 0 and p >= 0. Then see the comment of Robert Coquereaux (2014) at A193844. Conjecture: B(n+1,p) = A(n,p). (End)
T(n, k) = T(n, k-1) + T(n-1, k-1) for k>=1, T(n,0) = n. - Alois P. Heinz, Sep 12 2022
From G. C. Greubel, Sep 27 2022: (Start)
T(n, n-1) = A001792(n).
T(2*n-1, n-1) = A053220(n).
T(2*n+1, n-1) = 3*A001792(n).
T(m*n, n) = (2*m-1)*A001787(n), for m >= 1. (End)

Edited by N. J. A. Sloane, Dec 19 2008

More terms from R. J. Mathar, Jan 22 2009

A091613 Triangle: T(n,k) = number of compositions (ordered partitions) of n such that some part is repeated consecutively k times and no part is repeated consecutively more than k times.

Original entry on oeis.org

1, 1, 1, 3, 0, 1, 4, 3, 0, 1, 7, 6, 2, 0, 1, 14, 10, 5, 2, 0, 1, 23, 23, 11, 4, 2, 0, 1, 39, 50, 22, 10, 4, 2, 0, 1, 71, 99, 48, 22, 9, 4, 2, 0, 1, 124, 200, 105, 46, 21, 9, 4, 2, 0, 1, 214, 404, 223, 101, 46, 20, 9, 4, 2, 0, 1, 378, 805, 468, 218, 98, 45, 20, 9, 4, 2, 0, 1, 661, 1599, 979, 466, 213, 98, 44, 20, 9, 4, 2, 0, 1

Offset: 1

Christian G. Bower, Jan 23 2004

Cf. A232294 - A128695 = column 3. - Geoffrey Critzer, Mar 24 2014

			Triangle starts:
    1;
    1,   1;
    3,   0,   1;
    4,   3,   0,  1;
    7,   6,   2,  0,  1;
   14,  10,   5,  2,  0, 1;
   23,  23,  11,  4,  2, 0, 1;
   39,  50,  22, 10,  4, 2, 0, 1;
   71,  99,  48, 22,  9, 4, 2, 0, 1;
  124, 200, 105, 46, 21, 9, 4, 2, 0, 1;
  ...
In the partition 3+3+2+2+2+1+3+3+1, 2 is repeated consecutively 3 times, no part is repeated consecutively more than 3 times. (3 appears 4 times nonconsecutively.)

Alois P. Heinz, Rows n = 1..100, flattened

Row sums: A000079(n-1) (2^(n-1)).
Inverse: A091614.
Square: A091615.
Columns 1-6: A003242, A091616, A091617, A091618, A091619, A091620.
Convergent of columns: A034007.

b:= proc(n, l, k) option remember; `if`(n=0, 1, add(`if`(
      i=l, 0, add(b(n-i*j, i, k), j=1..min(k, n/i))), i=1..n))
    end:
T:= (n, k)-> b(n, 0, k)-b(n, 0, k-1):
seq(seq(T(n, k), k=1..n), n=1..14);  # Alois P. Heinz, Feb 08 2017

nn=15;Table[Take[Drop[Transpose[Map[PadRight[#,nn+1]&,Table[ CoefficientList[Series[1/(1-Sum[Sum[x^(j i),{i,1,k}]/Sum[x^(j i),{i,0,k}],{j,1,nn}])-1/(1-Sum[Sum[x^(j i),{i,1,k-1}]/Sum[x^(j i),{i,0,k-1}],{j,1,nn}]),{x,0,nn}],x],{k,1,nn}]]],1][[n]],n],{n,1,nn}]//Grid
(* or *)
Needs["Combinatorica`"];Table[Distribution[Map[Max,Map[Length,Map[Split, Level[Map[Permutations,IntegerPartitions[n,n]],{2}]],{2}]],Range[1,n]],{n,1,15}]//Grid (* Geoffrey Critzer, Mar 24 2014 *)
b[n_, l_, k_] := b[n, l, k] = If[n == 0, 1, Sum[If[i == l, 0,
     Sum[b[n - i*j, i, k], {j, 1, Min[k, n/i]}]], {i, 1, n}]];
T[n_, k_] := b[n, 0, k] - b[n, 0, k - 1];
Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 04 2021, after Alois P. Heinz *)

G.f. for column k: 1/(1 - Sum_{i>=1} (x^i + x^(2*i) + ... + x^(k*i))/( 1 + x^i + x^(2*i) + ... + x^(k*i)) ) - 1/(1 - Sum_{i>=1} (x^i + x^(2*i) + ... + x^((k-1)*i))/( 1 + x^i + x^(2*i) + ... + x^((k-1)*i))). - Geoffrey Critzer, Mar 24 2014

A168230 a(n) = n + 2 - a(n-1) for n>1; a(1) = 0.

