A014622 -id:A014622 - cp's OEIS frontend

A011784 Levine's sequence. First construct a triangle as follows. Row 1 is {1,1}; if row n is {r_1, ..., r_k} then row n+1 consists of {r_k 1's, r_{k-1} 2's, r_{k-2} 3's, etc.}; sequence consists of the final elements in each row.

1, 2, 2, 3, 4, 7, 14, 42, 213, 2837, 175450, 139759600, 6837625106787, 266437144916648607844, 508009471379488821444261986503540, 37745517525533091954736701257541238885239740313139682, 5347426383812697233786139576220450142250373277499130252554080838158299886992660750432

Offset: 1

Lionel Levine (levine(AT)ultranet.com)

Additional remarks.
The sequence is generated by this array, the final term in each row forming the sequence:
  1 1
  1 2
  1 1 2
  1 1 2 3
  1 1 1 2 2 3 4
  1 1 1 1 2 2 2 3 3 4 4 5 6 7
  1 1 1 1 1 1 1 2 2 2 2 2 2 3 3 3 3 3 4 4 4 4 5 5 5 5 6 6 6 7 7 7 8 8 9 9 10 10 11 12 13 14
  ...
where we start with the first row {1 1} and produce the rest of the array recursively as follows:
Suppose line n is {a_1, ..., a_k}; then line n+1 contains a_k 1's, a_{k-1} 2's, etc.
So the fifth line contains three 1's, two 2's, one 3 and one 4.
The sequence is 1,2,2,3,4,7,14,42,213,2837,175450,...,
where the n-th term a(n) is the sum of the elements in row n-2
= the number of elements in row n-1
= the last element in row n
= the number of 1's in row n+1
= ...
If the n-th row is r_{n,i} then
Sum_{i=1..f(n+1)} (a(n+1) - i + 1)*r_{n,i} ) = a(n+3)
Let {a( )} be the sequence; s(i,j) = j-th partial sum of the i-th row,
L(i) is the length of that row and S(i) = its sum. Then
L(i+1) = a(i+2) = S(i) = s(i,a(i+1));
L(i+2) = SUM(s(i,j));
L(i+3) = SUM(s(i,j)*(1+s(i,j))/2) (Allan Wilks).
Eric Rains and Bjorn Poonen have shown (June 1997) that the log of the n-th term is asymptotic to constant times phi^n, where phi = golden number.
This follows from the inequalities S(n) <= a(n)L(n) and S(n+1) >= ([L(n+1)/a(n)]+1) choose 2)*a(n). See N. J. A. Sloane et al., Scans of Notebook pages.
The n-th term is approximately exp(a*phi^n)/I, where phi = golden number, a = .05427 (last digit perhaps 6 or 8), I = .277 (last digit perhaps 6 or 8) (Colin Mallows).

			{1,1}, {1,2}, {1,1,2}, {1,1,2,3}, {1,1,1,2,2,3,4}, {1,1,1,1,2,2,2,3,3,4,4,5,6,7}.

Richard K. Guy, Unsolved Problems in Number Theory, Section E25.
R. K. Guy, What's left?, in The Edge of the Universe: Celebrating Ten Years of Math Horizons, Deanna Haunsperger, Stephen Kennedy (editors), 2006, p. 81.

Johan Claes, Table of n, a(n) for n = 1..19
Johnson Ihyeh Agbinya, Computer Board Games of Africa, (2004), see pages 113-114.
R. K. Guy, What's left?, Math Horizons, Vol. 5, No. 4 (April 1998), pp. 5-7.
Roland Miyamoto, Polynomial parametrisation of the canonical iterates to the solution of -gamma*g' = g^(-1), arXiv:2402.06618 [math.CO], 2024. See pp. 16-17.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane and Brady Haran, The Levine Sequence, Numberphile video (2021)
N. J. A. Sloane, Colin Mallows, and Bjorn Poonen, Discussion of A011784. [Scans of pages 150-155 and 164 of my notebook "Lattices 77", from June-July 1997.]

Cf. A012257, A014621, A014622.

a011784 = last . a012257_row  -- Reinhard Zumkeller, Aug 11 2014

(* This script is not suitable for computing more than 11 terms *) nmax = 11; ro = {{2, 1}}; a[1]=1; For[n=2, n <= nmax, n++, ro = Transpose[{Table[#[[2]], {#[[1]]}]& /@ Reverse[ro] // Flatten, Range[Total[ro[[All, 1]]]]}]; Print["a(", n, ") = ", a[n] = ro // Last // Last]]; Array[a, nmax] (* Jean-François Alcover, Feb 25 2016 *)
NestList[Flatten@ MapIndexed[ConstantArray[First@ #2, #1] &, Reverse@ #] &, {1, 1}, 10][[All, -1]] (* Michael De Vlieger, Jul 12 2017, same limitations as above *)

