A008314 -id:A008314 - cp's OEIS frontend

A100071 a(n) = n * binomial(n-1, floor((n-1)/2)) = n * max_{i=0..n} binomial(n-1, i).

0, 1, 2, 6, 12, 30, 60, 140, 280, 630, 1260, 2772, 5544, 12012, 24024, 51480, 102960, 218790, 437580, 923780, 1847560, 3879876, 7759752, 16224936, 32449872, 67603900, 135207800, 280816200, 561632400, 1163381400, 2326762800

Offset: 0

Paul Barry, Nov 02 2004

Old name: An inverse Chebyshev transform of n.
Hankel transform is (-1)^n*n*2^(n-1), A085750. This is the inverse binomial transform of -n. - Paul Barry, Jan 11 2007
Corollary 3 of the Farhi reference mentions this sequence. - Roger L. Bagula, Nov 08 2009
Number of UDUD's in all length n+3 left factors of Dyck paths (here U=(1,1) and D=(1,-1)). Example: a(2)=2 because in (UDUD)U, UDUUD, UDUUU, UUDDU, U(UDUD), UUDUU, UUUDD, UUUDU, UUUUD, and UUUUU we have a total of two UDUDs (shown between parentheses).  Also number of UUDD's in all length n+3 left factors of Dyck paths (here U=(1,1) and D=(1,-1)). Example: a(2)=2 because in UDUDU, UDUUD, UDUUU, (UUDD)U, UUDUD, UUDUU, U(UUDD), UUUDU, UUUUD, and UUUUU we have a total of two UUDDs (shown between parentheses). - Emeric Deutsch, Jun 19 2011
Apparently the number of long ascents in all symmetric Dyck (n+1)-paths. - David Scambler, Aug 17 2012
Beginning with the least positive term multiple of an odd prime p (which is a(p)), we have exactly p+1 consecutive terms multiple of p. - Vladimir Shevelev, Aug 17 2012
Apparently also the count of 'unmatched symbols' in the binary strings of length n (see A008314). - Wouter Meeussen, May 26 2013

T. D. Noe, Table of n, a(n) for n = 0..1000
Ruggero Bandiera and Florian Schaetz, Eulerian idempotent, pre-Lie logarithm and combinatorics of trees, arXiv:1702.08907 [math.CO], 2017. See p. 34.
F. Disanto, A. Frosini, and S. Rinaldi, Square involutions, J. Int. Seq., Vol. 14 (2011), Article 11.3.5.
Bakir Farhi, An identity involving the least common multiple of binomial coefficients and its application, The American Mathematical Monthly, Vol. 116, No. 9 (2009), pp. 836-839, see p. 838; arXiv preprint, arXiv:0906.2295 [math.NT], 2009.
Nikita Gogin and Mika Hirvensalo, On the Moments of Squared Binomial Coefficients, (2020).
D. Nečas, Elementary models of low-pressure plasma polymerisation into nanofibrous mats, Phys. Scr. (2025) Vol. 100, 055601. See p. 10.
Helmut Prodinger, Dispersed Dyck paths revisited, arXiv:2402.13026 [math.CO], 2024.

Cf. A134757, A008314.

[n*Binomial(n-1, Floor((n-1)/2)): n in [0..35]]; // Vincenzo Librandi, Sep 14 2015

swing := n -> n!/iquo(n,2)!^2:
A100071 := n -> swing(n)*(n/2)^(n-1 mod 2):
seq(A100071(i),i=0..30); # Peter Luschny, Aug 31 2011

Table[(Floor[n/2] + Ceiling[n/2] + 1)!/(Floor[n/2]!*Ceiling[n/2]!), {n, 1, 40}] (* Stefan Steinerberger, Nov 04 2008 *)
Table[If[n == 0, 0, n*Binomial[n - 1, Floor[(n - 1)/2]]], {n, 0, 30}] (* Roger L. Bagula, Nov 08 2009 *);
Table[ Tr[ Table[Count[match[-1 + 2*IntegerDigits[n, 2, k]], 0], {n, 2^(k - 1), 2^k - 1}]], {k, 16}] (* function 'match' see A008314; Wouter Meeussen, May 26 2013 *)

a(n) = n * binomial(n-1, (n-1)\2); \\ Michel Marcus, Sep 14 2015

def A100071(n):
    f = factorial(n)/factorial(n//2)^2
    return f if is_odd(n) else f*(n/2)
[A100071(n) for n in (0..50)]  # Peter Luschny, Aug 17 2012

