A081039 -id:A081039 - cp's OEIS frontend

A081038 3rd binomial transform of (1,2,0,0,0,0,0,0,...).

1, 5, 21, 81, 297, 1053, 3645, 12393, 41553, 137781, 452709, 1476225, 4782969, 15411789, 49424013, 157837977, 502211745, 1592728677, 5036466357, 15884240049, 49977243081, 156905298045, 491636600541, 1537671920841

Offset: 0

Paul Barry, Mar 03 2003

a(n) is the number of distinguished parts in all compositions of n+1 in which some (possibly all or none) of the parts have been distinguished. a(1) = 2 because we have: 2', 1'+1, 1+1', 1'+1' where we see 5's marking the distinguished parts. With offset=1, a(n) = Sum_{k=1..n} A200139(n,k)*k. - Geoffrey Critzer, Jan 12 2013
For n>=1, a(n-1) the number of ternary strings of length 2n containing the block 11..12 with n ones where no runs of length larger than n are permitted. - Marko Riedel, Mar 08 2016
Binomial transform of {A001787(n + 1)}{n >= 0}. - _Wolfdieter Lang, Oct 01 2019

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Jia Huang, Partially Palindromic Compositions, J. Int. Seq. (2023) Vol. 26, Art. 23.4.1. See p. 13.
Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021. See p. 67.
Index entries for linear recurrences with constant coefficients, signature (6,-9).

Cf. A001787, A001792, A081039, A081040, A086972, A200139.
First differences of A027471.
Cf. A269914, A269915, A269916, A269917.

[(2*n+3)*3^(n-1): n in [0..30]]; // Vincenzo Librandi, Jun 09 2011

A081038:=n->(2*n+3)*3^(n-1): seq(A081038(n), n=0..30); # Wesley Ivan Hurt, Mar 07 2016

LinearRecurrence[{6,-9},{1,5},40] (* Harvey P. Dale, Jun 22 2012 *)

G.f.: (1-x)/(1-3*x)^2.
a(n) = 6*a(n-1) - 9*a(n-2), with a(0)=1, a(1)=5.
a(n) = (2*n+3)*3^(n-1).
a(n) = Sum_{k=0..n} (k+1)*2^k*binomial(n, k).
a(n) = 2*A086972(n) - 1. - Lambert Herrgesell (zero815(AT)googlemail.com), Feb 10 2008
From Amiram Eldar, May 17 2022: (Start)
Sum_{n>=0} 1/a(n) = 9*(sqrt(3)*arctanh(1/sqrt(3)) - 1).
Sum_{n>=0} (-1)^n/a(n) = 9 - 3*sqrt(3)*Pi/2. (End)
E.g.f.: exp(3*x)*(1 + 2*x). - Stefano Spezia, Jan 31 2025

A081040 5th binomial transform of (1,4,0,0,0,0,...).

Original entry on oeis.org

1, 9, 65, 425, 2625, 15625, 90625, 515625, 2890625, 16015625, 87890625, 478515625, 2587890625, 13916015625, 74462890625, 396728515625, 2105712890625, 11138916015625, 58746337890625, 308990478515625, 1621246337890625

Offset: 0

Paul Barry, Mar 03 2003

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Silvana Ramaj, New Results on Cyclic Compositions and Multicompositions, Master's Thesis, Georgia Southern Univ., 2021. See p. 67.
Index entries for linear recurrences with constant coefficients, signature (10,-25).

Cf. A081039, A081041.

[(4*n+5)*5^(n-1): n in [0..25]]; // Vincenzo Librandi, Aug 06 2013

CoefficientList[Series[(1 - x) / (1 - 5 x)^2, {x, 0, 30}], x] (* Vincenzo Librandi, Aug 06 2013 *)
LinearRecurrence[{10,-25},{1,9},30] (* Harvey P. Dale, Jan 10 2021 *)

a(n)=(4*n+5)*5^(n-1) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 10*a(n-1) - 25*a(n-2), a(0)=1, a(1)=9.
a(n) = (4n+5)*5^(n-1).
a(n) = Sum_{k=0..n} (k+1)*4^k*binomial(n, k).
G.f.: (1-x)/(1-5*x)^2.
E.g.f.: exp(5*x)*(1 + 4*x). - Stefano Spezia, Jan 31 2025

A119733 Offsets of the terms of the nodes of the reverse Collatz function.

