A271527 a(n) = 1000^n + 1.

2, 1001, 1000001, 1000000001, 1000000000001, 1000000000000001, 1000000000000000001, 1000000000000000000001, 1000000000000000000000001, 1000000000000000000000000001, 1000000000000000000000000000001, 1000000000000000000000000000000001

Offset: 0

Author

Ilya Gutkovskiy, Apr 09 2016

Keywords

• nonn
• base
• easy

Comments

All terms in this sequence are palindromes (A002113).

Also, A062395 written in base 2 (see example).

a(n) minus one gives the number of nodes at n-th level of a 1000-ary tree.

More generally, the ordinary generating function for sequences of the form k^n + m, is (1 + m - (1 + k*m)*x)/((1 - x)*(1 - k*x)), and the exponential generating function is exp(k*x) + m*exp(x).

Examples

			a(n), n>0, is the binary representation of A062395(n)
n  ------------------------------------------
0  2........................................2
1  1001.....................................9
2  1000001.................................65
3  1000000001.............................513
4  1000000000001.........................4097
5  1000000000000001.....................32769
6  1000000000000000001.................262145
7  1000000000000000000001.............2097153
8  1000000000000000000000001.........16777217
9  1000000000000000000000000001.....134217729
		

Links

• Ilya Gutkovskiy, Examples of the ordinary generating function for the sequences of the form k^n + m
• Index entries for linear recurrences with constant coefficients, signature (1001,-1000)

Crossrefs

Cf. A000035, A000533, A007088, A007395, A007953, A057427, A060365, A062395, A152756.

Programs

Mathematica

Table[1000^n + 1, {n, 0, 11}]
LinearRecurrence[{1001, -1000}, {2, 1001}, 12]

PARI

x='x+O('x^99); Vec((2-1001*x)/((1-x)*(1-1000*x))) \\ Altug Alkan, Apr 09 2016

Python

for n in range(0,10**4):print(1000**n+1)
# Soumil Mandal, Apr 10 2016

Formula

G.f.: (2 - 1001*x)/((1 - x)*(1 - 1000*x)).

E.g.f.: exp(1000*x) + exp(x).

a(n) = 1001*a(n-1) - 1000*a(n-2).

a(n) = A060365(n) + 1.

a(n) = A000533(3n), n>0.

a(n) = A007088(A062395(n)).

A007953(a(n)) = A007395(n).

A000035(a(n)) = A057427(n).

Sum_{n>=0} 1/a(n) = 0.501000001999002...

Lim_{n->infinity} a(n + 1)/a(n) = 1000.