A243499 -id:A243499 - cp's OEIS frontend

A011965 Second differences of Bell numbers.

1, 2, 7, 27, 114, 523, 2589, 13744, 77821, 467767, 2972432, 19895813, 139824045, 1028804338, 7905124379, 63287544055, 526827208698, 4551453462543, 40740750631417, 377254241891064, 3608700264369193, 35613444194346451, 362161573323083920, 3790824599495473121

Offset: 0

N. J. A. Sloane

Number of partitions of n+3 with at least one singleton and with the smallest element in a singleton equal to 3. Alternatively, number of partitions of n+3 with at least one singleton and with the largest element in a singleton equal to n+1. - Olivier GERARD, Oct 29 2007

Out of the A005493(n) set partitions with a specific two elements clustered separately, number that have a different set of two elements clustered separately. - Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007

Olivier Gérard and Karol A. Penson, A budget of set partition statistics, in preparation, unpublished as of Sep 22 2011.

Chai Wah Wu, Table of n, a(n) for n = 0..570(terms 0..250 from Alois P. Heinz).
Martin Cohn, Shimon Even, Karl Menger, Jr. and Philip K. Hooper, On the Number of Partitionings of a Set of n Distinct Objects, Amer. Math. Monthly 69 (1962), no. 8, 782--785. MR1531841.
Martin Cohn, Shimon Even, Karl Menger, Jr. and Philip K. Hooper, On the Number of Partitionings of a Set of n Distinct Objects, Amer. Math. Monthly 69 (1962), no. 8, 782--785. MR1531841. [Annotated scanned copy]
Adam M. Goyt and Lara K. Pudwell, Avoiding colored partitions of two elements in the pattern sense, arXiv preprint arXiv:1203.3786 [math.CO], 2012.
Jocelyn Quaintance and Harris Kwong, A combinatorial interpretation of the Catalan and Bell number difference tables, Integers, 13 (2013), #A29.

Cf. A000110, A005493, A106436.

Similar recurrences: A006014, A090365, A124758, A217924, A243499, A284005, A290615, A329369, A341392, A347204, A347205, A350309.

[Bell(n+2) -2*Bell(n+1) + Bell(n): n in [0..40]]; // G. C. Greubel, Jan 07 2025

a:= n-> add((-1)^k*binomial(2,k)*combinat['bell'](n+k), k=0..2): seq(a(n), n=0..20);  # Alois P. Heinz, Sep 05 2008

Differences[BellB[Range[0, 30]], 2] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)

# requires python 3.2 or higher. Otherwise use def'n of accumulate in python docs.
from itertools import accumulate
A011965_list, blist, b = [1], [1, 2], 2
for _ in range(1000):
    blist = list(accumulate([b]+blist))
    b = blist[-1]
    A011965_list.append(blist[-3])
# Chai Wah Wu, Sep 02 2014

# or Sagemath
b=bell_number
print([b(n+2) -2*b(n+1) +b(n) for n in range(41)]) # G. C. Greubel, Jan 07 2025

a(n) = A005493(n) - A005493(n-1).
E.g.f.: exp(exp(x)-1)*(exp(2*x)-exp(x)+1). - Vladeta Jovovic, Feb 11 2003
a(n) = A000110(n) - 2*A000110(n-1) + A000110(n-2). - Andrey Goder (andy.goder(AT)gmail.com), Dec 17 2007
G.f.: G(0) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k+2*x-1) - x*(2*k+1)*(2*k+3)*(2*x*k+2*x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k+3*x-1)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 19 2012
G.f.: 1 - G(0) where G(k) = 1 - 1/(1-k*x-2*x)/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 17 2013
G.f.: 1 - 1/x + (1-x)^2/x/(G(0)-x) where G(k) = 1 - x*(k+1)/(1 - x/G(k+1) ); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 26 2013
G.f.: G(0)*(1-1/x) where G(k) = 1 - 1/(1-x*(k+1))/(1-x/(x-1/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Feb 07 2013
a(n) ~ n^2 * Bell(n) / LambertW(n)^2 * (1 - 2*LambertW(n)/n). - Vaclav Kotesovec, Jul 28 2021
Conjecture: a(n) = Sum_{k=0..2^n - 1} b(k) for n >= 0 where b(2n+1) = b(n) + b(A025480(n-1)), b(2n) = b(n - 2^f(n)) + b(2n - 2^f(n)) + b(A025480(n-1)) for n > 0 with b(0) = b(1) = 1 and where f(n) = A007814(n). Also b((4^n - 1)/3) = A141154(n+1). - Mikhail Kurkov, Jan 27 2022

A227184 a(n) = product of parts of the unordered partition encoded with the runlengths of binary expansion of n.