Original entry on oeis.org

0, 1, 2, 5, 4, 7, 10, 19, 8, 11, 14, 23, 20, 29, 38, 65, 16, 19, 22, 31, 28, 37, 46, 73, 40, 49, 58, 85, 76, 103, 130, 211, 32, 35, 38, 47, 44, 53, 62, 89, 56, 65, 74, 101, 92, 119, 146, 227, 80, 89, 98, 125, 116, 143, 170, 251, 152, 179, 206, 287, 260, 341, 422, 665, 64, 67

Offset: 0

William Entriken, Jun 14 2006

nonn

Create a binary tree starting with x. To follow 0 from the root, apply f(x)=2x. To follow 1, apply g(x)=(2x-1)/3. For example, starting with x, the string 010 {also known as f(g(f(x)))}, you would get (8x-2)/3. These expressions represent the reverse Collatz function and will provide numbers whose Collatz path may include x. These expressions will all be of the form (2^a*x-b)/3^c. This sequence concerns b. What makes b interesting is that if you draw the tree, each level of the tree will have the same sequence of values for b. The root of the tree x, can be written as (2^0*x-0)/3^0, which has the first value for b. Each subsequent level contains twice as many values of b.
This sequence is 0 followed by a permutation of A213539, and therefore consists of 0 plus the elements of A116640 multiplied by 2^k, where k >= 0. E.g., 1, 5, 7, 19 becomes 0, 2^0*1, 2^1*1, 2^0*5, 2^2*1, 2^0*7, 2^1*5, 2^0*19, ... - Joe Slater, Dec 19 2016
When this sequence is arranged as an irregular triangle the sum of each row a(2^k)...a(2^(k+1)-1) equals A081039(2^k). The cumulative sum from a(0) to a(2^k-1) equals A002697(k). - Joe Slater, Apr 12 2018

			a(1) = 1 = 2 * 0 + 3^0 since 0 written in binary contains no 1's.

Alois P. Heinz, Table of n, a(n) for n = 0..16383
Víctor Martín Chabrera, An algebraic fractal approach to Collatz Conjecture, Bachelor tesis, Universitat Politècnica de Catalunya (Barcelona, 2019), see lemma 6.1 with a(n) = r(A005836(n)).

Cf. A002697, A081039, A116623, A116640, A116641, A213539, A226383.

a:= proc(n) `if`(n=0, 0, `if`(irem(n, 2, 'r')=0, 0,
      3^add(i, i=convert(r, base, 2)))+2*a(r))
    end:
seq(a(n), n=0..127);  # Alois P. Heinz, Aug 13 2017

a[0] := 0; a[n_?OddQ] := 2a[(n - 1)/2] + 3^Plus@@IntegerDigits[(n - 1)/2, 2]; a[n_?EvenQ] := 2a[n/2]; Table[a[n], {n, 0, 65}] (* Alonso del Arte, Apr 21 2011 *)

a(n) = my(ret=0); if(n, for(i=0,logint(n,2), if(bittest(n,i), ret=3*ret+1<Kevin Ryde, Oct 22 2021

# call with n to get 2^n values
$depth=shift; sub funct { my ($i, $b, $c) = @_; if ($i < $depth) { funct($i+1, $b*2, $c); funct($i+1, 2*$b+$c, $c*3); } else { print "$b, "; } } funct(0, 0, 1); print " ";

from sympy.core.cache import cacheit
@cacheit
def a(n): return 0 if n==0 else 2*a((n - 1)//2) + 3**bin((n - 1)//2).count('1') if n%2 else 2*a(n//2)
print([a(n) for n in range(131)]) # Indranil Ghosh, Aug 13 2017

a(0) = 0, a(2*n + 1) = 2*a(n) + 3^wt(n) = 2*a(n) + A048883(n), a(2*n) = 2*a(n), where wt(n) = A000120(n) = the number 1's in the binary representation of n.
a(k) = [z^k] 1 + (1/(1-z)) * Sum_{s=0..n-1} 2^s*z^(2^s)*(1 - z^(2^s)) * Product_{r=s+1..n-1} (1 + 3*z^(2^r)), for 0 < k <= 2^n-1. - Wolfgang Hintze, Jul 28 2017
a(n) = Sum_{i=0..k} 2^e[i] * 3^i where binary expansion n = 2^e[0] + 2^e[1] + ... + 2^e[k] with descending e[0] > e[1] > ... > e[k] (A272011). [Martín Chabrera lemma 6.1, adapting index i] - Kevin Ryde, Oct 22 2021

A169585 A000004 preceded by 1, 3.

