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Search: a350406 -id:a350406
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A228960
a(n) = [x^n] (1 + x + x^3 + x^4)^n.
+10
5
1, 1, 4, 17, 51, 136, 393, 1233, 3865, 11851, 36301, 112520, 351352, 1098189, 3433704, 10758609, 33794505, 106344793, 335061790, 1056924667, 3338026857, 10554163533, 33402840615, 105809430024, 335444908176, 1064268538776, 3379009937161, 10735253448349, 34127137228747
(list; graph; refs; listen; history; text; internal format)
OFFSET
1,3
COMMENTS
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).
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..1000
FORMULA
L.g.f. L(x) satisfies:
(1) L(x) = log( (1/x)*Series_Reversion(x/((1+x)*(1+x^3))) ).
(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)).
(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)).
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.
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
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
From Peter Bala, Jun 15 2015: (Start)
a(n) = [x^(3*n)](1 + x + x^3 + x^4)^n.
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)
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
From Peter Bala, Apr 15 2023: (Start)
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.
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)
EXAMPLE
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 +...
Given G(x) = exp(L(x)), which is the g.f. of A198951:
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 +...
then the l.g.f. L(x) satisfies the series:
L(x) = (1 + x^2*G(x)^2)*x
+ (1 + 2^2*x^2*G(x)^2 + x^4*G(x)^4)*x^2/2
+ (1 + 3^2*x^2*G(x)^2 + 3^2*x^4*G(x)^4 + x^6*G(x)^6)*x^3/3
+ (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
+ (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 +...
The table of coefficients in (1 + x + x^3 + x^4)^n begins:
n=1: [1,(1), 0, 1, 1, 0, 0, 0, 0, 0, 0, ...];
n=2: [1, 2, (1), 2, 4, 2, 1, 2, 1, 0, 0, ...];
n=3: [1, 3, 3, (4), 9, 9, 6, 9, 9, 4, 3, ...];
n=4: [1, 4, 6, 8, (17), 24, 22, 28, 36, 28, 22, ...];
n=5: [1, 5, 10, 15, 30, (51), 60, 75, 105, 110, 100, ...];
n=6: [1, 6, 15, 26, 51, 96,(136), 180, 261, 326, 345, ...];
n=7: [1, 7, 21, 42, 84, 168, 273, (393), 588, 819, 987, ...];
n=8: [1, 8, 28, 64, 134, 280, 504, 792,(1233),1848, 2472, ...];
n=9: [1, 9, 36, 93, 207, 450, 876, 1494, 2439,(3865),5616, ...]; ...
the terms in parenthesis forms the initial terms of this sequence.
MAPLE
A228960 := proc(n)
(1+x+x^3+x^4)^n ;
coeftayl(%, x=0, n) ;
end proc: # R. J. Mathar, Sep 15 2013
MATHEMATICA
Table[Coefficient[(1 + x + x^3 + x^4)^n, x, n], {n, 1, 30}] (* Vaclav Kotesovec, Dec 27 2013 *)
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 *)
PROG
(PARI) {a(n)=polcoeff((1+x+x^3+x^4+x*O(x^n))^n, n)}
for(n=1, 30, print1(a(n), ", "))
(PARI) {a(n)=local(A=1/x*serreverse(x/(1+x+x^3+x^4+x*O(x^n)))); n*polcoeff(log(A), n)}
for(n=1, 30, print1(a(n), ", "))
(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)}
for(n=1, 30, print1(a(n), ", "))
(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)}
for(n=1, 30, print1(a(n), ", "))
CROSSREFS
Cf. A198951, A350383, A350406, A350407.
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Sep 10 2013
STATUS
approved
A350383
a(n) = [x^n] 1/(1 + x + x^2)^n.
+10
4
1, -1, 1, 2, -15, 49, -98, 48, 561, -2860, 8151, -12948, -9282, 149226, -594320, 1428952, -1448655, -5538975, 37450900, -122995950, 239589735, -37528755, -1886983020, 8939152560, -24579514050, 35197176924, 51580335366, -541312482256, 2033695030128, -4624358661240
(list; graph; refs; listen; history; text; internal format)
OFFSET
0,4
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..1000
FORMULA
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1+k,k) * binomial(n,3*k).