# This works, as with the others, up to 11.
lev2 <- function(x = 10, levprev= NULL){
x <- floor(x[1])
# levlen is the RLE values
levterm <-rep(1,x)
levlen[[1]] <- 2
for ( jl in 2:x) {
rk <- length(levlen[[jl-1]])
for (jrk in 1: rk) {
levlen[[jl]] <- c(levlen[[jl]], rep(jrk, times = levlen[[jl-1]][rk+1-jrk])) }
levterm[jl] <- length(levlen[[jl]]) }
return(invisible(list(levlen=levlen, levterm = levterm) ) ) }
# Carl Witthoft, Apr 01 2021

a(n+2) = n-th row sum of A012257; e.g., 5th row of A012257 is {1, 1, 1, 2, 2, 3, 4} and the sum of elements is 1+1+1+2+2+3+4=14=a(7) - Benoit Cloitre, Aug 06 2003

a(n) = A012257(n,a(n+1)). - Reinhard Zumkeller, Aug 11 2014

a(12) from Colin Mallows, a(13) from N. J. A. Sloane, a(14) and a(15) from Allan Wilks
a(16) from Johan Claes, Jun 09 2004
a(17) (an 85-digit number) from Johan Claes, Jun 18 2004
Edited by N. J. A. Sloane, Mar 08 2006
a(18) (a 137-digit number) from Johan Claes, Aug 19 2008

A210949 E.g.f. A(x) satisfies: A'(x) = 1/(1 - A(A(x))).

Original entry on oeis.org

1, 1, 4, 29, 309, 4383, 78121, 1684706, 42801222, 1255919755, 41918624013, 1572257236114, 65619165625383, 3022617826829288, 152615633802149416, 8397224009015443509, 500957609480739613321, 32261529179806961067634, 2234133327582388824135291

Offset: 1

Paul D. Hanna, Jul 22 2012

nonn

An unsigned version of A067146.
Equals row sums of triangle A277410.
Is this a duplicate of A014622, which is related to f(f(x))f'(x)=-1 ? - R. J. Mathar, May 13 2025

			E.g.f.: A(x) = x + x^2/2! + 4*x^3/3! + 29*x^4/4! + 309*x^5/5! + 4383*x^6/6! +...
Let G(x) = Integral A(x) dx, then A(x) = x + G(A(x)) where
G(x) = x^2/2! + x^3/3! + 4*x^4/4! + 29*x^5/5! + 309*x^6/6! + 4383*x^7/7! +...
Also,
A(x) = x + G(x) + d/dx G(x)^2/2! + d^2/dx^2 G(x)^3/3! + d^3/dx^3 G(x)^4/4! +...
log(A(x)/x) = G(x)/x + d/dx G(x)^2/(2!*x) + d^2/dx^2 G(x)^3/(3!*x) + d^3/dx^3 G(x)^4/(4!*x) +...
By definition, A'(x) = 1/(1 - A(A(x))), where
A(A(x)) = x + 2*x^2/2! + 11*x^3/3! + 101*x^4/4! + 1313*x^5/5! + 22235*x^6/6! + 466356*x^7/7! + 11710760*x^8/8! +...

Vaclav Kotesovec, Table of n, a(n) for n = 1..180

Cf. A277410, A067146, A279843, A279844, A279845, A280570, A280571, A280572, A280573, A280574, A280575.

m = 20; A[_] = 0;
Do[A[x_] = InverseSeries[Integrate[1 - A[x], x] + O[x]^m], {m}];
CoefficientList[A[x], x] * Range[0, m - 1]! // Rest (* Jean-François Alcover, Sep 30 2019 *)

{a(n)=local(A=x,G);for(i=1,n,G=intformal(A+x*O(x^n));A=serreverse(x-G));n!*polcoeff(A, n)}

{a(n)=local(A=x,G);for(i=1,n,G=intformal(A+x*O(x^n));A=x+subst(G,x,A+x*O(x^n))); n!*polcoeff(A, n)}

{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=x,G);for(i=0,n,G=intformal(A+x*O(x^n)); A=x+sum(m=1, n, Dx(m-1, G^m/m!))); n!*polcoeff(A, n)}

{Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D}
{a(n)=local(A=x,G);for(i=0,n,G=intformal(A+x*O(x^n)); A=x*exp(sum(m=1, n, Dx(m-1, G^m/x/m!)+x*O(x^n)))); n!*polcoeff(A, n)}
for(n=1, 25, print1(a(n), ", "))

{a(n) = local(A=x); for(i=1,n, A = serreverse(intformal(1-A +x*O(x^n)))); n!*polcoeff(A,n)}
for(n=1, 25, print1(a(n), ", "))

