A001792 a(n) = (n+2)*2^(n-1).

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

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy
• nice

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)my-deja.com), May 06 2001

The arithmetic mean of first n terms of the sequence is 2^(n-1). - 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(n-1) + 2^(n-2). - 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(i-j) then det(M_n) = (-1)^(n-1)*a(n-1). - 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

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(n-1) + 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 2-blocks 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(n-1) is the determinant of the n X n matrix with A[i, j] = n - |i-j|. - 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

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*I-A) = (X-(n+1)*2^(n-2))*(X-2^(n-2))^(n-1)*X^(2^n-n).

Eigenvector for (n+1)*2^(n-2) is V_i=|X_i|.

Sum_{k=1..2^n} |X_i intersect X_k|*|X_k| = (n+1)*2^(n-2)*|X_i|.

Eigenvectors for 2^(n-2) 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 m-th term in the x_binacci sequence (where the first term is one and the y-th term is the sum of x terms immediately preceding it) from 2^(m-2). - 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 k-th 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(n-1) 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 length-2n 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^(n-1) sequences since a zero must precede the initial one. Hence a(n) = 2^n + n*2^(n-1)=(n+2)2^(n-1). 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(n-1) = 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. (1-sqrt(1-4x/(1+(1-t)x)))/2 and inverse x(1-x)/(1+(t-1)x(1-x)). See A091867 for more info on this family. Here the interpolation is t=-3 (mod signs in the results).

Let C(x) = (1 - sqrt(1-4x))/2, an o.g.f. for the Catalan numbers A000108, with inverse Cinv(x) = x*(1-x) and P(x,t) = x/(1+t*x) with inverse P(x,-t).

Shifted o.g.f: G(x) = x*(1-x)/(1 - 4x*(1-x)) = 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. (End)

For n > 0, element a(n) of the sequence is equal to the gradients of the (n-1)-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

Number of self-avoiding paths connecting all the vertices of a convex (n+2)-gon. - Ivaylo Kortezov, Jan 19 2020

a(n-1) 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

Examples

			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 length-4 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 + ...
		

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. Tagi-Zade, Words with restrictions, pp. 67-74 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 and Nadir Murru, Colored compositions, Invert operator and elegant compositions with the "black tie", Discrete Math., Vol. 335 (2014), pp. 1-7. MR3248794.
• Marco Abrate, Stefano Barbero, Umberto Cerruti, and 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 and A. Pereira, The mu-permanent, a new graph labeling, and a known integer sequence, arXiv preprint arXiv:1609.04208 [math.CO], 2016.
• Félix Balado and Guénolé C. M. Silvestre, Runs of Ones in Binary Strings, arXiv:2302.11532 [math.CO], 2023. See pp. 6-7.
• Jean-Luc Baril and Nathanaël Hassler, Intervals in a family of Fibonacci lattices, Univ. de Bourgogne (France, 2024). See p. 7.
• Neil J. Calkin, A curious binomial identity, Discr. Math., Vol. 131, No. 1-3 (1994), pp. 335-337.
• Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs., Vol. 3 (2000), #00.1.5.
• Filippo Disanto and Simone Rinaldi, Symmetric convex permutominoes and involutions, PU. M. A., Vol. 22, No. 1 (2011), pp. 39-60.
• Frank Ellermann, Illustration of binomial transforms.
• David Eppstein, Non-crossing Hamiltonian Paths and Cycles in Output-Polynomial Time, arXiv:2303.00147 [cs.CG], 2023, pp. 2, 19.
• Guillermo Esteban, Clemens Huemer and Rodrigo I. Silveira, New production matrices for geometric graphs, arXiv:2003.00524 [math.CO], 2020.
• M. Hirschhorn, Calkin's binomial identity, Discr. Math., Vol. 159, No. 1-3 (1996), pp. 273-278.
• INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 146.
• Milan Janjić, Two Enumerative Functions
• Milan Janjić and Boris Petković, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013.
• Milan Janjić and Boris Petković, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
• C. W. Jones, J. C. P. Miller, J. F. C. Conn and R. C. Pankhurst, Tables of Chebyshev polynomials, Proc. Roy. Soc. Edinburgh. Sect. A., Vol. 62, No. 2 (1946), pp. 187-203.
• Sergey Kitaev and Jeffrey B. Remmel, A note on p-Ascent Sequences, Preprint, 2016.
• Sergey Kitaev and Jeffrey Remmel, p-Ascent Sequences, arXiv preprint arXiv:1503.00914 [math.CO], 2015.
• Ivaylo Kortezov, problem 8.4 ("Задача 8.4" in Bulgarian) in National Math Contest "Atanas Radev" 2020.
• Wolfdieter 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. 20-25.
• Donatella Merlini and Massimo Nocentini, Algebraic Generating Functions for Languages Avoiding Riordan Patterns, Journal of Integer Sequences, Vol. 21 (2018), Article 18.1.3.
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
• Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021.
• John Riordan and N. J. A. Sloane, Correspondence, 1974.
• N. J. A. Sloane, Transforms.
• I. Tasoulas, K. Manes, and A. Sapounakis, Hamiltonian intervals in the lattice of binary paths, Elect. J. Comb. (2024) Vol. 31, Issue 1, P1.39.
• Jun Wang and Zhizheng Zhang, On extensions of Calkin's binomial identities, Discrete Math., Vol. 274 (2004), 331-342.
• Index entries for linear recurrences with constant coefficients, signature (4,-4).
• Index entries for sequences related to Chebyshev polynomials.

