A193820 -id:A193820 - cp's OEIS frontend

A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

0, 1, 2, 4, 3, 4, 9, 7, 7, 8, 16, 16, 14, 15, 16, 25, 32, 30, 29, 31, 32, 36, 57, 62, 59, 60, 63, 64, 49, 93, 119, 121, 119, 123, 127, 128, 64, 142, 212, 240, 240, 242, 250, 255, 256, 81, 206, 354, 452, 480, 482, 492, 505, 511, 512, 100, 287, 560, 806, 932, 962, 974, 997, 1016, 1023, 1024

Offset: 1

Boris Putievskiy, Aug 15 2013

The third row is (n^4 - n^2 + 24*n + 24)/12.

For a closed-form formula for generalized Pascal's triangle see A228576. - Boris Putievskiy, Sep 04 2013

			The start of the sequence as a triangular array read by rows:
   0;
   1,  2;
   4,  3,  4;
   9,  7,  7,  8;
  16, 16, 14, 15, 16;
  25, 32, 30, 29, 31, 32;
  36, 57, 62, 59, 60, 63, 64;

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Rely Pellicer and David Alvo, Modified Pascal Triangle and Pascal Surfaces p.4
Index entries for triangles and arrays related to Pascal's triangle

Cf. We denote Pascal-like triangle with L(n) on the left border and R(n) on the right border by (L(n),R(n)). A007318 (1,1), A008949 (1,2^n), A029600 (2,3), A029618 (3,2), A029635 (1,2), A029653 (2,1), A037027 (Fibonacci(n),1), A051601 (n,n) n>=0, A051597 (n,n) n>0, A051666 (n^2,n^2), A071919 (1,0), A074829 (Fibonacci(n), Fibonacci(n)), A074909 (1,n), A093560 (3,1), A093561 (4,1), A093562 (5,1), A093563 (6,1), A093564 (7,1), A093565 (8,1), A093644 (9,1), A093645 (10,1), A095660 (1,3), A095666 (1,4), A096940 (1,5), A096956 (1,6), A106516 (3^n,1), A108561(1,(-1)^n), A132200 (4,4), A134636 (2n+1,2n+1), A137688 (2^n,2^n), A160760 (3^(n-1),1), A164844(1,10^n), A164847 (100^n,1), A164855 (101*100^n,1), A164866 (101^n,1), A172171 (1,9), A172185 (9,11), A172283 (-9,11), A177954 (int(n/2),1), A193820 (1,2^n), A214292 (n,-n), A227074 (4^n,4^n), A227075 (3^n,3^n), A227076 (5^n,5^n), A227550 (n!,n!), A228053 ((-1)^n,(-1)^n), A228074 (Fibonacci(n), n).

Cf. A000290 (row 1), A153056 (row 2), A000079 (column 1), A000225 (column 2), A132753 (column 3), A118885 (row sums of triangle array + 1), A228576 (generalized Pascal's triangle).

T:= function(n,k)
    if k=0 then return n^2;
    elif k=n then return 2^n;
    else return T(n-1,k-1) + T(n-1,k);
    fi;
  end;
Flat(List([0..12], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 12 2019

T:= proc(n, k) option remember;
      if k=0 then n^2
    elif k=n then 2^k
    else T(n-1, k-1) + T(n-1, k)
      fi
    end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 12 2019

T[n_, k_]:= T[n, k] = If[k==0, n^2, If[k==n, 2^k, T[n-1, k-1] + T[n-1, k]]]; Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Nov 12 2019 *)
Flatten[Table[Sum[i^2 Binomial[n-1-i, n-k-i], {i,1,n-k}] + Sum[2^i Binomial[n-1-i, k-i], {i,1,k}], {n,0,10}, {k,0,n}]] (* Greg Dresden, Aug 06 2022 *)

