[go: up one dir, main page]

login
Search: a318108 -id:a318108
     Sort: relevance | references | number | modified | created      Format: long | short | data
The Franel number a(n) = Sum_{k = 0..n} binomial(n,k)^3.
(Formerly M1971 N0781)
+10
137
1, 2, 10, 56, 346, 2252, 15184, 104960, 739162, 5280932, 38165260, 278415920, 2046924400, 15148345760, 112738423360, 843126957056, 6332299624282, 47737325577620, 361077477684436, 2739270870994736, 20836827035351596, 158883473753259752, 1214171997616258240
OFFSET
0,2
COMMENTS
Cusick gives a general method of deriving recurrences for the r-th order Franel numbers (this is the sequence of third-order Franel numbers), with floor((r+3)/2) terms.
This is the Taylor expansion of a special point on a curve described by Beauville. - Matthijs Coster, Apr 28 2004
An identity of V. Strehl states that a(n) = Sum_{k = 0..n} C(n,k)^2 * binomial(2*k,n). Zhi-Wei Sun conjectured that for every n = 2,3,... the polynomial f_n(x) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(2*k,n) * x^(n-k) is irreducible over the field of rational numbers. - Zhi-Wei Sun, Mar 21 2013
Conjecture: a(n) == 2 (mod n^3) iff n is prime. - Gary Detlefs, Mar 22 2013
a(p) == 2 (mod p^3) for any prime p since p | C(p,k) for all k = 1,...,p-1. - Zhi-Wei Sun, Aug 14 2013
a(n) is the maximal number of totally mixed Nash equilibria in games of 3 players, each with n+1 pure options. - Raimundas Vidunas, Jan 22 2014
This is one of the Apéry-like sequences - see Cross-references. - Hugo Pfoertner, Aug 06 2017
Diagonal of rational functions 1/(1 - x*y - y*z - x*z - 2*x*y*z), 1/(1 - x - y - z + 4*x*y*z), 1/(1 + y + z + x*y + y*z + x*z + 2*x*y*z), 1/(1 + x + y + z + 2*(x*y + y*z + x*z) + 4*x*y*z). - Gheorghe Coserea, Jul 04 2018
a(n) is the constant term in the expansion of ((1 + x) * (1 + y) + (1 + 1/x) * (1 + 1/y))^n. - Seiichi Manyama, Oct 27 2019
Diagonal of rational function 1 / ((1-x)*(1-y)*(1-z) - x*y*z). - Seiichi Manyama, Jul 11 2020
Named after the Swiss mathematician Jérôme Franel (1859-1939). - Amiram Eldar, Jun 15 2021
It appears that a(n) is equal to the coefficient of (x*y*z)^n in the expansion of (1 + x + y - z)^n * (1 + x - y + z)^n * (1 - x + y + z)^n. Cf. A036917. - Peter Bala, Sep 20 2021
REFERENCES
Matthijs Coster, Over 6 families van krommen [On 6 families of curves], Master's Thesis (unpublished), Aug 26 1983.
Jérôme Franel, On a question of Laisant, Intermédiaire des Mathématiciens, vol 1 1894 pp 45-47
H. W. Gould, Combinatorial Identities, Morgantown, 1972, (X.14), p. 56.
Murray Klamkin, ed., Problems in Applied Mathematics: Selections from SIAM Review, SIAM, 1990; see pp. 148-149.
John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 193.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..1000 (terms 0..100 from T. D. Noe)
Boris Adamczewski, Jason P. Bell, and Eric Delaygue, Algebraic independence of G-functions and congruences à la Lucas", arXiv preprint arXiv:1603.04187 [math.NT], 2016.
Prarit Agarwal and June Nahmgoong, Singlets in the tensor product of an arbitrary number of Adjoint representations of SU(3), arXiv:2001.10826 [math.RT], 2020.
Richard Askey, Orthogonal Polynomials and Special Functions, SIAM, 1975; see p. 43.
P. Barrucand, A combinatorial identity, Problem 75-4, SIAM Rev., Vol. 17 (1975), p. 168. Solution by D. R. Breach, D. McCarthy, D. Monk and P. E. O'Neil, SIAM Rev., Vol. 18 (1976), p. 303.