Original entry on oeis.org

0, 4, 1, 5, 2, 6, 3, 7, 4, 8, 5, 9, 6, 10, 7, 11, 8, 12, 9, 13, 10, 14, 11, 15, 12, 16, 13, 17, 14, 18, 15, 19, 16, 20, 17, 21, 18, 22, 19, 23, 20, 24, 21, 25, 22, 26, 23, 27, 24, 28, 25, 29, 26, 30, 27, 31, 28, 32, 29, 33, 30, 34, 31, 35, 32, 36, 33, 37, 34, 38, 35, 39, 36, 40, 37

Offset: 1

Vincenzo Librandi, Nov 21 2009

Interleaving of A001477 and A000027 without first three terms.
Binomial transform of 0, 4 followed by a signed version of A005009.
Inverse binomial transform of A034007 without first and third term.

			a(2) = 2+2-a(1) = 4-0 = 4; a(3) = 3+2-a(2) = 5-4 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A001477 (nonnegative integers), A000027 (positive integers), A168309 (repeat 4,-3), A005009 (7*2^n), A034007 (first differences of A045891).

[ n eq 1 select 0 else -Self(n-1)+n+2: n in [1..75] ];

a=3; Table[a=n-a, {n, 3, 200}] (* Vladimir Joseph Stephan Orlovsky, Nov 22 2009 *)
CoefficientList[Series[x (4 - 3 x) / ((1 + x) (1 - x)^2),{x, 0, 100}], x] (* Vincenzo Librandi, Sep 16 2013 *)
LinearRecurrence[{1,1,-1}, {0, 4, 1}, 50] (* G. C. Greubel, Jul 16 2016 *)
nxt[{n_,a_}]:={n+1,n+3-a}; NestList[nxt,{1,0},80][[All,2]] (* Harvey P. Dale, May 28 2021 *)

Vec(x^2*(4-3*x)/((1+x)*(1-x)^2) + O(x^100)) \\ Colin Barker, Nov 08 2014

G.f.: x^2*(4 - 3*x)/((1+x)*(1-x)^2).
a(n) = (7*(-1)^n + 2*n + 5)/4.
a(n) = a(n-2) + 1 for n>2; a(1)=0, a(2)=4.
a(n+1) - a(n) = A168309(n).
a(n) = a(n-1) + a(n-2) - a(n-3). - Colin Barker, Nov 08 2014
E.g.f.: (1/4)*(7 - 12*exp(x) + (5 + 2*x)*exp(2*x))*exp(-x). - G. C. Greubel, Jul 16 2016
Sum_{n>=2} (-1)^(n+1)/a(n) = 11/6. - Amiram Eldar, Feb 23 2023

Edited, three comments, four formulas, MAGMA program added by Klaus Brockhaus, Nov 22 2009

A159694 a(n) = 2*a(n-1) + 2^(n-1), for n > 0, with a(0)=6.

Original entry on oeis.org

6, 13, 28, 60, 128, 272, 576, 1216, 2560, 5376, 11264, 23552, 49152, 102400, 212992, 442368, 917504, 1900544, 3932160, 8126464, 16777216, 34603008, 71303168, 146800640, 301989888, 620756992, 1275068416, 2617245696, 5368709120

Offset: 0

Philippe Deléham, Apr 20 2009

Diagonal of triangles A062111, A152920.

			a(0) = 6,
a(1) = 2* 6 + 1 =  13,
a(2) = 2*13 + 2 =  28,
a(3) = 2*28 + 4 =  60,
a(4) = 2*60 + 8 = 128, ...