T(n,k) = if(k==0, n^2, if(k==n, 2^k, T(n-1, k-1) + T(n-1, k) )); \\ G. C. Greubel, Nov 12 2019

def funcL(n):
   q = n**2
   return q
def funcR(n):
   q = 2**n
   return q
for n in range (1,9871):
   t=int((math.sqrt(8*n-7) - 1)/ 2)
   i=n-t*(t+1)/2-1
   j=(t*t+3*t+4)/2-n-1
   sum1=0
   sum2=0
   for m1 in range (1,i+1):
      sum1=sum1+funcR(m1)*binomial(i+j-m1-1,i-m1)
   for m2 in range (1,j+1):
      sum2=sum2+funcL(m2)*binomial(i+j-m2-1,j-m2)
   sum=sum1+sum2

@CachedFunction
def T(n, k):
    if (k==0): return n^2
    elif (k==n): return 2^n
    else: return T(n-1, k-1) + T(n-1, k)
[[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 12 2019

T(n,0) = n^2, n>0; T(0,k) = 2^k; T(n, k) = T(n-1, k-1) + T(n-1, k) for n,k > 0. [corrected by G. C. Greubel, Nov 12 2019]
Closed-form formula for general case. Let L(m) and R(m) be the left border and the right border of Pascal like triangle, respectively. We denote binomial(n,k) by C(n,k).
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} R(m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} L(m2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} R(m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} L(m2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2); n>0.
Some special cases. If L(m)={b,b,b...} b*A000012, then the second sum takes form b*C(n+k-1,j). If L(m) is {0,b,2b,...} b*A001477, then the second sum takes form b*C(n+k,n-1). Similarly for R(m) and the first sum.
For this sequence L(m)=m^2 and R(m)=2^m.
As table read by antidiagonals T(n,k) = Sum_{m1=1..n} (2^m1)*C(n+k-m1-1, n-m1) + Sum_{m2=1..k} (m2^2)*C(n+k-m2-1, k-m2); n,k >=0.
As linear sequence a(n) = Sum_{m1=1..i} (2^m1)*C(i+j-m1-1, i-m1) + Sum_{m2=1..j} (m2^2)*C(i+j-m2-1, j-m2), where i=n-t*(t+1)/2-1, j=(t*t+3*t+4)/2-n-1, t=floor((-1+sqrt(8*n-7))/2).
As a triangular array read by rows, T(n,k) = Sum_{i=1..n-k} i^2*C(n-1-i, n-k-i) + Sum_{i=1..k} 2^i*C(n-1-i, k-i); n,k >=0. - Greg Dresden, Aug 06 2022

Cross-references corrected and extended by Philippe Deléham, Dec 27 2013

A128175 Binomial transform of A128174.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 4, 4, 3, 1, 8, 8, 7, 4, 1, 16, 16, 15, 11, 5, 1, 32, 32, 31, 26, 16, 6, 1, 64, 64, 63, 57, 42, 22, 7, 1, 128, 128, 127, 120, 99, 64, 29, 8, 1, 256, 256, 255, 247, 219, 163, 93, 37, 9, 1

Offset: 1

Gary W. Adamson, Feb 17 2007

Row sums = A045623: (1, 2, 5, 12, 28, 64, 144, ...).
A128176 = A128174 * A007318.
Riordan array ((1-x)/(1-2x),x/(1-x)). - Paul Barry, Oct 02 2010
Fusion of polynomial sequences p(n,x) = (x+1)^n and q(n,x) = x^n + x^(n-1) + ... + x + 1; see A193722 for the definition of fusion. - Clark Kimberling, Aug 04 2011

