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.

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A264960 Half-convolution of the central binomial coefficients A000984 with itself.

Original entry on oeis.org

1, 2, 10, 32, 146, 512, 2248, 8192, 35218, 131072, 556040, 2097152, 8815496, 33554432, 140107040, 536870912, 2230302098, 8589934592, 35541690568, 137438953472, 566823203656, 2199023255552, 9044910175520, 35184372088832, 144393718191496
Offset: 0

Views

Author

Peter Bala, Nov 29 2015

Keywords

Comments

The half-convolution of a sequence {s(n)}n>=0 with itself is defined by r(n) := Sum_{k = 0..floor(n/2)} s(k)*s(n-k). See A201204.

Crossrefs

Programs

  • GAP
    List([0..24],n->Sum([0..Int(n/2)],k->Binomial(2*k,k)*Binomial(2*n-2*k,n-k))); # Muniru A Asiru, Nov 25 2018
    
  • Magma
    [(&+[Binomial(2*k,k)*Binomial(2*n-2*k, n-k): k in [0..Floor(n/2)]]): n in [0..30]]; // G. C. Greubel, Nov 26 2018
    
  • Maple
    A264960:= n-> add(binomial(2*k,k)*binomial(2*n - 2*k, n - k),k = 0..floor(n/2)):
    seq(A264960(n),n = 0..24);
  • Mathematica
    a[n_] := Sum[Binomial[2k, k]*Binomial[2n - 2k, n - k], {k, 0, Floor[n/2]}]; Array[a, 30, 0] (* Amiram Eldar, Nov 25 2018 *)
  • PARI
    a(n) = sum(k = 0, n\2, binomial(2*k,k)*binomial(2*n - 2*k, n - k)); \\ Michel Marcus, Nov 30 2015
    
  • Sage
    [sum(binomial(2*k,k)*binomial(2*n-2*k, n-k) for k in (0..floor(n/2))) for n in range(30)] # G. C. Greubel, Nov 26 2018

Formula

a(n) = Sum_{k = 0..floor(n/2)} binomial(2*k,k)*binomial(2*n - 2*k, n - k).
a(2*n + 1) = 2^(4*n + 1) = A013776(n).
a(2*n) = (1/2)*(binomial(2*n,n)^2 + 16^n) = A112830(2*n,n).
O.g.f.: (1/2)*( 2/Pi*EllipticK(4*x) + 1/(1 - 4*x) ).
E.g.f.: (1/2)*( cosh(4*x) + sinh(4*x) + (BesselI(0,2*x))^2 ).
D-finite with recurrence: - (2*n-3)*n^2*a(n) + 4*(2*n-1)*(n-1)^2*a(n-1) + 16*(2*n-3)*(n-1)^2*a(n-2) - 64*(2*n-1)*(n-2)^2*a(n-3) = 0. - Georg Fischer, Nov 25 2022

A081393 a(n) is the smallest k such that number of non-unitary prime divisors of central binomial coefficient, A000984(k) = C(2*k,k) equals n.

Original entry on oeis.org

1, 3, 5, 14, 48, 74, 182, 314, 480, 774, 960, 1321, 1323, 1670, 3121, 3455, 3457, 3472, 3462, 3469, 8203, 9991, 12163, 15838, 15840, 17665, 18480, 18482, 19458, 19464, 36782, 19865, 36789, 40048, 43603, 43655, 47518, 61654, 61653, 61685, 61684, 87120, 92958, 93181, 93185, 93187, 93191
Offset: 0

Views

Author

Labos Elemer, Mar 27 2003

Keywords

Examples

			n=5: a(5)=74, C(148,74) has 5 non-unitary prime divisors: {2,3,5,7,11} and 74 is the smallest.
		

Crossrefs

Programs

Formula

a(n) = Min{k; A081387(k) = n}.

