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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A181147 a(n) = Sum_{k=0..n-1} (105k+44)*C(2k,k)^2*T(k)*(-1)^(n-1-k)/(2n*C(2n,n)), where T(k) (k=0,1,2,...) are central trinomial coefficients given by A002426.

Original entry on oeis.org

11, 23, 224, 1747, 16754, 162392, 1651206, 17126327, 181182446, 1943132842, 21080299228, 230802972664, 2546569337336, 28280754214358, 315824396838386, 3544003431783795, 39936833763112790, 451718158386620678
Offset: 1

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Author

Zhi-Wei Sun, Jan 24 2011

Keywords

Comments

On Jan 22 2011, Zhi-Wei Sun conjectured that a(n) is a positive integer for every n=1,2,3,... Let p > 3 be a prime. He also conjectured that a(p) == 5 + 6*(p/3)*(2-3^(p-1)) (mod p^2). Another conjecture of his states that Sum_{k=0..p-1} (-1)^k*binomial(2k,k)^2*T(k) is congruent to b(p) modulo p^2, where b(p)=0 if (p/15)=-1, b(p) = 4x^2-2p if p == 1,4 (mod 15) and p = x^2+15y^2 with x,y integers, and b(p) = 20x^2-2p if p == 2,8 (mod 15) and p=5x^2+3y^2 with x,y integers.

Examples

			For n=2 we have a(2) = (44*1^2*T(0)(-1) + (105+44)*2^2*T(1))/(2*2*binomial(4,2)) = 23.
		

Crossrefs

Programs

  • Maple
    T:= n-> `if`(n=0, 1, coeff ((x^2+x+1)^n, x, n)):
    a:= n-> add ((105*k+44) *binomial (2*k, k)^2 *T(k)*(-1)^(n-1-k),
                 k=0..n-1)/ (2*n*binomial (2*n,  n)):
    seq (a(n), n=1..30);
  • Mathematica
    T[k_]:=If[k>0,Coefficient[(x^2+x+1)^k,x^k],1]
    A[n_]:=Sum[(105k+44)Binomial[2k,k]^2*T[k](-1)^(n-1-k),{k,0,n-1}]/(2n*Binomial[2n,n])
    Table[A[n],{n,1,50}]

A186037 a(n) = log_2((1+A002426(n))/numerator((1+A002426(n))/2^n)).

Original entry on oeis.org

0, 1, 2, 3, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 2, 3, 2, 2, 1, 1, 1, 1, 3, 2, 1, 1, 3, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 3, 2, 1, 1, 2, 4, 3, 3, 1, 1, 1, 1, 3, 2, 3, 3, 1, 1, 2, 2, 1, 1, 1, 1, 2, 3, 2, 2, 1, 1, 1, 1, 3
Offset: 0

Views

Author

Paul Barry, Feb 11 2011

Keywords

Programs

  • Mathematica
    b[n_] := If[n < 0, 0, 3^n Hypergeometric2F1[1/2, -n, 1, 4/3]];
    a[n_] := Log[2, (1 + b[n])/Numerator[(1 + b[n])/2^n]]; Table[a[n], {n, 0, 50}] (* G. C. Greubel, Feb 27 2017 *)

A192670 Floor-Sqrt transform of central trinomial coefficients (A002426).

Original entry on oeis.org

1, 1, 1, 2, 4, 7, 11, 19, 33, 56, 94, 160, 271, 461, 785, 1337, 2279, 3890, 6644, 11357, 19426, 33246, 56930, 97535, 167177, 286665, 491745, 843842, 1448530, 2487302, 4272238, 7340067, 12614059, 21682694, 37279415, 64108676, 110268453, 189700207, 326408942, 561733093, 966869803
Offset: 0

Views

Author

Emanuele Munarini, Jul 07 2011

Keywords

Programs

  • Mathematica
    FSFromSeries[f_,x_,n_] := Map[Floor[Sqrt[#]]&,CoefficientList[Series[f,{x,0,n}],x]]
    FSFromSeries[1/Sqrt[1-2x-3x^2],x,100]

Formula

a(n) = floor(sqrt(trinomial(2*n,n))).

A212848 Least prime factor of n-th central trinomial coefficient (A002426).

Original entry on oeis.org

1, 1, 3, 7, 19, 3, 3, 3, 3, 43, 7, 3, 113, 73, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 7, 17, 3, 719, 7, 3, 3, 3, 3, 967, 9539, 3, 17, 47, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 19
Offset: 0

Views

Author

Jonathan Vos Post, May 28 2012

Keywords

Comments

A002426(n) is prime for n = 2, 3, 4, no more through 10^5. A002426 is semiprime iff A102445(n) = 2 (as is the case for n = 5, 6, 7, 9, 10, 12, 13).

