cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Prabha Sivaramannair

Prabha Sivaramannair's wiki page.

Prabha Sivaramannair has authored 8 sequences.

A377728 Convolution of Leonardo numbers with Jacobsthal numbers.

Original entry on oeis.org

0, 1, 2, 7, 16, 39, 86, 189, 402, 847, 1760, 3631, 7438, 15165, 30794, 62343, 125904, 253783, 510758, 1026685, 2061730, 4136991, 8295872, 16627167, 33311646, 66716029, 133582106, 267406999, 535206832, 1071049287, 2143127030, 4287918141, 8578528818, 17161414255
Offset: 0

Author

Prabha Sivaramannair, Nov 05 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{3, 0, -5, 1, 2}, {0, 1, 2, 7, 16}, 34] (* Amiram Eldar, Nov 07 2024 *)
  • Python
    from sympy import fibonacci
    def A377728(n): return 1-(fibonacci(n+2)<<2)+(m:=(4<>1 # Chai Wah Wu, Nov 09 2024

Formula

a(n) = Sum_{i=0..n} L(i)*J(n-i) where L = A001595 and J = A001045.
a(n) = (3*J(n+2) - 2*L(n+1) - 1)/2 where L = A001595 and J = A001045.
G.f.: -x*(x^2-x+1)/((x-1)*(2*x-1)*(x+1)*(x^2+x-1)). - Alois P. Heinz, Nov 05 2024
E.g.f.: 2*cosh(2*x) + sinh(x) + 2*sinh(2*x) - 2*exp(x/2)*(5*cosh(sqrt(5)*x/2) + 3*sqrt(5)*sinh(sqrt(5)*x/2))/5. - Stefano Spezia, Nov 06 2024

A372738 Binomial transform of A369795.

Original entry on oeis.org

1, 4, 28, 298, 4240, 75394, 1608688, 40045618, 1139279680, 36463487554, 1296712045648, 50724943433938, 2164652356532320, 100072984472662114, 4982304066392196208, 265770533884409878258, 15122101633293034668160, 914210942121577873619074, 58519992421072004957876368, 3954059527570115477197922578
Offset: 0

Author

Prabha Sivaramannair, May 11 2024

Keywords

Crossrefs

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[E^(2*x)/(1 + E^x - E^(3*x)), {x, 0, nmax}], x]*Range[0, nmax]! (* Vaclav Kotesovec, Jun 01 2024 *)
  • SageMath
    def a(n):
        if n==0:
            return 1
        else:
            return sum([(1-(-1)^j-(-2)^j)*binomial(n,j)*a(n-j) for j in [1,..,n]])
    list(a(n) for n in [0,..,20])

Formula

a(n) = Sum_{j=1..n} (1-(-1)^j-(-2)^j)*binomial(n,j)*a(n-j) for n > 0.
a(n) = 2^n + Sum_{j=1..n} (3^j-1)*binomial(n,j)*a(n-j).
a(n) = 1 + Sum_{j=1..n} (2^j-(-1)^j)*binomial(n,j)*a(n-j).
E.g.f.: exp(2*x)/(1 + exp(x) - exp(3*x)). - Vaclav Kotesovec, Jun 01 2024

A371984 Binomial transform of A371460.

Original entry on oeis.org

1, 3, 15, 117, 1227, 16053, 251955, 4613997, 96566667, 2273672133, 59482039395, 1711735382877, 53737315411707, 1827584253650613, 66936582030410835, 2626714554845111757, 109948916113808074347, 4889877314768678051493
Offset: 0

Author

Prabha Sivaramannair, Apr 15 2024

Keywords

Crossrefs

Cf. A371460.

Programs

  • Mathematica
    nn = 17; a[0] = 1; Do[Set[a[n], 2^n + Sum[(3^j - 2^j)*Binomial[n, j]*a[n - j], {j, n}]], {n, nn}]; Array[a, nn + 1, 0] (* Michael De Vlieger, Apr 19 2024 *)
  • SageMath
    def a(n):
        if n==0:
            return 1
        else:
            return sum([(1-(-2)^j)*binomial(n,j)*a(n-j) for j in [1,..,n]])
    list(a(n) for n in [0,..,20])
    
  • SageMath
    f= e^(2*x)/(1 + e^(2*x) - e^(3*x))
    print([(diff(f,x,i)).subs(x=0) for i in [0,..,20]])

Formula

a(0) = 1, a(n) = Sum_{j=1..n} (1-(-2)^j)*binomial(n,j)*a(n-j) for n > 0.
a(0) = 1, a(n) = 2^n + Sum_{j=1..n} (3^j-2^j)*binomial(n,j)*a(n-j) for n > 0.
E.g.f.: exp(2*x)/(1 + exp(2*x) - exp(3*x)).