P. Barrucand, Problem 75-4, A Combinatorial Identity, SIAM Rev., 17 (1975), 168. [Annotated scanned copy of statement of problem]
Arnaud Beauville, Les familles stables de courbes sur P_1 admettant quatre fibres singulières, Comptes Rendus, Académie Sciences Paris, Vol.. 294 (May 24 1982), pp. 657-660.
David Callan, A combinatorial interpretation for an identity of Barrucand, JIS, Vol. 11 (2008), Article 08.3.4.
Marc Chamberland and Armin Straub, Apéry Limits: Experiments and Proofs, arXiv:2011.03400 [math.NT], 2020.
Shaun Cooper, Apéry-like sequences defined by four-term recurrence relations, arXiv:2302.00757 [math.NT], 2023.
M. Coster, Email, Nov 1990
T. W. Cusick, Recurrences for sums of powers of binomial coefficients, J. Combin. Theory, Series A, Vol. 52, No. 1 (1989), pp. 77-83.
Eric Delaygue, Arithmetic properties of Apéry-like numbers, arXiv preprint arXiv:1310.4131 [math.NT], 2013.
Robert W. Donley Jr, Directed path enumeration for semi-magic squares of size three, arXiv:2107.09463 [math.CO], 2021.
Tomislav Došlic and Darko Veljan, Logarithmic behavior of some combinatorial sequences, Discrete Math., Vol. 308, No. 11 (2008), pp. 2182--2212. MR2404544 (2009j:05019) - From N. J. A. Sloane, May 01 2012
Carsten Elsner, On recurrence formulas for sums involving binomial coefficients, Fib. Q., VOl. 43, No. 1 (2005), pp. 31-45.
Jeff D. Farmer and Steven C. Leth, An asymptotic formula for powers of binomial coefficients, Math. Gaz., Vol. 89, No. 516 (2005), pp. 385-391.
Ofir Gorodetsky, New representations for all sporadic Apéry-like sequences, with applications to congruences, arXiv:2102.11839 [math.NT], 2021. See A p. 2.
Darij Grinberg, Introduction to Modern Algebra (UMN Spring 2019 Math 4281 Notes), University of Minnesota (2019).
S. Herfurtner, Elliptic surfaces with four singular fibres, Mathematische Annalen, 1991. Preprint.
Bradley Klee, Checking Weierstrass data, 2023.
Vaclav Kotesovec, Non-attacking chess pieces, 6ed, 2013, p. 282.
Amita Malik and Armin Straub, Divisibility properties of sporadic Apéry-like numbers, Research in Number Theory, Vol. 2, No. 5 (2016).
Guo-Shuai Mao and Yan Liu, On a congruence conjecture of Z.-W. Sun involving Franel numbers, arXiv:2111.08775 [math.NT], 2021.
Guo-Shuai Mao, On three conjectural congruences involving Domb numbers and Franel numbers, preprint on ResearchGate, April 2024.
Romeo Meštrović, Lucas' theorem: its generalizations, extensions and applications (1878--2014), arXiv preprint arXiv:1409.3820 [math.NT], 2014.
Marci A. Perlstadt, Some Recurrences for Sums of Powers of Binomial Coefficients, Journal of Number Theory, Vo. 27 (1987), pp. 304-309.
Juan Pla, Problem H-505, Advanced Problems and Solutions, The Fibonacci Quarterly, Vol. 33, No. 5 (1995), p. 473; Sum Formulae!, Solution to Problem H-505 by Paul S. Bruckman, ibid., Vol. 35, No. 1 (1997), pp. 93-95.
Armin Straub, and Wadim Zudilin, Sums of powers of binomials, their Apéry limits, and Franel's suspicions, arXiv:2112.09576 [math.NT], 2021.
Volker Strehl, Recurrences and Legendre transform, Séminaire Lotharingien de Combinatoire, B29b (1992), 22 pp.
Zhi-Hong Sun, Congruences for Apéry-like numbers, arXiv:1803.10051 [math.NT], 2018.
