A383254 -id:A383254 - cp's OEIS frontend

A385563 Expansion of 1/((1-x) * (1-5*x))^(3/2).

1, 9, 60, 360, 2055, 11403, 62132, 334260, 1781415, 9425295, 49581576, 259601004, 1353939405, 7038232425, 36484340400, 188665670880, 973545780195, 5014258620075, 25783103206100, 132378800689800, 678768332410245, 3476164133573505, 17782899991147500

Offset: 0

Seiichi Manyama, Aug 19 2025

Paolo Xausa, Table of n, a(n) for n = 0..1000

Partial sums of A383254.
Cf. A331516, A385716.
Cf. A385728, A385813.
Cf. A002212, A383949.

Module[{a, n}, RecurrenceTable[{a[n] == ((6*n+3)*a[n-1] - 5*(n+1)*a[n-2])/n, a[0] == 1, a[1] == 9}, a, {n, 0, 25}]] (* Paolo Xausa, Aug 21 2025 *)

my(N=30, x='x+O('x^N)); Vec(1/((1-x)*(1-5*x))^(3/2))

n*a(n) = (6*n+3)*a(n-1) - 5*(n+1)*a(n-2) for n > 1.
a(n) = (1/4)^n * Sum_{k=0..n} 5^k * (2*k+1) * (2*(n-k)+1) * binomial(2*k,k) * binomial(2*(n-k),n-k).
a(n) = Sum_{k=0..n} (2*k+1) * binomial(2*k,k) * binomial(n+2,n-k).
a(n) = Sum_{k=0..n} (-1)^k * 5^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(n+2,n-k).
a(n) = binomial(n+2,2) * A002212(n+1).
a(n) = ((n+2)/2) * Sum_{k=0..floor(n/2)} 3^(n-2*k) * binomial(n+1,n-2*k) * binomial(2*k+1,k).
a(n) = Sum_{k=0..n} (3/2)^k * (-5/6)^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(k,n-k).
a(n) ~ sqrt(n) * 5^(n + 3/2) / (4*sqrt(Pi)). - Vaclav Kotesovec, Aug 21 2025

A368070 a(n) is the number of sequences of binary words (w_1, ..., w_k) such that w_1 corresponds to the binary expansion of n (without leading zeros), for m = 1..k-1, w_{m+1} is obtained by removing one bit from w_m, and w_k is the empty word.

Original entry on oeis.org

1, 1, 2, 1, 3, 5, 3, 1, 4, 11, 16, 9, 6, 9, 4, 1, 5, 19, 40, 26, 35, 61, 40, 14, 10, 26, 35, 19, 10, 14, 5, 1, 6, 29, 78, 55, 99, 181, 132, 50, 64, 181, 272, 155, 111, 169, 78, 20, 15, 55, 111, 71, 90, 155, 99, 34, 20, 50, 64, 34, 15, 20, 6, 1, 7, 41, 133, 99

Offset: 0

Rémy Sigrist, Dec 10 2023

Leading zeros may appear in binary words w_2, ..., w_{k-1}.

a(n) gives the number of ways to erase the binary expansion of n bit by bit.

			For n = 5:
- the binary expansion of 5 is "101",
- we have the following appropriate sequences of binary words:
     ("101", "11", "1", "")
     ("101", "10", "1", "")
     ("101", "10", "0", "")
     ("101", "01", "1", "")
     ("101", "01", "0", "")
- hence a(5) = 5.

Rémy Sigrist, Table of n, a(n) for n = 0..8192
Rémy Sigrist, PARI program
Natalia L. Skirrow, bitwise permutations
Index entries for sequences related to binary expansion of n

See A060351 and A367019 for similar sequences.

Cf. A000225.