Original entry on oeis.org

1, 1, 1, 2, 4, 1, 2, 3, 9, 4, 1, 8, 6, 2, 3, 4, 16, 9, 4, 18, 16, 1, 8, 27, 12, 6, 2, 12, 8, 3, 4, 5, 25, 16, 9, 32, 36, 4, 18, 48, 81, 16, 1, 32, 54, 8, 27, 64, 20, 12, 6, 24, 24, 2, 12, 36, 15, 8, 3, 16, 10, 4, 5, 6, 36, 25, 16, 50, 64, 9, 32, 75, 144, 36, 4, 72

Offset: 0

Antti Karttunen, Jul 04 2013

a(0) = 1, as 0 is here considered to encode an empty partition {}, and the empty product is one.
Like A129594, this sequence is based on the fact that compositions (i.e., ordered partitions) can be mapped 1-to-1 to partitions by taking the partial sums of the list where one is subtracted from each composant except the first (originally explained by Marc LeBrun in his Jan 11 2006 post on SeqFan mailing list, with an additional twist involving factorization and prime exponents, cf. A129595). The example below show how this works.
Compare the scatterplot of this sequence to those of A002487, A243353, A243499 and A253552.

			8 has binary expansion "1000", whose runs have lengths [3,1] when arranged from the least significant to the most significant end. Taking partial sums of 3 and 0, we get 3 and 3, whose product is 9, thus a(8) = 9.
For 44, in binary "101100", the run lengths are [2,2,1,1] (from the least significant end), and subtracting one from all terms except the first one, we get [2,1,0,0], whose partial sums are [2,3,3,3], and 2*3*3*3 = 54, thus a(44)=54.

Antti Karttunen, Table of n, a(n) for n = 0..8192

For n>=1, a(n) gives the product of nonzero terms on row n of table A227189/A227739.
Cf. A227183 (gives the corresponding sums).
See also A167489 for a similar sequence, which gives the product of parts of the compositions (ordered partitions).
Cf. A243499, A003963, A243504 (other such product sequences) and A003188, A243353, A075157 (associated permutations mapping between these schemes).
Cf. also A002487, A243353, A253552.

Table[Function[b, Times @@ Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b]]@ Map[Length, Split@ Reverse@ IntegerDigits[n, 2]], {n, 0, 75}] // Flatten (* Michael De Vlieger, May 09 2017 *)

def A227184(n):
  '''Product of parts of the unique unordered partition encoded in the run lengths of the binary expansion of n.'''
  p = 1
  b = n%2
  i = 1
  while (n != 0):
    n >>= 1
    if ((n%2) == b): i += 1
    else:
      b = n%2
      p *= i
  return(p)

(define (A227184 n) (if (zero? n) 1 (apply * (binexp_to_ascpart n))))
(define (binexp_to_ascpart n) (let ((runlist (reverse! (binexp->runcount1list n)))) (PARTSUMS (cons (car runlist) (map -1+ (cdr runlist))))))
(define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
(define (PARTSUMS a) (cdr (reverse! (fold-left (lambda (psums n) (cons (+ n (car psums)) psums)) (list 0) a))))

Can be also obtained by mapping with an appropriate permutation from the products of parts of each partition computed for other enumerations similar to A227739:
a(n) = A243499(A003188(n)).
a(n) = A003963(A243353(n)).
a(n) = A243504(1+A075157(n)).

A217924 a(n) = n! * [x^n] exp(2*exp(x) - x - 2). Row sums of triangle A217537.