Examples

			a(9) = 43 because A002426(9) = 3139 = 43 * 73.
		

Crossrefs

Programs

  • Maple
    A002426:= gfun:-rectoproc({(n+2)*a(n+2)-(2*n+3)*a(n+1)-3*(n+1)*a(n) = 0, a(0)=1, a(1)=1},a(n),remember):
    lpf:= proc(n) local F;
        F:= map(proc(t) if t[1]::integer then t[1] else NULL fi end proc,
           ifactors(n, easy)[2]);
        if nops(F) > 0 then min(F)
        else min(numtheory:-factorset(n))
        fi
    end proc:
    lpf(1):= 1:
    map(lpf @ A002426, [$0..100]); # Robert Israel, Jun 20 2017
  • Mathematica
    a = b = 1; t = Join[{a, b}, Table[c = ((2 n - 1) b + 3 (n - 1) a)/n; a = b; b = c; c, {n, 2, 100}]]; Table[FactorInteger[n][[1, 1]], {n, t}] (* T. D. Noe, May 30 2012 *)
  • PARI
    a(n) = my(x=polcoeff((1 + x + x^2)^n, n)); if (x==1, 1, vecmin(factor(x)[,1])); \\ Michel Marcus, Jun 20 2017

Formula

a(n) = A020639(A002426(n)).

A216586 G.f.: exp( Sum_{n>=1} A002426(n)/2 * A002426(n) * x^n/n ), where A002426 is the central binomial coefficients and A002426 is the central trinomial coefficients.

Original entry on oeis.org

1, 1, 5, 28, 202, 1579, 13375, 118858, 1098458, 10453452, 101872926, 1012109860, 10218226307, 104570617520, 1082633236498, 11321654913838, 119438468577559, 1269787015989428, 13592294300856138, 146390465351654178, 1585337895099162317, 17253991887494062080
Offset: 0

Views

Author

Paul D. Hanna, Sep 09 2012

Keywords

Examples

			G.f.: A(x) = 1 + x + 5*x^2 + 28*x^3 + 202*x^4 + 1579*x^5 + 13375*x^6 +...
such that
log(A(x)) = 1*1*x + 3*3*x^2/2 + 10*7*x^3/3 + 35*19*x^4/4 + 126*51*x^5/5 + 462*141*x^6/6 +...+ A001700(n)*A002426(n)*x^n/n +...
		

Crossrefs

Programs

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

Formula

Self-convolution yields A216585.

A225604 G.f.: exp( Sum_{n>=1} A002426(n^2) * x^n/n ), where A002426 is the central trinomial coefficients.

Original entry on oeis.org

1, 1, 10, 1056, 1300253, 16436676927, 2026538428535847, 2377041996570919354629, 26137381916593225072659360863, 2668615348740645885804068311893052895, 2513426521807431879643802805359800329740903335, 21735453667359385540995804455408000917620356989063370267
Offset: 0

Views

Author

Paul D. Hanna, Aug 03 2013

Keywords

Examples

			G.f.: A(x) = A(x) = 1 + x + 10*x^2 + 1056*x^3 + 1300253*x^4 + 16436676927*x^5 +...
where
log(A(x)) = x + 19*x^2/2 + 3139*x^3/3 + 5196627*x^4/4 + 82176836301*x^5/5 +...+ A225602(n)*x^n/n +...
		

Crossrefs

Programs

  • PARI
    {A002426(n)=sum(k=0,n, binomial(n, k)*binomial(k, n-k))}
    {a(n)=polcoeff(exp(sum(m=1,n+1,A002426(m^2)*x^m/m) +x*O(x^n)),n)}
    for(n=0,20,print1(a(n),", "))

Formula

Logarithmic derivative yields A225602.

A231373 G.f. A(x) satisfies: A(x-x^2-x^3) = 1/sqrt(1-2*x-3*x^2), which is the g.f. the central trinomial coefficients (A002426).

