A333592 -id:A333592 - cp's OEIS frontend

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

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

A357671 a(n) = Sum_{k = 0..n} ( binomial(n+k-1,k) + binomial(n+k-1,k)^2 ).

Original entry on oeis.org

2, 4, 20, 166, 1812, 22504, 297362, 4067298, 56897300, 809019580, 11649254520, 169444978124, 2485270719570, 36707044807996, 545386321069862, 8144809732228666, 122177690210103060, 1839933274439787940, 27804610626798500372, 421476329345312885304, 6406685025104178888312

Offset: 0

Peter Bala, Oct 10 2022

Conjectures:
1) a(p) == 4 (mod p^5) for all primes p >= 7 (checked up to p = 499). Note that A000984(p) == 2 (mod p^3) and A333592(p) == 2 (mod p^3) for all primes p >= 5.
2) For r >= 2, and all primes p >= 5, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
3) More generally, let m be a positive integer and set u(n) = 2*m*Sum_{k = 0..m*n} binomial(n+k-1,k) + (m + 1)*Sum_{k = 0..m*n} binomial(n+k-1,k)^2. Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 1. See A357673 for the case m = 2.
4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).

			Examples of supercongruences:
a(19) - a(1) = 421476329345312885304 - 4 = (2^2)*(5^2)*(19^5)*1913*2383*373393 == 0 (mod 19^5).
a(25) - a(5) = 5375188503768910125546897504 - 22504 = (2^3)*(5^10)*1858537* 37019662696111 == 0 (mod 5^10).

Cf. A000984, A333592, A357509, A357565, A357566, A357672, A357673, A357674.

seq(add( binomial(n+k-1,k) + binomial(n+k-1,k)^2, k = 0..n ), n = 0..20);

a(n) = sum(k = 0, n, binomial(n+k-1,k) + binomial(n+k-1,k)^2); \\ Michel Marcus, Oct 24 2022

from math import comb
def A357671(n): return comb(n<<1,n)+sum(comb(n+k-1,k)**2 for k in range(n+1)) if n else 2 # Chai Wah Wu, Oct 28 2022

a(n) = A000984(n) + A333592(n).
a(p) == 4 (mod p^3) for all primes p >= 5.
a(n) ~ 2^(4*n) / (3*Pi*n).

A357566 a(n) = ( Sum_{k = 0..n} binomial(n+k-1,k)^2 )^3 * ( Sum_{k = 0..n} binomial(n+k-1,k)^3 )^2.

Original entry on oeis.org

1, 32, 3556224, 4816142496896, 14260946236464636800, 62923492736113950202540032, 355372959542696519903013302282592, 2376354966106399942850054560101358877184, 17973185649572984869873798116070605084766512000, 149319509846904520286037745483655872001727895961600000

Offset: 0

Peter Bala, Oct 16 2022

Conjectures:
1) a(p) == a(1) (mod p^5) for all odd primes p except p = 5 (checked up to p = 271).
2) a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 5.
3) More generally, let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+2) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^3 )^(2*m). Then the supercongruences u(p) == u(1) (mod p^5) hold for all primes p >= 5.
4) u(p^r) == u(p^(r-1)) (mod p^(3*r+3)) for r >= 2 and all primes p >= 5.

			a(7) - a(1) = 2376354966106399942850054560101358877184 - 32 = (2^5)*(7^5)*19*31*317*339247*25170329*2771351868561767 == 0 (mod 7^5).

Cf. A357565, A357671, A357672, A357673, A357674.

Cf. A333592.

seq((add(binomial(n+k-1,k)^2, k = 0..n))^3 * (add( binomial(n+k-1,k)^3, k = 0..n))^2, n = 0..20);

Table[Sum[Binomial[n+k-1,k]^2, {k, 0, n}]^3 * Sum[Binomial[n+k-1,k]^3, {k, 0, n}]^2, {n, 0, 10}] (* Vaclav Kotesovec, May 31 2025 *)

a(n) = sum(k = 0, n, binomial(n+k-1,k)^2)^3 * sum(k = 0, n, binomial(n+k-1,k)^3)^2; \\ Michel Marcus, Oct 25 2022

a(n) ~ 2^(24*n) / (1323*Pi^6*n^6). - Vaclav Kotesovec, May 31 2025

A357672 a(n) = Sum_{k = 0..n} binomial(n+k-1,k) * Sum_{k = 0..n} binomial(n+k-1,k)^2.

Original entry on oeis.org

1, 4, 84, 2920, 121940, 5607504, 273908712, 13947188112, 732102614100, 39332168075200, 2152235533317584, 119531412173662944, 6720552415489860584, 381775182057562837600, 21879043278489630349200, 1263402662473729731877920, 73438613319490294002441300, 4293679728171938162242298400

Offset: 0

Peter Bala, Oct 10 2022

Conjectures:
1) a(p) == 4 (mod p^5) for all odd primes p except p = 5 (checked up to p = 499). Note that both A000984(p) == 2 (mod p^3) and A333592(p) == 2 (mod p^3) for all primes p >= 5 and hence a(p) == 4 (mod p^3) for all primes p >= 5.
2) For r >= 2, and all primes p >= 3, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
3) More generally, let m be a positive integer and set u(n) = ( Sum_{k = 0..m*n} binomial(n+k-1,k) )^(2*m) * ( Sum_{k = 0..m*n} binomial(n+k-1,k)^2 )^(m+1). Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. See A357674 for the case m = 2.
4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).