Extensions

a(11)-a(21) from Michel Marcus, Sep 01 2019
a(22)-a(46) from Amiram Eldar, May 15 2023

A142962 Scaled convolution of (n^3)*A000984(n) with A000984(n).

Original entry on oeis.org

4, 26, 81, 184, 350, 594, 931, 1376, 1944, 2650, 3509, 4536, 5746, 7154, 8775, 10624, 12716, 15066, 17689, 20600, 23814, 27346, 31211, 35424, 40000, 44954, 50301, 56056, 62234, 68850, 75919, 83456, 91476, 99994, 109025, 118584, 128686, 139346, 150579
Offset: 1

Views

Author

Wolfdieter Lang, Sep 15 2008

Keywords

Comments

S(3,n) := Sum_{j=0..n} j^3*binomial(2*j,j)*binomial(2*(n-j),n-j).
a(n) = 2^3*S(3,n)/4^n, n >= 1.
O.g.f. for S(3,n) is G(k=3,x). See triangle A142963 for the general G(k,x) formula.
The author was led to compute such sums by a question asked by M. Greiter, Jun 27 2008.

Crossrefs

Cf. A142961 triangle: row k=3: [3, 5], with the row polynomial 3+5*n.
Cf. A049451 (scaled k=2 case).

Programs

  • Mathematica
    Rest@ CoefficientList[Series[x (4 + 10 x + x^2)/(1 - x)^4, {x, 0, 39}], x] (* Michael De Vlieger, Jul 02 2023 *)

Formula

a(n) = n^2*(3+5*n)/2.
a(n) = (2^3)*S(3,n)/4^n with the convolution S(3,n) defined above.
G.f.: x*(4+10*x+x^2)/(1-x)^4. - Joerg Arndt, Jul 02 2023

A152229 Eigentriangle, row sums = A000984.

Original entry on oeis.org

1, 1, 1, 3, 1, 2, 9, 3, 2, 6, 29, 9, 6, 6, 20, 97, 29, 18, 18, 20, 70, 333, 97, 58, 54, 60, 70, 252, 1165, 333, 194, 174, 180, 210, 252, 924, 4135, 1165, 666, 582, 580, 630, 756, 924, 3432, 14845, 4135, 2330, 1998, 1940, 2030, 2268, 2772, 3432, 12870
Offset: 0

Views

Author

Gary W. Adamson, Nov 29 2008

Keywords

Comments

Row sums = A000984: (1, 2, 6, 20, 70, 252,...), left border = A081696.
Sum of n-th row terms = rightmost term of next row.

Examples

			First few rows of the triangle =
1;
1, 1;
3, 1, 2;
9, 3, 2, 6;
29, 9, 6, 6, 20;
97, 29, 18, 18, 20, 70;
333, 97, 58, 54, 60, 70, 252;
1165, 333, 194, 174, 180, 210, 252, 924;
4135, 1165, 666, 582, 580, 630, 756, 924, 3432;
14845, 4135, 2330, 1998, 1940, 2030, 2268, 2772, 3432, 12870;
...
Row 3 = (9, 3, 2, 6) = termwise products of (9, 3, 1, 1) and (1, 1, 2, 6).
		

Crossrefs

Formula

Triangle read by rows, M*Q. M = an infinite lower triangular matrix with A081696: (1, 1, 3, 9, 29, 97, 333, 1165,...) in every column; and Q = a matrix with A000984 as the main diagonal (prefaced with a 1): (1, 1, 2, 6, 20, 70, 252,...) and the rest zeros.

A199813 G.f.: exp( Sum_{n>=1} A000984(n)*A000172(n) * x^n/n ), which involves central binomial coefficients (A000984) and Franel numbers (A000172).

