FORMULA
The sequence b(n) := [x^n] ( F(x)/F(-x) )^n, where F(x) = (x^7 - 1)/(x - 1), may satisfy the stronger supercongruences congruences b(p) == 2 (mod p^3) for prime p > 7 (checked up to p = 499). (End)
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Revision History for A318114
(Bold, blue-underlined text is an addition; faded, red-underlined text is aShowing entries 1-10 | older changes
STATUS
proposed
approved
STATUS
editing
proposed
FORMULA
a(n) = Sum_{i=0..n/7} (-1)^i*C(n,i)*C(2*n-7*i-1,n-7*i). - Vladimir Kruchinin, Mar 28 2019 [corrected by Peter Bala, Mar 31 2020]
a(n) = Sum_{i=0..n/7} (-1)^i*C(n,i)*C(2*n-7*i-1,n-7*i).
a(p) == 1 (mod p^2) for any prime p > 7.
a(p) == 1 (mod p^2) for any prime p > 7 (from Kruchnin's formula above). More generally, we may have a(p^k) == a(p^(k-1)) (mod p^(2*k)) for k >= 2 and any prime p.
STATUS
proposed
editing
STATUS
editing
proposed
FORMULA
a(n) = Sum_{i=0..n/57} (-1)^i*C(n,i)*C(2*n-57*i-1,n-57*i). - Vladimir Kruchinin, Mar 28 2019 [corrected by _Peter Bala_, Mar 31 2020]
a(n) = Sum_{k = 0..floor(n/7)} (-1)^k*C(n,k)*C(2*n-7*k-1,n-7*k).
a(p) == 1 (mod p^2) for any prime p > 7 (from Kruchnin's formula above). More generally, we may have a(p^k) == a(p^(k-1)) (mod p^(2*k)) for k >= 2 and any prime p.
FORMULA
a(n) = Sum_{i=0..n/5} (-1)^i*C(n,i)*C(2*n-5*i-1,n-5*i). - Vladimir Kruchinin, Mar 28 2019
From Peter Bala, Mar 31 2020: (Start)
a(n) = Sum_{k = 0..floor(n/7)} (-1)^k*C(n,k)*C(2*n-7*k-1,n-7*k).
a(p) == 1 (mod p^2) for any prime p > 7. More generally, we may have a(p^k) == a(p^(k-1)) (mod p^(2*k)) for k >= 2 and any prime p.
The sequence b(n) := [x^n] ( F(x)/F(-x) )^n, where F(x) = (x^7 - 1)/(x - 1), may satisfy the stronger supercongruences b(p) == 2 (mod p^3) for prime p > 7 (checked up to p = 499). (End)
STATUS
approved
editing
STATUS
editing
approved
LINKS
Alois P. Heinz, <a href="/A318114/b318114.txt">Table of n, a(n) for n = 0..1675</a>