A111776 -id:A111776 - cp's OEIS frontend

A081567 Second binomial transform of F(n+1).

1, 3, 10, 35, 125, 450, 1625, 5875, 21250, 76875, 278125, 1006250, 3640625, 13171875, 47656250, 172421875, 623828125, 2257031250, 8166015625, 29544921875, 106894531250, 386748046875, 1399267578125, 5062597656250, 18316650390625, 66270263671875, 239768066406250

Offset: 0

Paul Barry, Mar 22 2003

Binomial transform of F(2*n-1), index shifted by 1, where F is A000045. - corrected by Richard R. Forberg, Aug 12 2013
Case k=2 of family of recurrences a(n) = (2k+1)*a(n-1) - A028387(k-1)*a(n-2), a(0)=1, a(1)=k+1.
Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 10 and |s(i) - s(i-1)| = 1 for i = 1, 2, ..., 2*n+1, s(0) = 3, s(2*n+1) = 4.
a(n+1) gives the number of periodic multiplex juggling sequences of length n with base state <2>. - Steve Butler, Jan 21 2008
a(n) is also the number of idempotent order-preserving partial transformations (of an n-element chain) of waist n (waist(alpha) = max(Im(alpha))). - Abdullahi Umar, Sep 14 2008
Counts all paths of length (2*n+1), n>=0, starting at the initial node on the path graph P_9, see the Maple program. - Johannes W. Meijer, May 29 2010
Given the 3 X 3 matrix M = [1,1,1; 1,1,0; 1,1,3], a(n) = term (1,1) in M^(n+1). - Gary W. Adamson, Aug 06 2010
Number of nonisomorphic graded posets with 0 and 1 of rank n+2, with exactly 2 elements of each rank level between 0 and 1. Also the number of nonisomorphic graded posets with 0 of rank n+1, with exactly 2 elements of each rank level above 0. (This is by Stanley's definition of graded, that all maximal chains have the same length.) - David Nacin, Feb 26 2012
a(n) = 3^n a(n;1/3) = Sum_{k=0..n} C(n,k) * F(k-1) * (-1)^k * 3^(n-k), which also implies the Deleham formula given below and where a(n;d), n=0,1,...,d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers of Witula et al.). - Roman Witula, Jul 12 2012
The limiting ratio a(n)/a(n-1) is 1 + phi^2. - Bob Selcoe, Mar 17 2014
a(n) counts closed walks on K_2 containing 3 loops on the index vertex and 2 loops on the other. Equivalently the (1,1) entry of A^n where the adjacency matrix of digraph is A=(3,1; 1,2). - David Neil McGrath, Nov 18 2014

			a(4)=125: 35*(3 + (35 mod 10 - 10 mod 3)/(10-3)) = 35*(3 + 4/7) = 125. - _Bob Selcoe_, Mar 17 2014

R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pages 96-100.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Santiago Alzate, Oscar Correa, and Rigoberto Flórez, Fibonacci identities from Jordan Identities, arXiv:2009.02639 [math.NT], 2020.
Carolina Benedetti, Christopher R. H. Hanusa, Pamela E. Harris, Alejandro H. Morales, and Anthony Simpson, Kostant's partition function and magic multiplex juggling sequences, arXiv:2001.03219 [math.CO], 2020. See Table 1 p. 12.
S. Butler and R. Graham, Enumerating (multiplex) juggling sequences, arXiv:0801.2597 [math.CO], 2008.
P. E. Harris, E. Insko, and L. K. Williams, The adjoint representation of a Lie algebra and the support of Kostant's weight multiplicity formula, arXiv preprint arXiv:1401.0055 [math.RT], 2013.
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
A. Laradji and A. Umar, A. Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
A. Laradji and A. Umar, Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), #04.3.8.
Mircea Merca, A Note on Cosine Power Sums J. Integer Sequences, Vol. 15 (2012), Article 12.5.3.
D. Nacin, The Minimal Non-Koszul A(Gamma), arXiv preprint arXiv:1204.1534 [math.QA], 2012. - From _N. J. A. Sloane_, Oct 05 2012
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009) 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (5,-5).

a(n) = 5*A052936(n-1), n > 1.
Row sums of A114164.
Cf. A000045, A007051 (INVERTi transform), A007598, A028387, A030191, A039717, A049310, A081568 (binomial transform), A086351 (INVERT transform), A090041, A093129, A094441, A111776, A147748, A178381, A189315.

