

A001792


a(n) = (n+2)*2^(n1).
(Formerly M2739 N1100)


177



1, 3, 8, 20, 48, 112, 256, 576, 1280, 2816, 6144, 13312, 28672, 61440, 131072, 278528, 589824, 1245184, 2621440, 5505024, 11534336, 24117248, 50331648, 104857600, 218103808, 452984832, 939524096, 1946157056, 4026531840, 8321499136, 17179869184, 35433480192
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OFFSET

0,2


COMMENTS

Number of parts in all compositions (ordered partitions) of n + 1. For example, a(2) = 8 because in 3 = 2 + 1 = 1 + 2 = 1 + 1 + 1 we have 8 parts. Also number of compositions (ordered partitions) of 2n + 1 with exactly 1 odd part. For example, a(2) = 8 because the only compositions of 5 with exactly 1 odd part are 5 = 1 + 4 = 2 + 3 = 3 + 2 = 4 + 1 = 1 + 2 + 2 = 2 + 1 + 2 = 2 + 2 + 1.  Emeric Deutsch, May 10 2001
Binomial transform of natural numbers [1, 2, 3, 4, ...].
For n >= 1 a(n) is also the determinant of the n X n matrix with 3's on the diagonal and 1's elsewhere.  Ahmed Fares (ahmedfares(AT)mydeja.com), May 06 2001
The arithmetic mean of first n terms of the sequence is 2^(n1).  Amarnath Murthy, Dec 25 2001, corrected by M. F. Hasler, Dec 17 2016
Also the number of "winning paths" of length n across an n X n Hex board. Satisfies the recursion a(n) = 2a(n1) + 2^(n2).  David Molnar (molnar(AT)stolaf.edu), Apr 10 2002
Diagonal in A053218.  Benoit Cloitre, May 08 2002
Let M_n be the n X n matrix m_(i, j) = 1 + abs(ij) then det(M_n) = (1)^(n1)*a(n1).  Benoit Cloitre, May 28 2002
Absolute value of determinant of n X n matrix of form: [1 2 3 4 5 / 2 1 2 3 4 / 3 2 1 2 3 / 4 3 2 1 2 / 5 4 3 2 1].  Benoit Cloitre, Jun 20 2002
a(n) = A018804(2^n).  Matthew Vandermast, Mar 01 2003
a(n) = (1/4)A001787(n+2).  Emeric Deutsch, May 24 2003
Number of ones in all (n+1)bit integers (cf. A000120).  Ralf Stephan, Aug 02 2003
This sequence also emerges as a floretion force transform of powers of 2 (see program code). Define a(1) = 0 (as the sequence is returned by FAMP). Then a(n1) + A098156(n+1) = 2*a(n) (conjecture).  Creighton Dement, Mar 14 2005
This sequence gives the absolute value of the determinant of the Toeplitz matrix with first row containing the first n integers.  Paul Max Payton, May 23 2006
Equals sums of rows right of left edge of A102363 divided by three, + 2^K.  David G. Williams (davidwilliams(AT)paxway.com), Oct 08 2007
If X_1, X_2, ..., X_n are 2blocks of a (2n+1)set X then, for n >= 1, a(n) is the number of (n+1)subsets of X intersecting each X_i, (i = 1, 2, ..., n).  Milan Janjic, Nov 18 2007
Also, a(n1) is the determinant of the n X n matrix with A[i, j] = n  ij.  M. F. Hasler, Dec 17 2008
1/2 the number of permutations of 1 .. (n+2) arranged in a circle with exactly one local maximum.  R. H. Hardin, Apr 19 2009
The first corrector line for transforming 2^n offset 0 with a leading 1 into the Fibonacci sequence.  Al Hakanson (hawkuu(AT)gmail.com), Jun 01 2009
a(n) is the number of runs of consecutive 1's in all binary sequences of length (n+1).  Geoffrey Critzer, Jul 02 2009
a(n) = A164910(2^n).  Gary W. Adamson, Aug 30 2009
Let X_j (0 < j <= 2^n) all the subsets of N_n; m(i, j) := if {i} in X_j then 1 else 0. Let A = transpose(M).M; Then a(i, j) = (number of elements)X_i intersect X_j. Determinant(X*IA) = (X(n+1)*2^(n2))*(X2^(n2))^(n1)*X^(2^nn).
Eigenvector for (n+1)*2^(n2) is V_i=X_i.
Sum_{k=1..2^n} X_i intersect X_k*X_k = (n+1)*2^(n2)*X_i.
Eigenvectors for 2^(n2) are {line(M)[i]  line(M)[j], 1 <= i, j <= n).  CLARISSE Philippe (clarissephilippe(AT)yahoo.fr), Mar 24 2010
The sequence b(n) = 2*A001792(n), for n >= 1 with b(0) = 1, is an elephant sequence, see A175655. For the central square four A[5] vectors, with decimal values 187, 190, 250 and 442, lead to the b(n) sequence. For the corner squares these vectors lead to the companion sequence A134401.  Johannes W. Meijer, Aug 15 2010