G. C. Greubel, Table of n, a(n) for n = 0..1000
Milan Janjić and Boris Petković, A Counting Function, arXiv:1301.4550 [math.CO], 2013.
Milan Janjić and Boris Petković, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq., Vol. 17 (2014), Article 14.3.5.
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Cf. A000079, A001787, A001792, A034007, A045623.
Cf. A045891, A062111, A111297, A152920.
Seventh row of triangle A062111. - Klaus Brockhaus, Sep 27 2009

[(12+n)*2^(n-1): n in [0..30]]; // G. C. Greubel, Sep 27 2022

Table[(6 + n/2)*2^n, {n, 0, 30}] (* Amiram Eldar, Jan 19 2021 *)

[(12+n)*2^(n-1) for n in range(30)] # G. C. Greubel, Sep 27 2022

a(n) = Sum_{k=0..n} (k+6)*binomial(n,k).
From Klaus Brockhaus, Sep 27 2009: (Start)
a(n) = (6 + n/2)*2^n.
G.f.: (6 - 11*x)/(1-2*x)^2. (End)
From Amiram Eldar, Jan 19 2021: (Start)
Sum_{n>=0} 1/a(n) = 8192*log(2) - 3934820/693.
Sum_{n>=0} (-1)^n/a(n) = 11509636/3465 - 8192*log(3/2). (End)
E.g.f.: (6 + x)*exp(2*x). - G. C. Greubel, Sep 27 2022

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)

A159695 a(0)=7, a(n) = 2*a(n-1) + 2^(n-1) for n > 0.

Original entry on oeis.org

7, 15, 32, 68, 144, 304, 640, 1344, 2816, 5888, 12288, 25600, 53248, 110592, 229376, 475136, 983040, 2031616, 4194304, 8650752, 17825792, 36700160, 75497472, 155189248, 318767104, 654311424, 1342177280, 2751463424, 5637144576

Offset: 0

Philippe Deléham, Apr 20 2009

Diagonal of triangles A062111, A152920.

			a(0)=7, a(1) = 2*7 + 1 = 15, a(2) = 2*15 + 2 = 32, a(3) = 2*32 + 4 = 68, a(4) = 2*68 + 8 = 144, ...

G. C. Greubel, Table of n, a(n) for n = 0..3300
Index entries for linear recurrences with constant coefficients, signature (4, -4).

Cf. A000079, A001787, A001792, A045623, A045891, A034007, A111297, A159694.

[(14+n)*2^(n-1): n in [0..30]]; // G. C. Greubel, Jun 02 2018

LinearRecurrence[{4,-4}, {7,15}, 30] (* or *) Table[(14+n)*2^(n-1), {n, 0, 30}] (* G. C. Greubel, Jun 02 2018 *)
nxt[{n_,a_}]:={n+1,2a+2^n}; NestList[nxt,{0,7},30][[All,2]] (* Harvey P. Dale, Jan 01 2023 *)

for(n=0, 30, print1((14+n)*2^(n-1), ", ")) \\ G. C. Greubel, Jun 02 2018

a(n) = Sum_{k=0..n} (k+7)*binomial(n,k).
From R. J. Mathar, Apr 20 2009: (Start)
a(n) = (14+n)*2^(n-1).
a(n) = 4*a(n-1) - 4*a(n-2).
G.f.: (7-13*x)/(1-2x)^2. (End)
E.g.f.: (x+7)*exp(2*x). - G. C. Greubel, Jun 02 2018
From Amiram Eldar, Jan 19 2021: (Start)
Sum_{n>=0} 1/a(n) = 32768*log(2) - 204619418/9009.
Sum_{n>=0} (-1)^n/a(n) = 598484902/45045 - 32768*log(3/2). (End)

More terms from R. J. Mathar, Apr 20 2009

A079028 a(0) = 1, a(n) = (n + 4)*4^(n-1) for n >= 1.

Original entry on oeis.org

1, 5, 24, 112, 512, 2304, 10240, 45056, 196608, 851968, 3670016, 15728640, 67108864, 285212672, 1207959552, 5100273664, 21474836480, 90194313216, 377957122048, 1580547964928, 6597069766656, 27487790694400, 114349209288704, 474989023199232, 1970324836974592, 8162774324609024

Offset: 0

Benoit Cloitre, Feb 01 2003

a(n) = det(M(n)) where M(n) is the n X n matrix defined by m(i,i) = 5, m(i,j) = i/j.
Main diagonal of array defined by m(1,j) = j; m(i,1) = i and m(i,j) = m(i-1,j) + 3*m(i-1,j-1).
4th binomial transform of (1,1,0,0,0,0,...). - Paul Barry, Mar 07 2003
Number of independent vertex subsets of the graph obtained by attaching two pendant edges to each vertex of the complete graph K_n (see A235113). Example: a(1)=5; indeed, K_1 is the one vertex graph and after attaching two pendant vertices we obtain the path graph ABC; the independent vertex subsets are: empty, {A}, {B}, {C}, and {A,C}. - Emeric Deutsch, Jan 13 2014
Row sums of A235113.

