A014462 -id:A014462 - cp's OEIS frontend

A034868 Left half of Pascal's triangle.

1, 1, 1, 2, 1, 3, 1, 4, 6, 1, 5, 10, 1, 6, 15, 20, 1, 7, 21, 35, 1, 8, 28, 56, 70, 1, 9, 36, 84, 126, 1, 10, 45, 120, 210, 252, 1, 11, 55, 165, 330, 462, 1, 12, 66, 220, 495, 792, 924, 1, 13, 78, 286, 715, 1287, 1716, 1, 14, 91, 364, 1001, 2002, 3003, 3432, 1, 15

Offset: 0

N. J. A. Sloane

			1;
1;
1, 2;
1, 3;
1, 4,  6;
1, 5, 10;
1, 6, 15, 20;
...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle

Cf. A007318, A107430, A062344, A122366, A027306 (row sums).
Cf. A008619.
Cf. A225860.
Cf. A126257.
Cf. A034869 (right half), A014413, A014462, A265848.

a034868 n k = a034868_tabf !! n !! k
a034868_row n = a034868_tabf !! n
a034868_tabf = map reverse a034869_tabf
-- Reinhard Zumkeller, improved Dec 20 2015, Jul 27 2012

Flatten[ Table[ Binomial[n, k], {n, 0, 15}, {k, 0, Floor[n/2]}]] (* Robert G. Wilson v, May 28 2005 *)

for(n=0, 14, for(k=0, floor(n/2), print1(binomial(n, k),", ");); print();) \\ Indranil Ghosh, Mar 31 2017

import math
from sympy import binomial
for n in range(15):
    print([binomial(n, k) for k in range(int(math.floor(n/2)) + 1)]) # Indranil Ghosh, Mar 31 2017

from itertools import count, islice
def A034868_gen(): # generator of terms
    yield from (s:=(1,))
    for i in count(0):
        yield from (s:=(1,)+tuple(s[j]+s[j+1] for j in range(len(s)-1)) + ((s[-1]<<1,) if i&1 else ()))
A034868_list = list(islice(A034868_gen(),30)) # Chai Wah Wu, Oct 17 2023

T(n,k) = A034869(n,floor(n/2)-k), k = 0..floor(n/2). - Reinhard Zumkeller, Jul 27 2012

A034869 Right half of Pascal's triangle.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 6, 4, 1, 10, 5, 1, 20, 15, 6, 1, 35, 21, 7, 1, 70, 56, 28, 8, 1, 126, 84, 36, 9, 1, 252, 210, 120, 45, 10, 1, 462, 330, 165, 55, 11, 1, 924, 792, 495, 220, 66, 12, 1, 1716, 1287, 715, 286, 78, 13, 1, 3432, 3003, 2002, 1001, 364, 91, 14, 1

Offset: 0

N. J. A. Sloane

From R. J. Mathar, May 13 2006: (Start)
Also flattened table of the expansion coefficients of x^n in Chebyshev Polynomials T_k(x) of the first kind:
x^n is 2^(1-n) multiplied by the sum of floor(1+n/2) terms using only terms T_k(x) with even k if n even, only terms T_k(x) with odd k if n is odd and halving the coefficient a(..) in front of any T_0(x):
x^0=2^(1-0) a(0)/2 T_0(x)
x^1=2^(1-1) a(1) T_1(x)
x^2=2^(1-2) [a(2)/2 T_0(x)+a(3) T_2(x)]
x^3=2^(1-3) [a(4) T_1(x)+a(5) T_3(x)]
x^4=2^(1-4) [a(6)/2 T_0(x)+a(7) T_2(x) +a(8) T_4(x)]
x^5=2^(1-5) [a(9) T_1(x)+a(10) T_3(x) +a(11) T_5(x)]
x^6=2^(1-6) [a(12)/2 T_0(x)+a(13) T_2(x) +a(14) T_4(x) +a(15) T_6(x)]
x^7=2^(1-7) [a(16) T_1(x)+a(17) T_3(x) +a(18) T_5(x) +a(19) T_7(x)]" (End)
T(n,k) = A034868(n,floor(n/2)-k), k = 0..floor(n/2). - Reinhard Zumkeller, Jul 27 2012
Rows are binomial(r-1,(2r+1-(-1)^r)\4 -n ) where r is the row and n is the term. Columns are binomial(2m+c-3,m-1) where c is the column and m is the term. - Anthony Browne, May 17 2016

			The table starts:
  1
  1
  2 1
  3 1
  6 4 1
  ...