Equals partial sums of A045623: (1, 2, 5, 12, 28, ...); where A045623 = the convolution square of (1, 1, 2, 4, 8, 16, 32, ...).  Gary W. Adamson, Oct 26 2010
Equals (1, 2, 4, 8, 16, ...) convolved with (1, 1, 2, 4, 8, 16, ...); e.g., a(3) = 20 = (1, 1, 2, 4) dot (8, 4, 2, 1) = (8 + 4 + 4 + 4).  Gary W. Adamson, Oct 26 2010
This sequence seems to give the first x+1 nonzero terms in the sequence derived by subtracting the mth term in the x_binacci sequence (where the first term is one and the yth term is the sum of x terms immediately preceding it) from 2^(m2).  Dylan Hamilton, Nov 03 2010
Recursive formulas for a(n) are in many cases derivable from its property wherein delta^k(a(n))  a(n) = k*2^n where delta^k(a(n)) represents the kth forward difference of a(n). Provable with a difference table and a little induction.  Ethan Beihl, May 02 2011
Let f(n,k) be the sum of numbers in the subsets of size k of {1, 2, ..., n}. Then a(n1) is the average of the numbers f(n, 0), ... f(n, n). Example: (f(3, 1), f(3, 2), f(3, 3)) = (6, 12, 6), with average (6+12+6)/3.  Clark Kimberling, Feb 24 2012
a(n) is the number of length2n binary sequences that contain a subsequence of ones with length n or more. To derive this result, note that there are 2^n sequences where the initial one of the subsequence occurs at entry one. If the initial one of the subsequence occurs at entry 2, 3, ..., or n + 1, there are 2^(n1) sequences since a zero must precede the initial one. Hence a(n) = 2^n + n*2^(n1)=(n+2)2^(n1). An example is given in the example section below.  Dennis P. Walsh, Oct 25 2012
As the total number of parts in all compositions of n+1 (see the first line in Comments) the equivalent sequence for partitions is A006128. On the other hand, as the first differences of A001787 (see the first line in Crossrefs) the equivalent sequence for partitions is A138879.  Omar E. Pol, Aug 28 2013
a(n) is the number of spanning trees of the complete tripartite graph K_{n,1,1}.  James Mahoney, Oct 24 2013
a(n1) = denominator of the mean (2n/(n+1), after reduction), of the compositions of n; numerator is given by A022998(n).  Clark Kimberling, Mar 11 2014
From Tom Copeland, Nov 09 2014: (Start)
The shifted array belongs to an interpolated family of arrays associated to the Catalan A000108 (t=1), and Riordan, or Motzkin sums A005043 (t=0), with the interpolating o.g.f. (1sqrt(14x/(1+(1t)x)))/2 and inverse x(1x)/(1+(t1)x(1x)). See A091867 for more info on this family. Here the interpolation is t=3 (mod signs in the results).
Let C(x) = (1  sqrt(14x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1x) and P(x,t) = x/(1+t*x) with inverse P(x,t).
Shifted o.g.f: G(x) = x*(1x)/(1  4x*(1x)) = P[Cinv(x),4].
Inverse o.g.f: Ginv(x) = [1  sqrt(1  4*x/(1+4x))]/2 = C[P(x, 4)] (signed shifted A001700). Cf. A030528.
For n > 0, element a(n) of the sequence is equal to the gradients of the (n1)th row of Pascal triangle multiplied with the square of the integers from n+1,...,1. I.e., row 3 of Pascal's triangle 1,3,3,1 has gradients 1,2,0,2,1, so a(4) = 1*(5^2) + 2*(4^2) + 0*(3^2)  2*(2^2)  1*(1^2) = 48.  Jens Martin Carlsson, May 18 2017
Let T_n be the undirected graph formed by taking an edge with endpoints 'u' and 'v', and adding n vertices, all of which are adjacent to 'u' and 'v'. Then a(n) is the number of spanning trees of T_n.  Kevin Long, Sep 28 2018
Number of nonselfintersecting broken lines connecting all the vertices of a convex (n+2)gon.  Ivaylo Kortezov, Jan 19 2020
a(n1) is the total number of elements of subsets of {1,2,..,n} that contain n. For example, for n = 3, a(2) = 8, and the subsets of {1,2,3} that contain 3 are {3}, {1,3}, {2,3}, {1,2,3}, with a total of 8 elements.  Enrique Navarrete, Aug 01 2020