Original entry on oeis.org

1, 4, 38, 504, 8249, 154036, 3149326, 68741880, 1576163328, 37548785408, 922252542128, 23222906277952, 596981991939677, 15616173859832740, 414621835401615110, 11150969618415168280, 303278916800906999191, 8330190277527648516572, 230814933905555392525290
Offset: 0

Views

Author

Paul D. Hanna, Nov 10 2011

Keywords

Comments

Sum_{k=0..n} C(n,k)^2 = A000984(n) defines central binomial coefficients.
Sum_{k=0..n} C(n,k)^3 = A000172(n) defines Franel numbers.

Examples

			G.f.: A(x) = 1 + 4*x + 38*x^2 + 504*x^3 + 8249*x^4 + 154036*x^5 +...
where
log(A(x)) = 2*2*x + 6*10*x^2/2 + 20*56*x^3/3 + 70*346*x^4/4 + 252*2252*x^5/5 + 924*15184*x^6/6 +...+ A000984(n)*A000172(n)*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n,binomial(2*m, m)*sum(k=0, m, binomial(m, k)^3)*x^m/m)+x*O(x^n)),n)}

A216584 a(n) = A002426(n)*A000984(n); product of central trinomial coefficients and central binomial coefficients.

Original entry on oeis.org

1, 2, 18, 140, 1330, 12852, 130284, 1348776, 14247090, 152618180, 1654120468, 18096447096, 199536967084, 2214714164600, 24720932068200, 277289164574640, 3123590583844530, 35318969120870820, 400692715550057700, 4559427798654821400, 52020436064931914580
Offset: 0

Views

Author

Paul D. Hanna, Sep 08 2012

Keywords

Examples

			L.g.f.: L(x) = 2*x + 18*x^2/2 + 140*x^3/3 + 1330*x^4/4 + 12852*x^5/5 + 130284*x^6/6 + ...
where
exp(L(x)) = 1 + 2*x + 11*x^2 + 66*x^3 + 485*x^4 + 3842*x^5 + 32712*x^6 + ... + A216585(n)*x^n/n + ...
The central trinomial coefficients (A002426) begin:
[1, 1, 3, 7, 19, 51, 141, 393, 1107, 3139, 8953, 25653, 73789, ...];
The central binomial coefficients (A000984) begin:
[1, 2, 6, 20, 70, 252, 924, 3432, 12870, 48620, 184756, 705432, ...].
		

Crossrefs

Programs

  • Mathematica
    Table[Binomial[2*n, n]*Sum[ Binomial[n, 2*k]*Binomial[2*k, k], {k, 0, Floor[n/2]}], {n,0,50}] (* G. C. Greubel, Feb 27 2017 *)
  • PARI
    {a(n) = polcoeff((1+x+x^2)^n,n) * polcoeff((1+2*x+x^2)^n,n)}
    
  • PARI
    {a(n)=binomial(2*n,n)*sum(k=0,n\2,binomial(n,2*k)*binomial(2*k,k))}
    for(n=0,21,print1(a(n),", "))

Formula

a(n) = binomial(2*n, n) * Sum_{k=0..floor(n/2)} binomial(n, 2*k)*binomial(2*k, k).
Logarithmic derivative of A216585, after ignoring initial term a(0).
a(n) = [x^n*y^n] ( 1 + (x + y)^2 + (x + y)^4 )^n. - Peter Bala, Feb 17 2020
G.f.: hypergeom([1/2, 1/2],[1],16*x/(1+4*x))/sqrt(1+4*x). - Mark van Hoeij, May 13 2025

A216585 G.f.: exp( Sum_{n>=1} A000984(n)*A002426(n)*x^n/n ), where A000984 is the central binomial coefficients and A002426 is the central trinomial coefficients.

