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.

Showing 1-9 of 9 results.

A349554 a(n) = A054108(n) + 4*(-1)^n.

Original entry on oeis.org

1, 19, 51, 201, 723, 2709, 10161, 38459, 146297, 559135, 2145021, 8255579, 31861021, 123256499, 477823891, 1855782329, 7219352971, 28125910829, 109720617991, 428537256449, 1675561707271, 6557869020329, 25689734662771, 100720871774981, 395197661173123
Offset: 1

Views

Author

Clark Kimberling, Dec 30 2021

Keywords

Crossrefs

Cf. A000984, A054108.

Programs

  • Mathematica
    s[0] = 1; s[n_] := Binomial[2 n + 2, n + 1] - s[n - 1];
Table[s[n], {n, 0, 10}]; (* A054108 *)
a[n_] := s[n] + 4*(-1)^n;
Table[a[n], {n, 1, 30}] (* A349554 *)
  • PARI
    a(n) = (-1)^(n+1)*sum(k=0,n+1,(-1)^k*binomial(2*k,k)) + 4*(-1)^n; \\ Michel Marcus, Jan 19 2022
  • Python
    from math import comb
def A349554(n): return (1 if n % 2 else -1)*(sum((-1 if k % 2 else 1)*comb(2*k,k) for k in range(n+2))-4) # Chai Wah Wu, Jan 19 2022

Formula

a(n) = C(2(n+1),n+1) - a(n-1) with a(1) = 1. - Chai Wah Wu, Jan 19 2022

A082761 Trinomial transform of the Fibonacci numbers (A000045).

Original entry on oeis.org

1, 4, 20, 104, 544, 2848, 14912, 78080, 408832, 2140672, 11208704, 58689536, 307302400, 1609056256, 8425127936, 44114542592, 230986743808, 1209462292480, 6332826779648, 33159111507968, 173623361929216, 909103725543424, 4760128905543680, 24924358531088384, 130505635564355584
Offset: 0

Views

Author

Emanuele Munarini, May 21 2003

Keywords

Comments

Hankel transform of Sum_{k=0..n} (-1)^k*C(2k, k) (see A054108). - Paul Barry, Jan 13 2009
Hankel transform of A046748. - Paul Barry, Apr 14 2010
For positive n, a(n) equals the permanent of the (2n) X (2n) tridiagonal matrix with sqrt(2)'s along the three central diagonals. - John M. Campbell, Jul 12 2011
The limiting ratio is: Lim_{n -> oo} a(n)/a(n-1) = 1 + phi^3. - Bob Selcoe, Mar 18 2014
Invert transform of A052984. Invert transform is A083066. Binomial transform of A033887. Binomial transform is A163073. - Michael Somos, May 26 2014

Examples

			a(5) = 2848 = 5*(544) + 4 + 20 + 104.
G.f. = 1 + 4*x + 20*x^2 + 104*x^3 + 544*x^4 + 2848*x^5 + 14912*x^6 + ...

Links

Crossrefs

Cf. A000045, A001519, A027907, A033887, A046748, A052984, A054108, A069429, A083066, A147703, A163073.

Programs

  • Magma
    [2^n * Fibonacci(2*n+1): n in [0..40]]; // Vincenzo Librandi, Jul 15 2011
  • Mathematica
    a[ n_] := 2^n Fibonacci[ 2 n + 1]; (* Michael Somos, May 26 2014 *)
a[ n_] := If[ n < 0, SeriesCoefficient[ (2 - x) / (4 - 6 x + x^2), {x, 0, -1 - n}], SeriesCoefficient[ (1 - 2 x) / (1 - 6 x + 4 x^2), {x, 0, n}]]; (* Michael Somos, Oct 22 2017 *)
LinearRecurrence[{6,-4},{1,4},30] (* Harvey P. Dale, Jul 11 2014 *)
  • PARI
    a(n)=fibonacci(2*n+1)<Charles R Greathouse IV, Jul 15 2011
  • PARI
    {a(n) = if( n<0, n = -1 - n; 2^(-1-2*n), 1) * polcoeff( (1 - 2*x) / (1 - 6*x + 4*x^2) + x * O(x^n), n)}; /* Michael Somos, Oct 22 2017 */
  • SageMath
    [2^n*fibonacci(2*n+1) for n in range(41)] # G. C. Greubel, Jul 28 2024

