A329734 a(n) = (10^(1 + floor((n-1) * log_10 2)) + 1)^n, n >= 0.

1, 11, 121, 1331, 14641, 10510100501, 1061520150601, 107213535210701, 1008028056070056028008001, 1009036084126126084036009001, 1010045120210252210120045010001, 100110055016503300462046203300165005500110001, 1001200660220049507920924079204950220006600120001

Offset: 0

Author

Daniel Forgues, Nov 19 2019

Keywords

• nonn
• base
• easy

Comments

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.

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.

Examples

			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).
		

Links

• OEIS Wiki, Pascal's triangle rows.

Crossrefs

Cf. A007318, A092846 (uses number of digits of central binomial coefficients).

Programs

Mathematica

d[n_] := Floor[Log[10, 10*n]]; a[n_] := (10^(d[2^(n - 1)]) + 1)^n; Array[a, 12] (* Amiram Eldar, Nov 23 2019 *)

Formula

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).