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proposed

Cf. A003161, A183069, A361888, A361891

Conjecture: Let b(n) = a(2*n-1). Then the supercongruence b(n*p^rk) == b(n*p^(rk-1)) (mod p^(3*rk)) holds for positive integers n and r k and all primes p >= 5. See A183069. (End)

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) proposed the problem of showing that S(3,n) was always divisible by S(1,n). The present sequence is {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). See For other cases see A361888 (r = {S(5,n)/S(1,n)}) and A361890 A361891 ({S(r = 7,n)/ S(1,n)}).

Cf. A003161, A361888, A361890A361891

From Peter Bala, Mar 26 2023: (Start)

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) proposed the problem of showing that S(3,n) was always divisible by S(1,n). The present sequence is {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). See A361888 (r = 5) and A361890 (r = 7).

Conjecture: Let b(n) = a(2*n-1). Then the supercongruence b(n*p^r) == b(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5. See A183069. (End)

From Peter Bala, Mar 26 2023: (Start)

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) proposed the problem of showing that S(3,n) was always divisible by S(1,n). The present sequence is {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n).

Conjecture: Let b(n) = a(2*n-1). Then the supercongruence b(n*p^r) == b(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5. See A183069. (End)

Cf. A003161., A361888, A361890

Conjecture: Let b(n) = a(2*n-1). Then the supercongruence ab(2n*p - 1^r) == b(n*p^(r-1 )) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5 (checked up to p = 199). Indeed, for k >= 2, the congruence a(2*p^k - 1) == 1 (mod p^3) may also hold for all primes p >= 5See A183069. (End)

Conjecture: the supercongruence a(2*p - 1) == 1 (mod p^3) holds for all primes p >= 5 (checked up to p = 199). Indeed, for k >= 2, the supercongruence congruence a(2*p^k - 1) == 1 (mod p^3) may also hold for all primes p >= 5. (End)

nonn,changed,easy

From Peter Bala, Mar 26 2023: (Start)

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) proposed the problem of showing that S(3,n) was always divisible by S(1,n). The present sequence is {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n).

Conjecture: the supercongruence a(2*p - 1) == 1 (mod p^3) holds for all primes p >= 5 (checked up to p = 199). Indeed, for k >= 2, the supercongruence a(2*p^k - 1) == 1 (mod p^3) may also hold for all primes p >= 5. (End)

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