A186378 -id:A186378 - cp's OEIS frontend

A186375 a(n) equals the sum of the squares of the expansion coefficients for (x + y + 2*z)^n.

1, 6, 54, 588, 7110, 91476, 1224636, 16849944, 236523078, 3371140740, 48630906324, 708412918824, 10403176168476, 153813188724552, 2287366047735480, 34185974267420208, 513159651195396678, 7732530110414488932

Offset: 0

Paul D. Hanna, Feb 19 2011

nonn

Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials: (1 + 2*x^k + x^p)^n, (2 + x^k + x^p)^n, or (1 + x^k + 2*x^p)^n, for all p > n*k and fixed k > 0.

Rescaling the g.f. G(x) to T(u)=G(3*u/16) moves the singular point x=1/16 to u=1/3. Period function T(u) measures precession of the J-vector along an algebraic sphere curve with local cyclic C_3 symmetry. For precise definitions, pictures, a proof certificate, and more information, see A318245. - Bradley Klee, Aug 22 2018

			G.f.: A(x) = 1 + 6*x + 54*x^2/2!^2 + 588*x^3/3!^2 + 7110*x^4/4!^2 + ...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2]^2 *[Sum_{n>=0} (4x)^n/n!^2] where
[Sum_{n>=0} x^n/n!^2]^2 = 1 + 2*x + 6*x^2/2!^2 + 20*x^3/3!^2 + 70*x^4/4!^2 + ... + (2n)!/n!^2 *x^n/n!^2 + ...
a(4) =   256
       + 1024 + 1024
       + 576  + 2304 + 576
       + 64   + 576  + 576 + 64
       + 1    + 16   + 36  + 16 + 1  = 7110.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
E. Weisstein, Goursat's Surface, Mathworld--A Wolfram Web Resource.

Cf. A046816, A186376, A186377, A186378. Periods: A318245, A318417.

A186375 := n -> 4^n*hypergeom([1/2,-n,-n], [1,1], 1):
seq(simplify(A186375(n)), n=0..17); # Peter Luschny, May 24 2017

Table[Sum[Binomial[n,k]^2*Binomial[2k,k]*4^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 20 2012 *)
(* From Bradley Klee, Aug 22 2018: Start *)PyramidLevel[n_]:=If[n==0, {{1}}, Table[Coefficient[(2*x+y+z)^n,x^j*y^k*z^(n-j-k)]^2, {j,0,n}, {k,0,n-j}]]; a1[n_]:= Total[Flatten[PyramidLevel[n]]];
a1 /@ Range[0, 10]
RecurrenceTable[{4*(4*n-5)*(4*n-3)*a[n-2]-2*(10*n^2-10*n+3)*a[n-1]+n^2*a[n]==0, a[0]==1, a[1]==6},a,{n,0,1000}] (*  End *)
a[ n_] := If[ n < 0, 0, Block[ {x, y, z}, Expand[ (x + y + 2 z)^n] /. {t_Integer -> t^2, x -> 1, y -> 1, z -> 1}]]; (* Michael Somos, Aug 27 2018 *)

{a(n)=sum(k=0,n,binomial(n,k)^2*binomial(2*(n-k),n-k)*4^k)}

{a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)^2*sum(m=0,n,(2^2*x)^m/m!^2),n)}

{a(n)=local(V=Vec((1+2*x+x^(n+2))^n));V*V~}

(1) a(n) = Sum_{k=0..n} C(n,k)^2*C(2k,k)*4^(n-k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = B(x)^2 * B(2^2*x)
where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
Recurrence: n^2*a(n) = 2*(10*n^2-10*n+3)*a(n-1) - 4*(4*n-5)*(4*n-3)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(4*n+1/2)/(Pi*n). - Vaclav Kotesovec, Oct 20 2012
a(n) = 4^n*hypergeom([1/2,-n,-n], [1,1], 1). - Peter Luschny, May 24 2017
G.f.: G(x)=Sum_{n>=0}a(n)x^n, 6*(10*x-1)*G + (192*x^2-40*x+1)*G' + x*(16*x-1)*(4*x-1)*G''=0. - Bradley Klee, Aug 22 2018

Name edited by Bradley Klee, Aug 22 2018

A186376 a(n) is the sum of the squares of the coefficients of (x + 2y + 3z)^n.

