A370456 -id:A370456 - cp's OEIS frontend

A370092 a(0) = 1, a(n) = (-1)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.

1, 1, 3, 16, 105, 856, 8433, 96916, 1272225, 18789136, 308335713, 5565837916, 109603592145, 2338198823416, 53718370204593, 1322292130204516, 34718481333932865, 968552056638097696, 28609403248435931073, 892022330159009036716, 29276492753074019702385

Offset: 0

Prabha Sivaramannair, Feb 09 2024

nonn

Inverse binomial transform of A370456.

Conjecture: Let k > 2 be a positive integer. The sequence obtained by reducing a(n) modulo k is eventually periodic with the period dividing phi(k) = A000010(k). For example, modulo 10 we obtain the sequence [1, 1, 3, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, 3, 6, 5, 6, ...] with an apparent period of 4 beginning at a(2). See A000670 for a more general conjecture. - Peter Bala, Feb 16 2024

Cf. A370163, A370456.

a[0]=1;Table[(-1)^n+Sum[ (1-(-1)^j-  (-2) ^j) *Binomial[n,j]*a[n-j]/2,{j,1,n} ],{n,0,20}] (* James C. McMahon, Feb 10 2024 *)

seq(n)={my(p=exp(x + O(x*x^n))); Vec(serlaplace(2*p/(1 + p + p^2 - p^3)))} \\ Andrew Howroyd, Feb 10 2024

def a(m):
    if m==0:
        return 1
    else:
        return (-1)^m+1/2*sum([(1-(-2)^j-(-1)^j)*binomial(m,j)*a(m-j) for j in [1,..,m]])
list(a(m) for m in [0,..,20])

E.g.f.: 2*exp(x)/(1 + exp(x) + exp(2*x) - exp(3*x)).

A370163 a(0) = 2, a(n) = (-1)^n + (-2)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.

Original entry on oeis.org

2, 1, 5, 25, 161, 1321, 13025, 149605, 1963841, 29004721, 475975745, 8591917885, 169193833121, 3609452038921, 82924458549665, 2041207822721365, 53594538159184001, 1495143168658285921, 44164021453758342785, 1377005070100813288045, 45193800193226286112481

Offset: 0

Prabha Sivaramannair, Feb 26 2024

nonn

Inverse binomial transform of A370092 + A370456.

Cf. A370092, A370456.

seq(n)={my(p=exp(x + O(x*x^n))); Vec(serlaplace(2*(1 + p)/(1 + p + p^2 - p^3)))} \\ Andrew Howroyd, Feb 26 2024

def a(m):
    if m==0:
        return 2
    else:
        return (-1)^m+(-2)^m+1/2*sum([(1-(-2)^j-(-1)^j)*binomial(m,j)*a(m-j) for j in [1,..,m]])
list(a(m) for m in [0,..,20])

f=2*(1+e^x)/(1+e^x+e^(2*x)-e^(3*x))
print([(diff(f,x,i)).subs(x=0) for i in [0,..,20]])

E.g.f.: 2*(1 + exp(x))/(1 + exp(x) + exp(2*x) - exp(3*x)).

cp's OEIS Frontend

A370092 a(0) = 1, a(n) = (-1)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.

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A370163 a(0) = 2, a(n) = (-1)^n + (-2)^n + (1/2) * Sum_{j=1..n} (1-(-1)^j-(-2)^j) * binomial(n,j) * a(n-j) for n > 0.

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

SageMath

SageMath

Formula