A034670 Duplicate of A020516.
7, 127, 5461, 299593, 17895697, 1108378657, 69810262081, 4432676798593
Offset: 0
This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
For n=3, a(3)=S(4,2)=7, a Stirling number of the second kind, since there are 7 ways to partition {a,b,c,d} into 2 nonempty subsets, namely, {a}U{b,c,d}, {b}U{a,c,d}, {c}U{a,b,d}, {d}U{a,b,c}, {a,b}U{c,d}, {a,c}U{b,d}, and {a,d}U{b,c}. - _Dennis P. Walsh_, Mar 29 2011 From _Justin M. Troyka_, Aug 13 2011: (Start) Since a(3) = 7, there are 7 signed permutations of 4 that are equal to the bar of their reverse-complements and avoid {(-2,-1), (-1,+2), (+2,+1)}. These are: (+1,+2,-3,-4), (+1,+3,-2,-4), (+1,-3,+2,-4), (+2,+4,-1,-3), (+3,+4,-1,-2), (-3,+1,-4,+2), (-3,-4,+1,+2). (End) G.f. = x + 3*x^2 + 7*x^3 + 15*x^4 + 31*x^5 + 63*x^6 + 127*x^7 + ... For the Towers of Hanoi problem with 2 disks, the moves are as follows, so a(2) = 3. 12|_|_ -> 2|1|_ -> _|1|2 -> _|_|12 - _Allan Bickle_, Aug 07 2024
a000225 = (subtract 1) . (2 ^) a000225_list = iterate ((+ 1) . (* 2)) 0 -- Reinhard Zumkeller, Mar 20 2012
A000225 := n->2^n-1; [ seq(2^n-1,n=0..50) ]; A000225:=1/(2*z-1)/(z-1); # Simon Plouffe in his 1992 dissertation, sequence starting at a(1)
a[n_] := 2^n - 1; Table[a[n], {n, 0, 30}] (* Stefan Steinerberger, Mar 30 2006 *) Array[2^# - 1 &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *) NestList[2 # + 1 &, 0, 32] (* Robert G. Wilson v, Feb 28 2011 *) 2^Range[0, 20] - 1 (* Eric W. Weisstein, Jul 17 2017 *) LinearRecurrence[{3, -2}, {1, 3}, 20] (* Eric W. Weisstein, Sep 21 2017 *) CoefficientList[Series[1/(1 - 3 x + 2 x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)
A000225(n) = 2^n-1 \\ Michael B. Porter, Oct 27 2009
concat(0, Vec(x/((1-2*x)*(1-x)) + O(x^100))) \\ Altug Alkan, Oct 28 2015
def A000225(n): return (1<Chai Wah Wu, Jul 06 2022
def isMersenne(n): return n == sum([(1 - b) << s for (s, b) in enumerate((n+1).bits())]) # Peter Luschny, Sep 01 2019
[DivisorSigma(n, 22): n in [0..20]];
Total[#^Range[0, 20]&/@Divisors[22]] Table[(1 + 2^n) (1 + 11^n), {n, 0, 20}] (* Bruno Berselli, Apr 17 2014 *) LinearRecurrence[{36,-343,792,-484},{4,36,610,11988},30] (* Harvey P. Dale, May 21 2014 *)
makelist((1+2^n)*(1+11^n), n, 0, 20); /* Bruno Berselli, Apr 17 2014 */
17^Range[0,20]+1 (* Harvey P. Dale, Jul 19 2019 *)
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