A144463 -id:A144463 - cp's OEIS frontend

A174301 A symmetrical triangle: T(n,k) = binomial(n, k)*if(floor(n/2) greater than or equal to k then 4^k, otherwise 4^(n-k)).

1, 1, 1, 1, 8, 1, 1, 12, 12, 1, 1, 16, 96, 16, 1, 1, 20, 160, 160, 20, 1, 1, 24, 240, 1280, 240, 24, 1, 1, 28, 336, 2240, 2240, 336, 28, 1, 1, 32, 448, 3584, 17920, 3584, 448, 32, 1, 1, 36, 576, 5376, 32256, 32256, 5376, 576, 36, 1, 1, 40, 720, 7680, 53760, 258048, 53760, 7680, 720, 40, 1

Offset: 0

Roger L. Bagula, Mar 15 2010

Row sums are: {1, 2, 10, 26, 130, 362, 1810, 5210, 26050, 76490, ...}.

			Triangle begins:
  1;
  1,  1;
  1,  8,   1;
  1, 12,  12,    1;
  1, 16,  96,   16,     1;
  1, 20, 160,  160,    20,      1;
  1, 24, 240, 1280,   240,     24,     1;
  1, 28, 336, 2240,  2240,    336,    28,    1;
  1, 32, 448, 3584, 17920,   3584,   448,   32,   1;
  1, 36, 576, 5376, 32256,  32256,  5376,  576,  36,  1;
  1, 40, 720, 7680, 53760, 258048, 53760, 7680, 720, 40,  1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened

Cf. A144463, A144470.

T(2n,n) gives A098430.

[[Floor(n/2) ge k select 4^k*Binomial(n,k) else 4^(n-k)*Binomial(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 15 2019

Table[Binomial[n, m]*If[Floor[n/2]>=m , 4^m, 4^(n-m)], {n,0,10}, {m,0,n} ]//Flatten

{T(n,k) = binomial(n,k)*if(floor(n/2)>=k, 4^k, 4^(n-k))}; \\ G. C. Greubel, Apr 15 2019

def T(n,k):
   if floor(n/2)>=k: return 4^k*binomial(n,k)
   else: return 4^(n-k)*binomial(n,k)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 15 2019

T(n, m) = binomial(n, m)*if(floor(n/2) greater than or equal to m then 4^m, otherwise 4^(n-m)).

Edited by G. C. Greubel, Apr 15 2019

A174303 A symmetrical triangle: T(n,k) = A008292(n+1, k) * f(n,k), where f(n,k) = 2^k when floor(n/2) >= k, otherwise 2^(n-k).

Original entry on oeis.org

1, 1, 1, 1, 8, 1, 1, 22, 22, 1, 1, 52, 264, 52, 1, 1, 114, 1208, 1208, 114, 1, 1, 240, 4764, 19328, 4764, 240, 1, 1, 494, 17172, 124952, 124952, 17172, 494, 1, 1, 1004, 58432, 705872, 2499040, 705872, 58432, 1004, 1, 1, 2026, 191360, 3641536, 20965664, 20965664, 3641536, 191360, 2026, 1

Offset: 0

Roger L. Bagula, Mar 15 2010

Row sums are: {1, 2, 10, 46, 370, 2646, 29338, 285238, 4029658, ...}.

			Triangle begins as:
  1;
  1,    1;
  1,    8,     1;
  1,   22,    22,      1;
  1,   52,   264,     52,       1;
  1,  114,  1208,   1208,     114,      1;
  1,  240,  4764,  19328,    4764,    240,     1;
  1,  494, 17172, 124952,  124952,  17172,   494,    1;
  1, 1004, 58432, 705872, 2499040, 705872, 58432, 1004,   1;

G. C. Greubel, Rows n = 0..100 of triangle, flattened
Eric Weisstein's World of Mathematics, Eulerian Number

Cf. A008292, A144463, A144470, A174301.

Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[Floor(n/2) ge k select 2^k*Eulerian(n+1,k) else 2^(n-k)*Eulerian(n+1,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Apr 15 2019

Eulerian[n_, k_]:= Sum[(-1)^j*Binomial[n+1,j]*(k-j+1)^n, {j,0,k+1}];
Table[Eulerian[n+1,m]*If[Floor[n/2] >= m, 2^m, 2^(n-m)], {n,0,10}, {m,0,n} ]//Flatten (* modified by G. C. Greubel, Apr 15 2019 *)

{eulerian(n,k) = sum(j=0,k+1, (-1)^j*binomial(n+1,j)*(k-j+1)^n)};
for(n=0,10, for(k=0,n, print1(eulerian(n+1,k)*if(floor(n/2)>=k, 2^k, 2^(n-k)), ", "))) \\ G. C. Greubel, Apr 15 2019

def Eulerian(n,k): return sum((-1)^j*binomial(n+1,j)*(k-j+1)^n for j in (0..k+1))
def T(n,k):
   if floor(n/2)>=k: return 2^k*Eulerian(n+1,k)
   else: return 2^(n-k)*Eulerian(n+1,k)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Apr 15 2019

T(n,k) = Eulerian(n+1, k)*if(floor(n/2) greater than or equal to k then 2^m otherwise 2^(n-k)), where the Eulerian numbers are defined as A008292(n,k).

Edited by G. C. Greubel, Apr 15 2019

cp's OEIS Frontend

A174301 A symmetrical triangle: T(n,k) = binomial(n, k)*if(floor(n/2) greater than or equal to k then 4^k, otherwise 4^(n-k)).

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