

A000918


a(n) = 2^n  2.
(Formerly M1599 N0625)


150



1, 0, 2, 6, 14, 30, 62, 126, 254, 510, 1022, 2046, 4094, 8190, 16382, 32766, 65534, 131070, 262142, 524286, 1048574, 2097150, 4194302, 8388606, 16777214, 33554430, 67108862, 134217726, 268435454, 536870910, 1073741822, 2147483646, 4294967294, 8589934590, 17179869182, 34359738366, 68719476734, 137438953470
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OFFSET

0,3


COMMENTS

For n > 1, a(n) is the expected number of tosses of a fair coin to get n1 consecutive heads.  Pratik Poddar, Feb 04 2011
For n > 0, the number of nonempty proper subsets of an nelement set.  Ross La Haye, Feb 07 2004
Numbers j such that abs( Sum_{k=0..j} (1)^binomial(j, k)*binomial(j + k, j  k) ) = 1.  Benoit Cloitre, Jul 03 2004
For n > 2 this formula also counts edgerooted forests in a cycle of length n.  Woong Kook (andrewk(AT)math.uri.edu), Sep 08 2004
For n >= 1, conjectured to be the number of integers from 0 to (10^n)1 that lack 0, 1, 2, 3, 4, 5, 6 and 7 as a digit.  Alexandre Wajnberg, Apr 25 2005
Beginning with a(2) = 2, these are the partial sums of the subsequence of A000079 = 2^n beginning with A000079(1) = 2. Hence for n >= 2 a(n) is the smallest possible sum of exactly one prime, one semiprime, one triprime, ... and one product of exactly n1 primes. A060389 (partial sums of the primorials, A002110, beginning with A002110(1)=2) is the analog when all the almost primes must also be squarefree.  Rick L. Shepherd, May 20 2005
From the second term on (n >= 1), the binary representation of these numbers is a 0 preceded by (n  1) 1's. This pattern (0)111...1110 is the "opposite" of the binary 2^n+1: 1000...0001 (cf. A000051).  Alexandre Wajnberg, May 31 2005
The numbers 2^n  2 (n >= 2) give the positions of 0's in A110146. Also numbers k such that k^(k + 1) = 0 mod (k + 2).  Zak Seidov, Feb 20 2006
Number of surjections from an nelement set onto a twoelement set, with n >= 2.  Mohamed Bouhamida, Dec 15 2007
It appears that these are the numbers n such that 3*A135013(n) = n*(n + 1), thus answering Problem 2 on the Mathematical Olympiad Foundation of Japan, Final Round Problems, Feb 11 1993 (see link Japanese Mathematical Olympiad).
Let P(A) be the power set of an nelement set A and R be a relation on P(A) such that for all x, y of P(A), xRy if x is a proper subset of y or y is a proper subset of x and x and y are disjoint. Then a(n+1) = R.  Ross La Haye, Mar 19 2009
Apart the first term which is 1 the number of units of a(n) belongs to a periodic sequence: 0, 2, 6, 4.  Mohamed Bouhamida, Sep 04 2009
The permutohedron Pi_n has 2^n  2 facets [Pashkovich].  Jonathan Vos Post, Dec 17 2009
a(n) is the number of branches of a complete binary tree of n levels.  Denis Lorrain, Dec 16 2011
For n>=1, a(n) is the number of lengthn words on alphabet {1,2,3} such that the gap(w)=1. For a word w the gap g(w) is the number of parts missing between the minimal and maximal elements of w. Generally for words on alphabet {1,2,...,m} with g(w)=g>0 the e.g.f. is Sum_{k=g+2..m} (m  k + 1)*binomial((k  2),g)*(exp(x)  1)^(k  g). a(3)=6 because we have: 113, 131, 133, 311, 313, 331. Cf. A240506. See the Heubach/Mansour reference.  Geoffrey Critzer, Apr 13 2014
For n > 0, a(n) is the minimal number of internal nodes of a redblack tree of height 2*n2. See the Oct 02 2015 comment under A027383.  Herbert Eberle, Oct 02 2015
Actually this follows from the procedure for determining the multiplicity of prime p in C(n) given in A000108 by Franklin T. AdamsWatters: For p=2, the multiplicity is the number of 1 digits minus 1 in the binary representation of n+1. Obviously, the smallest k achieving "number of 1 digits" = k is 2^k1. Therefore C(2^k2) is divisible by 2^(k1) for k > 0 and there is no smaller m for which 2^(k1) divides C(m) proving the conjecture.  Peter Schorn, Feb 16 2020
For n >= 0, a(n) is the largest number you can write in bijective base2 (a.k.a. the dyadic system, A007931) with n digits.  Harald Korneliussen, May 18 2019
The terms of this sequence are also the sum of the terms in each row of Pascal's triangle other than the ones.  Harvey P. Dale, Apr 19 2020
For n > 1, binomial(a(n),k) is odd if and only if k is even.  Charlie Marion, Dec 22 2020
For n >= 2, a(n+1) is the number of n X n arrays of 0's and 1's with every 2 X 2 square having density exactly 2.  David desJardins, Oct 27 2022
For n >= 1, a(n+1) is the number of roots of unity in the unique degreen unramified extension of the 2adic field Q_2. Note that for each p, the unique degreen unramified extension of Q_p is the splitting field of x^(p^n)  x, hence containing p^n  1 roots of unity for p > 2 and 2*(2^n  1) for p = 2.  Jianing Song, Nov 08 2022