Zhi-Hong Sun, New congruences involving Apéry-like numbers, arXiv:2004.07172 [math.NT], 2020.
Zhi-Wei Sun, Congruences for Franel numbers, arXiv preprint arXiv:1112.1034 [math.NT], 2011.
Zhi-Wei Sun, Connections between p = x^2+3y^2 and Franel numbers, J. Number Theory, Vol. 133 (2013), pp. 2919-2928.
Zhi-Wei Sun, Conjectures involving arithmetical sequences, arXiv:1208.2683v9 [math.CO] 2013; Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H. Li and J. Liu), Proc. the 6th China-Japan Sem. (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258.
Zhi-Wei Sun, Congruences involving g_n(x) = Sum_{k= 0..n} C(n,k)^2 C(2k,k) x^k, arXiv preprint arXiv:1407.0967 [math.NT], 2014.
Raimundas Vidunas, MacMahon's master theorem and totally mixed Nash equilibria, arxiv 1401.5400 [math.CO], 2014.
Eric Weisstein's World of Mathematics, Binomial Sums.
Eric Weisstein's World of Mathematics, Franel Number.
Eric Weisstein's World of Mathematics, Schmidt's Problem.
Jin Yuan, Zhi-Juan Lu, Asmus L. Schmidt, On recurrences for sums of powers of binomial coefficients, J. Numb. Theory 128 (2008) 2784-2794
Don Zagier, Integral solutions of Apéry-like recurrence equations. See line A in sporadic solutions table of page 5.
Bao-Xuan Zhu, Higher order log-monotonicity of combinatorial sequences, arXiv preprint arXiv:1309.6025 [math.CO], 2013.
FORMULA
A002893(n) = Sum_{m = 0..n} binomial(n, m)*a(m) [Barrucand].
Sum_{k = 0..n} C(n, k)^3 = (-1)^n*Integral_{x = 0..infinity} L_k(x)^3 exp(-x) dx. - from Askey's book, p. 43
D-finite with recurrence (n + 1)^2*a(n+1) = (7*n^2 + 7*n + 2)*a(n) + 8*n^2*a(n-1) [Franel]. - Felix Goldberg (felixg(AT)tx.technion.ac.il), Jan 31 2001
a(n) ~ 2*3^(-1/2)*Pi^-1*n^-1*2^(3*n). - Joe Keane (jgk(AT)jgk.org), Jun 21 2002
O.g.f.: A(x) = Sum_{n >= 0} (3*n)!/n!^3 * x^(2*n)/(1 - 2*x)^(3*n+1). - Paul D. Hanna, Oct 30 2010
G.f.: hypergeom([1/3, 2/3], [1], 27 x^2 / (1 - 2x)^3) / (1 - 2x). - Michael Somos, Dec 17 2010
G.f.: Sum_{n >= 0} a(n)*x^n/n!^3 = [ Sum_{n >= 0} x^n/n!^3 ]^2. - Paul D. Hanna, Jan 19 2011
G.f.: A(x) = 1/(1-2*x)*(1+6*(x^2)/(G(0)-6*x^2)),
with G(k) = 3*(x^2)*(3*k+1)*(3*k+2) + ((1-2*x)^3)*((k+1)^2) - 3*(x^2)*((1-2*x)^3)*((k+1)^2)*(3*k+4)*(3*k+5)/G(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Dec 03 2011
In 2011 Zhi-Wei Sun found the formula Sum_{k = 0..n} C(2*k,n)*C(2*k,k)*C(2*(n-k),n-k) = (2^n)*a(n) and proved it via the Zeilberger algorithm. - Zhi-Wei Sun, Mar 20 2013
0 = a(n)*(a(n+1)*(-2048*a(n+2) - 3392*a(n+3) + 768*a(n+4)) + a(n+2)*(-1280*a(n+2) - 2912*a(n+3) + 744*a(n+4)) + a(n+3)*(+288*a(n+3) - 96*a(n+4))) + a(n+1)*(a(n+1)*(-704*a(n+2) - 1232*a(n+3) + 288*a(n+4)) + a(n+2)*(-560*a(n+2) - 1372*a(n+3) + 364*a(n+4)) + a(n+3)*(+154*a(n+3) - 53*a(n+4))) + a(n+2)*(a(n+2)*(+24*a(n+2) + 70*a(n+3) - 20*a(n+4)) + a(n+3)*(-11*a(n+3) + 4*a(n+4))) for all n in Z. - Michael Somos, Jul 16 2014
For r a nonnegative integer, Sum_{k = r..n} C(k,r)^3*C(n,k)^3 = C(n,r)^3*a(n-r), where we take a(n) = 0 for n < 0. - Peter Bala, Jul 27 2016
a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^2. - Peter Luschny, May 31 2017
From Gheorghe Coserea, Jul 04 2018: (Start)
a(n) = Sum_{k=0..floor(n/2)} (n+k)!/(k!^3*(n-2*k)!) * 2^(n-2*k).
