A110197 -id:A110197 - cp's OEIS frontend

A110198 Antidiagonal sums of number triangle A110197.

1, 2, 4, 9, 20, 46, 109, 262, 638, 1569, 3886, 9680, 24225, 60856, 153368, 387573, 981742, 2491934, 6336721, 16139616, 41166912, 105139773, 268841100, 688157430, 1763206441, 4521749642, 11605580290, 29809644693, 76621733444, 197074591420, 507193044993

Offset: 0

Paul Barry, Jul 15 2005

Partial sums of A051286.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Cf. A051286, A110197, A201631.

CoefficientList[Series[1/((1-x)*Sqrt[(1+x+x^2)*(1-3*x+x^2)]), {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 08 2014 *)

G.f.: 1/((1-x)*sqrt((1+x+x^2)*(1-3x+x^2))); a(n) = sum{k=0..floor(n/2), sum{i=0..n-2k, binomial(i+k, k)^2}}.
a(n) = sum{i=0..2n, A202411(i)}. - Peter Luschny, Jan 16 2012
Conjecture: n*a(n) +(-3*n+1)*a(n-1) +n*a(n-2) +(-n+2)*a(n-3) +(3*n-5)*a(n-4) +(-n+2)*a(n-5)=0. - R. J. Mathar, Nov 15 2012
a(n) ~ sqrt(100+45*sqrt(5)) * ((sqrt(5)+3)/2)^n / (10*sqrt(Pi*n)). - Vaclav Kotesovec, Feb 08 2014
Equivalently, a(n) ~ phi^(2*n + 3) / (2 * 5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 07 2021

A112029 a(n) = Sum_{k=0..n} binomial(n+k, k)^2.

Original entry on oeis.org

1, 5, 46, 517, 6376, 82994, 1119210, 15475205, 217994860, 3115374880, 45035696036, 657153097330, 9663914317396, 143050882063262, 2129448324373546, 31853280798384645, 478503774600509620, 7215090439396842572, 109154411037070011504, 1656268648035559711392

Offset: 0

N. J. A. Sloane, Nov 28 2005

Vincenzo Librandi, Table of n, a(n) for n = 0..200
F. Baldassarri, S. Bosch, B. Dwork, (eds), p-adic Analysis. Lecture Notes in Mathematics, vol. 1454, pp. 194 - 204, Springer, Berlin, Heidelberg.
Matthijs J. Coster, Supercongruences.
C. Elsner, On recurrence formulas for sums involving binomial coefficients, Fib. Q., 43,1 (2005), 31-45.
Vaclav Kotesovec, Asymptotic of generalized Apery sequences with powers of binomial coefficients, Nov 04 2012
Pedro J. Miana, Hideyuki Ohtsuka, and Natalia Romero, Sums of powers of Catalan triangle numbers, arXiv:1602.04347 [math.NT], 2016.

Cf. A001700, A110197, A112028, A173774, A176335, A219562, A219563, A219564, A333592.

[(&+[Binomial(n+j, j)^2: j in [0..n]]): n in [0..20]]; // G. C. Greubel, Jul 06 2021

f := 64*x^2/(16*x-1); S := sqrt(x)*sqrt(4-x);
H := ((10*x-5/8)*hypergeom([1/4,1/4],[1],f)-(21*x-21/8)*hypergeom([1/4,5/4],[1],f))/(S*(1-16*x)^(5/4));
ord := 30;
ogf := series(int(series(H,x=0,ord),x)/S,x=0,ord);
# Mark van Hoeij, Mar 27 2013

Table[Sum[Binomial[n+k,k]^2, {k,0,n}], {n,0,20}] (* Vaclav Kotesovec, Nov 23 2012 *)

a(n) = sum(k=0, n, binomial(n+k, k)^2); \\ Michel Marcus, Jul 07 2021

[sum(binomial(n+j, j)^2 for j in (0..n)) for n in (0..20)] # G. C. Greubel, Jul 06 2021

a(n) ~ 2^(4*n+2)/(3*Pi*n). - Vaclav Kotesovec, Nov 23 2012
Recurrence: 2*(2*n+1)*(21*n-13)*n^2*a(n) = (1365*n^4 - 1517*n^3 + 240*n^2 + 216*n - 64)*a(n-1) - 4*(n-1)*(2*n-1)^2*(21*n+8)*a(n-2). - Vaclav Kotesovec, Nov 23 2012
G.f.: see Maple code. - Mark van Hoeij, Mar 27 2013
a(p-1) == 1 (mod p^3) for all primes p >= 5. See the comments in A173774. - Peter Bala, Jul 12 2024
a(n-1) = 1/(4*n) * binomial(2*n, n)^2 * ( 1 + 3*((n - 1)/(n + 1))^3 + 5*((n - 1)*(n - 2)/((n + 1)*(n + 2)))^3 + 7*((n - 1)*(n - 2)*(n - 3)/((n + 1)*(n + 2)*(n + 3)))^3 + ... )  for n >= 1. - Peter Bala, Jul 22 2024
a(m*p^r - 1) == a(m*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and positive integers m and r. See Coster, Theorem 4. - Peter Bala, Nov 29 2024
a(n) = A110197(2n,n). - Alois P. Heinz, Mar 21 2025

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A110198 Antidiagonal sums of number triangle A110197.

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A112029 a(n) = Sum_{k=0..n} binomial(n+k, k)^2.

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