Original entry on oeis.org

1, 1, 3, 9, 35, 153, 755, 4105, 24323, 155513, 1064851, 7760745, 59895203, 487397849, 4166564147, 37298443977, 348667014723, 3395240969785, 34365336725715, 360837080222761, 3923531021460707, 44108832866004121, 511948390801374835, 6126363766802713481

Offset: 0

Peter Luschny, Oct 15 2012

nonn

The inverse binomial transform of a(n) is A194689.
A087981(n) = Sum_{k=0..n} (-1)^k*s(n+1,k+1)*a(k);
|A000023(n)| = |Sum_{k=0..n} (-1)^(n-k)*s(n,k)*a(k)|
where s(n,k) are the unsigned Stirling numbers of first kind.
a(n) is the number of inequivalent set partitions of {1,2,...,n} where two blocks are considered equivalent when one can be obtained from the other by an alternating (even) permutation. - Geoffrey Critzer, Mar 17 2013

			a(3)=9 because we have: {1,2,3}; {1,3,2}; {1}{2,3}; {1}{3,2}; {2}{1,3}; {2}{3,1}; {3}{1,2}; {3}{2,1}; {1}{2}{3}. - _Geoffrey Critzer_, Mar 17 2013

Vaclav Kotesovec, Table of n, a(n) for n = 0..556

Similar recurrences: A124758, A243499, A284005, A329369, A341392, A372205.

R:=PowerSeriesRing(Rationals(), 30);
Coefficients(R!(Laplace( Exp(2*Exp(x) -x-2) ))); // G. C. Greubel, Jan 09 2025

egf := exp(2*exp(x) - x - 2): ser := series(egf, x, 25):
seq(n!*coeff(ser, x, n), n = 0..23);  # Peter Luschny, Apr 22 2024

nn=23;Range[0,nn]!CoefficientList[Series[Exp[2 Exp[x]-x-2],{x,0,nn}],x]  (* Geoffrey Critzer, Mar 17 2013 *)
nmax = 25; CoefficientList[Series[1/(1 - x + ContinuedFractionK[-2*k*x^2 , 1 - (k + 1)*x, {k, 1, nmax}]), {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 25 2017 *)

a(n):=sum(sum(binomial(n,k-j)*2^j*(-1)^(k-j)*stirling2(n-k+j,j),j,0,k),k,0,n); /* Vladimir Kruchinin, Feb 28 2015 */

def A217924_list(n):
    T = A217537_triangle(n)
    return [add(T.row(n)) for n in range(n)]
A217924_list(24)

def A217924_list(prec):
    P. = PowerSeriesRing(QQ, prec)
    return P( exp(2*exp(x)-x-2) ).egf_to_ogf().list()
print(A217924_list(40)) # G. C. Greubel, Jan 09 2025

G.f.: 1/Q(0) where Q(k) = 1 + x*k - x/(1 - 2*x*(k+1)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 06 2013
E.g.f.: exp(2*exp(x) - x - 2). -  Geoffrey Critzer, Mar 17 2013
G.f.: 1/Q(0), where Q(k) = 1 - (k+1)*x - 2*(k+1)*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 03 2013
G.f.: T(0)/(1-x), where T(k) = 1 - 2*x^2*(k+1)/( 2*x^2*(k+1) - (1-x-x*k)*(1-2*x-x*k)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Oct 19 2013
a(n) = Sum_{k=0..n} Sum_{j=0..k} binomial(n,k-j)*2^j*(-1)^(k-j)*Stirling2(n-k+j,j). - Vladimir Kruchinin, Feb 28 2015
a(n) = exp(-2) * Sum_{k>=0} 2^k * (k - 1)^n / k!. - Ilya Gutkovskiy, Jun 27 2020
Conjecture: a(n) = Sum_{k=0..2^n-1} A372205(k). - Mikhail Kurkov, Nov 21 2021 [Rewritten by Peter Luschny, Apr 22 2024]
a(n) ~ 2 * n^(n-1) * exp(n/LambertW(n/2) - n - 2) / (sqrt(1 + LambertW(n/2)) * LambertW(n/2)^(n-1)). - Vaclav Kotesovec, Jun 26 2022

Name extended by a formula of Geoffrey Critzer by Peter Luschny, Apr 22 2024

A347205 a(2n+1) = a(n) for n >= 0, a(2n) = a(n) + a(n - 2^A007814(n)) for n > 0 with a(0) = 1.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, 2, 6, 3, 4, 1, 5, 4, 7, 3, 9, 5, 7, 2, 10, 6, 9, 3, 10, 4, 5, 1, 6, 5, 9, 4, 12, 7, 10, 3, 14, 9, 14, 5, 16, 7, 9, 2, 15, 10, 16, 6, 19, 9, 12, 3, 20, 10, 14, 4, 15, 5, 6, 1, 7, 6, 11, 5, 15, 9, 13, 4, 18, 12, 19, 7, 22, 10, 13

Offset: 0

Mikhail Kurkov, Aug 23 2021

Scatter plot might be called "Cypress forest on a windy day". - Antti Karttunen, Nov 30 2021

Antti Karttunen, Table of n, a(n) for n = 0..16384
J. Abate and W. Whitt, Brownian Motion and the Generalized Catalan Numbers, J. Int. Seq. 14 (2011) # 11.2.6.
Index entries for sequences with interesting graphs or plots

Cf. A000108, A006318, A007814, A059894, A108524, A115197, A116867.