I:=[1, 3]; [n le 2 select I[n] else 5*Self(n-1)-5*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 27 2012

with(GraphTheory):G:=PathGraph(9): A:= AdjacencyMatrix(G): nmax:=23; n2:=nmax*2+2: for n from 0 to n2 do B(n):=A^n; a(n):=add(B(n)[1,k],k=1..9); od: seq(a(2*n+1),n=0..nmax); # Johannes W. Meijer, May 29 2010

Table[MatrixPower[{{2,1},{1,3}},n][[2]][[2]],{n,0,44}] (* Vladimir Joseph Stephan Orlovsky, Feb 20 2010 *)
LinearRecurrence[{5,-5},{1,3},30] (* Vincenzo Librandi, Feb 27 2012 *)

Vec((1-2*x)/(1-5*x+5*x^2)+O(x^99)) \\ Charles R Greathouse IV, Mar 18 2014

def a(n, adict={0:1, 1:3}):
    if n in adict:
        return adict[n]
    adict[n]=5*a(n-1) - 5*a(n-2)
    return adict[n] # David Nacin, Mar 04 2012

a(n) = 5*a(n-1) - 5*a(n-2) for n >= 2, with a(0) = 1 and a(1) = 3.
a(n) = (1/2 - sqrt(5)/10) * (5/2 - sqrt(5)/2)^n + (sqrt(5)/10 + 1/2) * (sqrt(5)/2 + 5/2)^n.
G.f.: (1 - 2*x)/(1 - 5*x + 5*x^2).
a(n-1) = Sum_{k=1..n} binomial(n, k)*F(k)^2. - Benoit Cloitre, Oct 26 2003
a(n) = A090041(n)/2^n. - Paul Barry, Mar 23 2004
The sequence 0, 1, 3, 10, ... with a(n) = (5/2 - sqrt(5)/2)^n/5 + (5/2 + sqrt(5)/2)^n/5 - 2(0)^n/5 is the binomial transform of F(n)^2 (A007598). - Paul Barry, Apr 27 2004
From Paul Barry, Nov 15 2005: (Start)
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, j)*binomial(j+k, 2k);
a(n) = Sum_{k=0..n} Sum_{j=0..n} binomial(n, k+j)*binomial(k, k-j)2^(n-k-j);
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} binomial(n+k-j, n-k-j)*binomial(k, j)(-1)^j*2^(n-k-j). (End)
a(n) = A111776(n, n). - Abdullahi Umar, Sep 14 2008
a(n) = Sum_{k=0..n} A094441(n,k)*2^k. - Philippe Deléham, Dec 14 2009
a(n+1) = Sum_{k=-floor(n/5)..floor(n/5)} ((-1)^k*binomial(2*n, n+5*k)/2). -Mircea Merca, Jan 28 2012
a(n) = A030191(n) - 2*A030191(n-1). - R. J. Mathar, Jul 19 2012
G.f.: Q(0,u)/x - 1/x, where u=x/(1-2*x), Q(k,u) = 1 + u^2 + (k+2)*u - u*(k+1 + u)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Oct 07 2013
For n>=3: a(n) = a(n-1)*(3+(a(n-1) mod a(n-2) - a(n-2) mod a(n-3))/(a(n-2) - a(n-3))). - Bob Selcoe, Mar 17 2014
a(n) = sqrt(5)^(n-1)*(3*S(n-1, sqrt(5)) - sqrt(5)*S(n-2, sqrt(5))) with Chebyshev's S-polynomials (see A049310), where S(-1, x) = 0 and S(-2, x) = -1. This is the (1,1) entry of A^n with the matrix A=(3,1;1,2). See the comment by David Neil McGrath, Nov 18 2014. - Wolfdieter Lang, Dec 04 2014
Conjecture: a(n) = 2*a(n-1) + A039717(n). - Benito van der Zander, Nov 20 2015
a(n) = A189315(n+1) / 10. - Tom Copeland, Dec 08 2015
a(n) = A093129(n) + A030191(n-1). - Gary W. Adamson, Apr 24 2023
E.g.f.: exp(5*x/2)*(5*cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Jun 03 2024

