A094704 -id:A094704 - cp's OEIS frontend

A165155 a(n) = 100*a(n-1) + 11^(n-1) for n>0, a(0)=0.

0, 1, 111, 11221, 1123431, 112357741, 11235935151, 1123595286661, 112359548153271, 11235955029685981, 1123595505326545791, 112359550558592003701, 11235955056144512040711, 1123595505617589632447821, 112359550561793485956926031, 11235955056179728345526186341

Offset: 0

Mark Dols, Sep 05 2009

Generalization of A000225. - Mark Dols, Jan 28 2010

			From _Mark Dols_, Jan 28 2010: (Start)
As triangle:
  ........... 1
  ......... 1 1 1
  ....... 1 1 2 2 1
  ..... 1 1 2 3 4 3 1
  ... 1 1 2 3 5 7 7 4 1
  . 1 1 2 3 5 9 3 5 1 5 1
  1 1 2 3 5 9 5 2 8 6 6 6 1
(Mirrored version of A162741) (End)

G. C. Greubel, Table of n, a(n) for n = 0..495
Index entries for linear recurrences with constant coefficients, signature (111,-1100).

Cf. A000225, A021093, A094704, A162741, A164913.

[(1/89)*(100^n-11^n): n in [0..40]] // Vincenzo Librandi, Dec 05 2010

RecurrenceTable[{a[0]==0,a[n]==100a[n-1]+11^(n-1)},a,{n,40}] (* Harvey P. Dale, Feb 20 2016 *)

[(10^(2*n) - 11^n)/89 for n in range(41)] # G. C. Greubel, Feb 09 2023

G.f.: x/((1-11*x)*(1-100*x)). - R. J. Mathar, Nov 02 2016

E.g.f.: (1/89)*(exp(100*x) - exp(11*x)). - G. C. Greubel, Feb 09 2023

a(0) prepended by Bruno Berselli, Oct 02 2015

A164899 Binomial matrix (1,10^n) read by antidiagonals.

Original entry on oeis.org

1, 1, 10, 1, 11, 100, 1, 12, 110, 1000, 1, 13, 121, 1100, 10000, 1, 14, 133, 1210, 11000, 100000, 1, 15, 146, 1331, 12100, 110000, 1000000, 1, 16, 160, 1464, 13310, 121000, 1100000, 10000000, 1, 17, 175, 1610, 14641, 133100, 1210000, 11000000, 100000000

Offset: 1

Mark Dols, Aug 30 2009

			Matrix array, A(n, k), begins:
  1, 10, 100, 1000, ...
  1, 11, 110, 1100, ...
  1, 12, 121, 1210, ...
  1, 13, 133, 1331, ...
  1, 14, 146, 1464, ...
  1, 15, 160, 1610, ...
Antidiagonal triangle, T(n, k), begins as:
  1;
  1, 10;
  1, 11, 100;
  1, 12, 110, 1000;
  1, 13, 121, 1100, 10000;
  1, 14, 133, 1210, 11000, 100000;
  1, 15, 146, 1331, 12100, 110000, 1000000;

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

Cf. A001020, A007318, A021093, A164844.

Cf. A094704 (antidiagonal row sums).

function T(n,k) // T = A164899
  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 = A164899
    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

Sum_{k=1..n} T(n, k) = A094704(n).
As a triangle T(n,k) read by rows, T(n,1) = 1, T(n,n) = 10^(n-1), and T(n,k) = T(n-1, k) + T(n-2, k-1) otherwise. - Joerg Arndt, Dec 10 2016
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) = A(n-k+1, k) (antidiagonal triangle). (End)

cp's OEIS Frontend

A165155 a(n) = 100*a(n-1) + 11^(n-1) for n>0, a(0)=0.

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