A000270 -id:A000270 - cp's OEIS frontend

A005582 a(n) = n(n+1)(n+2)*(n+7)/24.

0, 2, 9, 25, 55, 105, 182, 294, 450, 660, 935, 1287, 1729, 2275, 2940, 3740, 4692, 5814, 7125, 8645, 10395, 12397, 14674, 17250, 20150, 23400, 27027, 31059, 35525, 40455, 45880, 51832, 58344, 65450, 73185, 81585, 90687, 100529, 111150, 122590, 134890

Offset: 0

N. J. A. Sloane

a(n) = number of Dyck (n+2)-paths with exactly 2 rows of peaks. A row of peaks is a maximal sequence of peaks all at the same height and 2 units apart. For example, UDUDUD ( = /\/\/\ ) contains exactly one row of peaks, as does UUUDDD, but UDUUDDUD has three and a(1)=2 counts UDUUDD, UUDDUD. - David Callan, Mar 02 2005
If X is an n-set and Y a fixed 2-subset of X then a(n-4) is equal to the number of (n-4)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
Let I=I_n be the n X n identity matrix and P=P_n be the incidence matrix of the cycle (1,2,3,...,n). Then, for n>=7, a(n-7) is the number of (0,1) n X n matrices A<=P^(-1)+I+P having exactly two 1's in every row and column with perA=16. - Vladimir Shevelev, Apr 12 2010
Row 2 of the convolution array A213550. - Clark Kimberling, Jun 20 2012
a(n-1) = risefac(n, 4)/4! - risefac(n, 2)/2! is for n >= 1 also the number of independent components of a symmetric traceless tensor of rank 4 and dimension n. Here risefac is the rising factorial. - Wolfdieter Lang, Dec 10 2015
Consider the array formed by the second polygonal numbers of increasing rank:
  A000217(-1-n): 0,  1,  3,  6, 10, 15, ...
  A000270(-1-n): 1,  4,  9, 16, 25, 36, ...
  A000326(-1-n): 2,  7, 15, 26, 40, 57, ...
  A000384(-1-n): 3, 10, 21, 36, 55, 78, ...
Then the antidiagonal sums yield this sequence. - Michael Somos, Nov 23 2021

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), Table 22.7, p. 797.
Vladimir S. Shevelyov (Shevelev), Extension of the Moser class of four-line Latin rectangles, DAN Ukrainy, 3(1992),15-19. [From Vladimir Shevelev, Apr 12 2010]
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
A. M. Yaglom and I. M. Yaglom: Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #51 (the case k=4) (First published: San Francisco: Holden-Day, Inc., 1964)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
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].
Richard K. Guy, Letter to N. J. A. Sloane, Feb 1988
F. T. Howard and Curtis Cooper, Some identities for r-Fibonacci numbers, Fibonacci Quart. 49 (2011), no. 3, 231-243.
Milan Janjic, Two Enumerative Functions
P. A. MacMahon, Properties of prime numbers deduced from the calculus of symmetric functions, Proc. London Math. Soc., 23 (1923), 290-316. = Coll. Papers, II, pp. 354-380. [See p. 301]
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
C. Rossiter, Depictions, Explorations and Formulas of the Euler/Pascal Cube.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Partial sums of A005581.

Cf. A000211, A052928, A128209, A176222.

[seq(binomial(n,4)+2*binomial(n,3), n=2..43)]; # Zerinvary Lajos, Jul 26 2006
seq((n+4)*binomial(n,4)/n, n=3..43); # Zerinvary Lajos, Feb 28 2007
A005582:=(-2+z)/(z-1)**5; # conjectured by Simon Plouffe in his 1992 dissertation