Recurrence: 3*(n-1)*n*(4*n - 7)*a(n) = -2*(n-1)*(28*n^2 - 63*n + 27)*a(n-1) - 3*(3*n - 5)*(3*n - 4)*(4*n - 3)*a(n-2). - Vaclav Kotesovec, Mar 18 2023
From Peter Bala, Apr 15 2023: (Start)
a(n) = (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1).
Conjecture: 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. A228960.
More generally, let k be a positive integer, m an integer and let f(x) = g(x)/h(x), where g(x) and h(x) are both finite products of cyclotomic polynomials. Then we conjecture that 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)
MAPLE
a := n -> (-1)^n*hypergeom([-n/3, 1/3 - n/3, 2/3 - n/3, n], [1/3, 2/3, 1], 1): seq(simplify(a(n)), n = 0..30); # Peter Bala, Apr 17 2023
MATHEMATICA
a[n_] := Coefficient[Series[1/(1 + x + x^2)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)
PROG
(PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n-1+k, k)*binomial(n, 3*k));
CROSSREFS
Cf. A002426, A027907, A110556, A165817, A228960, A350406, A350407.
KEYWORD
sign,easy
AUTHOR
Seiichi Manyama, Dec 29 2021
STATUS
approved
A350407
a(n) = [x^n] 1/(1 + x + x^2 + x^3 + x^4)^n.
+10
4
1, -1, 1, -1, 1, 4, -35, 146, -447, 1133, -2464, 4355, -4355, -9296, 77078, -314636, 1006145, -2738565, 6398155, -12175809, 14621376, 14828330, -179236815, 786257460, -2625667395, 7412386254, -17983703880, 36057065030, -49553122730, -14585596720
(list; graph; refs; listen; history; text; internal format)
OFFSET
0,6
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..1000
FORMULA
a(n) = Sum_{k=0..n} (-1)^(n-k) * binomial(n-1+k,k) * binomial(n,5*k).
Recurrence: 125*(n-3)*(n-2)*(n-1)*n*(2*n - 7)*(3*n - 11)*(3*n - 8)*(6*n - 23)*(6*n - 17)*(6*n - 11)*a(n) = -50*(n-3)*(n-2)*(n-1)*(3*n - 11)*(6*n - 23)*(6*n - 17)*(576*n^4 - 4896*n^3 + 14402*n^2 - 16875*n + 6250)*a(n-1) - 30*(n-3)*(n-2)*(6*n - 23)*(6*n - 5)*(7884*n^6 - 123516*n^5 + 791601*n^4 - 2652565*n^3 + 4894096*n^2 - 4707500*n + 1842500)*a(n-2) - 2*(n-3)*(3*n - 5)*(6*n - 11)*(6*n - 5)*(60192*n^6 - 1113552*n^5 + 8528546*n^4 - 34608379*n^3 + 78470893*n^2 - 94255700*n + 46855500)*a(n-3) - 5*(2*n - 5)*(3*n - 8)*(3*n - 5)*(5*n - 19)*(5*n - 18)*(5*n - 17)*(5*n - 16)*(6*n - 17)*(6*n - 11)*(6*n - 5)*a(n-4). - Vaclav Kotesovec, Mar 18 2023
From Peter Bala, Apr 16 2023: (Start)
a(n) = (-1)^n*hypergeom([-n/5, 1/5 - n/5, 2/5 - n/5, 3/5 - n/5, 4/5 - n/5, n], [1/5, 2/5, 3/5, 4/5, 1], 1).
Conjecture: 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 >= 3. (End)
MAPLE
a := n -> (-1)^n*hypergeom([-n/5, 1/5 - n/5, 2/5 - n/5, 3/5 - n/5, 4/5 - n/5, n], [1/5, 2/5, 3/5, 4/5, 1], 1): seq(simplify(a(n)), n = 0..30); # Peter Bala, Apr 16 2023
MATHEMATICA
a[n_] := Coefficient[Series[1/(1 + x + x^2 + x^3 + x^4)^n, {x, 0, n}], x, n]; Array[a, 30, 0] (* Amiram Eldar, Dec 29 2021 *)
PROG
(PARI) a(n) = sum(k=0, n, (-1)^(n-k)*binomial(n-1+k, k)*binomial(n, 5*k));
CROSSREFS
Cf. A187925, A228960, A350383, A350406.
KEYWORD
sign,easy
AUTHOR
Seiichi Manyama, Dec 29 2021
STATUS
approved
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