Original entry on oeis.org

1, 14, 294, 7292, 198310, 5717124, 171485916, 5290989816, 166688596998, 5335746337460, 172951272017524, 5662840724506056, 186960502253087836, 6215612039963043368, 207865952390729881080, 6987002286567227550192

Offset: 0

Paul D. Hanna, Feb 19 2011

nonn

a(n) equals the least sum of the squares of the coefficients in (1 + 2*x^k + 3*x^p)^n found at sufficiently large p for some fixed k>0.
Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials:
. (1 + 3*x^k + 2*x^p)^n, or
. (2 + x^k + 3*x^p)^n, or
. (3 + 2*x^k + x^p)^n, etc.,
for all p>(n+1)k and fixed k>0.
a(n) is the sum of the squares of the coefficients of (x + 2*y + 3*z)^n. - Michael Somos, Aug 25 2018

			G.f.: A(x) = 1 + 14*x + 294*x^2/2!^2 + 7292*x^3/3!^2 +...
The g.f. may be expressed as:
[Sum_{n>=0}x^n/n!^2]*[Sum_{n>=0}(4x)^n/n!^2]*[Sum_{n>=0}(9x)^n/n!^2].

Cf. A186375, A186377, A186378, A186391.

Table[Sum[Binomial[n,k]^2 * Sum[Binomial[k,j]^2 * 4^(k-j) * 9^j, {j,0,k}], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)
a[ n_] := If[ n < 0, 0, Block[ {x, y, z}, Expand[(x + y + 2 z)^n] /. {t_Integer -> t^2, x -> 1, y -> 2, z -> 3}]]; (* Michael Somos, Aug 25 2018 *)

{a(n)=local(V=Vec((1+2*x+3*x^(n+2))^n));V*V~}

{a(n)=sum(k=0,n,binomial(n,k)^2*sum(j=0,k,binomial(k,j)^2*4^(k-j)*9^j))}

{a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)*sum(m=0,n,(2^2*x)^m/m!^2)*sum(m=0,n,(3^2*x)^m/m!^2),n)}

(1) a(n) = Sum_{k=0..n} C(n,k)^2 *Sum_{j=0..k} C(k,j)^2*4^(k-j)*9^j.
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = B(x) * B(2^2*x) * B(3^2*x)
where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
Recurrence: n^2*(4*n - 7)*a(n) = 14*(16*n^3 - 44*n^2 + 34*n - 9)*a(n-1) - 196*(2*n - 3)^2*(4*n - 3)*a(n-2) + 144*(4*n - 9)*(4*n - 7)*(4*n - 3)*a(n-3). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 2^(2*n-1) * 3^(2*n+1) / (Pi*n). - Vaclav Kotesovec, Feb 12 2015

Name changed to match the definition given by Michael Somos. - Paul D. Hanna, Sep 05 2018

A186377 a(n) equals the least sum of the squares of the coefficients in (1 + 2*x^k + x^p + x^q)^n found at sufficiently large p and q>(n+1)p for some fixed k>0.

Original entry on oeis.org

1, 7, 79, 1129, 18559, 333577, 6365089, 126652183, 2598628543, 54577439833, 1167481074529, 25346459683783, 557042221952881, 12368307313680871, 277027947337574911, 6251808554314780009, 142015508983550880703

Offset: 0

Paul D. Hanna, Feb 19 2011

nonn

Equivalently, a(n) equals the sum of the squares of the coefficients in any one of the following polynomials:
. (2 + x^k + x^p + x^q)^n, or
. (1 + x^k + 2*x^p + x^q)^n, or
. (1 + x^k + x^p + 2*x^q)^n,
for all p>(n+1)k and q>(n+1)p and fixed k>0.