G.f. y=A(x) satisfies: 0 = x*(x + 1)*(8*x - 1)*y'' + (24*x^2 + 14*x - 1)*y' + 2*(4*x + 1)*y. (End)
a(n) = [x^n] (1 - x^2)^n*P(n,(1 + x)/(1 - x)), where P(n,x) denotes the n-th Legendre polynomial. See Gould, p. 56. - Peter Bala, Mar 24 2022
a(n) = (2^n/(4*Pi^2)) * Integral_{x,y=0..2*Pi} (1+cos(x)+cos(y)+cos(x+y))^n dx dy = (8^n/(Pi^2)) * Integral_{x,y=0..Pi} (cos(x)*cos(y)*cos(x+y))^n dx dy (Pla, 1995). - Amiram Eldar, Jul 16 2022
a(n) = Sum_{k = 0..n} m^(n-k)*binomial(n,k)*binomial(n+2*k,n)*binomial(2*k,k) at m = -4. Cf. A081798 (m = 1), A006480 (m = 0), A124435 (m = -1), A318109 (m = -2) and A318108 (m = -3). - Peter Bala, Mar 16 2023
From Bradley Klee, Jun 05 2023: (Start)
The g.f. T(x) obeys a period-annihilating ODE:
0=2*(1 + 4*x)*T(x) + (-1 + 14*x + 24*x^2)*T'(x) + x*(1 + x)*(-1 + 8*x)*T''(x).
The periods ODE can be derived from the following Weierstrass data:
g2 = (4/243)*(1 - 8*x + 240*x^2 - 464*x^3 + 16*x^4);
g3 = -(8/19683)*(1 - 12*x - 480*x^2 + 3080*x^3 - 12072*x^4 + 4128*x^5 +
64*x^6);
which determine an elliptic surface with four singular fibers. (End)
From Peter Bala, Oct 31 2024: (Start)
For n >= 1, a(n) = 2 * Sum_{k = 0..n-1} binomial(n, k)^2 * binomial(n-1, k). Cf. A361716.
For n >= 1, a(n) = 2 * hypergeom([-n, -n, -n + 1], [1, 1], -1). (End)
EXAMPLE
O.g.f.: A(x) = 1 + 2*x + 10*x^2 + 56*x^3 + 346*x^4 + 2252*x^5 + ...
