A245089 -id:A245089 - cp's OEIS frontend

A277640 a(n) is the integer r with |r| < prime(n)/2 such that (T(prime(n)^2)-T(prime(n)))/prime(n)^4 == r (mod prime(n)), where T(k) denotes the central trinomial coefficient A002426(k).

-2, 1, -3, -1, 7, -1, 6, 4, -15, -15, -13, 1, -23, 1, 8, -15, -22, 13, -33, 27, 25, 11, -17, 24, -32, -53, 31, 42, -19, 18, -35, 55, -5, 38, -29, 76, 34, 44, -71, -21, -13, 16, 46, 70, 92, 70, -39, 88, -84, -118, -120, 64, 107, 111, -56, 124, -13, -23

Offset: 3

Zhi-Wei Sun, Oct 25 2016

sign

Conjecture: (i) For any prime p > 3 and positive integer n, the number (T(p*n)-T(n))/(p*n)^2 is always a p-adic integer.
(ii) For any prime p > 3 and positive integer k, we have (T(p^k)-T(p^(k-1)))/p^(2k) == 1/6*(p^k/3)*B_{p-2}(1/3) (mod p), where (p^k/3) denotes the Legendre symbol and B_{p-2}(x) is the Bernoulli polynomial of degree p-2.
For any prime p > 3, the author has proved that (T(p*n)-T(n))/(p^2*n) is a p-adic integer for each positive integer n, and that T(p) == 1 + p^2/6*(p/3)*B_{p-2}(1/3) (mod p^3).

			a(3) = -2 since (T(prime(3)^2)-T(prime(3)))/prime(3)^4 = (T(25)-T(5))/5^4 = (82176836301-51)/5^4 = 131482938 is congruent to -2 modulo prime(3) = 5 with |-2| < 5/2.

Hao Pan and Zhi-Wei Sun, Supercongruences for central trinomial coefficients, arXiv:2012.05121 [math.NT], 2020.
Zhi-Wei Sun, Congruences involving generalized central trinomial coefficients, Sci. China Math. 57(2014), no.7, 1375-1400.
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Cf. A000040, A002426, A245089, A277860.

T[n_]:=T[n]=Sum[Binomial[n,2k]Binomial[2k,k],{k,0,n/2}]
rMod[m_,n_]:=Mod[Numerator[m]*PowerMod[Denominator[m],-1,n],n,-n/2]
Do[Print[n," ",rMod[(T[Prime[n]^2]-T[Prime[n]])/Prime[n]^4,Prime[n]]],{n,3,60}]

A245204 The unique integer r with |r| < prime(n)/2 such that E_{prime(n)-3}(1/4) == r (mod prime(n)), where E_m(x) denotes the Euler polynomial of degree m.

Original entry on oeis.org

1, 2, 2, 4, 1, 1, 5, 1, -2, -6, 10, 14, 5, 7, 7, -28, -12, 13, 14, 26, -21, -31, -13, -10, -11, -7, -6, 5, 2, -21, 2, 33, -15, -24, 34, 71, -15, 24, 9, 37, 73, -18, -84, -65, 9, -90, -65, -47, 97, -64, -100, -8, 41, 81, -81, -71, -65, -70, 113, 10, -80, 119, 57, 78, 20, 124, 167, -71, -48

Offset: 2

Zhi-Wei Sun, Jul 13 2014

sign

Conjecture: a(n) = 0 infinitely often. In other words, there are infinitely many odd primes p such that E_{p-3}(1/4) == 0 (mod p) (equivalently, p divides A001586(p-3)).
This seems reasonable in view of the standard heuristic arguments. The first n with a(n) = 0 is 171 with prime(171) = 1019. The next such a number n is greater than 2600 and hence prime(n) > 23321.
Zhi-Wei Sun made many conjectures on congruences involving E_{p-3}(1/4), see the reference.

			a(3) = 2 since E_{prime(3)-3}(1/4) = E_2(1/4) = -3/16 == 2 (mod prime(3)=5).