Table[n(n+1)(n+2)(n+7)/24,{n,0,40}] (* Harvey P. Dale, Jun 01 2012 *)

concat(0, Vec(x*(2-x)/(1-x)^5 + O(x^100))) \\ Altug Alkan, Dec 10 2015

a(n) = binomial(n+3, n-1) + binomial(n+2, n-1).
a(n) = binomial(n,4) + 2*binomial(n,3), n>=2. - Zerinvary Lajos, Jul 26 2006
From Colin Barker, Jan 28 2012: (Start)
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: x*(2-x)/(1-x)^5. (End)
a(n) = Sum_{k=1..n} ( Sum_{i=1..k} i(n-k+2) ). - Wesley Ivan Hurt, Sep 26 2013
a(n+1) = A127672(8+n, n), n >= 0, with the Chebyshev C-polynomial coefficients A127672(n, k). See the Abramowitz-Stegun reference.  - Wolfdieter Lang, Dec 10 2015
E.g.f.: (1/24)*x*(48 + 60*x + 16*x^2 + x^3)*exp(x). - G. C. Greubel, Jul 01 2017
Sum_{n>=1} 1/a(n) = 853/1225. - Amiram Eldar, Jan 02 2021
a(n) = A005587(-7-n) for all n in Z. - Michael Somos, Nov 23 2021

More terms from Larry Reeves (larryr(AT)acm.org), Jun 01 2000

A335391 Square array read by antidiagonals downwards: for n >= 2, T(k,n) is the number of permutations of [k+n] that differ in every position from both the identity permutation and a permutation consisting of k 1-cycles and one n-cycle.

Original entry on oeis.org

2, -1, 0, 0, 1, 2, 1, 0, 1, 4, 2, 3, 4, 7, 18, 13, 16, 19, 24, 35, 88, 80, 95, 114, 137, 168, 221, 530, 579, 672, 783, 916, 1077, 1280, 1589, 3708, 4738, 5397, 6164, 7061, 8114, 9359, 10860, 12979, 29666, 43387, 48704, 54773, 61720, 69697, 78888, 89527, 101976, 118663, 266992

Offset: 0

William P. Orrick, Jun 04 2020

The number of permutations of [k+n] that differ in every position from both the identity permutation and a permutation consisting of k 1-cycles and s cycles of lengths p_1, p_2, ... p_s, with p_j >= 2 and p_1+p_2+...+p_s = n, can be expressed as Sum T(k,p_1+-p_2+-...+-p_s), where the sum is over all 2^(s-1) choices of sign and where T(k,-n) = T(k,n) (Touchard).
The first of Touchard's formulas for T(k,n) involves A034807, the number of k-matchings of C_n (A213234 or A127677 with sign included) and A047920, the k-th differences of the factorial numbers.
A slightly different formula, due to Wyman and Moser in the k=0 case, involves A213234 and A000023.
The first column is twice A000166 (twice the number of derangements of [k]); the second column is A105926 (first differences of A000166); the third column is A331007 (with offset 2); the first row is A102761 (the ménage numbers); the second row is A000270.

			Array starts:
k/n |    0     1      2      3       4         5          6           7
-----------------------------------------------------------------------
0   |    2    -1      0      1       2        13         80         579
1   |    0     1      0      3      16        95        672        5397
2   |    2     1      4     19     114       783       6164       54773
3   |    4     7     24    137     916      7061      61720      602955
4   |   18    35    168   1077    8114     69697     671736     7172007
5   |   88   221   1280   9359   78888    749547    7913440    91815601
6   |  530  1589  10860  89527  837794   8741875  100478588  1260186153
7   | 3708 12979 101976 938181 9669196 110058257 1369406616 18475560567
There are T(1,3)=3 permutations that differ from 1234=(1)(2)(3)(4) and 1342=(1)(234) in every position: 2413, 3421, and 4123.

Mathematics Stack Exchange, Double derangement and the other unfamous kind of derangement.
J. Touchard, Sur un problème de permutations, Comptes Rendus Acad. Sci. Paris, 198 (1934) 631-633.
Max Wyman and Leo Moser, On the problème des ménages, Canadian Journal of Mathematics, 10 (1958) 468-480.

Cf. A034807, A213234, A127677, A047920, A000023, A000166, A105926, A331007, A102761, A000270.