Formula

a(n) = Sum_{k=0..2*n} A027907(n, k)*A000045(k+1).
From Paul Barry, Jul 16 2003: (Start)
Third binomial transform of (1, 1, 5, 5, 25, 25, ....).
a(n) = ((1+sqrt(5))(3+sqrt(5))^n-(1-sqrt(5))*(3-sqrt(5))^n)/(2*sqrt(5)). (End)
From R. J. Mathar, Nov 04 2008: (Start)
G.f.: (1-2*x)/(1-6*x+4*x^2).
a(n) = 6*a(n-1) - 4*a(n-2). (End)
a(n) = Sum_{k=0..n} A147703(n,k)*3^k. - Philippe Deléham, Nov 14 2008
For n>=2: a(n) = 5*a(n-1) + Sum_{i=1..n-2} a(i). - Bob Selcoe, Mar 18 2014
a(n) = a(-1-n) * 2^(2*n+1) for all n in Z. - Michael Somos, Mar 18 2014
a(n) = 2^n*Fibonacci(2*n+1), or 2^n*A001519(n+1). - Bob Selcoe, May 25 2014
From Michael Somos, May 26 2014: (Start)
a(n) - a(n-1) = A069429(n).
a(n+1) * a(n-1) - a(n)^2 = 4^n.
G.f.: 1 / (1 - 4*x / (1 - x / (1 - x))). (End)
E.g.f.: exp(3*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, May 24 2024

A360185 a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(2*n-4*k,n-2*k).

Original entry on oeis.org

1, 2, 5, 18, 65, 234, 859, 3198, 12011, 45422, 172745, 660010, 2531411, 9740590, 37585189, 145376930, 563495201, 2188229290, 8511640099, 33157034510, 129334888721, 505100839930, 1974764074999, 7728329887670, 30272839608101, 118682276550082, 465645693340003
Offset: 0

Views

Author

Seiichi Manyama, Jan 29 2023

Keywords

Links

Crossrefs

Cf. A054108, A360186.

Programs

  • Maple
    A360185 := proc(n)
    add((-1)^k*binomial(2*n-4*k,n-2*k),k=0..n/2) ;
end proc:
seq(A360185(n),n=0..70) ; # R. J. Mathar, Mar 12 2023
  • PARI
    a(n) = sum(k=0, n\2, (-1)^k*binomial(2*n-4*k, n-2*k));
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/(sqrt(1-4*x)*(1+x^2)))

Formula

G.f.: 1 / ( sqrt(1-4*x) * (1 + x^2) ).
a(n) ~ 2^(2*n + 4) / (17*sqrt(Pi*n)). - Vaclav Kotesovec, Jan 29 2023
D-finite with recurrence n*a(n) +2*(-2*n+1)*a(n-1) +n*a(n-2) +2*(-2*n+1)*a(n-3)=0. - R. J. Mathar, Mar 12 2023
a(n)+a(n-2) = A000984(n). - R. J. Mathar, Mar 12 2023

A360186 a(n) = Sum_{k=0..floor(n/3)} (-1)^k * binomial(2*n-6*k,n-3*k).

Original entry on oeis.org

1, 2, 6, 19, 68, 246, 905, 3364, 12624, 47715, 181392, 692808, 2656441, 10219208, 39423792, 152461079, 590861182, 2294182428, 8922674221, 34754402618, 135552346392, 529335200219, 2069344561102, 8097878381208, 31718268482881, 124341261876650
Offset: 0

Views

Author

Seiichi Manyama, Jan 29 2023

Keywords

Links

Crossrefs

Cf. A054108, A360185.
Cf. A000984, A360153.

Programs

  • Maple
    A360186 := proc(n)
    add((-1)^k*binomial(2*n-6*k,n-3*k),k=0..n/3) ;
end proc:
seq(A360186(n),n=0..70) ; # R. J. Mathar, Mar 12 2023
  • Mathematica
    Table[Sum[(-1)^k Binomial[2n-6k,n-3k],{k,0,Floor[n/3]}],{n,0,30}] (* Harvey P. Dale, Mar 05 2023 *)
  • PARI
    a(n) = sum(k=0, n\3, (-1)^k*binomial(2*n-6*k, n-3*k));
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/(sqrt(1-4*x)*(1+x^3)))

Formula

G.f.: 1 / ( sqrt(1-4*x) * (1 + x^3) ).
a(n) ~ 2^(2*n + 6) / (65*sqrt(Pi*n)). - Vaclav Kotesovec, Jan 29 2023
D-finite with recurrence n*a(n) +2*(-2*n+1)*a(n-1) +n*a(n-3) +2*(-2*n+1)*a(n-4)=0. - R. J. Mathar, Mar 12 2023
a(n)+a(n-3) = A000984(n). - R. J. Mathar, Mar 12 2023

A106188 Expansion of 1/((1-x^2)*sqrt(1-4*x)).