O.g.f.: A(x) = 1/(1-2*x) + 3!*x^2/(1-2*x)^4 + (6!/2!^3)*x^4/(1-2*x)^7 + (9!/3!^3)*x^6/(1-2*x)^10 + (12!/4!^3)*x^8/(1-2*x)^13 + ... - Paul D. Hanna, Oct 30 2010
Let g.f. A(x) = Sum_{n >= 0} a(n)*x^n/n!^3, then
A(x) = 1 + 2*x + 10*x^2/2!^3 + 56*x^3/3!^3 + 346*x^4/4!^3 + ... where
A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 + ...]^2. - Paul D. Hanna
MAPLE
A000172 := proc(n)
add(binomial(n, k)^3, k=0..n) ;
end proc:
seq(A000172(n), n=0..10) ; # R. J. Mathar, Jul 26 2014
A000172_list := proc(len) series(hypergeom([], [1, 1], x)^2, x, len);
seq((n!)^3*coeff(%, x, n), n=0..len-1) end:
A000172_list(21); # Peter Luschny, May 31 2017
MATHEMATICA
Table[Sum[Binomial[n, k]^3, {k, 0, n}], {n, 0, 30}] (* Harvey P. Dale, Aug 24 2011 *)
Table[ HypergeometricPFQ[{-n, -n, -n}, {1, 1}, -1], {n, 0, 20}] (* Jean-François Alcover, Jul 16 2012, after symbolic sum *)
a[n_] := Sum[ Binomial[2k, n]*Binomial[2k, k]*Binomial[2(n-k), n-k], {k, 0, n}]/2^n; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Mar 20 2013, after Zhi-Wei Sun *)
a[ n_] := SeriesCoefficient[ Hypergeometric2F1[ 1/3, 2/3, 1, 27 x^2 / (1 - 2 x)^3] / (1 - 2 x), {x, 0, n}]; (* Michael Somos, Jul 16 2014 *)
PROG
(PARI) {a(n)=polcoeff(sum(m=0, n, (3*m)!/m!^3*x^(2*m)/(1-2*x+x*O(x^n))^(3*m+1)), n)} \\ Paul D. Hanna, Oct 30 2010
(PARI) {a(n)=n!^3*polcoeff(sum(m=0, n, x^m/m!^3+x*O(x^n))^2, n)} \\ Paul D. Hanna, Jan 19 2011
(Haskell)
a000172 = sum . map a000578 . a007318_row
-- Reinhard Zumkeller, Jan 06 2013
(Sage)
def A000172():
x, y, n = 1, 2, 1
while True:
yield x
n += 1
x, y = y, (8*(n-1)^2*x + (7*n^2-7*n + 2)*y) // n^2
a = A000172()
[next(a) for i in range(21)] # Peter Luschny, Oct 12 2013
(PARI) A000172(n)={sum(k=0, (n-1)\2, binomial(n, k)^3)*2+if(!bittest(n, 0), binomial(n, n\2)^3)} \\ M. F. Hasler, Sep 21 2015
CROSSREFS
Cf. A002893, A052144, A005260, A096191, A033581, A189791. Second row of array A094424.
The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)
For primes that do not divide the terms of the sequences A000172, A005258, A002893, A081085, A006077, A093388, A125143, A229111, A002895, A290575, A290576, A005259 see A260793, A291275-A291284 and A133370 respectively.
Sum_{k = 0..n} C(n,k)^m for m = 1..12: A000079, A000984, A000172, A005260, A005261, A069865, A182421, A182422, A182446, A182447, A342294, A342295.
Column k=3 of A372307.
KEYWORD
nonn,easy,nice,changed
STATUS
approved
Number of effective multiple alignments of three equal-length sequences.
+10
6
1, 5, 67, 1109, 20251, 391355, 7847155, 161476565, 3387271675, 72114452255, 1553475100717, 33786532319435, 740681494769659, 16346552430326123, 362830907979309067, 8093356178498583509, 181311959402343288955, 4077310062938894133623, 91999289732199733092601
OFFSET
0,2
COMMENTS
This counts effective alignments rather than standard alignments, so that for example the following two alignments are equivalent:
-A A-
-T T-
C- -C
See Dress, Morgenstern and Stoye for more information.
LINKS
A. Bostan, S. Boukraa, J.-M. Maillard, and J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
A. Dress, B. Morgenstern and J. Stoye, On the number of standard and of effective multiple alignments, Applied Mathematics Letters, Vol. 11, No. 4, 1998, pp. 43-49.