T := proc(n,k) local t; t := proc(n, k) option remember;
   simplify((n + k)!*hypergeom([-n], [-n - k], -1)) end:
   if k = 0 then return 2*t(n, 0) fi;
   add((-1)^j*(2*k)/(2*k-j)*binomial(2*k-j, j)*t(n, k-j), j=0 ..k) end:
seq(lprint(seq(T(n, k), k=0..7)),n=0..7); # Peter Luschny, Jul 22 2020

f(k, n) = sum(j=0, k, (-1)^j*binomial(k, j)*(n+k-j)!);
T(k, n) = if (n==0, 2*f(k, 0), sum(j=0, n, (-1)^j*(2*n)/(2*n-j)*binomial(2*n-j, j)*f(k, n-j)));
matrix(7, 7,n, k, T(n-1,k-1))
\\ Michel Marcus, Jun 26 2020

def f(k,n):
    return sum((-1)^j*binomial(k,j)*factorial(n+k-j) for j in range(0,k+1))
def T(k,n):
    if n==0:
        return 2*f(k,0)
    else:
        return sum((-1)^j*(2*n)/(2*n-j)*binomial(2*n-j,j)*f(k,n-j) for j in range(0,n+1))

T(k,0) = 2*nu(k,k), T(k,n>0) = Sum_{j=0..n} A213234(2*n,j)*nu(k,k+n-j) = Sum_{j=0..n} (-1)^j*2*n/(2*n-j)*binomial(2*n-j,j)*nu(k,k+n-j) where nu(k,k+n) = A047920(k+n,k) = Sum_{j=0..k} (-1)^j*binomial(k,j)*(k+n-j)! (Touchard).
T(k,n) = 2*cos(2*n * arccos(1/2*sqrt(x))) = 2*Chebyshev_T(2*n,sqrt(x)/2), where, after expanding in powers of x, x^m gets replaced by nu(k,k+m) (Touchard).
T(k,n) = 2*(-1)^n*Sum_{j=0..n} (-1)^j*(Product_{r=0..j} n^2-r^2)/(2*j)!*nu(k,k+j) (Touchard).
T(k,n) = 2*Integral_{x=0..oo} e^(-x^2) * (x^2-1)^k * x * ((x+sqrt(x^2-4))^(2*n)+(x-sqrt(x^2-4))^(2*n)) / 2^(2*n) dx (Touchard).
T(k,0) = 2*Sum_{j=0..h} binomial(h,j)*k(j), T(k,n) = Sum_{i>=0} A213234(n,i)*Sum_{j=0..h} binomial(h,j)*k(n-2*i+j) = Sum_{i>=0} (-1)^i*n/(n-i)*binomial(n-i,i)*Sum_{j=0..h} binomial(h,j)*k(n-2*i+j) where k(n) = A000023(n) = n! * Sum_{i=0..n} (-2)^i / i! (k=0 case due to Wyman and Moser)
T(k+1,n+1) = T(k,n)+T(k,n+1)+T(k,n+2): This holds for all integers n if one defines T(k,-n) = T(k,n).
T(k,0) = 2*A000166(k).
T(k,1) = A105926(k).
T(k,2) = A331007(k+2).
T(0,n) = A102761(n).
T(1,n) = A000270(n).

cp's OEIS Frontend

A005582 a(n) = n(n+1)(n+2)*(n+7)/24.

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A335391 Square array read by antidiagonals downwards: for n >= 2, T(k,n) is the number of permutations of [k+n] that differ in every position from both the identity permutation and a permutation consisting of k 1-cycles and one n-cycle.

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cp's OEIS Frontend

A005582 a(n) = n*(n+1)*(n+2)*(n+7)/24.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

Extensions

A335391 Square array read by antidiagonals downwards: for n >= 2, T(k,n) is the number of permutations of [k+n] that differ in every position from both the identity permutation and a permutation consisting of k 1-cycles and one n-cycle.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Sage

Formula

A005582 a(n) = n(n+1)(n+2)*(n+7)/24.