REFERENCES

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
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. M. Stepin and A. T. TagiZade, Words with restrictions, pp. 6774 of Kvant Selecta: Combinatorics I, Amer. Math. Soc., 2001 (G_n on p. 70).


LINKS

T. D. Noe, Table of n, a(n) for n = 0..500
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Marco Abrate, Stefano Barbero, Umberto Cerruti, Nadir Murru, Colored compositions, Invert operator and elegant compositions with the "black tie", Discrete Math. 335 (2014), 17. MR3248794.
Marco Abrate, Stefano Barbero, Umberto Cerruti, Nadir Murru, Colored compositions, Invert operator and elegant compositions with the "black tie", arXiv:1409.6454 [math.NT], 2014.
Milica Andelic, C. M. da Fonseca, A. Pereira, The mupermanent, a new graph labeling, and a known integer sequence, arXiv preprint arXiv:1609.04208 [math.CO], 2016.
N. J. Calkin, A curious binomial identity, Discr. Math., 131 (1994), 335337.
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
F. Disanto and S. Rinaldi, Symmetric convex permutominoes and involutions, PU. M. A., Vol. 22 (2011), No. 1, pp. 3960.
F. Ellermann, Illustration of binomial transforms
Guillermo Esteban, Clemens Huemer, Rodrigo I. Silveira, New production matrices for geometric graphs, arXiv:2003.00524 [math.CO], 2020.
M. Hirschhorn, Calkin's binomial identity, Discr. Math., 159 (1996), 273278.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 146
Milan Janjic, Two Enumerative Functions
M. Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
M. Janjic, B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5
Jones, C. W.; Miller, J. C. P.; Conn, J. F. C.; Pankhurst, R. C.; Tables of Chebyshev polynomials Proc. Roy. Soc. Edinburgh. Sect. A. 62, (1946). 187203.
Sergey Kitaev, J. B. Remmel, A note on pAscent Sequences, Preprint, 2016.
S. Kitaev, J. Remmel, pAscent Sequences, arXiv preprint arXiv:1503.00914 [math.CO], 2015.
W. Lang, On generalizations of Stirling number triangles, J. Integer Seqs., Vol. 3 (2000), #00.2.4.
Maohua Le, Two Classes of Smarandache Determinants, Scientia Magna, vol 2, no 1 (2006), pp 2025.
Donatella Merlini, Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
National Math Contest "Atanas Radev", , in problem 8.4 ("Задача 8.4" in Bulgarian), Jan 2020.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.
Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.
John Riordan and N. J. A. Sloane, Correspondence, 1974
N. J. A. Sloane, Transforms
Jun Wang and Zhizheng Zhang, On extensions of Calkin's binomial identities, Discrete Math., 274 (2004), 331342.
Index entries for linear recurrences with constant coefficients, signature (4,4).
Index entries for sequences related to Chebyshev polynomials.