FORMULA
The recurrence is three-dimensional with the order of the three parameters immaterial. That is, a(i,j,k)=a(i,k,j)=a(j,i,k)=a(j,k,i)=a(k,i,j)=a(k,j,i). a(i, j, 0) = (i+j)! / i! / j! a(i, j, k) = a(i-1,j,k) + a(i,j-1,k) + a(i,j,k-1) - a(i-1,j-1,k-1).
a(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n,k)*binomial(n+2*k,n)*binomial(2*k,k). - Wadim Zudilin, Nov 26 2015
Diagonal of 1/(1 - x - y - z + x*y*z). - Mark van Hoeij, Dec 20 2013
G.f.: hypergeom([1/3, 2/3],[1],27*x/(1+x)^3)/(1+x). - Mark van Hoeij, Dec 20 2013
(3*n-1)*(n+1)^2*a(n+1)-(3*n+1)*(24*n^2+8*n-5)*a(n)+(9*n^3-3*n^2-4*n+2)*a(n-1)+(3*n+2)*(n-1)^2*a(n-2)=0. - Robert Israel, Nov 26 2015
0 = (2*x-1)*(x^3+3*x^2-24*x+1)*x*y'' + (6*x^4+8*x^3-57*x^2+48*x-1)*y' + (x+1)*(2*x^2-2*x+5)*y, where y is g.f. - Gheorghe Coserea, Jul 06 2016
From Peter Bala, Mar 16 2023: (Start)
(3*n - 4)*n^2*a(n) = (3*n - 2)*(24*n^2 - 40*n + 11)*a(n-1) - (9*n^3 - 30*n^2 + 29*n - 6)*a(n-2) - (3*n - 1)*(n - 2)^2*a(n-3) with a(0) = 1, a(1) = 5 and a(2) = 67.
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. (End)
EXAMPLE
a(1) = 5 because the five alignments are
A-- A- A- A- A
-C- C- -C -C C
--T -T T- -T T
MAPLE
G := series( hypergeom([1/3, 2/3], [1], 27*x/(1+x)^3)/(1+x), x=0, 31);
seq(coeff(G, x, i), i=0..30); # Mark van Hoeij, Dec 20 2013
MATHEMATICA
a[n_] := Sum[(-1)^(n-k) Binomial[n, k] Binomial[n+2k, n] Binomial[2k, k], {k, 0, n}];
Table[a[n], {n, 0, 18}] (* Jean-François Alcover, Sep 18 2018, after Wadim Zudilin *)
PROG
(PARI)
diag(expr, N=22, var=variables(expr)) = {
my(a = vector(N));
for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
for (n = 1, N, a[n] = expr;
for (k = 1, #var, a[n] = polcoef(a[n], n-1)));
return(a);
};
x='x; y='y; z='z; diag(1/(1 - x - y - z + x*y*z), 19)
(PARI) \\ system("wget http://www.jjj.de/pari/hypergeom.gpi");
read("hypergeom.gpi");
N = 20; x = 'x + O('x^N);
Vec(hypergeom([1/3, 2/3], [1], 27*x/(1+x)^3, N)/(1+x)) \\ Gheorghe Coserea, Jul 06 2016
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Lee A. Newberg, Dec 15 2006
EXTENSIONS
More terms from Mark van Hoeij, Dec 21 2013
STATUS
approved
a(n) = Sum_{k=0..n} (3*n-2*k)!/((n-k)!^3*k!)*(-2)^k.
+10
5
1, 4, 46, 652, 10186, 168304, 2884456, 50723824, 909192538, 16538659384, 304391739796, 5655971294824, 105929883322576, 1997228410630912, 37871584674309376, 721672204654077952, 13811327854028171098, 265324110145941691912, 5114208160758838538044, 98874597697991698311832, 1916741738060370782929036
OFFSET
0,2
COMMENTS
Diagonal of rational function 1/(1 - (x + y + z - 2*x*y*z)).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..766 (terms 0..100 from Gheorghe Coserea)
FORMULA
G.f. y=A(x) satisfies: 0 = x*(x - 1)*(4*x - 1)*(8*x^2 + 20*x - 1)*y'' + (96*x^4 + 64*x^3 - 120*x^2 + 42*x - 1)*y' + 4*(2*x + 1)*(4*x^2 - 2*x + 1)*y.
From Peter Bala, Mar 16 2023: (Start)
n^2*(3*n - 4)*a(n) = (3*n - 2)*(21*n^2 - 35*n + 10)*a(n-1) - 4*(9*n^3 - 30*n^2 + 29*n - 6)*a(n-2) - 8*(3*n - 1)*(n - 2)^2*a(n-3) with a(0) = 1, a(1) = 4 and a(2) = 46.