Similar recurrences: A124758, A243499, A284005, A329369, A341392.

a[0] = 1; a[n_] := a[n] = If[OddQ[n], a[(n - 1)/2], a[n/2] + a[n/2 - 2^IntegerExponent[n/2, 2]]]; Array[a, 100, 0] (* Amiram Eldar, Sep 06 2021 *)

a(n) = if (n==0, 1, if (n%2, a(n\2), a(n/2) + a(n/2 - 2^valuation(n/2, 2)))); \\ Michel Marcus, Sep 09 2021

a(2n+1) = a(n) for n >= 0.
a(2n) = a(n) + a(n - 2^A007814(n)) = a(2*A059894(n)) for n > 0 with a(0) = 1.
Sum_{k=0..2^n - 1} a(k) = A000108(n+1) for n >= 0.
a((4^n - 1)/3) = A000108(n) for n >= 0.
a(2^m*(2^n - 1)) = binomial(n + m, n) for n >= 0, m >= 0.
Generalization:
b(2n+1, p, q) = b(n, p, q) for n >= 0.
b(2n, p, q) = p*b(n, p, q) + q*b(n - 2^A007814(n), p, q) = for n > 0 with b(0, p, q) = 1.
Conjectured formulas: (Start)
Sum_{k=0..2^n - 1} b(k, 2, 1) = A006318(n) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 2, 2) = A115197(n) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 3, 1) = A108524(n+1) for n >= 0.
Sum_{k=0..2^n - 1} b(k, 3, 3) = A116867(n) for n >= 0.
b((4^n - 1)/3, p, q) is generalized Catalan number C(p, q; n). (End)
Conjecture: a(n) = T(n, wt(n)+1), a(2n) = Sum_{k=1..wt(n)+1} T(n, k) where T(2n+1, k) = T(n, k) for 1 <= k <= wt(n)+1, T(2n+1, wt(n)+2) = T(n, wt(n)+1), T(2n, k) = Sum_{i=1..k} T(n, i) for 1 <= k <= wt(n)+1 with T(0, 1) = 1. - Mikhail Kurkov, Dec 13 2024

A090365 Shifts 1 place left under the INVERT transform of the BINOMIAL transform of this sequence.

Original entry on oeis.org

1, 1, 3, 11, 47, 225, 1177, 6625, 39723, 251939, 1681535, 11764185, 86002177, 655305697, 5193232611, 42726002123, 364338045647, 3215471252769, 29331858429241, 276224445794785, 2682395337435723, 26832698102762435, 276221586866499839, 2923468922184615897

Offset: 0

Paul D. Hanna, Nov 26 2003

nonn

The Hankel transform of this sequence is A000178(n+1); example: det([1,1,3; 1,3,11; 3,11,47]) = 12. - Philippe Deléham, Mar 02 2005
a(n) appears to be the number of indecomposable permutations (A003319) of [n+1] that avoid both of the dashed patterns 32-41 and 41-32. - David Callan, Aug 27 2014
This is true: A nonempty permutation avoids 32-41 and 41-32 if and only if all its components do so. So if A(x) denotes the g.f. for indecomposable {32-41,41-32}-avoiders, then F(x):=1/(1-A(x)) is the g.f. for all {32-41,41-32}-avoiders. From A074664, F(x)=1/x(1-1/B(x)) where B(x) is the o.g.f. for the Bell numbers. Solve for A(x). - David Callan, Jul 21 2017
The Hankel transform of this sequence without the a(0)=1 term is also A000178(n+1). - Michael Somos, Oct 02 2024

Alois P. Heinz, Table of n, a(n) for n = 0..574

Cf. A090366, A090367.
Cf. A074664.
Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A329369, A341392, A347204, A347205.