			G.f.: A(x) = 1 + 7*x + 79*x^2/2!^2 + 1129*x^3/3!^2 + 18559*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2]^3 *[Sum_{n>=0} (4x)^n/n!^2] where
[Sum_{n>=0} x^n/n!^2]^3 = 1 + 3*x + 15*x^2/2!^2 + 93*x^3/3!^2 + 639*x^4/4!^2 + 4653*x^5/5!^2 +...+ A002893(n)*x^n/n!^2 +...

Cf. A186375, A186376, A186378.

Table[Sum[Binomial[n,k]^2 * 4^(n-k) *Sum[Binomial[k,j]^2 * Binomial[2j,j], {j,0,k}], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)

{a(n)=local(V=Vec((1+2*x+x^(n+2)+x^(n^2+2*n+3))^n));V*V~}

{a(n)=sum(k=0,n,binomial(n,k)^2*4^(n-k)*sum(j=0,k,binomial(k,j)^2*binomial(2*j,j)))}

{a(n)=n!^2*polcoeff(sum(m=0,n,x^m/m!^2)^3*sum(m=0,n,(2^2*x)^m/m!^2),n)}

(1) a(n) = Sum_{k=0..n} C(n,k)^2 *4^(n-k) *Sum_{j=0..k} C(k,j)^2*C(2j,j).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = B(x)^3 * B(2^2*x)
where B(x) = Sum_{n>=0} x^n/n!^2 = BesselI(0, 2*sqrt(x)).
Recurrence: (n-1)*n^3*(3*n - 5)*a(n) = 2*(n-1)*(54*n^4 - 174*n^3 + 192*n^2 - 99*n + 20)*a(n-1) - 2*(441*n^5 - 2604*n^4 + 6102*n^3 - 7107*n^2 + 4111*n - 940)*a(n-2) + 2*(n-2)^2*(726*n^3 - 3076*n^2 + 4188*n - 1655)*a(n-3) - 225*(n-3)^2*(n-2)^2*(3*n - 2)*a(n-4). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 5^(2*n+2) / (2^(7/2) * Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Feb 12 2015

A186391 a(n) equals the least sum of the squares of the coefficients in (1 + x^k + x^(2k) + x^p)^n found at sufficiently large p for some fixed k>0.

Original entry on oeis.org

1, 4, 32, 340, 4096, 52704, 705956, 9717488, 136443904, 1945097296, 28063032832, 408836809088, 6004266204964, 88779091937488, 1320294416004736, 19733192546306640, 296219343194357760, 4463668854432401280

Offset: 0

Paul D. Hanna, Feb 19 2011

nonn

Equivalently, a(n) equals the sum of the squares of the coefficients in the polynomial: (1+x+x^2 + x^p)^n for all p>2(n+1).
...
More generally, let B(x) = Sum_{n>=0} b(n)*x^n/n!^2 such that b(n) is the least sum of the squares of the coefficients in (F(x^k) + t*x^p)^n where F(x) is a finite polynomial in x with degree d and p>(n+1)dk for some fixed k>0,
then B(x) = [Sum_{n>=0} (t^2*x)^n/n!^2]*[Sum_{n>=0} c(n)/n!^2] where c(n) equals the sum of the squares of the coefficients in the polynomial F(x)^n.

			G.f.: A(x) = 1 + 4*x + 32*x^2/2!^2 + 340*x^3/3!^2 + 4096*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2] * C(x) where
C(x)= 1 + 3*x + 19*x^2/2!^2 + 141*x^3/3!^2 + 1107*x^4/4!^2 + 8953*x^5/5!^2 + 73789*x^6/6!^2 +...+ A082758(n)*x^n/n!^2 +...

Cf. A082758, A186392, A186378.

