A202817 The right-hand half-triangle of A202691.
0, 1, 1, 2, 1, 0, 8, 10, 11, 11, 40, 32, 22, 11, 0, 256, 296, 328, 350, 361, 361
Offset: 1
Examples
Triangle begins: 0 1 1 2 1 0 8 10 11 11 40 32 22 11 0 256 296 328 350 361 361
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
Triangle begins: 0 1 1 2 1 0 8 10 11 11 40 32 22 11 0 256 296 328 350 361 361
Triangle begins:
1;
1, 0;
-2, -3, -3;
-8, -6, -3, 0;
40, 48, 54, 57, 57;
256, 216, 168, 114, 57, 0;
T:= (n,m) -> add(add(4^i*euler(2*i)*binomial(2*k,2*i)*binomial(n-m,2*k-m),i=0..k),k=floor(m/2)..floor(n/2)): seq(seq(T(n,m),m=0..n),n=0..10); # Robert Israel, Apr 08 2015 # Second program, about 100 times faster than the first for the first 100 rows. Triangle := proc(len) local s, A, n, k; A := Array(0..len-1,[1]); lprint(A[0]); for n from 1 to len-1 do if n mod 2 = 1 then s := 0 else s := 2^(3*n+1)*(Zeta(0,-n,1/8)-Zeta(0,-n,5/8)) fi; A[n] := s; for k from n-1 by -1 to 0 do s := s + A[k]; A[k] := s od; lprint(seq(A[k], k=0..n)); od end: Triangle(100); # Peter Luschny, Apr 08 2015
T[n_, m_] := Sum[4^i EulerE[2i] Binomial[2k, 2i] Binomial[n-m, 2k-m], {k, Floor[m/2], n/2}, {i, 0, k}];
Table[T[n, m], {n, 0, 8}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 12 2019 *)
T(n,m):=(sum((sum(4^i*euler(2*i)*binomial(2*k,2*i),i,0,k))*binomial(n-m,2*k-m),k,floor(m/2),n/2));
def triangle(len):
L = [1]; print(L)
for n in range(1,len):
if is_even(n):
s = 2^(3*n+1)*(hurwitz_zeta(-n,1/8)-hurwitz_zeta(-n,5/8))
else: s = 0
L.append(s)
for k in range(n-1,-1,-1):
s = s + L[k]; L[k] = s
print(L)
triangle(7) # Peter Luschny, Apr 08 2015
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