PARI
```
\\ See Links section.
```

def A368070(n):
  m=0
  r=[1]
  for k in range(n.bit_length()):
    if m!=(m:=n>>k&1): r=r[::-1]
    for j in range(k): r[j+1]+=r[j]
    r.insert(0,0)
  return sum(r) # Natalia L. Skirrow, Apr 20 2025

from fractions import Fraction as frac
inte=lambda p: [0]+[frac(c,i+1) for i,c in enumerate(p)]
from math import factorial as fact
def A368070(n):
  r=[1]
  for k in range(n.bit_length()):
    i=inte(r)
    r=i if n>>k&1 else [sum(i)]+[-c for c in i[1:]]
  return int(fact(n.bit_length()+1)*sum(inte(r)))
#without the multiplication, this is the probability that a sequence of real numbers in [0,1] satisfies the chain of inequalities. # Natalia L. Skirrow, Apr 20 2025

a(n) = 1 iff n belongs to A000225.
a(2^k) = k + 1 for any k >= 0.
a(n) >= A367019(n).
a(n) <= A383254(n). (See comment there) - Natalia L. Skirrow, Apr 20 2025

A383503 Expansion of 1/sqrt( (1-x) * (1-x-4*x^3)^3 ).

Original entry on oeis.org

1, 2, 3, 10, 23, 42, 97, 218, 435, 918, 1977, 4062, 8393, 17590, 36303, 74614, 154211, 317334, 650505, 1335054, 2736453, 5595950, 11439475, 23370270, 47681965, 97217882, 198110199, 403383026, 820820215, 1669405626, 3393344257, 6893850650, 13999109715, 28414742790

Offset: 0

Seiichi Manyama, May 05 2025

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A383254, A383499.

Cf. A217615, A377213.

R:=PowerSeriesRing(Rationals(), 35); Coefficients(R!( 1/Sqrt( (1-x) * (1-x-4*x^3)^3 ))); // Vincenzo Librandi, May 06 2025

Table[Sum[(2*k+1)*Binomial[2*k,k]*Binomial[n-2*k+1,k+1],{k,0,Floor[n/3]}],{n,0,35}] (* Vincenzo Librandi, May 06 2025 *)

a(n) = sum(k=0, n\3, (2*k+1)*binomial(2*k, k)*binomial(n-2*k+1, k+1));

a(n) = Sum_{k=0..floor(n/3)} (2*k+1) * binomial(2*k,k) * binomial(n-2*k+1,k+1).

A383499 Expansion of 1/sqrt( (1-x) * (1-x-4*x^2)^3 ).

Original entry on oeis.org

1, 2, 9, 22, 71, 186, 537, 1434, 3957, 10586, 28603, 76266, 203767, 540986, 1435533, 3796050, 10026015, 26422350, 69544765, 182759750, 479731113, 1257750486, 3294264627, 8619879726, 22535782953, 58869786162, 153671378139, 400861115498, 1045005290059, 2722601576322

Offset: 0

Seiichi Manyama, May 05 2025

nonn

Cf. A383254, A383503.

Cf. A026569, A377204.

a(n) = sum(k=0, n\2, (2*k+1)*binomial(2*k, k)*binomial(n-k+1, k+1));

a(n) = Sum_{k=0..floor(n/2)} (2*k+1) * binomial(2*k,k) * binomial(n-k+1,k+1).

a(n) ~ sqrt(n/Pi) * ((1 + sqrt(17))/2)^(n + 5/2) / 17^(3/4). - Vaclav Kotesovec, May 05 2025

cp's OEIS Frontend

A385563 Expansion of 1/((1-x) * (1-5*x))^(3/2).

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A368070 a(n) is the number of sequences of binary words (w_1, ..., w_k) such that w_1 corresponds to the binary expansion of n (without leading zeros), for m = 1..k-1, w_{m+1} is obtained by removing one bit from w_m, and w_k is the empty word.

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A383503 Expansion of 1/sqrt( (1-x) * (1-x-4*x^3)^3 ).

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Magma

Mathematica

PARI

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A383499 Expansion of 1/sqrt( (1-x) * (1-x-4*x^2)^3 ).

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Author

Keywords

Crossrefs

Programs

PARI

Formula