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%I #29 Apr 17 2023 11:01:41
%S 1,1,4,17,51,136,393,1233,3865,11851,36301,112520,351352,1098189,
%T 3433704,10758609,33794505,106344793,335061790,1056924667,3338026857,
%U 10554163533,33402840615,105809430024,335444908176,1064268538776,3379009937161,10735253448349,34127137228747
%N a(n) = [x^n] (1 + x + x^3 + x^4)^n.
%C Equals the Logarithmic derivative of A198951, where the g.f. of A198951 satisfies: G(x) = (1 + x*G(x))*(1 + x^3*G(x)^3).
%H G. C. Greubel, <a href="/A228960/b228960.txt">Table of n, a(n) for n = 1..1000</a>
%F L.g.f. L(x) satisfies:
%F (1) L(x) = log( (1/x)*Series_Reversion(x/((1+x)*(1+x^3))) ).
%F (2) L(x) = Sum_{n>=1} x^n/n * Sum_{k=0..n} C(n,k)^2 * x^(2*k) * exp(2*k*L(x)).
%F (3) L(x) = Sum_{n>=1} x^n/n * (1 - x^2*exp(2*L(x)))^(2*n+1) * Sum_{k>=0} C(n+k,k)^2 * x^(2*k) * exp(2*k*L(x)).
%F Conjecture: 3*n*(1661*n-1820) *(3*n-1) *(3*n-2) *a(n) +(1927258*n^4 -17091925*n^3 +50171975*n^2 -59448794*n +24442440) *a(n-1) +12*(-705605*n^4 +6363374*n^3 -20228575*n^2 +27219817*n -13185456) *a(n-2) +54*(n-2) *(250934*n^3 -1892927*n^2 +4367836*n -3055963) *a(n-3) -486*(n-2)*(n-3) *(26090*n-34343) *(2*n-7) *a(n-4)=0. - _R. J. Mathar_, Sep 15 2013.
%F Recurrence (of order 3): 3*n*(3*n-2)*(3*n-1)*(238*n^3 - 1302*n^2 + 2285*n - 1293)*a(n) = 2*(13328*n^6 - 92904*n^5 + 249452*n^4 - 329211*n^3 + 224408*n^2 - 74649*n + 9360)*a(n-1) - 18*(n-1)*(2380*n^5 - 15400*n^4 + 36710*n^3 - 39398*n^2 + 18345*n - 2880)*a(n-2) + 162*(n-2)*(n-1)*(2*n-5)*(238*n^3 - 588*n^2 + 395*n - 72)*a(n-3). - _Vaclav Kotesovec_, Dec 27 2013
%F a(n) ~ c*d^n/sqrt(n), where d = 1/81*((2144134 + 520506*sqrt(17))^(2/3) - 2036 + 112*(2144134 + 520506*sqrt(17))^(1/3))*(2144134 + 520506 * sqrt(17))^(-1/3) = 3.23407602060970245... is the root of the equation -324 + 180*d - 112*d^2 + 27*d^3 = 0 and c = 1/102*sqrt(17)*sqrt((34102 + 8262*sqrt(17))^(1/3)*((34102+8262*sqrt(17))^(2/3) + 136 + 136*(34102 + 8262*sqrt(17))^(1/3)))/((34102 + 8262*sqrt(17))^(1/3)*sqrt(Pi)) = 0.3061270429417747... - _Vaclav Kotesovec_, Dec 27 2013
%F From _Peter Bala_, Jun 15 2015: (Start)
%F a(n) = [x^(3*n)](1 + x + x^3 + x^4)^n.
%F a(n) = Sum_{k = 0..floor(n/3)} binomial(n,k)*binomial(n,3*k). Applying Maple's sumrecursion command to this formula gives the above recurrence of Kotesovec. (End)
%F a(n) = hypergeom([1/3-n/3, 2/3-n/3, -n, -n/3], [1/3, 2/3, 1], 1). - _Vladimir Reshetnikov_, Oct 04 2016
%F From _Peter Bala_, Apr 15 2023: (Start)
%F Conjecture 1: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for positive integers n and r and all primes p >= 5. Cf. A350383.
%F Conjecture 2: let k be a positive integer, m an integer and let f(x) = g(x)/h(x), where both g(x) and h(x) are finite products of cyclotomic polynomials. Then the same supercongruences hold, except for a finite number of primes p depending on f(x), for the sequence {a_(k,m,f)(n): n >= 0} defined by a_(k,m,f)(n) = [x^(k*n)] f(x)^(m*n). (End)
%e L.g.f.: L(x) = x + x^2/2 + 4*x^3/3 + 17*x^4/4 + 51*x^5/5 + 136*x^6/6 +...
