A108625 Square array, read by antidiagonals, where row n equals the crystal ball sequence for the A_n lattice.
1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 13, 19, 7, 1, 1, 21, 55, 37, 9, 1, 1, 31, 131, 147, 61, 11, 1, 1, 43, 271, 471, 309, 91, 13, 1, 1, 57, 505, 1281, 1251, 561, 127, 15, 1, 1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1, 1, 91, 1405, 6637, 12559, 11253, 5321, 1415, 217, 19, 1
Offset: 0
Examples
Square array begins:
1, 1, 1, 1, 1, 1, 1, ... A000012;
1, 3, 5, 7, 9, 11, 13, ... A005408;
1, 7, 19, 37, 61, 91, 127, ... A003215;
1, 13, 55, 147, 309, 561, 923, ... A005902;
1, 21, 131, 471, 1251, 2751, 5321, ... A008384;
1, 31, 271, 1281, 4251, 11253, 25493, ... A008386;
1, 43, 505, 3067, 12559, 39733, 104959, ... A008388;
1, 57, 869, 6637, 33111, 124223, 380731, ... A008390;
1, 73, 1405, 13237, 79459, 350683, 1240399, ... A008392;
1, 91, 2161, 24691, 176251, 907753, 3685123, ... A008394;
1, 111, 3191, 43561, 365751, 2181257, ... ... A008396;
...
As a triangle:
[0] 1
[1] 1, 1
[2] 1, 3, 1
[3] 1, 7, 5, 1
[4] 1, 13, 19, 7, 1
[5] 1, 21, 55, 37, 9, 1
[6] 1, 31, 131, 147, 61, 11, 1
[7] 1, 43, 271, 471, 309, 91, 13, 1
[8] 1, 57, 505, 1281, 1251, 561, 127, 15, 1
[9] 1, 73, 869, 3067, 4251, 2751, 923, 169, 17, 1
...
Inverse binomial transform of rows yield rows of triangle A063007:
1;
1, 2;
1, 6, 6;
1, 12, 30, 20;
1, 20, 90, 140, 70;
1, 30, 210, 560, 630, 252; ...
Product of the g.f. of row n and (1-x)^(n+1) generates the symmetric triangle A008459:
1;
1, 1;
1, 4, 1;
1, 9, 9, 1;
1, 16, 36, 16, 1;
1, 25, 100, 100, 25, 1;
...
Links
- Alois P. Heinz, Antidiagonals n = 0..140, flattened
- R. Bacher, P. de la Harpe, and B. Venkov, Séries de croissance et séries d'Ehrhart associées aux réseaux de racines, C. R. Acad. Sci. Paris, 325 (Series 1) (1997), 1137-1142.
- Joseph T. Iosue, T. C. Mooney, Adam Ehrenberg, and Alexey V. Gorshkov, Projective toric designs, difference sets, and quantum state designs, arXiv:2311.13479 [quant-ph], 2023. See page 6.
- Armin Straub, Multivariate Apéry numbers and supercongruences of rational functions, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; arXiv preprint, arXiv:1401.0854 [math.NT], 2014.
- A. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3). An informal report. Math. Intelligencer 1 (1978/79), no 4, 195-203.
- Eric Weisstein's World of Mathematics, Apéry number.
Crossrefs
Programs
-
Magma
T:= func< n,k | (&+[Binomial(n,j)^2*Binomial(n+k-j,k-j): j in [0..k]]) >; // array A108625:= func< n,k | T(n-k,k) >; // antidiagonals [A108625(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
-
Maple
T := (n,k) -> binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1): seq(seq(simplify(T(n,k)),k=0..n),n=0..10); # Peter Luschny, Feb 10 2018
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Mathematica
T[n_, k_]:= HypergeometricPFQ[{-n, -k, n+1}, {1, 1}, 1] (* Michael Somos, Jun 03 2012 *) -
PARI
T(n,k)=sum(i=0,k,binomial(n,i)^2*binomial(n+k-i,k-i))
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SageMath
def T(n,k): return sum(binomial(n,j)^2*binomial(n+k-j, k-j) for j in range(k+1)) # array def A108625(n,k): return T(n-k, k) # antidiagonals flatten([[A108625(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023
Formula
T(n, k) = Sum_{i=0..k} C(n, i)^2 * C(n+k-i, k-i).
G.f. for row n: (Sum_{i=0..n} C(n, i)^2 * x^i)/(1-x)^(n+1).
Sum_{k=0..n} T(n-k, k) = A108626(n) (antidiagonal sums).
From Peter Bala, Jul 23 2008 (Start):
O.g.f. row n: 1/(1 - x)*Legendre_P(n,(1 + x)/(1 - x)).
G.f. for square array: 1/sqrt((1 - x)*((1 - t)^2 - x*(1 + t)^2)) = (1 + x + x^2 + x^3 + ...) + (1 + 3*x + 5*x^2 + 7*x^3 + ...)*t + (1 + 7*x + 19*x^2 + 37*x^3 + ...)*t^2 + ... . Cf. A142977.
Main diagonal is A005258.