Original entry on oeis.org

1, 2, 7, 22, 77, 274, 1001, 3706, 13871, 52326, 198627, 757758, 2902783, 11158358, 43019383, 166275878, 644099773, 2499882098, 9719235073, 37845145898, 147565763893, 576103020338, 2251664727613, 8809533747938, 34499268410713
Offset: 0

Views

Author

Paul Barry, Apr 24 2005

Keywords

Comments

Diagonal sums of number triangle A106187.

Examples

			1 + 2*x + 7*x^2 + 22*x^3 + 77*x^4 + 274*x^5 + 1001*x^6 + 3706*x^7 + 13871*x^8 + ...

Links

Crossrefs

Cf. A006134, A054108. Convolution of A000984 and A059841.

Programs

  • Mathematica
    CoefficientList[Series[1/((1-x^2)*Sqrt[1-4*x]), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 03 2014 *)
  • PARI
    x='x+O('x^50); Vec(1/((1-x^2)*sqrt(1-4*x))) \\ G. C. Greubel, Mar 16 2017

Formula

G.f. 1 / ((1 - x^2) * sqrt(1 - 4*x)).
a(n)=sum{k=0..floor(n/2), binomial(2(n-2k), n-2k)}.
PSUMSIGN transform of A006134. a(n+1) + a(n) = A006134(n). a(n) = Sum_{k=0..n} (-1)^k * binomial(2 * (n-k), n-k). - Michael Somos, Jun 20 2012
First difference is A054108. a(n+1) - a(n) = A054108(n). - Michael Somos, Jun 20 2012
D-finite with recurrence: n*a(n)+2*(1-2*n)*a(n-1) -n*a(n-2) +2*(2*n-1)*a(n-3)=0. - R. J. Mathar, Nov 09 2012
a(n) ~ 2^(2*n+4) / (15*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 03 2014

A103245 Triangle read by rows: T(n,k) = binomial(2n+1, n-k)*Fibonacci(2k+1), 0 <= k <= n.

Original entry on oeis.org

1, 3, 2, 10, 10, 5, 35, 42, 35, 13, 126, 168, 180, 117, 34, 462, 660, 825, 715, 374, 89, 1716, 2574, 3575, 3718, 2652, 1157, 233, 6435, 10010, 15015, 17745, 15470, 9345, 3495, 610, 24310, 38896, 61880, 80444, 80920, 60520, 31688, 10370, 1597, 92378
Offset: 0

Views

Author

Emeric Deutsch, Mar 19 2005

Keywords

Examples

			Triangle begins:
    1;
    3,   2;
   10,  10,   5;
   35,  42,  35,  13;
  126, 168, 180, 117,  34;

References

  • S. G. Guba, Problem No. 174, Issue No. 4, July-August 1965, p. 73 of Matematika v Skole.

Links

Crossrefs

Column 0 is A001700.
Column 1 is A024483.
T(n, n) = A001519(n+1) (the odd-indexed Fibonacci numbers).
Row sums are the powers of 5 (A000351).
Alternating row sums yield A054108.

Programs

  • Maple
    with(combinat): T:=(n,k)->binomial(2*n+1,n-k)*fibonacci(2*k+1): for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
  • Mathematica
    Table[Binomial[2 n + 1, n - k] Fibonacci[2 k + 1], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 01 2019 *)

Formula

T(n, k) = binomial(2n+1, n-k)*Fibonacci(2k+1), 0 <= k <= n.

A227995 Alternate partial sums of the binomial coefficients C(4*n,n).

Original entry on oeis.org

1, 3, 25, 195, 1625, 13879, 120717, 1063323, 9454977, 84688303, 762972225, 6906366907, 62762167561, 572251392039, 5232480571761, 47961608620959, 440565328458621, 4054586252967027, 37377503512616413, 345083448151227987, 3190232694060946333, 29529002023029712547
Offset: 0

Views

Author

Vaclav Kotesovec, Aug 06 2013

Keywords

Comments

Generally (for p>1), alternate partial sums of the binomial coefficients C(p*n,n) is asymptotic to (1/(1+(p-1)^(p-1)/p^p)) * sqrt(p/(2*Pi*n*(p-1))) * (p^p/(p-1)^(p-1))^n.