G.f.: 2*x*(1 - sqrt(1 - 4*x^2))/(sqrt(1 - 4*x^2)*(sqrt(1 - 4*x^2) + 2*x - 1)^2).
G.f.: (1/sqrt(1 - 4*x^2))*x*c(x^2)/(1 - x*c(x^2))^2.
a(n) = Sum_{k = 0..floor(n/2)} binomial(n,k)*(n - 2*k).
Sum_{k = 0..floor(n/2)} binomial(n-k,k)*(-1)^k*a(n-2k) = 1.
From Paul Barry, Jan 11 2007: (Start)
a(n) = n*binomial(n-1, floor((n-1)/2));
a(n) = Sum_{k = 0..n} binomial(n,k)*2^(n-k)*binomial(2*k-2, k-1)*(-1)^(k-1). (End)
Starting (1, 2, 6, 12, ...), = inverse binomial transform of A134757: (1, 3, 11, 37, 123, 401, ...). - Gary W. Adamson, Nov 08 2007
a(n) = a(n-1)*n/floor(n/2) for n > 0. - Reinhard Zumkeller, Jan 20 2008
G.f.: x/((1 - 2*x)*sqrt(1 - 4*x^2)). - Paul Barry, Apr 25 2008
a(n) = (floor(n/2) + ceiling(n/2) + 1)!/(floor(n/2)! * ceiling(n/2)!). - Stefan Steinerberger, Nov 04 2008
a(n) = A056040(n)*(n/2)^((n-1) mod 2). - Peter Luschny, Aug 31 2011
Asymptotic: a(n) ~ b(n) where b(n) = ceiling(2^(n-1)*sqrt(2*n-(-1)^n)/sqrt(Pi)). b(n) is also a lower bound of a(n) and an upper bound of 2^(n-1). With corollary 3 from Bakir Farhi (see reference) lcm(1,2,...,n) >= a(n) >= b(n) >= 2^(n-1). - Peter Luschny, Aug 17 2012
a(n) = n for n < 3, a(n) = 4*a(n-2) + 2*a(n-1)/(n-1) for n >= 3. - Alexander R. Povolotsky, Aug 17 2012
E.g.f.: x*(BesselI(0,2*x) + BesselI(1,2*x)). - Peter Luschny, Aug 19 2012
a(n) = (-1)^(n*(n+1)/2) * Sum_{k = 0..n} (-1)^k*k*binomial(n,k)^2. - Peter Bala, Jul 25 2016
a(n) = n!/(floor((n-1)/2)!*ceiling((n-1)/2)!). See the Bandiera link. - Michel Marcus, Feb 28 2017
D-finite with recurrence (-n+1)*a(n) + 2*a(n-1) + 4*(n-1)*a(n-2) = 0. - R. J. Mathar, Aug 09 2017
From Amiram Eldar, Mar 10 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi/sqrt(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(3*sqrt(3)). (End)

Name changed, using part of a comment from Paul Barry, by Peter Luschny, Aug 17 2012

A122366 Triangle read by rows: T(n,k) = binomial(2*n+1,k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 5, 10, 1, 7, 21, 35, 1, 9, 36, 84, 126, 1, 11, 55, 165, 330, 462, 1, 13, 78, 286, 715, 1287, 1716, 1, 15, 105, 455, 1365, 3003, 5005, 6435, 1, 17, 136, 680, 2380, 6188, 12376, 19448, 24310, 1, 19, 171, 969, 3876, 11628, 27132, 50388, 75582, 92378, 1, 21