Recurrence relations:
Row n entries: (k+1)^2*T(n,k+1) = (2*k^2+2*k+n^2+n+1)*T(n,k) - k^2*T(n,k-1), k = 1,2,3,... ;
Column k entries: (n+1)^2*T(n+1,k) = (2*k+1)*(2*n+1)*T(n,k) + n^2*T(n-1,k), n = 1,2,3,... ;
Main diagonal entries: (n+1)^2*T(n+1,n+1) = (11*n^2+11*n+3)*T(n,n) + n^2*T(n-1,n-1), n = 1,2,3,... .
Series acceleration formulas for zeta(2):
Row n: zeta(2) = 2*(1 - 1/2^2 + 1/3^2 - ... + (-1)^(n+1)/n^2) + (-1)^n*Sum_{k >= 1} 1/(k^2*T(n,k-1)*T(n,k));
Column k: zeta(2) = 1 + 1/2^2 + 1/3^2 + ... + 1/k^2 + 2*Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,k)*T(n,k));
Main diagonal: zeta(2) = 5 * Sum_{n >= 1} (-1)^(n+1)/(n^2*T(n-1,n-1)*T(n,n)).
Conjectural result for superdiagonals: zeta(2) = 1 + 1/2^2 + ... + 1/k^2 + Sum_{n >= 1} (-1)^(n+1) * (5*n^2 + 6*k*n + 2*k^2)/(n^2*(n+k)^2*T(n-1,n+k-1)*T(n,n+k)), k = 0,1,2... .
Conjectural result for subdiagonals: zeta(2) = 2*(1 - 1/2^2 + ... + (-1)^(k+1)/k^2) + (-1)^k*Sum_{n >= 1} (-1)^(n+1)*(5*n^2 + 4*k*n + k^2)/(n^2*(n+k)^2*T(n+k-1,n-1)*T(n+k,n)), k = 0,1,2... .
Conjectural congruences: the main superdiagonal numbers S(n) := T(n,n+1) appear to satisfy the supercongruences S(m*p^r - 1) = S(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers m and r. If p is prime of the form 4*n + 1 we can write p = a^2 + b^2 with a an odd number. Then calculation suggests the congruence S((p-1)/2) == 2*a^2 (mod p). (End)
From Michael Somos, Jun 03 2012: (Start)
T(n, k) = hypergeom([-n, -k, n + 1], [1, 1], 1).
T(n, n-1) = A208675(n).
T(n+1, n) = A108628(n). (End)
T(n, k) = binomial(n, k)*hypergeom([-k, k - n, k - n], [1, -n], 1). - Peter Luschny, Feb 10 2018
From Peter Bala, Jun 23 2023: (Start)
T(n, k) = Sum_{i = 0..k} (-1)^i * binomial(n, i)*binomial(n+k-i, k-i)^2.
T(n, k) = binomial(n+k, k)^2 * hypergeom([-n, -k, -k], [-n - k, -n - k], 1). (End)
From Peter Bala, Jun 28 2023; (Start)
T(n,k) = the coefficient of (x^n)*(y^k)*(z^n) in the expansion of 1/( (1 - x - y)*(1 - z ) - x*y*z ).
T(n,k) = B(n, k, n) in the notation of Straub, equation 24.
The supercongruences T(n*p^r, k*p^r) == T(n*p^(r-1), k*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and k.
The formula T(n,k) = hypergeom([n+1, -n, -k], [1, 1], 1) allows the table indexing to be extended to negative values of n and k; clearly, we find that T(-n,k) = T(n-1,k) for all n and k. It appears that T(n,-k) = (-1)^n*T(n,k-1) for n >= 0, while T(n,-k) = (-1)^(n+1)*T(n,k-1) for n <= -1 [added Sep 10 2023: these follow from the identities immediately below]. (End)
T(n,k) = Sum_{i = 0..n} (-1)^(n+i) * binomial(n, i)*binomial(n+i, i)*binomial(k+i, i) = (-1)^n * hypergeom([n + 1, -n, k + 1], [1, 1], 1). - Peter Bala, Sep 10 2023
From G. C. Greubel, Oct 05 2023: (Start)
Let t(n,k) = T(n-k, k) (antidiagonals).
t(n, k) = Hypergeometric3F2([k-n, -k, n-k+1], [1,1], 1).
T(n, 2*n) = A363867(n).
T(3*n, n) = A363868(n).
T(2*n, 2*n) = A363869(n).
T(n, 3*n) = A363870(n).
T(2*n, 3*n) = A363871(n). (End)
T(n, k) = Sum_{i = 0..n} binomial(n, i)*binomial(n+i, i)*binomial(k, i). - Peter Bala, Feb 26 2024
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n, k) = A005259(n), the Apéry numbers associated with zeta(3). - Peter Bala, Jul 18 2024
From Peter Bala, Sep 21 2024: (Start)
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*T(n, k) = binomial(2*n, n) = A000984(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(n-1, n-k) = A376458(n).
Sum_{k = 0..n} (-1)^(n+k)*binomial(n, k)*binomial(n+k, k)*T(i, k) = A143007(n, i). (End)
From Peter Bala, Oct 12 2024: (Start)
Conjecture: for positive integer m, Sum_{k = 0..n} (-1)^(n+k) * binomial(n, k) * T(m*n, k) = ((m+1)*n)!/( ((m-1)*n)!*n!^2) (verified up to m = 10 using the MulZeil procedure in Doron Zeilberger's MultiZeilberger package). (End)
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