Original entry on oeis.org

1, 1, 4, 16, 71, 327, 1550, 7490, 36720, 182028, 910330, 4585318, 23233722, 118315318, 605088690, 3105994302, 15994906965, 82602799485, 427662046960, 2219130114108, 11538302709769, 60102637378353, 313591732265662, 1638671208390738, 8574718477933404, 44926247350136232
Offset: 0

Views

Author

Paul D. Hanna, Nov 08 2013

Keywords

Examples

			G.f.: A(x) = 1 + x + 4*x^2 + 16*x^3 + 71*x^4 + 327*x^5 + 1550*x^6 +...
where A(x-x^2-x^3)^2 = 1/(1-2*x-3*x^2):
A(x-x^2-x^3) = 1 + x + 3*x^2 + 7*x^3 + 19*x^4 + 51*x^5 + 141*x^6 + 393*x^7 + 1107*x^8 +...+ A002426(n)*x^n +...
The square of the g.f. begins (cf. A038112):
A(x)^2 = 1 + 2*x + 9*x^2 + 40*x^3 + 190*x^4 + 924*x^5 + 4578*x^6 +...
such that A(x)^2 = d/dx x*G(x) where G(x) is the g.f. of A001002:
G(x) = 1 + x + 3*x^2 + 10*x^3 + 38*x^4 + 154*x^5 + 654*x^6 +...
and satisfies G(x-x^2-x^3) = 1/(1-x-x^2).
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[Sqrt[D[InverseSeries[Series[x - x^2 - x^3, {x, 0, 30}], x], x]], {x, 0, 30}], x] (* Vaclav Kotesovec, Mar 31 2014 *)
  • PARI
    {a(n)=local(G=serreverse(x-x^2-x^3+x^2*O(x^n)),A);A=sqrt(deriv(G));polcoeff(A,n)}
    for(n=0,30,print1(a(n),", "))
    
  • PARI
    {Dx(n, F)=local(D=F); for(i=1, n, D=deriv(D)); D} \\ = d^n/dx^n F
    {a(n)=local(A2=x); A2=1+sum(m=1, n+1, Dx(m, x^(2*m)*(1+x +x*O(x^n))^m/m!)); polcoeff(sqrt(A2), n)}
    for(n=0,30,print1(a(n),", "))

Formula

Self-convolution yields A038112.
G.f. A(x) satisfies:
(1) A(x) = sqrt( Sum_{n>=0} d^n/dx^n x^(2*n)*(1+x)^n/n! ).
(2) A(x) = sqrt((1+x)*(5-27*x)*A(x)^6 - 1)/2, from a formula by Mark van Hoeij in A038112.
(3) A(x) = sqrt( d/dx x*G(x) ) where G(x) = Series_Reversion(x-x^2-x^3)/x is the g.f. of A001002.
(4) A(x) = 1/sqrt(1 - 2*x*G(x) - 3*x^2*G(x)^2) where G(x) = Series_Reversion(x-x^2-x^3)/x is the g.f. of A001002.
Sum_{k=0..n} a(k)*a(n-k) = Sum_{k=0..n} C(n+k, k)*C(k, n-k), from a formula by Paul Barry in A038112.
Recurrence: 25*(n-2)*(n-1)*n*a(n) = 110*(n-2)*(n-1)*(2*n-3)*a(n-1) - (n-2)*(214*n^2 - 856*n + 717)*a(n-2) - 33*(2*n-5)*(18*n^2 - 90*n + 113)*a(n-3) - 81*(n-3)*(3*n-11)*(3*n-7)*a(n-4). - Vaclav Kotesovec, Nov 10 2013
a(n) ~ 3^(3/4) * GAMMA(3/4) * (27/5)^n / (2*10^(1/4)*Pi*n^(3/4)). - Vaclav Kotesovec, Dec 29 2013

A329475 a(n) = Sum_{k=0..n} C(n,k)^2*T(k)*T(n-k), where T(k) = A002426(k) is the coefficient of x^k in the expansion of (x^2+x+1)^k.

Original entry on oeis.org

1, 2, 10, 68, 586, 5252, 49204, 475400, 4723786, 47937812, 494786260, 5177188040, 54794164660, 585565913480, 6309889976680, 68484312535568, 747985368753226, 8214968193003860, 90669516557975524, 1005156080857529768, 11187435500257898836, 124964856185950621832
Offset: 0

Views

Author

Zhi-Wei Sun, Nov 13 2019

Keywords

Comments

The author introduced this sequence in arXiv:1911.05456 and made the following conjecture.
Conjecture: Let p be an odd prime and let S = Sum_{k=0..p-1}a(k)/(-4)^k. If p == 1 (mod 12) and p = x^2 + 9*y^2 with x and y integers, then S == 4*x^2-2*p (mod p^2). If p == 5 (mod 12) and p = x^2 + y^2 with x == y (mod 3), then S == 4*x*y (mod p^2). If p == 3 (mod 4), then S == 0 (mod p^2).
Note that if p > 3 is a prime, then a(p-1) == Sum_{k=0..p-1} T(k)*T(p-1-k) == Legendre(p/3)*Sum_{k=0..p-1}T(k)^2/(-3)^k == 1 (mod p) by (1.7) and (2.3) of the author's 2014 paper in Sci. China Math.

