A034869 -id:A034869 - cp's OEIS frontend

A111417 a(n) = A034869(n) - A008311(n).

0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 0, 0, 0, 35, 0, 0, 0, 0, 0, 0, 0, 0, 0, 126, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 462, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1716, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 6435, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 24310, 0

Offset: 0

Philippe Deléham, Nov 13 2005

a(n) = A001700(m) if n = (m+1)*(m+2) and a(n) = 0 otherwise.

A034868 Left half of Pascal's triangle.

Original entry on oeis.org

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

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

A008311 Triangle of expansions of powers of x in terms of Chebyshev polynomials T_n (x).

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

This triangle is the right half of Pascal's triangle (A007318), but with each number along the center of Pascal's triangle (except the 1 at the top) divided by 2. - Benjamin Schak (schak(AT)math.upenn.edu), Dec 02 2005

For n>=2 found in A002378, a(n)=A034869(n)/2, for all others a(n)=A034869(n). - R. J. Mathar, May 13 2006

			Triangle begins:
1;
-, 1;
1, -, 1;
-, 3, -, 1;
3, -, 4, -, 1;
-, 10, -, 5, -, 1;
...
From _Philippe Deléham_, Mar 09 2013: (Start)
cos(x)      = 1*cos(x),
2*cos(x)^2  = 1 + cos(2x),
4*cos(x)^3  = 3*cos(x) + cos(3x),
8*cos(x)^4  = 3 + 4*cos(2x) + cos(4x),
16*cos(x)^5 = 10*cos(x) + 5*cos(3x) + cos(5x), etc. (End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.

Vincenzo Librandi, Table of n, a(n) for n = 0..5775
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
H. J. Brothers, Pascal's Prism: Supplementary Material.
Index entries for sequences related to Chebyshev polynomials.

With zeros: A100257.

printf("1,") ; for n from 1 to 20 do for j from n mod 2 to n by 2 do if j = 0 then printf("%d,",binomial(n,(n-j)/2)/2) ; else printf("%d,",binomial(n,(n-j)/2)) ; fi ; od ; od ; # R. J. Mathar, May 13 2006

row[n_] := If[n == 0, {1}, Table[If[j == 0, Binomial[n, (n - j)/2]/2, Binomial[n, (n - j)/2]], {j, Mod[n, 2], n, 2}]];
Table[row[n], {n, 0, 15}] // Flatten (* Jean-François Alcover, May 05 2017, after R. J. Mathar *)

Sum_{k, 0<=k}T(n,k)*cos(kx) = 2^(n-1)*cos(x)^n. - Philippe Deléham, Mar 09 2013

Corrected by Philippe Deléham, Nov 12 2005

More terms from R. J. Mathar, May 13 2006

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).

A067802 Triangle read by rows: T(n, k) = binomial(2n+1, n-k)^2(2k+1)/(2n+1).

Original entry on oeis.org

1, 3, 1, 20, 15, 1, 175, 189, 35, 1, 1764, 2352, 720, 63, 1, 19404, 29700, 12375, 1925, 99, 1, 226512, 382239, 196625, 44044, 4212, 143, 1, 2760615, 5010005, 3006003, 869505, 124215, 8085, 195, 1, 34763300, 66745536, 45048640, 15767024, 2998800, 299200, 14144, 255, 1

Offset: 0

Henry Bottomley, Feb 07 2002

			Triangle starts:
  [0]      1
  [1]      3,      1
  [2]     20,     15,      1
  [3]    175,    189,     35,     1
  [4]   1764,   2352,    720,    63,    1
  [5]  19404,  29700,  12375,  1925,   99,   1
  [6] 226512, 382239, 196625, 44044, 4212, 143, 1

First column is A000891.

Cf. A034869, A039599, A002894 (row sums).

T := (n, k) -> binomial(2*n+1, n-k)^2*(2*k+1)/(2*n+1):
seq(seq(T(n, k), k = 0..n), n = 0..8);  # Peter Luschny, Dec 07 2024

T(n, k) = A034869(2n+1, k) * A039599(n, k).

A111505 Right half of Pascal's triangle (A007318) with zeros.

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 0, 3, 1, 0, 0, 6, 4, 1, 0, 0, 0, 10, 5, 1, 0, 0, 0, 20, 15, 6, 1, 0, 0, 0, 0, 35, 21, 7, 1, 0, 0, 0, 0, 70, 56, 28, 8, 1, 0, 0, 0, 0, 0, 126, 84, 36, 9, 1, 0, 0, 0, 0, 0, 252, 210, 120, 45, 10, 1, 0, 0, 0, 0, 0, 0, 462, 330, 165

Offset: 0

Philippe Deléham, Nov 16 2005

A034869 is the version without zeros.

			Triangle begins:
1;
0, 1;
0, 2, 1;
0, 0, 3, 1;
0, 0, 6, 4, 1;
0, 0, 0, 10, 5, 1;
0, 0, 0, 20, 15, 6, 1;
0, 0, 0, 0, 35, 21, 7, 1;
0, 0, 0, 0, 70, 56, 28, 8, 1;
0, 0, 0, 0, 0, 126, 84, 36, 9, 1;
0, 0, 0, 0, 0, 252, 210, 120, 45, 10, 1;
0, 0, 0, 0, 0, 0, 462, 330, 165, 55, 11, 1;
0, 0, 0, 0, 0, 0, 924, 792, 495, 220, 66, 12, 1;
0, 0, 0, 0, 0, 0, 0, 1716, 1287, 715, 286, 78, 13, 1;
0, 0, 0, 0, 0, 0, 0, 3432, 3003, 2002, 1001, 364, 91, 14, 1;

Cf. A007318, A034869, A094527, A111418.

Sum_{n, n>=k} T(n, k) = A001700(k).
Sum_{k =0..2*n} T(2*n, k) = A032443(n).
Sum_{k=0..2*n+1} T(2*n+1, k) = 4^n = A000302(n).
Sum_{k=0..2*n} T(2*n, k)^2 = A036910(n).
Sum_{k=0..2*n+1} T(2*n+1, k)^2 = C(4*n+1, 2*n) = A002458(n) . Paul D. Hanna

cp's OEIS Frontend

A111417 a(n) = A034869(n) - A008311(n).

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A034868 Left 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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A008311 Triangle of expansions of powers of x in terms of Chebyshev polynomials T_n (x).

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

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A067802 Triangle read by rows: T(n, k) = binomial(2*n+1, n-k)^2*(2*k+1)/(2*n+1).

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A111505 Right half of Pascal's triangle (A007318) with zeros.

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A067802 Triangle read by rows: T(n, k) = binomial(2n+1, n-k)^2(2k+1)/(2n+1).