%e Given G(x) = exp(L(x)), which is the g.f. of A198951:
%e G(x) = 1 + x + x^2 + 2*x^3 + 6*x^4 + 16*x^5 + 39*x^6 + 99*x^7 + 271*x^8 + 763*x^9 + 2146*x^10 +...+ A198951(n)*x^n +...
%e then the l.g.f. L(x) satisfies the series:
%e L(x) = (1 + x^2*G(x)^2)*x
%e + (1 + 2^2*x^2*G(x)^2 + x^4*G(x)^4)*x^2/2
%e + (1 + 3^2*x^2*G(x)^2 + 3^2*x^4*G(x)^4 + x^6*G(x)^6)*x^3/3
%e + (1 + 4^2*x^2*G(x)^2 + 6^2*x^4*G(x)^4 + 4^2*x^6*G(x)^6 + x^8*G(x)^8)*x^4/4
%e + (1 + 5^2*x^2*G(x)^2 + 10^2*x^4*G(x)^4 + 10^2*x^6*G(x)^6 + 5^2*x^8*G(x)^8 + x^10*G(x)^10)*x^5/5 +...
%e The table of coefficients in (1 + x + x^3 + x^4)^n begins:
%e n=1: [1,(1), 0, 1, 1, 0, 0, 0, 0, 0, 0, ...];
%e n=2: [1, 2, (1), 2, 4, 2, 1, 2, 1, 0, 0, ...];
%e n=3: [1, 3, 3, (4), 9, 9, 6, 9, 9, 4, 3, ...];
%e n=4: [1, 4, 6, 8, (17), 24, 22, 28, 36, 28, 22, ...];
%e n=5: [1, 5, 10, 15, 30, (51), 60, 75, 105, 110, 100, ...];
%e n=6: [1, 6, 15, 26, 51, 96,(136), 180, 261, 326, 345, ...];
%e n=7: [1, 7, 21, 42, 84, 168, 273, (393), 588, 819, 987, ...];
%e n=8: [1, 8, 28, 64, 134, 280, 504, 792,(1233),1848, 2472, ...];
%e n=9: [1, 9, 36, 93, 207, 450, 876, 1494, 2439,(3865),5616, ...]; ...
%e the terms in parenthesis forms the initial terms of this sequence.
%p A228960 := proc(n)
%p (1+x+x^3+x^4)^n ;
%p coeftayl(%,x=0,n) ;
%p end proc: # _R. J. Mathar_, Sep 15 2013
%t Table[Coefficient[(1 + x + x^3 + x^4)^n,x,n],{n,1,30}] (* _Vaclav Kotesovec_, Dec 27 2013 *)
%t Table[HypergeometricPFQ[{1/3 - n/3, 2/3 - n/3, -n, -n/3}, {1/3, 2/3, 1}, 1], {n, 20}] (* _Vladimir Reshetnikov_, Oct 04 2016 *)
%o (PARI) {a(n)=polcoeff((1+x+x^3+x^4+x*O(x^n))^n, n)}
%o for(n=1,30,print1(a(n),", "))
%o (PARI) {a(n)=local(A=1/x*serreverse(x/(1+x+x^3+x^4+x*O(x^n)))); n*polcoeff(log(A), n)}
%o for(n=1,30,print1(a(n),", "))
%o (PARI) {a(n)=local(A=x); for(i=1, n, A=sum(m=1, n, sum(j=0, m, binomial(m, j)^2*x^(2*j)*exp(2*j*A+x*O(x^n)))*x^m/m)); n*polcoeff(A, n)}
%o for(n=1,30,print1(a(n),", "))
%o (PARI) {a(n)=local(A=1+x); for(i=1, n, A=sum(m=1, n, (1-x^2*exp(2*A))^(2*m+1)*sum(j=0, n\2, binomial(m+j, j)^2*x^(2*j)*exp(2*j*A+x*O(x^n)))*x^m/m)); n*polcoeff(A, n, x)}
%o for(n=1,30,print1(a(n),", "))
%Y Cf. A198951, A350383, A350406, A350407.
%K nonn,easy
%O 1,3
%A _Paul D. Hanna_, Sep 10 2013
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