A159695 -id:A159695 - cp's OEIS frontend

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

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

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)

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

A159697 a(0)=9, a(n) = 2*a(n-1) + 2^(n-1) for n > 0.

Original entry on oeis.org

9, 19, 40, 84, 176, 368, 768, 1600, 3328, 6912, 14336, 29696, 61440, 126976, 262144, 540672, 1114112, 2293760, 4718592, 9699328, 19922944, 40894464, 83886080, 171966464, 352321536, 721420288, 1476395008, 3019898880

Offset: 0

Philippe Deléham, Apr 20 2009

Diagonal of triangles A062111, A152920.

			a(0)=9, a(1) = 2*9 + 1 = 19, a(2) = 2*19 + 2 = 40, a(3) = 2*40 + 4 = 84, a(4) = 2*84 + 8 = 176, ...

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, A159696.

I:=[9,19]; [n le 2 select I[n] else 4*Self(n-1) - 4*Self(n-2): n in [1..30]]; // G. C. Greubel, Jun 02 2018

RecurrenceTable[{a[0]==9,a[n]==2a[n-1]+2^(n-1)},a,{n,30}] (* or *) LinearRecurrence[{4,-4},{9,19},30] (* Harvey P. Dale, Mar 24 2013 *)

Vec((9-17*x)/(1-2*x)^2 + O(x^40)) \\ Michel Marcus, Sep 29 2015

a(n) = Sum_{k=0..n} (k+9)*binomial(n,k).
From R. J. Mathar, Apr 20 2009: (Start)
a(n) = (18+n)*2^(n-1).
a(n) = 4*a(n-1) - 4*a(n-2).
G.f.: (9-17*x)/(1-2*x)^2. (End)
a(0)=9, a(1)=19, a(n) = 4*a(n-1) - 4*a(n-2). - Harvey P. Dale, Mar 24 2013
a(n) = 2*A079862(n-10). - Michel Marcus, Sep 29 2015
E.g.f.: (x+9)*exp(2*x). - G. C. Greubel, Jun 02 2018

More terms from Vincenzo Librandi, Apr 30 2009

A327916 Triangle T(k, n) read by rows: Array A(k, n) = 2^k(k + 1 + 2n), k >= 0, n >= 0, read by antidiagonals upwards.

Original entry on oeis.org

1, 4, 3, 12, 8, 5, 32, 20, 12, 7, 80, 48, 28, 16, 9, 192, 112, 64, 36, 20, 11, 448, 256, 144, 80, 44, 24, 13, 1024, 576, 320, 176, 96, 52, 28, 15, 2304, 1280, 704, 384, 208, 112, 60, 32, 17, 5120, 2816, 1536, 832, 448, 240, 128, 68, 36, 19, 11264, 6144, 3328, 1792, 960, 512, 272, 144, 76, 40, 21