FORMULA

a(n) = (n+2)*2^(n1).
G.f.: (1  x)/(1  2*x)^2 = 2F1(1,3;2;2x).
a(n) = 4*a(n1)  4*a(n2).
G.f. (1 + (12*x)^(2))/(x*2^2).  Wolfdieter Lang
a(n) = Sum_{k=0..n+2} binomial(n+2, 2k)*k.  Paul Barry, Mar 06 2003
With a leading 0, this is ((n+1)2^n  0^n)/4 = Sum_{m=0..n} binomial(n  1, m  1)*m, the binomial transform of A004526(n+1).  Paul Barry, Jun 05 2003
a(n) = Sum_{k=0..n} binomial(n, k)*(k + 1).  Lekraj Beedassy, Jun 24 2004
a(n) = A000244(n)  A066810(n).  Ross La Haye, Apr 29 2006
Row sums of triangle A130585.  Gary W. Adamson, Jun 06 2007
Equals A125092 * [1/1, 1/2, 1/3, ...].  Gary W. Adamson, Nov 16 2007
a(n) = (n+1)*2^n  n*2^(n1). Equals A128064 * A000079.  Gary W. Adamson, Dec 28 2007
G.f.: F(3, 1; 2; 2x).  Paul Barry, Sep 03 2008
a(n) = 1 + Sum_{k=1..n} (n  k + 4)2^(n  k  1). This follows from the result that the number of parts equal to k in all compositions of n is (n  k + 3)2^(n  k  2) for 0 < k < n.  Geoffrey Critzer, Sep 21 2008
a(n) = 2^(n1) + 2 a(n1) ; a(n1) = det(n  i  j)_{i, j = 1..n}.  M. F. Hasler, Dec 17 2008
a(n) = 2*a(n1) + 2^(n1).  Philippe Deléham, Apr 19 2009
a(n) = Sum_{i=1..2^n} gcd(i, 2^n) = A018804(2^n). So we have: 2^0 * phi(2^n) + ... + 2^n * phi(2^0) = (n + 2)*2^(n1), where phi is the Euler totient function.  Jeffrey R. Goodwin, Nov 11 2011
a(n) = Sum_{j=0..n} Sum_{i=0..n} binomial(n, i + j).  Yalcin Aktar, Jan 17 2012
Eigensequence of an infinite lower triangular matrix with 2^n as the left border and the rest 1's.  Gary W. Adamson, Jan 30 2012
G.f.: 1 + 2*x*U(0) where U(k) = 1 + (k + 1)/(2  8*x/(4*x + (k + 1)/U(k + 1))); (continued fraction, 3  step).  Sergei N. Gladkovskii, Oct 19 2012
a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n,j).  Peter Luschny, Dec 03 2013
a(n) = Hyper2F1([n, 2], [1], 1).  Peter Luschny, Aug 02 2014
G.f.: 1 / (1  3*x / (1 + x / (3  4*x))).  Michael Somos, Aug 26 2015
a(n) = A053120(2+n, n), n >= 0, the negative of the third (sub)diagonal of the triangle of Chebyshev's T polynomials.  Wolfdieter Lang, Nov 26 2019


EXAMPLE

a(0) = 1, a(1) = 2*1 + 1 = 3, a(2) = 2*3 + 2 = 8, a(3) = 2*8 + 4 = 20, a(4) = 2*20 + 8 = 48, a(5) = 2*48 + 16 = 112, a(6) = 2*112 + 32 = 256, ...  Philippe Deléham, Apr 19 2009
a(2) = 8 since there are 8 length4 binary sequences with a subsequence of ones of length 2 or more, namely, 1111, 1110, 1101, 1011, 0111, 1100, 0110, and 0011.  Dennis P. Walsh, Oct 25 2012
G.f. = 1 + 3*x + 8*x^2 + 20*x^3 + 48*x^4 + 112*x^5 + 256*x^6 + 576*x^7 + ...


