OFFSET
1,2
COMMENTS
LINKS
Clark Kimberling, Antidiagonals n = 1..60, flattened
FORMULA
Rows: T(n,k) = 2*T(n,k-1) + T(n,k-2) - 2*T(n,k-3) - T(n,k-4).
Columns: T(n,k) = T(n-1,k) + T(n-2,k).
G.f. for row n: f(x)/g(x), where f(x) = F(n+1) + F(n+2)*x + F(n)*x^2 and g(x) = (1 - x - x^2)^2.
T(n, k) = (k*Lucas(n+k+2) - Fibonacci(k)*Lucas(n-1))/5. - G. C. Greubel, Jul 08 2019
EXAMPLE
Northwest corner (the array is read by falling antidiagonals):
1....4....10....22....45....88....167
2....7....17....37....75....146...276
3....11...27....59....120...234...443
5....18...44....96....195...380...719
8....29...71....155...315...614...1162
13...47...115...251...510...994...1881
MATHEMATICA
(* First program *)
b[n_]:= Fibonacci[n+1]; c[n_]:= Fibonacci[n+1];
T[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
TableForm[Table[T[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[T[n-k+1, k], {n, 12}, {k, n, 1, -1}]] (* A213587 *)
r[n_]:= Table[T[n, k], {k, 40}] (* columns of antidiagonal triangle *)
Table[T[n, n], {n, 1, 40}] (* A213588 *)
s[n_]:= Sum[T[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A213589 *)
(* Second program *)
Table[((n-k+1)*LucasL[n+3] - Fibonacci[n-k+1]*LucasL[k-1])/5, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Jul 08 2019 *)
PROG
(PARI) lucas(n) = fibonacci(n+1) + fibonacci(n-1);
t(n, k) = ((n-k+1)*lucas(n+3) - fibonacci(n-k+1)*lucas(k-1))/5;
for(n=1, 12, for(k=1, n, print1(t(n, k), ", "))) \\ G. C. Greubel, Jul 08 2019
(Magma) [[((n-k+1)*Lucas(n+3) - Fibonacci(n-k+1)*Lucas(k-1))/5: k in [1..n]]: n in [1..12]]; // G. C. Greubel, Jul 08 2019
(Sage) [[((n-k+1)*lucas_number2(n+3, 1, -1) - fibonacci(n-k+1)* lucas_number2(k-1, 1, -1))/5 for k in (1..n)] for n in (1..12)] # G. C. Greubel, Jul 08 2019
(GAP) Flat( List([1..12], n-> List([1..n], k-> ((n-k+1)*Lucas(1, -1, n+3)[2] - Fibonacci(n-k+1)*Lucas(1, -1, k-1)[2])/5 ))); # G. C. Greubel, Jul 08 2019
CROSSREFS
KEYWORD
AUTHOR
Clark Kimberling, Jun 19 2012
STATUS
approved