REFERENCES

H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 212.
Ralph P. Grimaldi, Discrete and Combinatorial Mathematics: An Applied Introduction, Fifth Edition, AddisonWesley, 2004, p. 134. [From Mohammad K. Azarian, October 27 2011]
S. Heubach and T. Mansour, Combinatorics of Compositions and Words, Chapman and Hall, 2009 page 86, Exercise 3.16.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.


LINKS

P. A. Piza, Kummer numbers, Mathematics Magazine, 21 (1947/1948), 257260.
P. A. Piza, Kummer numbers, Mathematics Magazine, 21 (1947/1948), 257260. [Annotated scanned copy]


FORMULA

G.f.: 1/(1  2*x)  2/(1  x), e.g.f.: (e^x  1)^2  1.  Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 07 2001
G.f.: (3*x  1)/((2*x  1)*(x  1)).  Simon Plouffe in his 1992 dissertation for the sequence without the leading 1
G.f.: U(0)  1, where U(k) = 1 + x/(2^k + 2^k/(x  1  x^2*2^(k + 1)/(x*2^(k + 1)  (k + 1)/U(k + 1) ))); (continued fraction, 3rd kind, 4step).  Sergei N. Gladkovskii, Dec 01 2012
a(n) = Sum_{k=1..n1} binomial(n, k) for n > 0.  Dan McCandless, Nov 14 2015
a(n+1) = 2*(n + Sum_{j=1..n1} (nj)*2^(j1)), n >= 1. This is the number of the rationals k/2, k = 1..2*n for n >= 1 and (2*k+1)/2^j for j = 2..n, n >= 2, and 2*k+1 < n(j1). See the example for n = 3 below. Motivated by the proposal A287012 by Mark Rickert.  Wolfdieter Lang, Jun 14 2017


EXAMPLE

a(4) = 14 because the 14 = 6 + 4 + 4 rationals (in lowest terms) for n = 3 are (see the Jun 14 2017 formula above): 1/2, 1, 3/2, 2, 5/2, 3; 1/4, 3/4, 5/4, 7/4; 1/8, 3/8, 5/8, 7/8.  Wolfdieter Lang, Jun 14 2017


MAPLE

seq(2^n2, n=0..20) ;
ZL := [S, {S=Prod(B, B), B=Set(Z, 1 <= card)}, labeled]: [1, seq(combstruct[count](ZL, size=n), n=1..28)]; # Zerinvary Lajos, Mar 13 2007


MATHEMATICA



PROG

(Sage) [gaussian_binomial(n, 1, 2)1 for n in range(0, 29)] # Zerinvary Lajos, May 31 2009
(Haskell)
a000918 = (subtract 2) . (2 ^)
a000918_list = iterate ((subtract 2) . (* 2) . (+ 2)) ( 1)


CROSSREFS



KEYWORD

sign,easy


AUTHOR



EXTENSIONS



STATUS

approved