Original entry on oeis.org

1, 2, 11, 66, 485, 3842, 32712, 291568, 2697610, 25679316, 250190125, 2484270622, 25062816127, 256275246582, 2650947762450, 27697861115740, 291943603838698, 3101066786857876, 33167191013319532, 356924515784037128, 3862299973917286526, 42003704374124712172
Offset: 0

Views

Author

Paul D. Hanna, Sep 09 2012

Keywords

Examples

			G.f.: A(x) = 1 + 2*x + 11*x^2 + 66*x^3 + 485*x^4 + 3842*x^5 + 32712*x^6 +...
such that
log(A(x)) = 2*1*x + 6*3*x^2/2 + 20*7*x^3/3 + 70*19*x^4/4 + 252*51*x^5/5 + 924*141*x^6/6 +...+ A000984(n)*A002426(n)*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {a(n)=polcoeff(exp(sum(m=1,n+1,binomial(2*m,m)*polcoeff((1+x+x^2)^m,m)*x^m/m+x*O(x^n))),n)}
    for(n=0,30,print1(a(n),", "))

Formula

Logarithmic derivative yields A216584.

A227325 a(n) = A000272(n+1) * A000984(n).

Original entry on oeis.org

1, 2, 18, 320, 8750, 326592, 15529668, 899678208, 61556811030, 4862000000000, 435644983598396, 43678490079264768, 4846281282497517772, 589650705050503577600, 78074729079345703125000, 11177395284330167371038720, 1720546364146510165684599270
Offset: 0

Views

Author

Reinhard Zumkeller, Jul 07 2013

Keywords

Crossrefs

Central terms of triangle A088956.

Programs

  • Haskell
    a227325 n = a000272 (n + 1) * a000984 n

Formula

a(n) = A088956(2*n, n) = (n+1)^(n-1) * C(2*n, n).

Extensions

Typo in formula fixed by Zak Seidov, Jul 08 2013

A289800 p-INVERT of the central binomial coefficients (A000984), where p(S) = 1 - S - S^2.

Original entry on oeis.org

1, 4, 17, 75, 336, 1517, 6879, 31276, 142439, 649431, 2963266, 13528285, 61785007, 282257992, 1289734455, 5894167695, 26939918564, 123142940445, 562928407213, 2573477722376, 11765383864555, 53790586563231, 245933621620228, 1124446028551665, 5141224466008849
Offset: 0

Views

Author

Clark Kimberling, Aug 12 2017

Keywords

Comments

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).
See A289780 for a guide to related sequences.

Crossrefs

Programs

  • Mathematica
    z = 60; s = x/Sqrt[1 - 4 x]; p = 1 - s - s^2;
    Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A000984 *)
    Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1] (* A289800 *)

A292441 Largest m such that m^2 divides A000984(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 6, 2, 2, 3, 2, 2, 2, 2, 10, 30, 12, 3, 6, 10, 10, 6, 2, 2, 60, 30, 42, 42, 28, 2, 4, 4, 4, 21, 14, 14, 6, 2, 2, 10, 140, 14, 126, 6, 60, 90, 12, 84, 84, 210, 30, 18, 12, 6, 36, 4, 4, 6, 4, 4, 12, 12, 132, 132, 440, 55, 330, 10, 10, 90, 30, 30, 180
Offset: 0

Views

Author

Seiichi Manyama, Sep 16 2017

Keywords

Comments

a(n) is the product of p^floor(m(n,p)/2) over primes pRobert Israel, Sep 17 2017
Granville and Ramaré show that A006530(a(n)) > sqrt(n/5) if n >= 2082.
In particular a(n) -> infinity as n -> infinity. - Robert Israel, Sep 18 2017

Examples

			binomial(10,5)/7           =  252/7   = 36 = a(5)^2.
binomial(12,6)/(3*7*11)    =  924/231 =  4 = a(6)^2.
binomial(14,7)/(2*3*11*13) = 3432/858 =  4 = a(7)^2.
		

Crossrefs

Programs

  • Maple
    A000188:= n -> mul(t[1]^floor(t[2]/2), t = ifactors(n)[2]):
    seq(A000188(binomial(2*n,n)),n=0..100); # Robert Israel, Sep 17 2017

Formula

a(n) > 1 for n > 4.
a(n) = A000188(A000984(n)). - Robert Israel, Sep 17 2017
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