Examples

			a(1) = 2 since Sum_{k=0,1} C(1,k)^2*T(k)*T(1-k) = C(1,0)^2*T(0)*T(1) + C(1,1)^2*T(1)*T(0) = 2*T(0)*T(1) = 2*1*1 = 2.
		

Crossrefs

Programs

  • Mathematica
    T[0]=1; T[1]=1; T[n_]:=T[n]=((2n-1)T[n-1]+3*(n-1)*T[n-2])/n;
    a[n_]:=a[n]=Sum[Binomial[n,k]^2*T[k]*T[n-k],{k,0,n}];
    Table[a[n],{n,0,21}]

Formula

a(n) ~ (3/2)*12^n/(n*Pi)^(3/2) as n tends to the infinity.

A179100 a(n) = (1/n) * Sum_{k=0..n-1} (8k+5) T_k^2, where T_0, T_1, ... are central trinomial coefficients given by A002426.

Original entry on oeis.org

5, 9, 69, 407, 2997, 22005, 169389, 1325889, 10573677, 85386881, 697013325, 5739021051, 47599593941, 397234035333, 3332690347437, 28089543969855, 237711099004461, 2018856328439841, 17200553934626253, 146966002696538271
Offset: 1

Views

Author

Zhi-Wei Sun, Jun 29 2010

Keywords

Comments

On Jun 17 2010, Zhi-Wei Sun conjectured that a(n) is an integer for every n=1,2,3,... and that a(p) == 3(p/3) (mod p) for any prime p, where (p/3) is the Legendre symbol. He also observed that Sum_{k=0..n-1} (2k+1) T_k*3^{n-1-k} = n * Sum_{k=0..n-1} C(n-1,k)*(-1)^(n-1-k)*(k+1)*C(2k,k).

Examples

			For n=3 we have a(3) = (5*T_0^2 + 13*T_1^2 + 21*T_2^2)/3 = (5 + 13 + 21*9)/3 = 69.
		

Crossrefs

Programs

  • Mathematica
    TT[n_]:=Sum[Binomial[n,2k]Binomial[2k,k],{k,0,Floor[n/2]}] SS[n_]:=Sum[(8k+5)*TT[k]^2,{k,0,n-1}]/n Table[SS[n],{n,1,50}]

A308482 Composites c such that T_{c-1} == (c/3)*3^(c-1) (mod c), where T_i denotes the i-th central trinomial coefficient (A002426) and (/) denotes the Kronecker symbol.

Original entry on oeis.org

4, 9, 20, 25, 27, 40, 49, 80, 81, 121, 169, 189, 243, 272, 289, 361, 369, 400, 416, 470, 529, 544, 567, 729, 841, 961, 1071, 1323, 1369, 1539, 1681, 1849, 2000, 2187, 2209, 2809, 2889, 3213, 3481, 3721, 4489, 4617, 5041, 5329, 6241, 6561, 6889, 7749, 7921, 8667
Offset: 1

Views

Author

Felix Fröhlich, May 30 2019

Keywords

Comments

Composites satisfying a weaker version of an analog to a congruence satisfied by all primes > 3 (cf. Cao, Sun, 2015, Theorem 1.1 (i); cf. Sun, 2011, Remark to Conjecture A69).
Up to 9000, 189 is the only composite satisfying the congruence modulo c^2. Do any other such composites exist?

Crossrefs

Cf. A002426.

Programs

  • Mathematica
    aQ[n_] := CompositeQ[n] && Divisible[3^(n-1)*(Hypergeometric2F1[1/2, 1-n, 1, 4/3] - JacobiSymbol[n,3]) ,n]; Select[Range[1000], aQ] (* Amiram Eldar, Jul 10 2019 *)
  • PARI
    t(n) = sum(k=0, floor(n/2), binomial(n, k)*binomial(n-k, k))
    is(n) = Mod(t(n-1), n)==kronecker(n, 3)*3^(n-1)
    forcomposite(c=1, , if(is(c), print1(c, ", ")))

Extensions

a(45)-a(50) from Amiram Eldar, Jul 10 2019
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