Offset: 0

Reinhard Zumkeller, Aug 30 2006

Sum of n-th row = A000302(n) = 4^n.
Central terms give A052203.
Reversal of A111418. - Philippe Deléham, Mar 22 2007
Coefficient triangle for the expansion of one half of odd powers of 2*x in terms of Chebyshev's T-polynomials: ((2*x)^(2*n+1))/2 = Sum_{k=0..n} a(n,k)*T(2*(n-k)+1,x) with Chebyshev's T-polynomials. See A053120. - Wolfdieter Lang, Mar 07 2007
The signed triangle T(n,k)*(-1)^(n-k) appears in the formula ((2*sin(phi))^(2*n+1))/2 = Sum_{k=0..n} ((-1)^(n-k))*a(n,k)*sin((2*(n-k)+1)*phi). - Wolfdieter Lang, Mar 07 2007
The signed triangle T(n,k)*(-1)^(n-k) appears therefore in the formula (4-x^2)^n = Sum_{k=0..n} ((-1)^(n-k))*a(n,k)*S(2*(n-k),x) with Chebyshev's S-polynomials. See A049310 for S(n,x). - Wolfdieter Lang, Mar 07 2007
From Wolfdieter Lang, Sep 18 2012: (Start)
The triangle T(n,k) appears also in the formula F(2*l+1)^(2*n+1) = (1/5^n)*Sum_{k=0..n} T(n,k)*F((2*(n-k)+1)*(2*l+1)), l >= 0, n >= 0, with F=A000045 (Fibonacci).
The signed triangle Ts(n,k):=T(n,k)*(-1)^k appears also in the formula
  F(2*l)^(2*n+1) = (1/5^n)*Sum_{k=0..n} Ts(n,k)*F((2(n-k)+1)*2*l), l >= 0, n >= 0, with F=A000045 (Fibonacci).
This is Lemma 2 of the K. Ozeki reference, p. 108, written for odd and even indices separately.
(End)

			.......... / 1 \ .......... =A062344(0,0)=A034868(0,0),
......... / 1 . \ ......... =T(0,0)=A034868(1,0),
........ / 1 2 . \ ........ =A062344(1,0..1)=A034868(2,0..1),
....... / 1 3 ... \ ....... =T(1,0..1)=A034868(3,0..1),
...... / 1 4 6 ... \ ...... =A062344(2,0..2)=A034868(4,0..2),
..... / 1 5 10 .... \ ..... =T(2,0..2)=A034868(5,0..2),
.... / 1 6 15 20 ... \ .... =A062344(3,0..3)=A034868(6,0..3),
... / 1 7 21 35 ..... \ ... =T(3,0..3)=A034868(7,0..3),
.. / 1 8 28 56 70 .... \ .. =A062344(4,0..4)=A034868(8,0..4),
. / 1 9 36 84 126 ..... \ . =T(4,0..4)=A034868(9,0..4).
Row n=2:[1,5,10] appears in the expansion ((2*x)^5)/2 = T(5,x)+5*T(3,x)+10*T(1,x).
Row n=2:[1,5,10] appears in the expansion ((2*cos(phi))^5)/2 = cos(5*phi)+5*cos(3*phi)+10*cos(1*phi).
The signed row n=2:[1,-5,10] appears in the expansion ((2*sin(phi))^5)/2 = sin(5*phi)-5*sin(3*phi)+10*sin(phi).
The signed row n=2:[1,-5,10] appears therefore in the expansion (4-x^2)^2 = S(4,x)-5*S(2,x)+10*S(0,x).
Triangle T(n,k) starts:
  n\k 0  1   2   3    4     5     6     7     8     9  ...
  0   1
  1   1  3
  2   1  5  10
  3   1  7  21  35
  4   1  9  36  84  126
  5   1 11  55 165  330   462
  6   1 13  78 286  715  1287  1716
  7   1 15 105 455 1365  3003  5005  6435
  8   1 17 136 680 2380  6188 12376 19448 24310
  9   1 19 171 969 3876 11628 27132 50388 75582 92378
  ...  - _Wolfdieter Lang_, Sep 18 2012
Row n=2, with F(n)=A000045(n) (Fibonacci number), l >= 0, see a comment above:
F(2*l)^5   = (1*F(10*l) - 5*F(6*l) + 10*F(2*l))/25,
F(2*l+1)^5 = (1*F(10*l+5) + 5*F(6*l+3) + 10*F(2*l+1))/25.
- _Wolfdieter Lang_, Sep 19 2012