Table[Sum[Binomial[n,k]^2 * Sum[Binomial[2*k-j,j] * Binomial[2*k, j], {j,0,k}], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)

{a(n)=local(V=Vec((1+x+x^2+x^(2*n+3))^n)); V*V~}

{a(n)=sum(k=0,n,binomial(n,k)^2*sum(j=0, k, binomial(2*k-j, j)*binomial(2*k, j)))}

{a(n)=n!^2*polcoeff(sum(m=0, n, x^m/m!^2+x*O(x^n)) *sum(m=0, n, sum(k=0, m, binomial(2*m-k, k)*binomial(2*m, k))*x^m/m!^2), n)}

(1) a(n) = Sum_{k=0..n} C(n,k)^2*A082758(k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = [Sum_{n>=0} x^n/n!^2]*[Sum_{n>=0} A082758(n)*x^n/n!^2]
where A082758(n) is the sum of the squares of the trinomial coefficients in (1+x+x^2)^n.
Recurrence: n^2*(2*n-1)^2*(240*n^5 - 2720*n^4 + 12016*n^3 - 25824*n^2 + 26966*n - 10939)*a(n) = 4*(6000*n^9 - 81920*n^8 + 470480*n^7 - 1486720*n^6 + 2841766*n^5 - 3403995*n^4 + 2557086*n^3 - 1163574*n^2 + 291045*n - 30429)*a(n-1) - 24*(6720*n^9 - 103040*n^8 + 678448*n^7 - 2511512*n^6 + 5740528*n^5 - 8358208*n^4 + 7691502*n^3 - 4262963*n^2 + 1269092*n - 151182)*a(n-2) + 4*(96000*n^9 - 1633280*n^8 + 11993280*n^7 - 49661760*n^6 + 127015344*n^5 - 206357088*n^4 + 210522120*n^3 - 127943860*n^2 + 41074746*n - 5150007)*a(n-3) - 64*(n-3)^2*(4*n - 13)*(4*n - 11)*(240*n^5 - 1520*n^4 + 3536*n^3 - 3696*n^2 + 1686*n - 261)*a(n-4). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 2^(4*n + 1/2) / (sqrt(3) * Pi * n). - Vaclav Kotesovec, Feb 12 2015

A186392 a(n) equals the least sum of the squares of the coefficients in ((1 + x^k)^3 + x^p)^n found at sufficiently large p for some fixed k>0.

Original entry on oeis.org

1, 21, 1005, 57117, 3515661, 227676321, 15287457741, 1054718889525, 74310865827597, 5323117605120273, 386421018984886905, 28357462296640927845, 2099749565250183356973, 156648556486910137353777

Offset: 0

Paul D. Hanna, Feb 19 2011

nonn

Equivalently, a(n) equals the sum of the squares of the coefficients in the polynomial: ((1+x)^3 + x^p)^n for all p>3(n+1).
...
More generally, let B(x) = Sum_{n>=0} b(n)*x^n/n!^2 such that b(n) is the least sum of the squares of the coefficients in (F(x^k) + t*x^p)^n where F(x) is a finite polynomial in x with degree d and p>(n+1)dk for some fixed k>0,
then B(x) = [Sum_{n>=0} (t^2*x)^n/n!^2]*[Sum_{n>=0} c(n)/n!^2] where c(n) equals the sum of the squares of the coefficients in the polynomial F(x)^n.

			G.f.: A(x) = 1 + 21*x + 1005*x^2/2!^2 + 57117*x^3/3!^2 + 3515661*x^4/4!^2 +...
The g.f. may be expressed as:
A(x) = [Sum_{n>=0} x^n/n!^2] * C(x) where
C(x)= 1 + 20*x + 924*x^2/2!^2 + 48620*x^3/3!^2 + 2704156*x^4/4!^2 +...+ (6n)!/(3n)!^2*x^n/n!^2 +...

Cf. A186378, A186391.

