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A271527
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a(n) = 1000^n + 1.
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1
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2, 1001, 1000001, 1000000001, 1000000000001, 1000000000000001, 1000000000000000001, 1000000000000000000001, 1000000000000000000000001, 1000000000000000000000000001, 1000000000000000000000000000001, 1000000000000000000000000000000001
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OFFSET
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0,1
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COMMENTS
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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).
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LINKS
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FORMULA
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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).
Sum_{n>=0} 1/a(n) = 0.501000001999002...
Lim_{n->infinity} a(n + 1)/a(n) = 1000.
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EXAMPLE
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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
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MATHEMATICA
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Table[1000^n + 1, {n, 0, 11}]
LinearRecurrence[{1001, -1000}, {2, 1001}, 12]
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PROG
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(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)
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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