A371460 Binomial transform of A355409.

Original entry on oeis.org

1, 2, 10, 80, 838, 10952, 171910, 3148280, 65890198, 1551389192, 40586247910, 1167964662680, 36666464437558, 1247011549249832, 45672691012357510, 1792280373542404280, 75021202465129000918, 3336499249170658956872, 157116438405334017308710, 7809681380575733223237080, 408621675981135189773468278
Offset: 0

Author

Prabha Sivaramannair, Mar 24 2024

Keywords

Crossrefs

Cf. A355409.

Programs

  • SageMath
    def a(n):
        if n==0:
            return 1
        else:
            return (-1)^n + sum([(1-(-2)^j)*binomial(n,j)*a(n-j) for j in [1,..,n]])
    list(a(n) for n in [0,..,20])
    
  • SageMath
    f= e^(x)/(1 + e^(2*x) - e^(3*x))
    print([(diff(f,x,i)).subs(x=0) for i in [0,..,20]])

Formula

a(0) = 1, a(n) = (-1)^n + Sum_{j=1..n} (1-(-2)^j)*binomial(n,j)*a(n-j) for n > 0.
a(0) = 1, a(n) = 1 + Sum_{j=1..n} (3^j-2^j)*binomial(n,j)*a(n-j) for n > 0.
E.g.f.: exp(x)/(1 + 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

Author

Prabha Sivaramannair, Feb 26 2024

Keywords

Comments

Inverse binomial transform of A370092 + A370456.

Crossrefs

Programs

  • PARI
    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
  • SageMath
    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])
    
  • SageMath
    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]])
    

Formula

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

A370456 a(0) = 1, a(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

1, 2, 6, 29, 192, 1577, 15516, 178229, 2339952, 34559057, 567117876, 10237161629, 201592448712, 4300618438937, 98803485774636, 2432074390036229, 63857242954421472, 1781444969999245217, 52620896463516221796, 1640684857196257578029, 53847865360369426418232
Offset: 0

Author

Prabha Sivaramannair, Feb 23 2024

Keywords

Comments

Binomial transform of A370092.

Crossrefs

Programs

  • PARI
    seq(n)={my(p=exp(x + O(x*x^n))); Vec(serlaplace(2*p^2/(1 + p + p^2 - p^3)))} \\ Andrew Howroyd, Feb 23 2024
  • SageMath
    def a(m):
        if m==0:
            return 1
        else:
            return 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])
    

Formula

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

A369795 Binomial transform of A355408.

Original entry on oeis.org

1, 3, 21, 225, 3201, 56913, 1214361, 30229545, 860016801, 27525472353, 978858962601, 38291126920665, 1634047719138801, 75542860973042193, 3761030066169432441, 200624240375801784585, 11415336789685550907201, 690117422445926970890433, 44175435307592982599575881
Offset: 0

Author

Prabha Sivaramannair, Feb 01 2024

Keywords

Crossrefs

Cf. A355408.

Programs

  • Mathematica
    nmax = 20; CoefficientList[Series[E^x/(1 + E^x - E^(3*x)), {x, 0, nmax}], x]*
    Range[0, nmax]! (* Vaclav Kotesovec, Feb 01 2024*)
  • SageMath
    def a(m):
        if m==0:
            return 1
        else:
            return 1+sum([(3^j-1)*binomial(m,j)*a(m-j) for j in [1,..,m]])
    list(a(m) for m in [1,..,50])

Formula

a(n) = 1 + Sum_{k=1..n} (3^k - 1) * binomial(n,k) * a(n-k) for n > 0.
E.g.f.: exp(x)/(1 + exp(x) - exp(3*x)). - Vaclav Kotesovec, Feb 01 2024

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.

Original entry on oeis.org

1, 1, 3, 16, 105, 856, 8433, 96916, 1272225, 18789136, 308335713, 5565837916, 109603592145, 2338198823416, 53718370204593, 1322292130204516, 34718481333932865, 968552056638097696, 28609403248435931073, 892022330159009036716, 29276492753074019702385
Offset: 0

Author

Prabha Sivaramannair, Feb 09 2024

Keywords

Comments

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

Crossrefs

Programs

  • Mathematica
    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 *)
  • PARI
    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
  • SageMath
    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])
    

Formula

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