Table[Sum[Binomial[n,k]^2 * Binomial[6*k,3*k], {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Feb 11 2015 *)

{a(n)=sum(k=0,n,binomial(n,k)^2*(6*k)!/(3*k)!^2)}

{a(n)=n!^2*polcoeff(sum(m=0, n, (6*m)!/(3*m)!^2*x^m/m!^2)*sum(m=0, n, x^m/m!^2+x*O(x^n)), n)}

{a(n)=local(V=Vec(((1+x)^3+x^(3*n+4))^n)); V*V~}

(1) a(n) = Sum_{k=0..n} C(n,k)^2*C(6k,3k).
Let g.f. A(x) = Sum_{n>=0} a(n)*x^n/n!^2, then
(2) A(x) = [Sum_{n>=0} x^n/n!^2]*[Sum_{n>=0} (6n)!/(3n)!^2 *x^n/n!^2]
where (6n)!/(3n)!^2 is the sum of the squares of the coefficients in (1+x)^(3n).
Recurrence: n^2*(3*n-2)^2*(3*n-1)^2*(11664000*n^8 - 185457600*n^7 + 1268251200*n^6 - 4872227850*n^5 + 11501613345*n^4 - 17086352076*n^3 + 15601848563*n^2 - 8008019030*n + 1769497668)*a(n) = (125656272000*n^14 - 2377737892800*n^13 + 20176977398400*n^12 - 101636310193650*n^11 + 339033082048335*n^10 - 790997589023868*n^9 + 1328807234532186*n^8 - 1629746362828908*n^7 + 1463401419459585*n^6 - 955188456600918*n^5 + 445022508698326*n^4 - 143136353903096*n^3 + 29975298427288*n^2 - 3656735400000*n + 197437737600)*a(n-1) - 2*(2060573904000*n^14 - 41192487921600*n^13 + 371601043200000*n^12 - 2002776983698050*n^11 + 7194382158658545*n^10 - 18189133596160956*n^9 + 33303386556391095*n^8 - 44736636269608884*n^7 + 44153839934527497*n^6 - 31729853553647838*n^5 + 16260230748029395*n^4 - 5728259846949480*n^3 + 1303148021418356*n^2 - 170502613376352*n + 9707820967872)*a(n-2) + 18*(n-2)^2*(645497424000*n^12 - 11574979185600*n^11 + 91670927428800*n^10 - 422559236833650*n^9 + 1257324245932095*n^8 - 2530974275757936*n^7 + 3511780338639909*n^6 - 3357925240298748*n^5 + 2175657267448355*n^4 - 921464025234426*n^3 + 239290954736149*n^2 - 33846317262624*n + 1986410906748)*a(n-3) - 81*(n-3)^2*(n-2)^2*(140399568000*n^10 - 1953070315200*n^9 + 11498171428800*n^8 - 37500795421650*n^7 + 74520854931765*n^6 - 93517328384172*n^5 + 74333125575977*n^4 - 36536327729802*n^3 + 10492652783974*n^2 - 1569370920528*n + 93953212632)*a(n-4) + 321489*(n-4)^2*(n-3)^2*(n-2)^2*(11664000*n^8 - 92145600*n^7 + 296640000*n^6 - 504146250*n^5 + 489706095*n^4 - 274985196*n^3 + 85944305*n^2 - 13448002*n + 818220)*a(n-5). - Vaclav Kotesovec, Feb 12 2015
a(n) ~ 3^(4*n + 5/2) / (16 * Pi * n). - Vaclav Kotesovec, Feb 12 2015

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A186375 a(n) equals the sum of the squares of the expansion coefficients for (x + y + 2*z)^n.

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A186376 a(n) is the sum of the squares of the coefficients of (x + 2*y + 3*z)^n.

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A186377 a(n) equals the least sum of the squares of the coefficients in (1 + 2*x^k + x^p + x^q)^n found at sufficiently large p and q>(n+1)p for some fixed k>0.

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A186391 a(n) equals the least sum of the squares of the coefficients in (1 + x^k + x^(2k) + x^p)^n found at sufficiently large p for some fixed k>0.

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A186392 a(n) equals the least sum of the squares of the coefficients in ((1 + x^k)^3 + x^p)^n found at sufficiently large p for some fixed k>0.

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A186376 a(n) is the sum of the squares of the coefficients of (x + 2y + 3z)^n.