T. J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2nd ed., Wiley, New York, 1990, pp. 54-55, Ex. 1.5.31.

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
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].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 795.
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
K. Ozeki, On Melham's sum, The Fibonacci Quart. 46/47 (2008/2009), no. 2, 107-110.
Index entries for sequences related to Chebyshev polynomials.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A062344.

Odd numbered rows of A008314. Even numbered rows of A008314 are A127673.

a122366 n k = a122366_tabl !! n !! k
a122366_row n = a122366_tabl !! n
a122366_tabl = f 1 a007318_tabl where
   f x (_:bs:pss) = (take x bs) : f (x + 1) pss
-- Reinhard Zumkeller, Mar 14 2014

T[_, 0] = 1;
T[n_, k_] := T[n, k] = T[n-1, k-1] 2n(2n+1)/(k(2n-k+1));
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 01 2021 *)

T(n,0)=1; T(n,k) = T(n-1,k-1)*2*n*(2*n+1)/(k*(2*n-k+1)) for k > 0.
T(n,0)=1; for n > 0: T(n,1)=n+2; for n > 1: T(n,n) = T(n-1,n-2) + 3*T(n-1,n-1), T(n,k) = T(n-1,k-2) + 2*T(n-1,k-1) + T(n-1,k), 1 < k < n.
T(n,n) = A001700(n).
T(n,k) = A034868(2*n+1,k) = A007318(2*n+1,k), 0 <= k <= n;
G.f.: (2*y)/((y-1)*sqrt(1-4*x*y)-4*x*y^2+(1-4*x)*y+1). - Vladimir Kruchinin, Oct 30 2020

Chebyshev and trigonometric comments from Wolfdieter Lang, Mar 07 2007.

Typo in comments fixed, thanks to Philippe Deléham, who indicated this.

A014462 Triangular array formed from elements to left of middle of Pascal's triangle.

Original entry on oeis.org

1, 1, 1, 3, 1, 4, 1, 5, 10, 1, 6, 15, 1, 7, 21, 35, 1, 8, 28, 56, 1, 9, 36, 84, 126, 1, 10, 45, 120, 210, 1, 11, 55, 165, 330, 462, 1, 12, 66, 220, 495, 792, 1, 13, 78, 286, 715, 1287, 1716, 1, 14, 91, 364, 1001, 2002, 3003, 1, 15, 105, 455, 1365, 3003, 5005, 6435, 1, 16

Offset: 1

Mohammad K. Azarian

Coefficients for Pontryagin classes of projective spaces. See p. 3 of the Wilson link. Aerated to become a lower triangular matrix with alternating zeros on the diagonal, this matrix appparently becomes the reverse, or mirror, of A117178. - Tom Copeland, May 30 2017

			Array begins:
  1;
  1;
  1,  3;
  1,  4;
  1,  5, 10;
  1,  6, 15;
  1,  7, 21,  35;
  1,  8, 28,  56;
  1,  9, 36,  84, 126;
  1, 10, 45, 120, 210;
  1, 11, 55, 165, 330, 462;

Reinhard Zumkeller, Rows n = 1..200 of triangle, flattened
Jon Maiga, Computer-generated formulas for A014462, Sequence Machine.
D. Wilson, Genera: From cobordism to K-theory, Jul 2016.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A014413, A034868, A058622 (row sums).
Cf. A001791 (a half-diagonal and diagonal sums).
Cf. A007318, A000037, A008314.
Cf. A117178.

a014462 n k = a014462_tabf !! (n-1) !! (k-1)
a014462_row n = a014462_tabf !! (n-1)
a014462_tabf = map reverse a014413_tabf
-- Reinhard Zumkeller, Dec 24 2015

More terms from James Sellers

A127673 One half of even powers of 2*x in terms of Chebyshev's T-polynomials.