Reinhard Zumkeller, Rows n=0..150 of triangle, flattened

Cf. A007318, A008619 (row lengths).
Cf. A110654.
Cf. A034868 (left half), A014413, A014462, A027306 (row sums).
Columns k=0-1-2-3-4 give: A001405, A037955, A037956, A037957, A037958.

a034869 n k = a034869_tabf !! n !! k
a034869_row n = a034869_tabf !! n
a034869_tabf = [1] : f 0 [1] where
   f 0 us'@(_:us) = ys : f 1 ys where
                    ys = zipWith (+) us' (us ++ [0])
   f 1 vs@(v:_) = ys : f 0 ys where
                  ys = zipWith (+) (vs ++ [0]) ([v] ++ vs)
-- Reinhard Zumkeller, improved Dec 21 2015, Jul 27 2012

for n from 0 to 60 do for j from n mod 2 to n by 2 do print( binomial(n,(n-j)/2) ); od; od; # R. J. Mathar, May 13 2006
# Second program:
egf:= k-> BesselI(2*k, 2*x) + BesselI(2*k+1, 2*x):
A034869:= (n, k)-> n! * coeff(series(egf(k), x, n+1), x, n):
seq(print(seq(A034869(n, k), k=0..iquo(n, 2))), n=0..14); # Mélika Tebni, Sep 05 2024

Table[Binomial[n, k], {n, 0, 14}, {k, Ceiling[n/2], n}] // Flatten (* Michael De Vlieger, May 19 2016 *)

for(n=0, 14, for(k=ceil(n/2), n, print1(binomial(n, k),", ");); print();) \\ Indranil Ghosh, Mar 31 2017

import math
from sympy import binomial
for n in range(15):
    print([binomial(n, k) for k in range(math.ceil(n/2), n + 1)]) # Indranil Ghosh, Mar 31 2017

E.g.f. of column k: BesselI(2*k,2*x) + BesselI(2*k+1,2*x). - Mélika Tebni, Sep 05 2024

Keyword fixed and example added by Franklin T. Adams-Watters, May 27 2010

A014413 Triangular array formed from elements to right of middle of Pascal's triangle.

Original entry on oeis.org

1, 1, 3, 1, 4, 1, 10, 5, 1, 15, 6, 1, 35, 21, 7, 1, 56, 28, 8, 1, 126, 84, 36, 9, 1, 210, 120, 45, 10, 1, 462, 330, 165, 55, 11, 1, 792, 495, 220, 66, 12, 1, 1716, 1287, 715, 286, 78, 13, 1, 3003, 2002, 1001, 364, 91, 14, 1, 6435, 5005, 3003, 1365, 455, 105, 15, 1

Offset: 1

Mohammad K. Azarian

			Triangle begins as:
    1;
    1;
    3,   1;
    4,   1;
   10,   5,   1;
   15,   6,   1;
   35,  21,   7,  1;
   56,  28,   8,  1;
  126,  84,  36,  9,  1;
  210, 120,  45, 10,  1;
  462, 330, 165, 55, 11, 1;
  792, 495, 220, 66, 12, 1;
  ...

Reinhard Zumkeller, Rows n = 1..200 of triangle, flattened

Cf. A014462.

Cf. A007318, A034868, A034869, A265848.

a014413 n k = a014413_tabf !! (n-1) !! (k-1)
a014413_row n = a014413_tabf !! (n-1)
a014413_tabf = [1] : f 1 [1] where
   f 0 us'@(_:us) = ys : f 1 ys where
                    ys = zipWith (+) us' (us ++ [0])
   f 1 vs@(v:_) = ys' : f 0 ys where
                  ys@(_:ys') = zipWith (+) (vs ++ [0]) ([v] ++ vs)
-- Reinhard Zumkeller, Dec 24 2015

Table[Binomial[n,k],{n,15},{k,Ceiling[(n+1)/2],n}]//Flatten  (* Stefano Spezia, Jan 16 2025 *)

More terms from James Sellers

A265848 Pascal's triangle, right and left halves interchanged.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 3, 1, 1, 3, 4, 1, 1, 4, 6, 10, 5, 1, 1, 5, 10, 15, 6, 1, 1, 6, 15, 20, 35, 21, 7, 1, 1, 7, 21, 35, 56, 28, 8, 1, 1, 8, 28, 56, 70, 126, 84, 36, 9, 1, 1, 9, 36, 84, 126, 210, 120, 45, 10, 1, 1, 10, 45, 120, 210, 252, 462, 330, 165, 55, 11, 1, 1, 11, 55, 165, 330, 462

Offset: 0

Reinhard Zumkeller, Dec 24 2015

Concatenations of rows of A014413 and A034868.

Alternative mirrored variant: concatenation of A034869 and A014462.

			.   0:                                1
.   1:                              1   1
.   2:                            1   1   2
.   3:                          3   1   1   3
.   4:                        4   1   1   4   6
.   5:                     10   5   1   1   5   10
.   6:                   15   6   1   1   6   15  20
.   7:                 35   21  7   1   1   7   21   35
.   8:              56   28   8   1   1   8   28  56   70
.   9:           126   84   36  9   1   1   9   36   84   126
.  10:        210   120  45  10   1   1   10  45  120  210  252
.  11:     462   330  165   55  11  1   1   11  55  165   330  462
.  12:  792   495  220   66  12   1   1   12  66  220  495  792   924  .

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
Index entries for triangles and arrays related to Pascal's triangle.