bintr:= proc(p) proc(n) add(p(k) *binomial(n,k), k=0..n) end end:
invtr:= proc(p) local b;
           b:= proc(n) option remember; local i;
                `if`(n<1, 1, add(b(n-i) *p(i-1), i=1..n+1))
               end;
        end:
b:= invtr(bintr(a)):
a:= n-> `if`(n<0, 0, b(n-1)):
seq(a(n), n=0..30);  # Alois P. Heinz, Jun 28 2012

a[n_] := Module[{A, B}, A = 1+x; For[k=1, k <= n, k++, B = (A /. x -> x/(1 - x))/(1-x) + O[x]^n // Normal; A = 1 + x*A*B]; SeriesCoefficient[A, {x, 0, n}]]; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Oct 23 2016, adapted from PARI *)

{a(n)=local(A); if(n<0,0,A=1+x+x*O(x^n); for(k=1,n,B=subst(A,x, x/(1-x))/(1-x)+x*O(x^n); A=1+x*A*B);polcoeff(A,n,x))}

G.f.: A(x) satisfies A(x) = 1/(1 - A(x/(1-x))*x/(1-x) ).
a(n) = Sum_{k = 0..n} A085838(n, k). - Philippe Deléham, Jun 04 2004
G.f.: 1/x-1-1/(B(x)-1) where B(x) = g.f. for A000110 the Bell numbers. - Vladeta Jovovic, Aug 08 2004
a(n) = Sum_{k=0..n} A094456(n,k). - Philippe Deléham, Nov 07 2007
G.f.: 1/(1-x/(1-2x/(1-x/(1-3x/(1-x/(1-4x/(1-x/(1-5x/(1-... (continued fraction). - Paul Barry, Feb 25 2010
From Sergei N. Gladkovskii, Jan 06 2012  - May 12 2013: (Start)
Continued fractions:
G.f.: 1 - x/(G(0)+x); G(k) = x - 1 + x*k + x*(x-1+x*k)/G(k+1).
G.f.: 1/x - 1/2 + (x^2-4)/(4*U(0)-2*x^2+8) where U(k) = k*(2*k+3)*x^2 + x - 2 - (2-x+2*k*x)*(2+3*x+2*k*x)*(k+1)*x^2/U(k+1).
G.f.: 1/x+1/(U(0)-1) where U(k) = -x*k + 1 - x - x^2*(k+1)/U(k+1).
G.f.: (1 - U(0))/x - 1 where U(k) = 1 - x*(k+2) - x^2*(k+1)/U(k+1).
G.f.: (1 - U(0))/x where U(k) = 1 - x*(k+1)/(1-x/U(k+1)).
G.f.: 1/x + 1/( G(0)-1) where G(k) = 1 - x/(1 - x*(2*k+1)/(1 - x/(1 - x*(2*k+2)/ G(k+1) ))).
G.f.:1/x + 1/( G(0) - 1 ) where G(k) = 1 - x/(1 - x*(k+1)/G(k+1) ).
G.f.: (1 - Q(0))/x where Q(k) = 1 + x/(x*k - 1 )/Q(k+1).
G.f.: 1/x - 1/x/Q(0), where Q(k) = 1 + x/(1 - x + x*(k+1)/(x - 1/Q(k+1))).
(End)
Conjecture: a(n) = b(2^(n-1) - 1) for n > 0 with a(0) = 1 where b(n) = b((n - 2^f(n))/2) + b(floor((2n - 2^f(n))/2)) + b(A025480(n-1)) for n > 0 with b(0) = 1 and where f(n) = A007814(n). - Mikhail Kurkov, Jan 11 2022

A006014 a(n+1) = (n+1)a(n) + Sum_{k=1..n-1} a(k)a(n-k).

Original entry on oeis.org

1, 2, 7, 32, 178, 1160, 8653, 72704, 679798, 7005632, 78939430, 965988224, 12762344596, 181108102016, 2748049240573, 44405958742016, 761423731533286, 13809530704348160, 264141249701936818, 5314419112717217792, 112201740111374359516, 2480395343617443024896

Offset: 1

N. J. A. Sloane

			G.f. = x + 2*x^2 + 7*x^3 + 32*x^4 + 178*x^5 + 1160*x^6 + 8653*x^7 + 72704*x^8 + ...