Original entry on oeis.org

1, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 0

Klaus Brockhaus, Dec 02 2009

Inverse binomial transform of A016777; second inverse binomial transform of A053220; third inverse binomial transform of A027471 without first term; fourth inverse binomial transform of A081039.

Cf. A000004 (zero sequence), A016777 (3*n+1), A053220 ((3*n-1)*2^(n-2)), A027471 ((n-1)*3^(n-2)), A081039 ((3*n+4)*4^(n-1), a(0)=1, a(1)=7), A130706 (1, 2, 0, 0, 0, ...), A166926 (1, 2, 4, 0, 0, 0, ...), A130779 (1, 1, 2, 0, 0, 0, ...).

PARI
```
{concat([1, 3], vector(103))}
```

a(0) = 1, a(1) = 3, a(n) = 0 for n > 1.
G.f.: 1+3*x.
a(n) = 3^n mod 9. - Ridouane Oudra, Apr 09 2025

A380747 Array read by ascending antidiagonals: A(n,k) = [x^n] (1 - x)/(1 - k*x)^2.

Original entry on oeis.org

1, -1, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 8, 5, 1, 0, 1, 20, 21, 7, 1, 0, 1, 48, 81, 40, 9, 1, 0, 1, 112, 297, 208, 65, 11, 1, 0, 1, 256, 1053, 1024, 425, 96, 13, 1, 0, 1, 576, 3645, 4864, 2625, 756, 133, 15, 1, 0, 1, 1280, 12393, 22528, 15625, 5616, 1225, 176, 17, 1

Offset: 0

Stefano Spezia, Jan 31 2025

			The array begins as:
   1, 1,   1,    1,     1,     1, ...
  -1, 1,   3,    5,     7,     9, ...
   0, 1,   8,   21,    40,    65, ...
   0, 1,  20,   81,   208,   425, ...
   0, 1,  48,  297,  1024,  2625, ...
   0, 1, 112, 1053,  4864, 15625, ...
   0, 1, 256, 3645, 22528, 90625, ...
   ...

Cf. A000012 (k=1 or n=0), A000567 (n=2), A001792 (k=2), A007778, A060747 (n=1), A081038 (k=3), A081039 (k=4), A081040 (k=5), A081041 (k=6), A081042 (k=7), A081043 (k=8), A081044 (k=9), A081045 (k=10), A103532, A154955, A380748 (antidiagonal sums).

A[0,0]:=1; A[1,0]:=-1; A[n_,k_]:=((k-1)*n+k)k^(n-1); Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten (* or *)
A[n_,k_]:=SeriesCoefficient[(1-x)/(1-k*x)^2,{x,0,n}]; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten (* or *)
A[n_,k_]:=n!SeriesCoefficient[Exp[k*x](1+(k-1)*x),{x,0,n}]; Table[A[n-k,k],{n,0,10},{k,0,n}]//Flatten

A(n,k) = ((k - 1)*n + k)*k^(n-1) with A(0,0) = 1.
A(n,k) = n! * [x^n] exp(k*x)*(1 + (k - 1)*x).
A(n,0) = A154955(n+1).
A(3,n) = A103532(n-1) for n > 0.
A(n,n) = A007778(n) for n > 0.

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A081038 3rd binomial transform of (1,2,0,0,0,0,0,0,...).

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A081040 5th binomial transform of (1,4,0,0,0,0,...).

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A119733 Offsets of the terms of the nodes of the reverse Collatz function.

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A169585 A000004 preceded by 1, 3.

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A380747 Array read by ascending antidiagonals: A(n,k) = [x^n] (1 - x)/(1 - k*x)^2.

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