E.g.f. A(x) satisfies:
(1) A(x) = Series_Reversion( Integral 1 - A(x) dx ).
(2) A''(x) = 1 / ( (1 - A(A(x)))^3 * (1 - A(A(A(x)))) ).
Let G(x) = Integral A(x) dx with G(0)=0, then the e.g.f. A(x) satisfies:
(3) A(x) = x + G(A(x)) or, equivalently, A(x - G(x)) = x.
(4) A(x) = x + Sum_{n>=1} d^(n-1)/dx^(n-1) G(x)^n / n!.
(5) A(x) = x*exp( Sum_{n>=1} d^(n-1)/dx^(n-1) G(x)^n/(n!*x) ).
a(n) = Sum_{k=0..n-1} A277410(n,k).

A014621 Triangle of numbers arising from analysis of Levine's sequence A011784.

Original entry on oeis.org

1, 1, 3, 1, 15, 10, 3, 1, 105, 105, 55, 30, 10, 3, 1, 945, 1260, 910, 630, 350, 168, 76, 30, 10, 3, 1, 10395, 17325, 15750, 12880, 9135, 5789, 3381, 1806, 910, 434, 196, 76, 30, 10, 3, 1, 135135, 270270, 294525, 275275, 228375, 172200, 120960, 78519, 48006, 28336, 16065, 8609, 4461, 2166, 1018, 470, 196, 76, 30, 10, 3, 1

Offset: 1

Colin Mallows

			Triangle begins:
    1;
    1;
    3,    1;
   15,   10,   3,   1;
  105,  105,  55,  30,  10,   3,  1;
  945, 1260, 910, 630, 350, 168, 76, 30, 10, 3, 1;
10395, 17325, 15750, 12880, 9135, 5789, 3381, 1806, 910, 434, 196, 76, 30,
  10, 3, 1;
135135, 270270, 294525, 275275, 228375, 172200, 120960, 78519, 48006, 28336, 16065, 8609, 4461, 2166, 1018, 470, 196, 76, 30, 10, 3, 1;
2027025, 4729725, 5990985, 6276270, 5853925, 4996530, 3999765, 2997225, 2115960, 1432725, 938644, 593646, 364551, 215940, 123639, 68886, 37276, 19485, 9959, 4911, 2301, 1063, 470, 196, 76, 30, 10, 3, 1;

Roland Miyamoto, Rows n = 1..50 of triangle, flattened (rows n = 1..6 from _Colin Mallows_)
Roland Miyamoto, Comments on A014621, Oct 14 2022.
Roland Miyamoto and J. W. Sander, Solving the iterative differential equation -gamma*g' = g^{-1}, in: H. Maier, J. & R. Steuding (eds.), Number Theory in Memory of Eduard Wirsing, Springer, 2023, pp. 223-236, alternative link.
Roland Miyamoto, Python3 program a014621.py implementing below formulae.
Roland Miyamoto, Polynomial parametrisation of the canonical iterates to the solution of -gamma*g' = g^(-1), arXiv:2402.06618 [math.CO], 2024.

Cf. A011784, A014622 (row sums), A144006.

Python
```
# See Miyamoto link.
```

From Roland Miyamoto, Nov 20 2022: (Start)
The n-th row contains 1 + (n-1)*(n-2)/2 numbers a(n,k), where n >= 1 and k = 0..(n-1)*(n-2)/2.
Let f be a solution to the iterative differential equation f(f(x))*f'(x) = -1 defined on some nonnegative interval and let tau=f(tau) be a fixed point of f. Then the n-th derivative of f at tau is
  f^{(n)}(tau) = Sum_{k=0..(n-1)*(n-2)/2} (-1)^(n+k)*a(n,k)*tau^(2-3*n-k).
Thus, a(n,k) can be calculated recursively using the equations
  0 = (f ° f * f')^{(n)} = Sum_{k=0..n} binomial(n,k) (f ° f)^{(n-k)}*f^{(k+1)} for n=1,2,... (End)

More terms from Roland Miyamoto, Nov 20 2022

Offset corrected by Max Alekseyev, Sep 19 2023

A144006 Triangle, read by rows of coefficients of x^ny^k for k=0..n(n-1)/2 for n>=0, defined by e.g.f.: A(x,y) = 1 + Series_Reversion( Integral A(-xy,y) dx ), with leading zeros in each row suppressed.