MAPLE

A001792 := n> (n+2)*2^(n1);
spec := [S, {B=Set(Z, 0 <= card), S=Prod(Z, B, B)}, labeled]: seq(combstruct[count](spec, size=n)/4, n=2..30); # Zerinvary Lajos, Oct 09 2006
A001792:=(3+4*z)/(2*z1)^2; # Simon Plouffe in his 1992 dissertation, which gives the sequence without the initial 1
G(x):=1/exp(2*x)*(1x): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n1], x) od: x:=0: seq(abs(f[n]), n=0..28 ); # Zerinvary Lajos, Apr 17 2009
a := n > hypergeom([n, 2], [1], 1);
seq(round(evalf(a(n), 32)), n=0..31); # Peter Luschny, Aug 02 2014


MATHEMATICA

matrix[n_Integer /; n >= 1] := Table[Abs[p  q] + 1, {q, n}, {p, n}]; a[n_Integer /; n >= 1] := Abs[Det[matrix[n]]] (* Josh Locker (joshlocker(AT)macfora.com), Apr 29 2004 *)
g[n_, m_, r_] := Binomial[n  1, r  1] Binomial[m + 1, r] r; Table[1 + Sum[g[n, k  n, r], {r, 1, k}, {n, 1, k  1}], {k, 1, 29}] (* Geoffrey Critzer, Jul 02 2009 *)
a[n_] := (n + 2)*2^(n  1); a[Range[0, 40]] (* Vladimir Joseph Stephan Orlovsky, Feb 09 2011 *)
LinearRecurrence[{4, 4}, {1, 3}, 40] (* Harvey P. Dale, Aug 29 2011 *)
CoefficientList[Series[(1  x) / (1  2 x)^2, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 10 2014 *)
b[i_]:=i; a[n_]:=Abs[Det[ToeplitzMatrix[Array[b, n], Array[b, n]]]]; Array[a, 40] (* Stefano Spezia, Sep 25 2018 *)


PROG

(Octave, MATLAB) abs(det(toeplitz(1:n))) % Paul Max Payton, May 23 2006
(PARI) A001792(n)=(n+2)<<(n1) \\ M. F. Hasler, Dec 17 2008
(Haskell)
a001792 n = a001792_list !! n
a001792_list = scanl1 (+) a045623_list
 Reinhard Zumkeller, Jul 21 2013
(MAGMA) [(n+2)*2^(n1): n in [0..40]]; // Vincenzo Librandi, Nov 10 2014
(GAP) List([0..35], n>(n+2)*2^(n1)); # Muniru A Asiru, Sep 25 2018
(Python) for n in range(0, 40): print(int((n+2)*2**(n1)), end=' ') # Stefano Spezia, Oct 16 2018


CROSSREFS

First differences of A001787.
a(n) = A049600(n, 1), a(n) = A030523(n + 1, 1).
Cf. A053113.
Row sums of triangles A008949 and A055248.
a(n) = A039991(n+2, 2).
Cf. A001787, A130584, A125092, A128064, A000079, A164910, A045623, A000120, A053120.
Cf. A000108, A005043, A091867, A001700, A030528.
If the exponent E in a(n) = Sum_{m=0..n} (Sum_{k=0..m} C(n,k))^E is 1, 2, 3, 4, 5 we get A001792, A003583, A007403, A294435, A294436 respectively.
Sequence in context: A151975 A049610 A168150 * A018795 A018794 A018793
Adjacent sequences: A001789 A001790 A001791 * A001793 A001794 A001795


KEYWORD

nonn,easy,nice


AUTHOR

N. J. A. Sloane


STATUS

approved