A109814 a(n) is the largest k such that n can be written as sum of k consecutive positive integers.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 2, 1, 3, 4, 2, 3, 2, 4, 5, 1, 2, 4, 2, 5, 6, 4, 2, 3, 5, 4, 6, 7, 2, 5, 2, 1, 6, 4, 7, 8, 2, 4, 6, 5, 2, 7, 2, 8, 9, 4, 2, 3, 7, 5, 6, 8, 2, 9, 10, 7, 6, 4, 2, 8, 2, 4, 9, 1, 10, 11, 2, 8, 6, 7, 2, 9, 2, 4, 10, 8, 11, 12, 2, 5, 9, 4, 2, 8, 10, 4, 6, 11, 2, 12, 13, 8, 6, 4, 10, 3, 2, 7, 11, 8, 2, 12

Offset: 1

David W. Wilson

nonn

n is the sum of at most a(n) consecutive positive integers. As suggested by David W. Wilson, Aug 15 2005: Suppose n is to be written as sum of k consecutive integers starting with m, then 2n = k(2m + k - 1). Only one of the factors is odd. For each odd divisor d of n there is a unique corresponding k = min(d,2n/d). a(n) is the largest among those k. - Jaap Spies, Aug 16 2005
The numbers that can be written as a sum of k consecutive positive integers are those in column k of A141419 (as a triangle). - Peter Munn, Mar 01 2019
The numbers that cannot be written as a sum of two or more consecutive positive integers are the powers of 2. So a(n) = 1 iff n = 2^k for k >= 0. - Bernard Schott, Mar 03 2019

			Examples provided by _Rainer Rosenthal_, Apr 01 2008:
1 = 1     ---> a(1) = 1
2 = 2     ---> a(2) = 1
3 = 1+2   ---> a(3) = 2
4 = 4     ---> a(4) = 1
5 = 2+3   ---> a(5) = 2
6 = 1+2+3 ---> a(6) = 3
a(15) = 5: 15 = 15 (k=1), 15 = 7+8 (k=2), 15 = 4+5+6 (k=3) and 15 = 1+2+3+4+5 (k=5). - _Jaap Spies_, Aug 16 2005

Donovan Johnson, Table of n, a(n) for n = 1..10000
K. S. Brown's Mathpages, Partitions into Consecutive Integers
A. Heiligenbrunner, Sum of adjacent numbers (in German).
Jaap Spies, Problem C NAW 5/6 nr. 2 June 2005, July 2005 (solution to problem below).
Jaap Spies, Sage program for computing A109814
Universitaire Wiskunde Competitie, Problem C, Nieuw Archief voor Wiskunde, 5/6, no. 2, Problems/UWC, Jun 2005, pp. 181-182.

Cf. A001227, A111774, A111775, A141419, A138591.

Cf. A000079 (powers of 2), A000217 (triangular numbers).

A109814:= proc(n) local m, k, d; m := 0; for d from 1 by 2 to n do if n mod d = 0 then k := min(d, 2*n/d): fi; if k > m then m := k fi: od; return(m); end proc; seq(A109814(i),i=1..150); # Jaap Spies, Aug 16 2005

a[n_] := Reap[Do[If[OddQ[d], Sow[Min[d, 2n/d]]], {d, Divisors[n]}]][[2, 1]] // Max; Table[a[n], {n, 1, 102}]

from sympy import divisors
def a(n): return max(min(d, 2*n//d) for d in divisors(n) if d&1)
print([a(n) for n in range(1, 103)]) # Michael S. Branicky, Dec 23 2022

[sloane.A109814(n) for n in range(1,20)]
# Jaap Spies, Aug 16 2005

From Reinhard Zumkeller, Apr 18 2006: (Start)
a(n)*(a(n)+2*A118235(n)-1)/2 = n;
a(A000079(n)) = 1;
a(A000217(n)) = n. (End)

Edited by N. J. A. Sloane, Aug 23 2008 at the suggestion of R. J. Mathar

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A081567 Second binomial transform of F(n+1).

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