Revision History for A333593
(Underlined text is an addition;
strikethrough text is a deletion.)
Showing entries 1-10
| older changes
|
| |
| |
| |
|
|
#28 by Michael De Vlieger at Mon May 01 10:00:53 EDT 2023
| |
| |
| |
|
|
#27 by Joerg Arndt at Mon May 01 09:33:51 EDT 2023
| |
| |
| |
|
|
#26 by Michel Marcus at Mon May 01 04:00:35 EDT 2023
| |
| |
| |
|
|
#25 by Michel Marcus at Mon May 01 04:00:32 EDT 2023
|
|
| LINKS
|
P. Peter Bala, <a href="/A333593/a333593.pdf">Notes on A333593</a>
|
|
| STATUS
|
proposed
editing
|
| |
| |
|
|
#24 by Peter Bala at Mon May 01 03:59:58 EDT 2023
| |
| |
| |
|
|
#23 by Peter Bala at Sun Apr 30 17:40:19 EDT 2023
|
|
| COMMENTS
|
The sequence b(n) = C(2*n,n) of central binomial coefficients satisfies the congruencessupercongruences 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.
|
|
| STATUS
|
approved
editing
|
| |
| |
|
|
#22 by Peter Bala at Tue Mar 28 11:00:29 EDT 2023
|
|
| NAME
|
a(n) = Sum_{k = 0..n} (-1)^(n+ + k) * )*binomial(n+ + k- - 1,, k)^2.
|
|
| COMMENTS
|
The sequence { b(n) = C(2*n,n): n >= 1} ) of central binomial coefficients satisfies the supercongruencescongruences 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 Meštrović. 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.
|
|
| LINKS
|
Peter P. Bala, <a href="/A333593/a333593.pdf">Notes on A333593</a>
Romeo Meštrović, <R. Mestrovic, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.
|
|
| FORMULA
|
It appears that a(n) = Sum_{k = 0..n} (-1)^k*binomial(2*n-k-1,n-1)^2 for n >= 1. - Peter Bala, Mar 28 2023
|
|
| KEYWORD
|
nonn,easy,changed
|
|
| STATUS
|
editing
approved
|
| |
| |
|
|
#21 by Peter Bala at Tue Mar 28 10:51:14 EDT 2023
|
|
| NAME
|
a(n) = Sum_{k = 0..n} (-1)^(n + +k)*) * binomial(n + +k - -1, ,k)^2.
|
|
| COMMENTS
|
The sequence {b(n) = C(2*n,n): n) >= 1} of central binomial coefficients satisfies the congruencessupercongruences 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. Meštrović. 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.
|
|
| LINKS
|
P. Peter Bala, <a href="/A333593/a333593.pdf">Notes on A333593</a>
R. Mestrovic, <Romeo Meštrović, <a href="http://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011)</a>, arXiv:1111.3057 [math.NT], 2011.
|
|
| FORMULA
|
It appears that a(n) = Sum_{k = 0..n} (-1)^k*binomial(2*n-k-1,n-1)^2 for n >= 1. - Peter Bala, Mar 28 2023
|
|
| STATUS
|
approved
editing
|
| |
| |
|
|
#20 by N. J. A. Sloane at Wed Oct 06 14:35:58 EDT 2021
| |
| |
| |
|
|
#19 by N. J. A. Sloane at Wed Oct 06 14:35:55 EDT 2021
|
|
| COMMENTS
|
The sequence b(n) = C(2*n,n) of central binomial coefficients satisfies the congruences 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 supercongruencescongruences. Some examples of the supercongruencescongruences are given below. For a proof of the particular case of these supercongruencescongruences, 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 supercongruencescongruences for all prime p >= 5.
|
|
| STATUS
|
approved
editing
|
| |
| |
|
|