Cf. A014413, A014462, A034868, A034869, A007318, A001405, A037952, A000079 (row sums), A001142 (row products).

a265848 n k = a265848_tabl !! n !! k
a265848_row n = a265848_tabl !! n
a265848_tabl = zipWith (++) ([] : a014413_tabf) a034868_tabf

row[n_] := Binomial[n, Join[Range[Floor[n/2] + 1, n], Range[0, Floor[n/2]]]]; Array[row, 12, 0] // Flatten (* Amiram Eldar, May 13 2025 *)

T(n,k) = A007318(n, (k + floor((n+2)/2)) mod (n+1)).
T(n,k) = if k <= [(n+1)/2] then A014413(n,k+1) else A034868(n,k-[(n+1)/2]).
T(n,0) = A037952(n) for n > 0.
T(n,n) = A001405(n).

A117178 Riordan array (c(x^2)/sqrt(1-4x^2), xc(x^2)), c(x) the g.f. of A000108.

Original entry on oeis.org

1, 0, 1, 3, 0, 1, 0, 4, 0, 1, 10, 0, 5, 0, 1, 0, 15, 0, 6, 0, 1, 35, 0, 21, 0, 7, 0, 1, 0, 56, 0, 28, 0, 8, 0, 1, 126, 0, 84, 0, 36, 0, 9, 0, 1, 0, 210, 0, 120, 0, 45, 0, 10, 0, 1, 462, 0, 330, 0, 165, 0, 55, 0, 11, 0, 1, 0, 792, 0, 495, 0, 220, 0, 66, 0, 12, 0, 1

Offset: 0

Paul Barry, Mar 01 2006

Row sums are A058622(n+1). Diagonal sums are A001791(n+1), with interpolated zeros. Inverse is A117179.

De-aerated and rows reversed, this matrix apparently becomes A014462. The nonzero antidiagonals are embedded in several entries and apparently contain partial sums of previous nonzero antidiagonals. - Tom Copeland, May 30 2017

			Triangle begins
    1;
    0,  1;
    3,  0,  1;
    0,  4,  0,  1;
   10,  0,  5,  0,  1;
    0, 15,  0,  6,  0,  1;
   35,  0, 21,  0,  7,  0,  1;
    0, 56,  0, 28,  0,  8,  0,  1;
  126,  0, 84,  0, 36,  0,  9,  0,  1;

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

Cf. A001791, A014462, A052203, A058622 (row sums), A117179, A224274.

[(1+(-1)^(n-k))*Binomial(n+1, Floor((n-k)/2))/2: k in [0..n], n in [0..15]]; // G. C. Greubel, Aug 08 2022

T[n_, k_]:= Binomial[n+1, (n-k)/2]*(1+(-1)^(n-k))/2;
Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Aug 08 2022 *)

def A117178(n,k): return (1 + (-1)^(n-k))*binomial(n+1, (n-k)//2)/2
flatten([[A117178(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Aug 08 2022

T(n,k) = C(n+1, (n-k)/2) * (1 + (-1)^(n-k))/2.
Column k has e.g.f. Bessel_I(k,2x) + Bessel_I(k+2, 2x).
From G. C. Greubel, Aug 08 2022: (Start)
Sum_{k=0..n} T(n, k) = A058622(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = ((1+(-1)^n)/2) * A001791((n+2)/2).
T(2*n, n) = ((1+(-1)^n)/2) * A052203(n/2).
T(2*n+1, n) = ((1-(-1)^n)/2) * A224274((n+1)/2).
T(2*n-1, n-1) = ((1+(-1)^n)/2) * A224274(n/2). (End)

A014463 Triangular array formed from elements to left of middle of rows of Pascal's triangle that are not 1.

Original entry on oeis.org

3, 4, 5, 10, 6, 15, 7, 21, 35, 8, 28, 56, 9, 36, 84, 126, 10, 45, 120, 210, 11, 55, 165, 330, 462, 12, 66, 220, 495, 792, 13, 78, 286, 715, 1287, 1716, 14, 91, 364, 1001, 2002, 3003, 15, 105, 455, 1365, 3003, 5005, 6435, 16, 120, 560, 1820, 4368, 8008, 11440

Offset: 1

Mohammad K. Azarian

A014462 without the first column.

			3;
4;
5, 10;
6, 15;
7, 21, 35;
8, 28, 56;
9, 36, 84, 126;
10, 45, 120, 210;
11, 55, 165, 330, 462;

Cf. A014411.

More terms from James Sellers

Offset corrected by Mohammad K. Azarian, Nov 19 2008

cp's OEIS Frontend

A034868 Left half of Pascal's triangle.

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A034869 Right half of Pascal's triangle.

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A014413 Triangular array formed from elements to right of middle of Pascal's triangle.

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A265848 Pascal's triangle, right and left halves interchanged.

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A117178 Riordan array (c(x^2)/sqrt(1-4*x^2), x*c(x^2)), c(x) the g.f. of A000108.

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A014463 Triangular array formed from elements to left of middle of rows of Pascal's triangle that are not 1.

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A117178 Riordan array (c(x^2)/sqrt(1-4x^2), xc(x^2)), c(x) the g.f. of A000108.