D. E. Knuth, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Michael De Vlieger, Table of n, a(n) for n = 1..449
Jimmy Devillet and Bruno Teheux, Associative, idempotent, symmetric, and order-preserving operations on chains, arXiv:1805.11936 [math.RA], 2018.
E. Duchi, V. Guerrini, S. Rinaldi, and G. Schaeffer, Fighting fish. J. Phys. A, Math. Theor. 50, No. 2, Article ID 024002, 16 p. (2017), chapter 4.

Similar recurrences: A124758, A243499, A284005, A329369, A341392.

Cf. A007814, A025480, A073017.

Nest[Append[#1, #1[[-1]] (#2 + 1) + Total@ Table[#1[[k]] #1[[#2 - k]], {k, #2 - 1}]] & @@ {#, Length@ #} &, {1}, 17] (* Michael De Vlieger, Aug 22 2018 *)
(* or *)
a[1] = 1; a[n_] := a[n] = n a[n-1] + Sum[a[k] a[n-1-k], {k, n-2}]; Array[a, 18] (* Giovanni Resta, Aug 23 2018 *)

{a(n) = local(A); if( n<1, 0, A = vector(n); A[1] = 1; for( k=2, n, A[k] = k * A[k-1] + sum( j=1, k-2, A[j] * A[k-1-j])); A[n])} /* Michael Somos, Jul 24 2011 */

from sage.all import *
@CachedFunction
def a(n): # a = A006014
    if n<5: return (pow(5,n-1) + 3)//4
    else: return n*a(n-1) + 2*sum(a(k)*a(n-k-1) for k in range(1,(n//2))) + (n%2)*pow(a((n-1)//2),2)
print([a(n) for n in range(1,61)]) # G. C. Greubel, Jan 10 2025

G.f. A(x) satisfies A(x) = x * (1 + A(x) + A(x)^2 + x * A'(x)). - Michael Somos, Jul 24 2011
Conjecture: a(n) = Sum_{k=0..2^(n-1) - 1} b(k) for n > 0 where b(2n+1) = b(n), b(2n) = b(n) + b(n - 2^f(n)) + b(2n - 2^f(n)) + b(A025480(n-1)) for n > 0 with b(0) = b(1) = 1 and where f(n) = A007814(n). - Mikhail Kurkov, Nov 19 2021
a(n) ~ cosh(sqrt(3)*Pi/2) * n! / Pi = A073017 * n!. - Vaclav Kotesovec, Nov 21 2024

A253552 Permutation of natural numbers: a(n) = A252752(A005940(n+1)) - 1.

Original entry on oeis.org

1, 3, 2, 6, 4, 5, 7, 10, 11, 8, 16, 9, 37, 12, 29, 15, 22, 17, 46, 13, 106, 23, 67, 14, 301, 47, 154, 18, 352, 38, 121, 21, 56, 30, 92, 24, 211, 57, 191, 19, 596, 122, 436, 31, 991, 80, 277, 20, 1177, 327, 1226, 58, 2776, 173, 631, 25, 7751, 380, 1432, 48, 3241, 138, 497, 28, 79, 68, 232, 39, 529, 107, 379, 32, 1486, 233, 862, 69, 1954, 212, 781, 26

Offset: 1

Antti Karttunen, Jan 03 2015

Antti Karttunen, Table of n, a(n) for n = 1..8192
Index entries for sequences that are permutations of the natural numbers

Inverse: A253551.
Cf. A005940, A252752.
Differs from A249725 for the first time at n=13, where a(13) = 37, while A249725(13) = 22.
Cf. also A243499 & A253563 (for similar scatterplots).

(define (A253552 n) (+ -1 (A252752 (A005940 (+ 1 n)))))

a(n) = A252752(A005940(n+1)) - 1.

A290615 Number of maximal independent vertex sets (and minimal vertex covers) in the n-triangular honeycomb bishop graph.