Original entry on oeis.org

1, 1, 1, 1, 4, 3, 1, 6, 15, 10, 1, 8, 28, 56, 35, 1, 10, 45, 120, 210, 126, 1, 12, 66, 220, 495, 792, 462, 1, 14, 91, 364, 1001, 2002, 3003, 1716, 1, 16, 120, 560, 1820, 4368, 8008, 11440, 6435, 1, 18, 153, 816, 3060, 8568, 18564, 31824, 43758, 24310, 1, 20, 190

Offset: 0

Wolfdieter Lang, Mar 07 2007

See A122366 for one half of odd powers of 2*x in terms of Chebyshev's T-polynomials.
This is, for n >= 1, the left half of Pascal's triangle for even rows with the central coefficients divided by 2.
The signed version of this triangle, b(n,k) := a(n,k)*(-1)^(n-k), appears in the formula (1/2)*(2*sin(phi))^(2*n) = (Sum_{k=0..n-1} b(n,k)*cos(2*(n-k)*phi)) + a(n,n).
Correspondingly, (1/2)*(4*(1-x^2))^n = (Sum_{k=0..n-1} b(n,k)*T(2*(n-k),x)) + a(n,n).
The proofs follow from Euler's formula 2*x = 2*cos(phi) = exp(i*phi) + exp(-i*phi) or 2*sqrt(1-x^2) = 2*sin(phi) = (exp(i*phi) - exp(-i*phi))/i and the binomial formula.

			[1/2];
[ 1, 1];
[ 1, 4,  3];
[ 1, 6, 15, 10];
[ 1, 8, 28, 56, 35];
...
Row n=3: [1, 6, 15, 20/2 = 10] appears in ((2*x)^6)/2 = 1*T(6,x) + 6*T(4,x) + 15*T(2,x) + 10.
Row n=3: [1, 6, 15, 20/2 = 10] appears in ((2*cos(phi))^6)/2 = 1*cos(6*phi) + 6*cos(4*phi) + 15*cos(2*phi) + 10.
The signed row n=3, [-1, 6, -15, +20/2 = 10], appears in ((4*(1-x^2))^3)/2 = -1*T(6,x) + 6*T(4,x) - 15*T(2,x) + 10).
The signed row n=3, [-1, 6, -15, +20/2 = 10], appears in ((2*sin(phi))^6)/2 = -1*cos(6*phi) + 6*cos(4*phi) - 15*cos(2*phi) + 10.

Theodore J. Rivlin, Chebyshev polynomials: from approximation theory to algebra and number theory, 2. ed., Wiley, New York, 1990. pp. 54-55, Ex. 1.5.31.

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].
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, p. 795.
Index entries for sequences related to Chebyshev polynomials.
Index entries for triangles and arrays related to Pascal's triangle

Cf. A122366, A008314, A008311.

a(n,k) = binomial(2*n,k), k=0..n-1 and a(n,n) = binomial(2*n,n)/2, n >= 1. Instead of a(0,0)=1 one should take 1/2.

A286030 Irregular triangle T(n,k) read by rows: Let S be a 3-member set of integers {f,g,h} where f >= g >= h >= 0 and f+g+h = n. Let S(n,k) be an irregular triangle composed of all S listed in reverse lexicographic order by row n. Then T(n,k) = n!Q/(3f!g!h!), where Q is the number of permutations of S(n,k). (See "Comments" and "Examples" for additional explanation.)