			First few rows of the triangle:
   1;
   1,  1;
   2,  2,  1;
   4,  4,  3,  1;
   8,  8,  7,  4,  1;
  16, 16, 15, 11,  5,  1;
  32, 32, 31, 26, 16,  6,  1;
  64, 64, 63, 57, 42, 22,  7,  1;
  ...
From _Paul Barry_, Oct 02 2010: (Start)
Production matrix is
  1, 1;
  1, 1, 1;
  0, 0, 1, 1;
  0, 0, 0, 1, 1;
  0, 0, 0, 0, 1, 1;
  0, 0, 0, 0, 0, 1, 1;
  0, 0, 0, 0, 0, 0, 1, 1;
  0, 0, 0, 0, 0, 0, 0, 1, 1;
  0, 0, 0, 0, 0, 0, 0, 0, 1, 1;
  ...
Matrix logarithm is
  0;
  1, 0;
  1, 2, 0;
  1, 1, 3, 0;
  1, 1, 1, 4, 0;
  1, 1, 1, 1, 5, 0;
  1, 1, 1, 1, 1, 6, 0;
  1, 1, 1, 1, 1, 1, 7, 0;
  1, 1, 1, 1, 1, 1, 1, 8, 0;
  1, 1, 1, 1, 1, 1, 1, 1, 9,  0;
  1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 0;
  ... (End)
.
First few rows of the array:
  1, 1,  2,  4,  8,  16, ...
  1, 2,  4,  8, 16,  32, ...
  1, 3,  7, 15, 31,  63, ...
  1, 4, 11, 26, 57, 120, ...
  1, 5, 16, 42, 99, 219, ...
  ...

Cf. A045623, A128176, A007318.

A193820 := (n,k) -> `if`(k=0 or n=0, 1, A193820(n-1,k-1)+A193820(n-1,k));
A128175 := (n,k) -> A193820(n-1,n-k);
seq(print(seq(A128175(n,k),k=0..n)),n=0..10); # Peter Luschny, Jan 22 2012

z = 10; a = 1; b = 1;
p[n_, x_] := (a*x + b)^n
q[0, x_] := 1
q[n_, x_] := x*q[n - 1, x] + 1; q[n_, 0] := q[n, x] /. x -> 0;
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, 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}]]   (* A193820 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]]  (* A128175 *)
(* Clark Kimberling, Aug 06 2011 *)
(* function dotTriangle[] is defined in A128176 *)
a128175[r_] := dotTriangle[Binomial, If[EvenQ[#1 + #2], 1, 0]&, r]
TableForm[a128174[7]] (* triangle *)
Flatten[a128174[9]] (* data *) (* Hartmut F. W. Hoft, Mar 15 2017 *)

A007318 * A128174 as infinite lower triangular matrices.
Antidiagonals of an array in which the first row = (1, 1, 2, 4, 8, 16, ...); and (n+1)-th row = partial sums of n-th row.
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(4 + 4*x + 3*x^2/2! + x^3/3!) = 4 + 8*x + 15*x^2/2! + 26*x^3/3! + 42*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 22 2014
T(n, k) = Sum_{i=0..floor((n-k)/2)} binomial(n-1, k-1+2*i). - Werner Schulte, Mar 05 2025
T(n, k) = binomial(n-1, k-1)*hypergeom([1, (k-n)/2, (1+k-n)/2], [(1+k)/2, k/2], 1). - Stefano Spezia, Mar 07 2025

A117670 Triangle read by rows: partial sums of the Pascal triangle minus 1.

Original entry on oeis.org

1, 2, 3, 3, 6, 7, 4, 10, 14, 15, 5, 15, 25, 30, 31, 6, 21, 41, 56, 62, 63, 7, 28, 63, 98, 119, 126, 127, 8, 36, 92, 162, 218, 246, 254, 255, 9, 45, 129, 255, 381, 465, 501, 510, 511, 10, 55, 175, 385, 637, 847, 967, 1012, 1022, 1023