Original entry on oeis.org

1, 2, 5, 14, 45, 164, 661, 2906, 13829, 70736, 386397, 2242118, 13759933, 88975628, 604202693, 4296191090, 31904681877, 246886692680, 1986631886029, 16592212576862, 143589971363981, 1285605080403332, 11891649654471285, 113491862722958474, 1116236691139398565

Offset: 1

Eric W. Weisstein, Aug 07 2017

nonn

From Andrew Howroyd, Aug 09 2017: (Start)
See A146304 for algorithm and PARI code to produce this sequence.
The total number of independent vertex sets is given by Bell(n+1) where Bell=A000110.
A bishop can move along two axes in the triangular honeycomb grid.
Equivalently, the number of arrangements of non-attacking rooks on an n X n right triangular board with every square controlled by at least one rook. (End)

Andrew Howroyd, Table of n, a(n) for n = 1..150
Max Alekseyev, Subsequences of odd powers, answer to question on Mathoverflow.
Max Alekseyev, Generating function for partial sums of the sequence, answer to question on Mathoverflow.
Peter Taylor, Subsequence of the cubes, answer to question on Mathoverflow.
Eric Weisstein's World of Mathematics, Maximal Independent Vertex Set
Eric Weisstein's World of Mathematics, Minimal Vertex Cover

Row sums of A290724.
Cf. A000110 (independent vertex sets), A007814, A146304.
Similar recurrences: A124758, A243499, A284005, A329369, A341392.

Table[Sum[k! StirlingS2[m, k] StirlingS2[n + 1 - m, k + 1], {m, 0, n}, {k, 0, Min[m, n - m]}], {n, 20}] (* Eric W. Weisstein, Feb 01 2024 *)

{ A290615(n) = sum(m=0, n, sum(k=0, min(m,n-m), k! * stirling(m,k,2) * stirling(n+1-m,k+1,2) )); } \\ Max Alekseyev, Oct 14 2021

Conjecture: a(n) = Sum_{k=0..2^(n-1) - 1} b(k) for n > 0 where b(2n+1) = b(n - 2^f(n)), b(2n) = b(n) + b(2n - 2^f(n)) for n > 0 with b(0) = b(1) = 1 and where f(n) = A007814(n). Also b((4^n - 1)/3) = (floor((n+1)/2)!)^3. - Mikhail Kurkov, Sep 18 2021

a(n) = Sum_{m=0..n} Sum_{k=0..min(m,n-m)} k! * S(m,k) * S(n+1-m,k+1), where S(,) are Stirling numbers of second kind. - Max Alekseyev, Oct 14 2021

Terms a(10) and beyond from Andrew Howroyd, Aug 09 2017

A347204 a(n) = a(f(n)/2) + a(floor((n+f(n))/2)) for n > 0 with a(0) = 1 where f(n) = A129760(n).

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 10, 15, 5, 9, 13, 20, 17, 27, 37, 52, 6, 11, 16, 25, 21, 34, 47, 67, 26, 43, 60, 87, 77, 114, 151, 203, 7, 13, 19, 30, 25, 41, 57, 82, 31, 52, 73, 107, 94, 141, 188, 255, 37, 63, 89, 132, 115, 175, 235, 322, 141, 218, 295, 409, 372, 523, 674

Offset: 0

Mikhail Kurkov, Aug 23 2021 [verification needed]

Modulo 2 binomial transform of A243499(n).

Antti Karttunen, Table of n, a(n) for n = 0..8191
Kevin Ryde, PARI/GP Code

Cf. A000110, A000120, A002720, A005493, A005494, A007814, A008277, A035009, A045379, A053645, A129760, A143494, A143495, A196834, A265649, A295989.

Similar recurrences: A124758, A243499, A284005, A329369, A341392.

function a = A347204(max_n)
    a(1) = 1;
    a(2) = 2;
    for nloop = 3:max_n
        n = nloop-1;
        s = 0;
        for k = 0:floor(log2(n))-1
            s = s + a(1+A053645(n)-2^k*(mod(floor(n/(2^k)),2)));
        end
        a(nloop) = 2*a(A053645(n)+1) + s;
    end
end
function a_n = A053645(n)
    a_n = n - 2^floor(log2(n));
end % Thomas Scheuerle, Oct 25 2021

f[n_] := BitAnd[n, n - 1]; a[0] = 1; a[n_] := a[n] = a[f[n]/2] + a[Floor[(n + f[n])/2]]; Array[a, 100, 0] (* Amiram Eldar, Nov 19 2021 *)

f(n) = bitand(n, n-1); \\ A129760
a(n) = if (n<=1, n+1, if (n%2, a(n\2)+a(n-1), a(f(n/2)) + a(n/2+f(n/2)))); \\ Michel Marcus, Oct 25 2021