Original entry on oeis.org

1, 1, 1, 3, -1, 15, -10, 3, -1, 105, -105, 55, -30, 10, -3, 1, 945, -1260, 910, -630, 350, -168, 76, -30, 10, -3, 1, 10395, -17325, 15750, -12880, 9135, -5789, 3381, -1806, 910, -434, 196, -76, 30, -10, 3, -1, 135135, -270270, 294525, -275275, 228375

Offset: 0

Paul D. Hanna, Sep 10 2008

Comment from Lucas Larsen, Aug 20 2024: (Start)
The nonzero entries in the n-th row appear to be the nonzero coefficients (up to sign) in the following:
 Let c be a fixed point in (0,oo) and f a smooth function such that f(c) = c and f(f'(x)) = x in a neighborhood of c. Then the n-th derivative of f evaluated at c can be written as a Laurent polynomial in c with the (descending) coefficients in question.
For instance:
 f'(c) = c
 f''(c) = c^(-1)
 f'''(c) = -c^(-4)
 f''''(c) = 3c^(-7) + c^(-8)
(End)

			Triangle begins (without suppressing leading zeros):
1;
1;
0, 1;
0,0, 3, -1;
0,0,0, 15, -10, 3, -1;
0,0,0,0, 105, -105, 55, -30, 10, -3, 1;
0,0,0,0,0, 945, -1260, 910, -630, 350, -168, 76, -30, 10, -3, 1;
0,0,0,0,0,0, 10395, -17325, 15750, -12880, 9135, -5789, 3381, -1806, 910, -434, 196, -76, 30, -10, 3, -1;
0,0,0,0,0,0,0, 135135, -270270, 294525, -275275, 228375, -172200, 120960, -78519, 48006, -28336, 16065, -8609, 4461, -2166, 1018, -470, 196, -76, 30, -10, 3, -1; ...

Cf. A144005, A067146.

Generates A014621, A014622 and A014623, which are related to Levine's sequence A011784.

{T(n,k)=local(A=1+x*O(x^n)); for(i=0,n,A=1+serreverse(intformal(subst(A,x,-x*y))));n!*polcoeff(polcoeff(A,n,x),k,y)}

#This is only correct if the observation in the comment from 2024/08/20 is true.
def T(n,k):
    if 0 <= n <= 1:
        return 1 if k == 0 else 0
    c = {(-1,):1} #Polynomial in infinitely many variables (function iterates)
    for _ in range(n-1):
        cnext = {}
        for key, value in c.items():
            key += (0,)
            for i, ni in enumerate(key):
                term = tuple(nj-2 if j==i else nj-1 if j<=i+1 else nj
                             for j,nj in enumerate(key))
                cnext[term] = cnext.get(term,0) + value*ni
                if cnext[term] == 0:
                    del cnext[term]
        c = cnext
    pairs = {} #Reduction to single variable (evaluation at fixpoint)
    for key, value in c.items():
        s = -sum(key)
        pairs[s] = pairs.get(s,0) + value
    _, row = zip(*sorted(pairs.items())) #Coefficients
    if 0 <= k-n+1 < len(row): #Correcting number of leading 0s
        return (-1)**(n+k+1)*abs(row[k-n+1]) #Correcting signs
    else:
        return 0
# Lucas Larsen, Aug 22 2024

E.g.f. satisfies: A(x,y) = 1 + Series_Reversion[Integral A(-x*y,y) dx].
T(n,k) = [x^n*y^k] n!*A(x,y) for k=0..n(n-1)/2, n>=0.
Row sums equal A144005.
A067146(n) = Sum_{k=0..n(n-1)/2} (-1)^k*T(n,k).
This is a signed version of table A014621 because setting f((1+x)/y):=A(-x*y,y)/y for fixed y>0 implies f(f(x))*f'(x)=-1 and f(1/y)=1/y, as in the second formula of A014621. Therefore, the row sums form A014623 and the unsigned row sums form A014622. - Roland Miyamoto, Jun 03 2024

cp's OEIS Frontend

A011784 Levine's sequence. First construct a triangle as follows. Row 1 is {1,1}; if row n is {r_1, ..., r_k} then row n+1 consists of {r_k 1's, r_{k-1} 2's, r_{k-2} 3's, etc.}; sequence consists of the final elements in each row.

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A210949 E.g.f. A(x) satisfies: A'(x) = 1/(1 - A(A(x))).

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A014621 Triangle of numbers arising from analysis of Levine's sequence A011784.

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A144006 Triangle, read by rows of coefficients of x^n*y^k for k=0..n(n-1)/2 for n>=0, defined by e.g.f.: A(x,y) = 1 + Series_Reversion( Integral A(-x*y,y) dx ), with leading zeros in each row suppressed.

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A144006 Triangle, read by rows of coefficients of x^ny^k for k=0..n(n-1)/2 for n>=0, defined by e.g.f.: A(x,y) = 1 + Series_Reversion( Integral A(-xy,y) dx ), with leading zeros in each row suppressed.