A164899 -id:A164899 - cp's OEIS frontend

A164915 Inverse of binomial matrix (10^n,1) A164899. (See A164899 for companion sequence.)

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

Mark Dols, Aug 31 2009

Alternate sum and difference of diagonal integers generates A164913.

			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;

G. C. Greubel, Antidiagonals n = 1..50, flattened

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

Cf. A128834, A164881, A164899, A164913.

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

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 *)

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

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)

A164925 Array, binomial(j-i,j), read by rising antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, -1, 0, 1, 1, -2, 0, 0, 1, 1, -3, 1, 0, 0, 1, 1, -4, 3, 0, 0, 0, 1, 1, -5, 6, -1, 0, 0, 0, 1, 1, -6, 10, -4, 0, 0, 0, 0, 1, 1, -7, 15, -10, 1, 0, 0, 0, 0, 1, 1, -8, 21, -20, 5, 0, 0, 0, 0, 0, 1, 1, -9, 28, -35, 15, -1, 0, 0, 0, 0, 0, 1, 1, -10, 36, -56, 35, -6, 0, 0, 0, 0, 0, 0, 1

Offset: 0

Mark Dols, Aug 31 2009

Inverse of A052509, or A004070???

			Array, A(n, k), begins as:
  1,  1,  1,   1,  1,   1,  1,  1,  1, ...
  1,  0,  0,   0,  0,   0,  0,  0,  0, ...
  1, -1,  0,   0,  0,   0,  0,  0,  0, ...
  1, -2,  1,   0,  0,   0,  0,  0,  0, ...
  1, -3,  3,  -1,  0,   0,  0,  0,  0, ...
  1, -4,  6,  -4,  1,   0,  0,  0,  0, ...
  1, -5, 10, -10,  5,  -1,  0,  0,  0, ...
  1, -6, 15, -20, 15,  -6,  1,  0,  0, ...
  1, -7, 21, -35, 35, -21,  7, -1,  0, ...
Antidiagonal triangle, T(n, k), begins as:
  1;
  1,  1;
  1,  0,  1;
  1, -1,  0,  1;
  1, -2,  0,  0,  1;
  1, -3,  1,  0,  0,  1;
  1, -4,  3,  0,  0,  0,  1;
  1, -5,  6, -1,  0,  0,  0,  1;
  1, -6, 10, -4,  0,  0,  0,  0,  1;

G. C. Greubel, Antidiagonals n = 0..50, flattened

Cf. A004070, A008346, A010892, A052509, A052992.

Cf. A109466, A130595, A164899, A164915, A164965.

A164925:= func< n,k | k eq 0 or k eq n select 1 else Binomial(2*k-n,k) >;
[A164925(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Feb 10 2023

T[n_, k_]:= If[k==0 || k==n, 1, Binomial[2*k-n, k]];
Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 10 2023 *)

{A(i, j) = if( i<0, 0, if(i==0 || j==0, 1, binomial(j-i, j)))}; /* Michael Somos, Jan 25 2012 */

def A164925(n,k): return 1 if (k==0 or k==n) else binomial(2*k-n, k)
flatten([[A164925(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Feb 10 2023

Sum_{k=0..n} T(n, k) = A164965(n). - Mark Dols, Sep 02 2009
From G. C. Greubel, Feb 10 2023: (Start)
A(n, k) = binomial(k-n, k), with A(0, k) = A(n, 0) = 1 (array).
T(n, k) = binomial(2*k-n, k), with T(n, 0) = T(n, n) = 1 (antidiagonal triangle).
Sum_{k=0..n} (-1)^k*T(n, k) = A008346(n).
Sum_{k=0..n} (-2)^k*T(n, k) = (-1)^n*A052992(n). (End)

Edited by Michael Somos, Jan 26 2012

Offset changed by G. C. Greubel, Feb 10 2023

cp's OEIS Frontend

A164915 Inverse of binomial matrix (10^n,1) A164899. (See A164899 for companion sequence.)

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