%I #29 Jan 24 2020 23:27:04
%S 1,11,121,1331,14641,10510100501,1061520150601,107213535210701,
%T 1008028056070056028008001,1009036084126126084036009001,
%U 1010045120210252210120045010001,100110055016503300462046203300165005500110001,1001200660220049507920924079204950220006600120001
%N a(n) = (10^(1 + floor((n-1) * log_10 2)) + 1)^n, n >= 0.
%C a(n), n >= 1, gives all the binomial coefficients (without overlap, since binomial(n,k) < 2^(n-1) for n >= 3) of the n-th row of Pascal's triangle (A007318). The k-th "digit", 0 <= k <= n, of the base 10^(d(2^(n-1))) representation of a(n) gives binomial(n,k), albeit with leading zeros: binomial(n,k) = (a(n) div (10^(d(2^(n-1))))^k) mod 10^(d(2^(n-1))), where div means integer division.
%C A092846 uses number of digits of binomial(n, floor(n/2)) instead [optimal], but requires the evaluation of central binomial coefficients. a(n) uses number of digits of 2^(n-1) [not optimal], but does not require the evaluation of any binomial coefficient.
%H OEIS Wiki, <a href="/wiki/Pascal_triangle#Pascal's_triangle_rows">Pascal's triangle rows</a>.
%F a(0) = 1; a(n) = (10^(d(2^(n-1))) + 1)^n, n >= 1, where d(2^(n-1)) = 1 + floor((n-1) * log_10 2) = A034887(n-1) is the number of [decimal] digits of 2^(n-1).
%e a(9) = (10^3 + 1)^9 = 1009036084126126084036009001 = 1:009:036:084:126:126:084:036:009:001 (base 10^3, since 2^8 has 3 decimal digits).
%t d[n_] := Floor[Log[10, 10*n]]; a[n_] := (10^(d[2^(n - 1)]) + 1)^n; Array[a, 12] (* _Amiram Eldar_, Nov 23 2019 *)
%Y Cf. A007318, A092846 (uses number of digits of central binomial coefficients).
%K nonn,base,easy
%O 0,2
%A _Daniel Forgues_, Nov 19 2019