A014473 -id:A014473 - cp's OEIS frontend

A323211 Level 1 of Pascal's pyramid. T(n, k) triangle read by rows for n >= 0 and 0 <= k <= n.

1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 3, 2, 1, 1, 2, 4, 4, 2, 1, 1, 2, 5, 7, 5, 2, 1, 1, 2, 6, 11, 11, 6, 2, 1, 1, 2, 7, 16, 21, 16, 7, 2, 1, 1, 2, 8, 22, 36, 36, 22, 8, 2, 1, 1, 2, 9, 29, 57, 71, 57, 29, 9, 2, 1, 1, 2, 10, 37, 85, 127, 127, 85, 37, 10, 2, 1

Offset: 0

Peter Luschny, Feb 11 2019

Pascal's pyramid is defined by recurrence. P(0) is Pascal's triangle. Now assume P(n-1) already constructed. Then P(n) is found by the steps: (1) Add 1 to each term of P(n-1). (2) Add at the left and at the right side a diagonal consisting all of 1s and complement the top with the rows 1 and 1, 1. A similar construction starting from the Pascal's triangle and subtracting 1 from all terms leads to A014473.

			Triangle starts:
                                1
                              1,  1
                            1,  2,  1
                          1,  2,  2,  1
                        1,  2,  3,  2,  1
                      1,  2,  4,  4,  2,  1
                    1,  2,  5,  7,  5,  2,  1
                 1,  2,  6,  11, 11,  6,  2,  1
               1,  2,  7,  16,  21, 16,  7,  2,  1
             1,  2,  8,  22, 36, 36,  22,  8,  2,  1
           1,  2,  9,  29, 57,  71,  57, 29,  9,  2,  1

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

Differs from A323231 only in the second term.
Row sums are A323227.
Cf. A000045, A014473, A323230.

A323211:= func< n,k | n le 1 select 1 else 1 + Binomial(n-2,k-1) >;
[A323211(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Sep 26 2024

T := (n, k) -> `if`(n=1, 1, binomial(n-2, k-1) + 1):
seq(seq(T(n, k), k=0..n), n=0..10);
# Alternative:
T := proc(n, k) option remember;
if k = n then return 1 fi; if k < 2 then return k+1 fi;
T(n-1, k-1) + T(n-1, k) - 1 end:
seq(seq(T(n, k), k=0..n), n=0..10);

A323211[n_, k_]:= If[n<2, 1, Binomial[n-2, k-1] +1];
Table[A323211[n,k], {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 26 2024 *)

def A323211(n,k): return 1 if (n<2) else 1 + binomial(n-2,k-1)
flatten([[A323211(n,k) for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Sep 26 2024

T(n, k) = binomial(n-2, k-1) + 1 if n != 1 else 1.
G.f.: (1 + 3*y + y^2 + x^4*y^2*(1 + y)^2 + x^2*y*(2 + 5*y + 2*y^2) - x^3*y*(1 + 4*y + 4*y^2 + y^3) - x*(1 + 5*y + 5*y^2 + y^3)/((1 - x)*(1 + y)^2*(1 - x*y)*(1 - x - x*y)). - Stefano Spezia, Sep 26 2024
From G. C. Greubel, Sep 26 2024: (Start)
T(n, n-k) = T(n, k) (symmetry).
T(2*n, n) = A323230(n).
Sum_{k=0..n} (-1)^k*T(n, k) = (n+1 mod 2) - [n=2].
Sum_{k=0..floor(n/2)} T(n-k, k) = Fibonacci(n-2) + (1/4)*(2*n + 3 + (-1)^n) +[n=0] - [n=1]. (End)

A259525 First differences of A007318, when Pascal's triangle is seen as flattened list.

Original entry on oeis.org

0, 0, 0, 1, -1, 0, 2, 0, -2, 0, 3, 2, -2, -3, 0, 4, 5, 0, -5, -4, 0, 5, 9, 5, -5, -9, -5, 0, 6, 14, 14, 0, -14, -14, -6, 0, 7, 20, 28, 14, -14, -28, -20, -7, 0, 8, 27, 48, 42, 0, -42, -48, -27, -8, 0, 9, 35, 75, 90, 42, -42, -90, -75, -35, -9, 0, 10, 44, 110

Offset: 0

Reinhard Zumkeller, Jul 18 2015

sign

A214292 gives first differences per row in Pascal's triangle.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A014473, A163866, A214292.