Offset: 0

Wolfdieter Lang, Oct 03 2019

The array A(k, n) arises from the following Pascal-type triangles PTodd(k), k >= 0 based on the positive odd integers A005408.
For example, the Pascal-type triangle PTodd(k), for k = 3 is
   1     3     5     7
      4     8    12
        12    20
           32
Taken upside-down such triangles become so-called addition towers of height k+1 (Rechenturm in German elementary schools; thanks to my correspondent Bennet D.), starting with any k+1 numbers. Here the positive odd numbers are used.
The sequence s of the final number of these Pascal-type triangles PT(k), for k >= 0, begins 1, 4, 12, 32, ...; s(k) = (k+1)*2^k = A001787(k+1), for k >= 0.
For k -> infinity the left-aligned row sequences build the array A(k, n), with k >= 0 and n >= 0, namely A(k, n) = 2^k*(k + 2*n + 1); this array begins:
  k\n  0   1   2   3   4    5 ...
  -------------------------------
  0:   1   3   5   7   9   11 ... {A005408(n)}
  1:   4   8  12  16  20   24 ... {A008586(n+1)}
  2:  12  20  28  36  44   52 ... {A017113(n+1)}
  3:  32  48  64  80  96  112 ... {A008598(n+2)}
  4:  80 112 144 176 208  240 ... {16*A005408(n+2)}
  5: 192 256 320 384 448  512 ... {A152691(n+3)}
  6: 448 576 704 832 960 1088 ... {64*A005408(n+3)}
  ...
The sequence s, the first (n=0) column of A, is always the binomial transform of the first (k=0) row in A.
A(k, n) = Sum_{j=0..k} binomial(k, j)*(2*(n+j)+1) = 2^k*(k + 1 + 2*n), for k >= 0 and n >= 0.
The corresponding antidiagonal-upwards read triangle is T(k, n) = A(k-n, n) = 2^(k-n)*(k + n + 1), n >= 0, k = 0..n.
If the nonnegative integers A001477 are used as k = 0 row of the array Anneg(k, n) = 2^(k-1)*(2*n + k), for k >= 0, n >= 0, with the triangle Tnneg(k, n) = Anneg(k-n, n) = (n + k)*2^(k-n-1), k >= 0, n = 0..k, then the s sequence is snneg(k) = Tnneg(k, 0) = k*2^{k-1} = A001787(k), the binomial transform of the sequence{A001477(n)}_{n>=0}. The triangle Tnneg begins [0], [1, 1], [4, 3, 2], [12, 8, 5, 3], [32, 20, 12, 7, 4], ... . See A062111 and the row-reversed triangle A152920 for other versions.

			The triangle T(k, n) begins:
   k\n    0    1    2    3   4   5   6   7  8  9 10 ...
  -----------------------------------------------------
   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: 11264 6144 3328 1792 960 512 272 144 76 40 21
  ...

Column sequences without leading zeros are for n=0..9: A001787(n+1), A001792(n+1), A045623(n+2), A045891(n+3), A034007(n+4), A111297(n+3), A159694(n+1), A159695(n+1), A159696(n+1), A159697(n+1).
The sequence of (sub)diagonal k, for k >= 0, is the row k sequence of array A: {(k + 2*n + 1)*2^k}_{k >= 0}.
Row sums: A213569(k+1), k >= 0 (see the J. M. Bergot comments there).
Cf. A006211, A152920.

Table[2^#*(# + 1 + 2 n) &[k - n], {k, 0, 10}, {n, 0, k}] // Flatten (* Michael De Vlieger, Oct 03 2019 *)

Array A(k, n) = Sum_{j=0..k} binomial(k, j)*(2*(n+j) + 1) = 2^k*(k + 1+ 2*n), for k >= 0 and n >= 0.
Triangle T(k, n) =  A(k-n, n) = 2^(k-n)*(k + n + 1), n >= 0, k = 0..n.
Recurrence: T(k, 0) = (k+1)*2^k = A001787(k+1), for k >= 0, and  T(k, n) = T(k, n-1) - T(k-1, n-1), for n >= 1, k >= 1, with T(k, n) = 0 if k < n.
O.g.f. for row polynomials: G(z,x) = Sum_{n=0..k} R(k, x)*z^n =
(1  + x*z*(1 - 4*z))/((1 - 2*z)^2*(1 - x*z)^2).
T(k, 0) = Sum_{n=0..k} binomial(k,n)*T(n, n), k >= 0 (binomial transform).

Definition corrected by Georg Fischer, Jul 13 2023

cp's OEIS Frontend

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

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A152920 Triangle read by rows: triangle A062111 reversed.

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

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

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A159697 a(0)=9, a(n) = 2*a(n-1) + 2^(n-1) for n > 0.

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A327916 Triangle T(k, n) read by rows: Array A(k, n) = 2^k*(k + 1 + 2*n), k >= 0, n >= 0, read by antidiagonals upwards.

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A327916 Triangle T(k, n) read by rows: Array A(k, n) = 2^k(k + 1 + 2n), k >= 0, n >= 0, read by antidiagonals upwards.