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. 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.

Programs

GAP

List([0..35],n->(n+2)*2^(n-1)); # Muniru A Asiru, Sep 25 2018

Haskell

a001792 n = a001792_list !! n
a001792_list = scanl1 (+) a045623_list
-- Reinhard Zumkeller, Jul 21 2013

Magma

[(n+2)*2^(n-1): n in [0..40]]; // Vincenzo Librandi, Nov 10 2014

Maple

A001792 := n-> (n+2)*2^(n-1);
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*z-1)^2; # Simon Plouffe in his 1992 dissertation, which gives the sequence without the initial 1
G(x):=1/exp(2*x)*(1-x): f[0]:=G(x): for n from 1 to 54 do f[n]:=diff(f[n-1],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 *)
a[n_]:=Hypergeometric2F1[2,-n+1,1,-1];Array[a,32] (* Giorgos Kalogeropoulos, Jan 04 2022 *)

PARI

A001792(n)=(n+2)<<(n-1) \\ M. F. Hasler, Dec 17 2008

Python

for n in range(0,40): print(int((n+2)*2**(n-1)), end=' ') # Stefano Spezia, Oct 16 2018

Formula

a(n) = (n+2)*2^(n-1).

G.f.: (1 - x)/(1 - 2*x)^2 = 2F1(1,3;2;2x).

a(n) = 4*a(n-1) - 4*a(n-2).

G.f. (-1 + (1-2*x)^(-2))/(x*2^2). - Wolfdieter Lang

a(n) = A018804(2^n). - Matthew Vandermast, Mar 01 2003

a(n) = Sum_{k=0..n+2} binomial(n+2, 2k)*k. - Paul Barry, Mar 06 2003

a(n) = (1/4)*A001787(n+2). - Emeric Deutsch, May 24 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^(n-1). 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^(n-1) + 2 a(n-1) ; a(n-1) = det(n - |i - j|){i, j = 1..n}. - _M. F. Hasler, Dec 17 2008

a(n) = 2*a(n-1) + 2^(n-1). - Philippe Deléham, Apr 19 2009

a(n) = A164910(2^n). - Gary W. Adamson, Aug 30 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^(n-1), 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

From Amiram Eldar, Jan 12 2021: (Start)

Sum_{n>=0} 1/a(n) = 8*log(2) - 4.

Sum_{n>=0} (-1)^n/a(n) = 4 - 8*log(3/2). (End)

E.g.f.: exp(2*x)*(1 + x). - Stefano Spezia, Jun 11 2021