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A164915
Inverse of binomial matrix (10^n,1) A164899. (See A164899 for companion sequence.)
2
1, 1, 10, 1, 9, 100, 1, 8, 90, 1000, 1, 7, 81, 900, 10000, 1, 6, 73, 810, 9000, 100000, 1, 5, 66, 729, 8100, 90000, 1000000, 1, 4, 60, 656, 7290, 81000, 900000, 10000000, 1, 3, 55, 590, 6561, 72900, 810000, 9000000, 100000000
OFFSET
1,3
COMMENTS
Alternate sum and difference of diagonal integers generates A164913.
LINKS
FORMULA
From G. C. Greubel, Feb 10 2023: (Start)
A(n, k) = A(n-1, k) - A(n-1, k-1), with A(n, 1) = 1 and A(1, k) = 10^(k-1) (array).
T(n, k) = T(n-1, k) - T(n-2, k-1), with T(n, 1) = 1 and T(n, n) = 10^(n-1) (antidiagonal triangle).
Sum_{k=1..n} T(n, k) = (1/273)*(3*10^(n+1) - 15*A057079(n+1) - 12*A128834(n)).
Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (1/109)*(4*Fibonacci(n) + 5*LucasL(n) + (-10)^(n+1)). (End)
EXAMPLE
Matrix array, A(n, k), begins:
1, 10, 100, 1000, 10000, 100000, ...
1, 9, 90, 900, 9000, 90000, ...
1, 8, 81, 810, 8100, 81000, ...
1, 7, 73, 729, 7290, 72900, ...
1, 6, 66, 656, 6561, 65610, ...
1, 5, 60, 590, 5905, 59049, ...
1, 4, 55, 530, 5315, 53144, ...
Antidiagonal triangle, T(n, k), begins as:
1;
1, 10;
1, 9, 100;
1, 8, 90, 1000;
1, 7, 81, 900, 10000;
1, 6, 73, 810, 9000, 100000;
1, 5, 66, 729, 8100, 90000, 1000000;
MATHEMATICA
T[n_, k_]:= T[n, k]= If[k==n, 10^(n-1), If[k==1, 1, T[n-1, k] - T[n-2, k -1]]];
Table[T[n, k], {n, 15}, {k, n}]//Flatten (* G. C. Greubel, Feb 10 2023 *)
PROG
(Magma)
function T(n, k) // T = A164915
if k eq n then return 10^(n-1);
elif k eq 1 then return 1;
else return T(n-1, k) - T(n-2, k-1);
end if; return T;
end function;
[T(n, k): k in [1..n], n in [1..15]]; // G. C. Greubel, Feb 10 2023
(SageMath)
def T(n, k): # T = A164915
if (k==n): return 10^(n-1)
elif (k==1): return 1
else: return T(n-1, k) - T(n-2, k-1)
flatten([[T(n, k) for k in range(1, n+1)] for n in range(1, 16)]) # G. C. Greubel, Feb 10 2023
KEYWORD
sign,tabl
AUTHOR
Mark Dols, Aug 31 2009
STATUS
approved