Offset: 1

Arie Bos, Jul 06 2008, Jul 08 2008

Imagine that you are in a building with floors starting at floor 1, the lowest floor and you have a large number of eggs. For each floor in the building, you want to know whether or not an egg dropped from that floor will break.
If an egg breaks when dropped from floor i, then all eggs are guaranteed to break when dropped from any floor j > i. Likewise, if an egg doesn't break when dropped from floor i, then all eggs are guaranteed to never break when dropped from any floor j <= i.
a(n,k) is the maximum number of floors where you can determine whether or not an egg will break when dropped from any floor, with the following restrictions: you may drop a maximum of n eggs (one at a time, from any floors of your choosing) and you may break a maximum of k eggs.
Each row of the triangle is the running sum of the corresponding row with the first 1 omitted of Pascal's triangle (A007318), see A008949, A054143, A193820.
The k-th entry in the n-th row is the number of possible combinations of on/off switches after k attempts to turn on a switch in a set of n distinguishable switches. An attempt to turn on the same switch twice does not result in a new combination. See example. - Sergei Viznyuk, Jun 24 2012
T(n,k) is the number of nonempty subsets of the n-set with at most k elements, see example. - Joerg Arndt, May 04 2014

			Triangle a(n,k) begins:
n\k  1   2    3    4    5    6    7     8     9    10 ...
1:   1
2:   2   3
3:   3   6    7
4:   4  10   14   15
5:   5  15   25   30   31
6:   6  21   41   56   62   63
7:   7  28   63   98  119  126  127
8:   8  36   92  162  218  246  254   255
9:   9  45  129  255  381  465  501   510   511
10: 10  55  175  385  637  847  967  1012  1022  1023
...  Reformatted and extended by _Wolfdieter Lang_, Feb 07 2013
From _Sergei Viznyuk_, Jun 24 2012: (Start)
For example, we have n=3 distinguishable switches A,B,C (third row above). We attempt k=2 times to turn on a switch at random. The possible resulting combinations are:
A=on, B=off, C=off (the same A switch was turned on 2 times)
A=off, B=on, C=off (the same B switch was turned on 2 times)
A=off, B=off, C=on (the same C switch was turned on 2 times)
A=on, B=on, C=off  (switches A and B were turned on)
A=on, B=off, C=on  (switches A and C were turned on)
A=off, B=on, C=on  (switches B and C were turned on)
Thus, we have 6 different combinations, which is the number 6 at row n=3 column k=2 in the sequence above.
(End)
From _Joerg Arndt_, May 04 2014: (Start)
There are T(4,2) = 10 subsets of {0, 1, 2, 3}:
01:    1...    { 0 }
02:    11..    { 0, 1 }
03:    111.    { 0, 1, 2 }
04:    11.1    { 0, 1, 3 }
05:    1.1.    { 0, 2 }
06:    1.11    { 0, 2, 3 }
07:    1..1    { 0, 3 }
08:    .1..    { 1 }
09:    .11.    { 1, 2 }
10:    .111    { 1, 2, 3 }
11:    .1.1    { 1, 3 }
12:    ..1.    { 2 }
13:    ..11    { 2, 3 }
14:    ...1    { 3 }
(End)

Susanne Wienand, Table of n, a(n) for n = 1..1830
Google code.jam, Problem C. Egg Drop
Sergei Viznyuk, C-Program
Sergei Viznyuk, Local copy of C-Program

Table[Sum[Binomial[n, m], {m, k}], {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Nov 25 2015 *)

tabl(nrows) = {for (n=1, nrows, for (k=1, n, print1(sum(m=1,k,binomial(n,m)), ", ");); print(););} \\ Michel Marcus, May 21 2013

a(n,1) = n ; a(n,n) = 2^n-1; a(n+1,k+1) = 1 + a(n,k) + a(n,k-1), 0 < k < n.

a(n,k) = sum(binomial(n,m),m=1..k), 1 <= k <= n. (see the running sum comment above). - Wolfdieter Lang, Feb 07 2013

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A228196 A triangle formed like Pascal's triangle, but with n^2 on the left border and 2^n on the right border instead of 1.

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A128175 Binomial transform of A128174.

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A117670 Triangle read by rows: partial sums of the Pascal triangle minus 1.

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