Cf. A000108, A021499, A047171, A074331.

a259525 n = a259525_list !! n
a259525_list = zipWith (-) (tail pascal) pascal
                           where pascal = concat a007318_tabl

[k eq n select 0 else (n-2*k-1)*Binomial(n,k+1)/(n-k): k in [0..n], n in [0..14]]; // G. C. Greubel, Apr 25 2024

Table[If[k==n, 0, ((n-2*k-1)/(n-k))*Binomial[n,k+1]], {n,0,12}, {k,0, n}]//Flatten (* G. C. Greubel, Apr 25 2024 *)

flatten([[binomial(n,k+1) -binomial(n,k) +int(k==n) for k in range(n+1)] for n in range(15)]) # G. C. Greubel, Apr 25 2024

From G. C. Greubel, Apr 25 2024: (Start)
If viewed as a triangle then:
T(n, k) = binomial(n, k+1) - binomial(n, k), with T(n, n) = 0.
T(n, n-k) = - T(n, k), for 0 <= k < n.
T(2*n, n) = [n=0] - A000108(n).
Sum_{k=0..n} T(n, k) = 0 (row sums).
Sum_{k=0..floor(n/2)} T(n, k) = A047171(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A021499(n).
Sum_{k=0..floor(n/2)} T(n-k, k) = A074331(n-1). (End)

A177767 Triangle read by rows: T(n,k) = binomial(n - 1, k - 1), 1 <= k <= n, and T(n,0) = A153881(n+1), n >= 0.

Original entry on oeis.org

1, -1, 1, -1, 1, 1, -1, 1, 2, 1, -1, 1, 3, 3, 1, -1, 1, 4, 6, 4, 1, -1, 1, 5, 10, 10, 5, 1, -1, 1, 6, 15, 20, 15, 6, 1, -1, 1, 7, 21, 35, 35, 21, 7, 1, -1, 1, 8, 28, 56, 70, 56, 28, 8, 1, -1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1, -1, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1

Offset: 0

Roger L. Bagula, May 13 2010

Row sums yield A000225 preceded by 1.

Except for signs, this is A135225.

			Triangle begins:
   1;
  -1, 1;
  -1, 1, 1;
  -1, 1, 2,  1;
  -1, 1, 3,  3,  1;
  -1, 1, 4,  6,  4,   1;
  -1, 1, 5, 10, 10,   5,   1;
  -1, 1, 6, 15, 20,  15,   6,  1;
  -1, 1, 7, 21, 35,  35,  21,  7,  1;
  -1, 1, 8, 28, 56,  70,  56, 28,  8, 1;
  -1, 1, 9, 36, 84, 126, 126, 84, 36, 9, 1;
   ...

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

Cf. A000225, A000071, A007318, A014473, A097805, A131026, A135225.

A177767:= func< n,k | k eq n select 1 else  k eq 0 select -1 else Binomial(n-1, k-1) >;
[A177767(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Apr 22 2024

Flatten[Table[If[n == 0, {1}, CoefficientList[x*(1 + x)^( n - 1) - 1, x]], {n, 0, 10}]]

T(n, k) := if k = 0 then 2*floor(1/(n + 1)) - 1 else binomial(n - 1, k - 1)$
create_list(T(n, k), n, 0, 12, k, 0, n); /* Franck Maminirina Ramaharo, Oct 23 2018 */

flatten([[binomial(n-1, k-1) - int(k==0) + 2*int(n==0) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Apr 22 2024

Row n = coefficients in the expansion of x*(1 + x)^(n - 1) - 1, n > 0.
From Franck Maminirina Ramaharo, Oct 23 2018: (Start)
G.f.: (1 - 3*y + (2 + x)*y^2)/(1 - (2 + x)*y + (1 + x)*y^2).
E.g.f.: (2 + x - (1 + x)*exp(y) + x*exp((1 + x)*y))/(1 + x). (End)
From G. C. Greubel, Apr 22 2024: (Start)
Sum_{k=0..n} (-1)^k*T(n, k) = A153881(n+1) - [n=1].
Sum_{k=0..floor(n/2)} T(n-k, k) = A000071(n-1) + [n=0].
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = -A131026(n-1) + [n=0]. (End)

Edited and new name by Franck Maminirina Ramaharo, Oct 23 2018

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A323211 Level 1 of Pascal's pyramid. T(n, k) triangle read by rows for n >= 0 and 0 <= k <= n.

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A259525 First differences of A007318, when Pascal's triangle is seen as flattened list.

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A177767 Triangle read by rows: T(n,k) = binomial(n - 1, k - 1), 1 <= k <= n, and T(n,0) = A153881(n+1), n >= 0.

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