F. Disanto, A. Frosini, R. Pinzani and S. Rinaldi, A closed formula for the number of convex permutominoes, arXiv:math/0702550 [math.CO], 2007.
Index entries for linear recurrences with constant coefficients, signature (8,-16).

Cf. A001792, A006234, A081105, A006234, A235113.

Cf. A002697, A034007, A079861, A045891, A087449.

LinearRecurrence[{8, -16}, {1, 5}, 22] (* Jean-François Alcover, Nov 06 2018 *)

[lucas_number2(n, 4, 0)*n/2^10 for n in range(4, 26)] # Zerinvary Lajos, Mar 13 2009

a(n) = 8*a(n-1)-16*a(n-2), a(0) = 1, a(1) = 5. - Paul Barry, Mar 07 2003
G.f.: (1 - 3*x)/(1 - 4*x)^2. - Philippe Deléham, Dec 11 2008
From Amiram Eldar, Jan 14 2021: (Start)
Sum_{n>=0} 1/a(n) = 1024*log(4/3) - 880/3.
Sum_{n>=0} (-1)^n/a(n) = 688/3 - 1024*log(5/4). (End)
E.g.f.: exp(4*x)*(1 + x). - Stefano Spezia, Mar 05 2023

More terms from Stefano Spezia, Mar 05 2023

A159696 a(0)=8, a(n) = 2*a(n-1) + 2^(n-1) for n > 0.

Original entry on oeis.org

8, 17, 36, 76, 160, 336, 704, 1472, 3072, 6400, 13312, 27648, 57344, 118784, 245760, 507904, 1048576, 2162688, 4456448, 9175040, 18874368, 38797312, 79691776, 163577856, 335544320, 687865856, 1409286144, 2885681152, 5905580032

Offset: 0

Philippe Deléham, Apr 20 2009

Diagonal of triangles A062111, A152920.

			a(0)=8, a(1) = 2*8 + 1 = 17, a(2) = 2*17 + 2 = 36, a(3) = 2*36 + 4 = 76, a(4) = 2*76 + 8 = 160, ...

G. C. Greubel, Table of n, a(n) for n = 0..3300
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Cf. A000079, A001787, A001792, A045623, A045891, A034007, A111297, A159694, A159695.

[(16+n)*2^(n-1): n in [0..30]]; // G. C. Greubel, Jun 02 2018

LinearRecurrence[{4,-4}, {8,17}, 30] (* or *) Table[(16+n)*2^(n-1), {n,0,30}] (* G. C. Greubel, Jun 02 2018 *)

for(n=0, 30, print1((16+n)*2^(n-1), ", ")) \\ G. C. Greubel, Jun 02 2018

a(n) = Sum_{k=0..n} (k+8)*binomial(n,k).
From R. J. Mathar, Apr 20 2009: (Start)
a(n) = (16+n)*2^(n-1).
a(n) = 4*a(n-1) - 4*a(n-2).
G.f.: (8-15*x)/(1-2*x)^2. (End)
E.g.f.: (x+8)*exp(2*x). - G. C. Greubel, Jun 02 2018

More terms from R. J. Mathar, Apr 20 2009

cp's OEIS Frontend

A045891 First differences of A045623.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

A062111 Upper-right triangle resulting from binomial transform calculation for nonnegative integers.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath

Formula

A152920 Triangle read by rows: triangle A062111 reversed.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

SageMath

Formula

Extensions

A091613 Triangle: T(n,k) = number of compositions (ordered partitions) of n such that some part is repeated consecutively k times and no part is repeated consecutively more than k times.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A168230 a(n) = n + 2 - a(n-1) for n>1; a(1) = 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A159694 a(n) = 2*a(n-1) + 2^(n-1), for n > 0, with a(0)=6.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

SageMath