Original entry on oeis.org

1, 1, 2, 1, 6, 2, 1, 8, 6, 12, 1, 10, 20, 20, 30, 1, 12, 30, 30, 20, 120, 30, 1, 14, 42, 42, 70, 210, 140, 210, 1, 16, 56, 56, 112, 336, 70, 560, 420, 560, 1, 18, 72, 72, 168, 504, 252, 1008, 756, 630, 2520, 560

Offset: 1

Bob Selcoe, Apr 30 2017

See "Example" below for the starting construction of S(n,k) and T(n,k).
To understand S(n,k) and Q, consider example S(5,k), i.e., f+g+h = 5, and S(n,k) are listed in reverse lexicographic order. So S(5,k) = {5,0,0}, {4,1,0}, {3,2,0}, {3,1,1}, {2,2,1} k=1..5, respectively. Q is the number of permutations of S(n,k). So Q=3 when S(n,k) = {5,0,0}, {3,1,1} and {2,2,1}; and Q=6 when S(n,k) = {4,1,0} and {3,2,0}.
In general, by definition: Q=1 when all members of S(n,k) are equal, Q=3 when S(n,k) contains a pair, and Q=6 when none of the members of S(n,k) is equal.
Suppose three equally-matched players are playing a tournament of n games; and for each game there is one winner and two losers. Then S(n,k) is the "overall win record" (where player order does not matter) after n games. Let p be the probability that any S(n,k) occurs after n games. Then p = T(n,k)/3^(n-1). (See also "Example" section.)
Generally, when S(n,k) is a z-member set {f,g,h,i..,y}, then Q is the number of permutations of S(n,k), T(n,k) = n!*Q/(z*f!*g!*h!..*y!) and p = T(n,k)/z^(n-1). So when z=2 we get A008314. (Observation prompted by query from Linda Rogers.)
For triangle T(n,k):
Row sums are 3^(n-1).
Row lengths are A001399(n).
Final terms in each row are A199127(n).
For n >= 3: T(n,2) = 2*n.
For n >= 5: T(n,3) = T(n,4) = A002378(n-1) (oblong numbers).
For n >= 6: T(n,6) = A007531(n).
For n >= 8: T(n,9) = A033487(n-3).

			Triangle T(n,k) begins:
n/k 1    2    3    4    5   6    7     8    9    10    11    12    13    14
1:  1
2:  1,   2
3:  1,   6,   2
4:  1,   8,   6,  12
5:  1,  10,  20,  20,  30
6:  1,  12,  30,  30,  20, 120,  30
7:  1,  14,  42,  42,  70, 210, 140,  210
8:  1,  16,  56,  56, 112, 336,  70,  560,  420, 560
9:  1,  18,  72,  72, 168, 504, 252, 1008,  756, 630, 2520, 560
10: 1,  20,  90,  90, 240, 720, 420, 1680, 1260, 252, 2520, 5040, 3150, 4200
Triangle S(n,k) begins:
n/k    1        2        3        4        5        6        7
1:  {1,0,0}
2:  {2,0,0}  {1,1,0}
3:  {3,0,0}  {2,1,0}  {1,1,1}
4:  {4,0,0}  {3,1,0}  {2,2,0}  {2,1,1}
5:  {5,0,0}  {4,1,0}  {3,2,0}  {3,1,1}  {2,2,1}
6:  {6,0,0}  {5,1,0}  {4,2,0}  {4,1,1}  {3,3,0}  {3,2,1}  {2,2,2}
T(4,3) = 6 because n=4 and S(4,3) = {2,2,0}; so Q=3 and 3*4!/(3*2!*2!*0!) = 6. Therefore p = 6/27 = 2/9 that the overall win record = {2,2,0} after playing 4 tournament games.

Nicolas Behr, Pawel Sobocinski, Rule Algebras for Adhesive Categories, arXiv:1807.00785 [cs.LO], 2018, also LIPIcs 27th EACSL Annual Conference on Computer Science Logic (CSL 2018), Vol. 119, pp. 11:1-11:21.

Cf. A001399, A002378, A007531, A008314, A033487, A199127.

Cf. A000041 (partition numbers).

cp's OEIS Frontend

A100071 a(n) = n * binomial(n-1, floor((n-1)/2)) = n * max_{i=0..n} binomial(n-1, i).

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A122366 Triangle read by rows: T(n,k) = binomial(2*n+1,k), 0 <= k <= n.

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A014462 Triangular array formed from elements to left of middle of Pascal's triangle.

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A127673 One half of even powers of 2*x in terms of Chebyshev's T-polynomials.

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