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Inverse of binomial matrix (10^n,1) A164899. (See A164899 for companion sequence.)
2

%I #21 Mar 29 2023 17:43:15

%S 1,1,10,1,9,100,1,8,90,1000,1,7,81,900,10000,1,6,73,810,9000,100000,1,

%T 5,66,729,8100,90000,1000000,1,4,60,656,7290,81000,900000,10000000,1,

%U 3,55,590,6561,72900,810000,9000000,100000000

%N Inverse of binomial matrix (10^n,1) A164899. (See A164899 for companion sequence.)

%C Alternate sum and difference of diagonal integers generates A164913.

%H G. C. Greubel, <a href="/A164915/b164915.txt">Antidiagonals n = 1..50, flattened</a>

%F From _G. C. Greubel_, Feb 10 2023: (Start)

%F 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).

%F 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).

%F Sum_{k=1..n} T(n, k) = (1/273)*(3*10^(n+1) - 15*A057079(n+1) - 12*A128834(n)).

%F Sum_{k=1..n} (-1)^(k-1)*T(n, k) = (1/109)*(4*Fibonacci(n) + 5*LucasL(n) + (-10)^(n+1)). (End)

%e Matrix array, A(n, k), begins:

%e 1, 10, 100, 1000, 10000, 100000, ...

%e 1, 9, 90, 900, 9000, 90000, ...

%e 1, 8, 81, 810, 8100, 81000, ...

%e 1, 7, 73, 729, 7290, 72900, ...

%e 1, 6, 66, 656, 6561, 65610, ...

%e 1, 5, 60, 590, 5905, 59049, ...

%e 1, 4, 55, 530, 5315, 53144, ...

%e Antidiagonal triangle, T(n, k), begins as:

%e 1;

%e 1, 10;

%e 1, 9, 100;

%e 1, 8, 90, 1000;

%e 1, 7, 81, 900, 10000;

%e 1, 6, 73, 810, 9000, 100000;

%e 1, 5, 66, 729, 8100, 90000, 1000000;

%t 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]]];

%t Table[T[n, k], {n,15}, {k,n}]//Flatten (* _G. C. Greubel_, Feb 10 2023 *)

%o (Magma)

%o function T(n,k) // T = A164915

%o if k eq n then return 10^(n-1);

%o elif k eq 1 then return 1;

%o else return T(n-1,k) - T(n-2,k-1);

%o end if; return T;

%o end function;

%o [T(n,k): k in [1..n], n in [1..15]]; // _G. C. Greubel_, Feb 10 2023

%o (SageMath)

%o def T(n,k): # T = A164915

%o if (k==n): return 10^(n-1)

%o elif (k==1): return 1

%o else: return T(n-1,k) - T(n-2,k-1)

%o flatten([[T(n,k) for k in range(1,n+1)] for n in range(1,16)]) # _G. C. Greubel_, Feb 10 2023

%Y Cf. A000032, A000045, A001019, A007318, A057079.

%Y Cf. A128834, A164881, A164899, A164913.

%K sign,tabl

%O 1,3

%A _Mark Dols_, Aug 31 2009