			Examples of supercongruences:
a(17) - a(1) = 4293679728171938162242298400 - 4 = (2^2)*(17^5)*3457* 218688360593678551 == 0 (mod 17^5).
a(5^2) - a(5) = (2^4)*(3^2)*(5^9)*7*7229*102559*465516030080883405648119 == 0 (mod 5^9).

Cf. A000984, A333592, A357565, A357566, A357671, A357673, A357674.

seq(add(binomial(n+k-1,k), k = 0..n) * add( binomial(n+k-1,k)^2, k = 0..n), n = 0..20);

a(n) = sum(k = 0, n, binomial(n+k-1,k)) * sum(k = 0, n, binomial(n+k-1,k)^2); \\ Michel Marcus, Oct 24 2022

a(n) = A000984(n) * A333592(n).

a(n) ~ 2^(6*n)/(3*(Pi*n)^(3/2)).

A333593 a(n) = Sum_{k = 0..n} (-1)^(n + k)*binomial(n + k - 1, k)^2.

Original entry on oeis.org

1, 0, 6, 72, 910, 12000, 163086, 2266544, 32043726, 459167040, 6651400756, 97214919648, 1431514320886, 21213380196736, 316072831033350, 4731683468079072, 71128104013487310, 1073134004384407680, 16243463355081280080, 246585461357885877920

Offset: 0

Peter Bala, Mar 27 2020

It is known that Sum_{k = 0..2*n} (-1)^(n+k)*C(2*n,k)^2 = C(2*n,n). Here, we consider by analogy Sum_{k = 0..n} (-1)^(n+k)*C(-n,k)^2, where C(-n,k) = (-1)^k * C(n+k-1,k) for integer n and nonnegative integer k.
The sequence b(n) = C(2*n,n) of central binomial coefficients satisfies the supercongruences b(n*p^k) = b(n*p^(k-1)) ( mod p^(3*k) ) for all prime p >= 5 and any positive integers n and k -  see Mestrovic. We conjecture that the present sequence also satisfies these congruences. Some examples of the congruences are given below. For a proof of the particular case of these congruences, a(p) == 0 ( mod p^3 ) for prime p >= 5, see the Bala link. The sequence a(p)/(2*p^3) for prime p >= 5 begins [48, 304, 36519504, 4827806144, 109213719151680, 17975321574419440, ...]. Cf. A034602.
More generally, calculation suggests that for positive integer A and integer B, the sequence a(A,B;n) := Sum_{k = 0..A*n} (-1)^(n+k)* C(-B*n,k)^2 = Sum_{k = 0..A*n} (-1)^(n+k)*C(B*n+k-1,k)^2 may also satisfy the above congruences for all prime p >= 5.

			Examples of congruences:
a(11)  = 97214919648 = (2^5)*3*(7^2)*(11^3)*15527 == 0 ( mod 11^3 ).
a(2*7) - a(2) = 316072831033350 - 6 = (2^13)*3*(7^3)*11*691*4933 == 0 ( mod 7^3 ).
a(5^2) - a(5) = 3164395891098711251676512000 - 12000 = (2^5)*(5^6)*29* 124891891*1747384859327 == 0 ( mod 5^6 ).

Seiichi Manyama, Table of n, a(n) for n = 0..500
Peter Bala, Notes on A333593
R. Mestrovic, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Cf. A000984, A034602, A333592.

seq( add( (-1)^(n+k)*binomial(n+k-1,k)^2, k = 0..n ), n = 0..25);

Table[Binomial[2*n-1, n]^2 * HypergeometricPFQ[{1, -n, -n}, {1 - 2 n, 1 - 2 n}, -1], {n, 0, 20}] (* Vaclav Kotesovec, Mar 28 2020 *)

a(n) = sum(k=0, n, (-1)^(n+k)*binomial(n+k-1, k)^2); \\ Michel Marcus, Mar 29 2020

a(n) ~ 2^(4*n) / (5*Pi*n). - Vaclav Kotesovec, Mar 28 2020

cp's OEIS Frontend

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

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A357671 a(n) = Sum_{k = 0..n} ( binomial(n+k-1,k) + binomial(n+k-1,k)^2 ).

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A357566 a(n) = ( Sum_{k = 0..n} binomial(n+k-1,k)^2 )^3 * ( Sum_{k = 0..n} binomial(n+k-1,k)^3 )^2.

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A357672 a(n) = Sum_{k = 0..n} binomial(n+k-1,k) * Sum_{k = 0..n} binomial(n+k-1,k)^2.

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A333593 a(n) = Sum_{k = 0..n} (-1)^(n + k)*binomial(n + k - 1, k)^2.

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