PARI
```
\\ Also see links.
```

A129760(n) = bitand(n, n-1);
memoA347204 = Map();
A347204(n) = if (n<=1, n+1, my(v); if(mapisdefined(memoA347204,n,&v), v, v = if(n%2, A347204(n\2)+A347204(n-1), A347204(A129760(n/2)) + A347204(n/2+A129760(n/2))); mapput(memoA347204,n,v); (v))); \\ (Memoized version of Michel Marcus's program given above) - Antti Karttunen, Nov 20 2021

a(n) = a(n - 2^f(n)) + (1 + f(n))*a((n - 2^f(n))/2) for n > 0 with a(0) = 1 where f(n) = A007814(n).
a(2n+1) = a(n) + a(2n) for n >= 0.
a(2n) = a(n - 2^f(n)) + a(2n - 2^f(n)) for n > 0 with a(0) = 1 where f(n) = A007814(n).
a(n) = 2*a(f(n)) + Sum_{k=0..floor(log_2(n))-1} a(f(n) - 2^k*T(n,k)) for n > 1 with a(0) = 1, a(1) = 2, and where f(n) = A053645(n), T(n,k) = floor(n/2^k) mod 2.
Sum_{k=0..2^n - 1} a(k) = A035009(n+1) for n >= 0.
a((4^n - 1)/3) = A002720(n) for n >= 0.
a(2^n - 1) = A000110(n+1),
a(2*(2^n - 1)) = A005493(n),
a(2^2*(2^n - 1)) = A005494(n),
a(2^3*(2^n - 1)) = A045379(n),
a(2^4*(2^n - 1)) = A196834(n),
a(2^m*(2^n-1)) = T(n,m+1) is the n-th (m+1)-Bell number for n >= 0, m >= 0 where T(n,m) = m*T(n-1,m) + Sum_{k=0..n-1} binomial(n-1,k)*T(k,m) with T(0,m) = 1.
a(n) = Sum_{j=0..2^A000120(n)-1} A243499(A295989(n,j)) for n >= 0. Also A243499(n) = Sum_{j=0..2^f(n)-1} (-1)^(f(n)-f(j)) a(A295989(n,j)) for n >= 0 where f(n) = A000120(n). In other words, a(n) = Sum_{j=0..n} (binomial(n,j) mod 2)*A243499(j) and A243499(n) = Sum_{j=0..n} (-1)^(f(n)-f(j))*(binomial(n,j) mod 2)*a(j) for n >= 0 where f(n) = A000120(n).
Generalization:
b(n, x) = (1/x)*b((n - 2^f(n))/2, x) + (-1)^n*b(floor((2n - 2^f(n))/2), x) for n > 0 with b(0, x) = 1 where f(n) = A007814(n).
Sum_{k=0..2^n - 1} b(k, x) = (1/x)^n for n >= 0.
b((4^n - 1)/3, x) = (1/x)^n*n!*L_{n}(x) for n >= 0 where L_{n}(x) is the n-th Laguerre polynomial.
b((8^n - 1)/7, x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A265649(n, k) for n >= 0.
b(2^n - 1, x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A008277(n+1, k+1),
b(2*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A143494(n+2, k+2),
b(2^2*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*A143495(n+3, k+3),
b(2^m*(2^n - 1), x) = (1/x)^n*Sum_{k=0..n} (-x)^k*T(n+m+1, k+m+1, m+1) for n >= 0, m >= 0 where T(n,k,m) is m-Stirling numbers of the second kind.

cp's OEIS Frontend

A011965 Second differences of Bell numbers.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Python

Python

Formula

A227184 a(n) = product of parts of the unordered partition encoded with the runlengths of binary expansion of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python

Scheme

Formula

A217924 a(n) = n! * [x^n] exp(2*exp(x) - x - 2). Row sums of triangle A217537.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Maxima

Sage

SageMath

Formula

Extensions

A347205 a(2n+1) = a(n) for n >= 0, a(2n) = a(n) + a(n - 2^A007814(n)) for n > 0 with a(0) = 1.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A090365 Shifts 1 place left under the INVERT transform of the BINOMIAL transform of this sequence.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A006014 a(n+1) = (n+1)*a(n) + Sum_{k=1..n-1} a(k)*a(n-k).

Views

Author

Keywords

Examples

References

Links

Crossrefs

Programs

Mathematica

PARI

A006014 a(n+1) = (n+1)a(n) + Sum_{k=1..n-1} a(k)a(n-k).