Links

Crossrefs

Cf. A054108(n-1) (p=2), A188676 (p=3), A227996 (p=5).

Programs

  • Mathematica
    Table[Sum[Binomial[4*k, k]*(-1)^(n-k), {k, 0, n}], {n, 0, 20}]
  • PARI
    for(n=0,50, print1(sum(k=0,n, binomial(4*k,k)), ", ")) \\ G. C. Greubel, Apr 03 2017

Formula

Recurrence: 3*n*(3*n-2)*(3*n-1)*a(n) = (229*n^3 - 357*n^2 + 170*n - 24)*a(n-1) + 8*(2*n-1)*(4*n-3)*(4*n-1)*a(n-2).
a(n) ~ 2^(8*n+17/2)/(283*sqrt(Pi*n)*3^(3*n+1/2)).

A227996 Alternate partial sums of the binomial coefficients C(5*n,n).

Original entry on oeis.org

1, 4, 41, 414, 4431, 48699, 545076, 6179444, 70725241, 815437894, 9456840276, 110196725574, 1289162119401, 15131911395879, 178121845513281, 2101890841202799, 24856330289305726, 294500697587787599, 3495147445120811176, 41542892270532317969, 494440478133277365001
Offset: 0

Views

Author

Vaclav Kotesovec, Aug 06 2013

Keywords

Comments

Generally (for p>1), alternate partial sums of the binomial coefficients C(p*n,n) is asymptotic to (1/(1+(p-1)^(p-1)/p^p)) * sqrt(p/(2*Pi*n*(p-1))) * (p^p/(p-1)^(p-1))^n.

Links

Crossrefs

Cf. A054108(n-1) (p=2), A188676 (p=3), A227995 (p=4).

Programs

  • Mathematica
    Table[Sum[Binomial[5*k, k]*(-1)^(n-k), {k, 0, n}], {n, 0, 20}]
  • PARI
    for(n=0,50, print1(sum(k=0,n, binomial(5*k,k)), ", ")) \\ G. C. Greubel, Apr 03 2017

Formula

Recurrence: 8*n*(2*n-1)*(4*n-3)*(4*n-1)*a(n) = (2869*n^4 - 5866*n^3 + 4199*n^2 - 1226*n + 120)*a(n-1) + 5*(5*n-4)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-2).
a(n) ~ 5^(5*n+11/2)/(3381*sqrt(Pi*n)*2^(8*n+3/2)).

A104551 Expansion of x/((1-x)*sqrt(1+4*x^2)).

Original entry on oeis.org

0, 1, 1, -1, -1, 5, 5, -15, -15, 55, 55, -197, -197, 727, 727, -2705, -2705, 10165, 10165, -38455, -38455, 146301, 146301, -559131, -559131, 2145025, 2145025, -8255575, -8255575, 31861025, 31861025, -123256495, -123256495, 477823895, 477823895, -1855782325, -1855782325, 7219352975
Offset: 0

Views

Author

Paul Barry, Mar 14 2005

Keywords

Comments

A transformation of the Fibonacci numbers A000045 by the Riordan array (1/sqrt(1+4*x^2), (sqrt(1+4*x^2)-1)/(2*x)).

Links

Crossrefs

Cf. A000045, A054108.

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( x/((1-x)*Sqrt(1+4*x^2)) )); // G. C. Greubel, Jan 01 2023
  • Mathematica
    CoefficientList[Series[x/((1-x)*Sqrt[1+4*x^2]), {x,0,40}], x] (* G. C. Greubel, Jan 01 2023 *)
  • SageMath
    def A104551_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( x/((1-x)*sqrt(1+4*x^2)) ).list()
A104551_list(40) # G. C. Greubel, Jan 01 2023

Formula

G.f.: x/((1-x)*sqrt(1+4*x^2)).
a(n) = Sum_{k=0..n} (sin(Pi*k/2)+cos(Pi*k)/2+1/2)*C(k-1,(k-1)/2)*(1-(-1)^k)/2.
D-finite with recurrence: (n-1)*a(n) = (n-1)*a(n-1) - 4*(n-2)*a(n-2) + 4*(n-2)*a(n-3). - R. J. Mathar, Feb 20 2015
Showing 1-9 of 9 results.

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