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A213584
Rectangular array: (row n) = b**c, where b(h) = F(h+1), c(h) = n-1+h, where F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.
5
1, 4, 2, 10, 7, 3, 21, 16, 10, 4, 40, 32, 22, 13, 5, 72, 59, 43, 28, 16, 6, 125, 104, 78, 54, 34, 19, 7, 212, 178, 136, 97, 65, 40, 22, 8, 354, 299, 231, 168, 116, 76, 46, 25, 9, 585, 496, 386, 284, 200, 135, 87, 52, 28, 10, 960, 816, 638, 473, 337, 232, 154, 98, 58, 31, 11
OFFSET
1,2
COMMENTS
Principal diagonal: A213585.
Antidiagonal sums: A213586.
Row 1, (1,2,3,5,...)**(1,2,3,4,...): A001891.
Row 2, (1,2,3,5,...)**(2,3,4,5,...): A023550.
Row 3, (1,2,3,5,...)**(3,4,5,6,...): A023554.
For a guide to related arrays, see A213500.
LINKS
Clark Kimberling, Antidiagonals n = 1..60
FORMULA
T(n,k) = 3*T(n,k-1) - 2*T(n,k-2) - T(n,k-3) + T(n,k-4).
G.f. for row n: f(x)/g(x), where f(x) = n + x - (n - 1)*x and g(x) = (1 - x - x^2)*(1 - x)^2.
T(n, k) = Fibonacci(k+4) + n*Fibonacci(k+3) - 2*(n+k) - 3. - G. C. Greubel, Jul 08 2019
EXAMPLE
Northwest corner (the array is read by falling antidiagonals):
1...4....10...21...40....72
2...7....16...32...59....104
3...10...22...43...78....136
4...13...28...54...97....168
5...16...34...65...116...200
6...19...40...76...135...232
MATHEMATICA
(* First program *)
b[n_]:= Fibonacci[n+1]; c[n_]:= n;
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}]] (* A213584 *)
r[n_]:= Table[T[n, k], {k, 40}] (* columns of antidiagonal triangle *)
d = Table[T[n, n], {n, 1, 40}] (* A213585 *)
s[n_]:= Sum[T[i, n+1-i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A213586 *)
(* Second program *)
Table[Fibonacci[n-k+5] + k*Fibonacci[n-k+4] -2*n-5, {n, 12}, {k, n}]//Flatten (* G. C. Greubel, Jul 08 2019 *)
PROG
(PARI) t(n, k) = fibonacci(n-k+5) + k*fibonacci(n-k+4) -(2*n+5);
for(n=1, 12, for(k=1, n, print1(t(n, k), ", "))) \\ G. C. Greubel, Jul 08 2019
(Magma) [[Fibonacci(n-k+5) + k*Fibonacci(n-k+4) -(2*n+5): k in [1..n]]: n in [1..12]]; // G. C. Greubel, Jul 08 2019
(Sage) [[fibonacci(n-k+5) + k*fibonacci(n-k+4) -(2*n+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-> Fibonacci(n-k+5) + k*Fibonacci(n-k+4) -(2*n+5)))) # G. C. Greubel, Jul 08 2019
CROSSREFS
KEYWORD
nonn,tabl,easy
AUTHOR
Clark Kimberling, Jun 18 2012
STATUS
approved