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. (End)
a(n) ~ (1 + sqrt(3))^(3*n + 1) / (2*Pi*sqrt(3)*n). - Vaclav Kotesovec, Mar 17 2023
EXAMPLE
A(x) = 1 + 4*x + 46*x^2 + 652*x^3 + 10186*x^4 + 168304*x^5 + 2884456*x^6 + ...
PROG
(PARI)
a(n) = sum(k=0, n, (3*n-2*k)!/((n-k)!^3*k!)*(-2)^k);
vector(21, n, a(n-1))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Gheorghe Coserea, Sep 20 2018
STATUS
approved
Triangle read by rows: T(n,k) = (3*n - 2*k)!/((n-k)!^3*k!).
+10
4
1, 6, 1, 90, 24, 1, 1680, 630, 60, 1, 34650, 16800, 2520, 120, 1, 756756, 450450, 92400, 7560, 210, 1, 17153136, 12108096, 3153150, 369600, 18900, 336, 1, 399072960, 325909584, 102918816, 15765750, 1201200, 41580, 504, 1, 9465511770, 8779605120, 3259095840, 617512896, 63063000, 3363360, 83160, 720, 1
OFFSET
0,2
COMMENTS
Diagonal of rational function R(x,y,z,t) = 1/(1 - (x + y + z + t*x*y*z)) with respect to x,y,z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t).
Annihilating differential operator: x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*Dx^2 + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x - 1)*Dx + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6).
LINKS
Gheorghe Coserea, Rows n=0..100, flattened
FORMULA
Let P_n(t) = Sum_{k=0..n} T(n,k)*t^k. Then A000172(n) = P_n(-4), A318108(n) = P_n(-3), A318109(n) = P_n(-2), A124435(n) = P_n(-1), A006480(n) = P_n(0), A081798(n) = P_n(1).
G.f. y = Sum_{n>=0} P_n(t)*x^n satisfies:
0 = x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*y'' + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x - 1)*y' + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6)*y.
EXAMPLE
A(x;t) = 1 + (6 + t)*x + (90 + 24*t + t^2)*x^2 + (1680 + 630*t + 60*t^2 + t^3)*x^3 + ...
Triangle starts:
n\k [0] [1] [2] [3] [4] [5] [6] [7]
[0] 1;
[1] 6, 1;
[2] 90, 24, 1;
[3] 1680, 630, 60, 1;
[4] 34650, 16800, 2520, 120, 1;
[5] 756756, 450450, 92400, 7560, 210, 1;
[6] 17153136, 12108096, 3153150, 369600, 18900, 336, 1;
[7] 399072960, 325909584, 102918816, 15765750, 1201200, 41580, 504, 1;
[8] ...
PROG
(PARI)
T(n, k) = (3*n - 2*k)!/((n-k)!^3*k!);
concat(vector(10, n, vector(n, k, T(n-1, k-1))))
/* test:
P(n, v='t) = subst(Polrev(vector(n+1, k, T(n, k-1)), 't), 't, v);
diag(expr, N=22, var=variables(expr)) = {
my(a = vector(N));
for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
for (n = 1, N, a[n] = expr;
for (k = 1, #var, a[n] = polcoef(a[n], n-1)));
return(a);
};
apply_diffop(p, s) = { \\ apply diffop p (encoded as Pol in Dx) to Ser s
s=intformal(s);
sum(n=0, poldegree(p, 'Dx), s=s'; polcoef(p, n, 'Dx) * s);
};
\\ diagonal property:
x='x; y='y; z='z; t='t;
diag(1/(1 - (x+y+z + t*x*y*z)), 11, [x, y, z]) == vector(11, n, P(n-1))
\\ annihilating diffop:
y = Ser(vector(101, n, P(n-1)), 'x);
p=x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*Dx^2 + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x - 1)*Dx + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6);
0 == apply_diffop(p, y)
*/
KEYWORD
nonn,tabl
AUTHOR
Gheorghe Coserea, Sep 18 2018
STATUS
approved

Search completed in 0.012 seconds