Zhi-Wei Sun, Table of n, a(n) for n = 2..1300
Zhi-Wei Sun, Super congruences and Euler numbers, arXiv:1001.4453 [math.NT].
Zhi-Wei Sun, Super congruences and Euler numbers, Sci. China Math. 54(2011), 2509-2535.

Cf. A000040, A001586, A122045, A198245, A245089, A245206.

rMod[m_,n_]:=Mod[Numerator[m]*PowerMod[Denominator[m],-1,n],n,-n/2]
a[n_]:=rMod[EulerE[Prime[n]-3,1/4],Prime[n]]
Table[a[n],{n,2,70}]

A277860 a(n) = Sum_{k=0..n-1} binomial(4k, 2k+1)binomial(2k, k)48^(n-1-k).

Original entry on oeis.org

0, 8, 720, 50400, 3220000, 196885920, 11756961216, 692835631488, 40536961717824, 2363784447147552, 137716866109432896, 8030173585594013568, 469162781054378536320, 27486632292027996114560, 1615617140290621588826880, 95302760085090826490672640

Offset: 1

Zhi-Wei Sun, Nov 02 2016

nonn

Conjecture: For any prime p > 3 and positive integer n, we have
(Sum_{k=0..p*n-1} binomial(4k, 2k+1)*binomial(2k, k)/48^k - (p/3)*Sum_{r=0..n-1} binomial(4r, 2r+1)*binomial(2r, r)/48^r)*48^n/((p*n)^2*binomial(4n, 2n)*binomial(2n, n)) == (5/3)*B_{p-2}(1/3) (mod p), where (p/3) is the Legendre symbol and B_{p-2}(x) is the Bernoulli polynomial of degree p-2.
This conjecture with n = 1 gives the congruence a(p) == (5/12)*p^2*B_{p-2}(1/3) (mod p^3) for any prime p > 3.

			a(1) = 0 since binomial(4*0, 2*0+1)*binomial(2*0, 0)*48^(1-1-0) = 0.
a(2) = 8 since Sum_{k=0..1} binomial(4k, 2k+1)*binomial(2k, k)*48^(2-1-k) = binomial(4, 2+1)*binomial(2, 1)*48^0 = 8.

Zhi-Wei Sun, Table of n, a(n) for n = 1..100
Zhi-Wei Sun, Super congruences and Euler numbers, Sci. China Math. 54(2011), no.12, 2509--2535.
Zhi-Wei Sun, New series for some special values of L-functions, Nanjing Univ. J. Math. Biquarterly 32(2015), no.2, 189-218.
Zhi-Wei Sun, Supercongruences involving Lucas sequences, arXiv:1610.03384 [math.NT], 2016.

Cf. A000984, A001448, A245089, A277640.

a[n_]:=Sum[Binomial[4k,2k+1]Binomial[2k,k]48^(n-1-k),{k,0,n-1}]
Table[a[n],{n,1,16}]

a(n) ~ 2^(6*n - 9/2) / (Pi*n). - Vaclav Kotesovec, Nov 06 2021

A245206 Odd primes p with E_{p-3}(1/4) == 0 (mod p), where E_n(x) denotes the Euler polynomial of degree n.

Original entry on oeis.org

1019

Offset: 1

Zhi-Wei Sun, Jul 13 2014

The conjecture in A245204 asserts that the current sequence contains infinitely many primes.

Our computation shows that the second term should be greater than prime(2600) = 23321.

			a(1) = 1019 since 1019 is a prime with E_{1019-3}(1/4) == 88*1019 (mod 1019^2).

Zhi-Wei Sun, Super congruences and Euler numbers, Sci. China Math., Vol. 54, No. 12 (2011), 2509-2535; arXiv preprint, arXiv:1001.4453 [math.NT], 2010-2011.
Index entries for one-term sequences.

Cf. A000040, A001586, A122045, A198245, A245089, A245204.

rMod[m_,n_]:=Mod[Numerator[m]*PowerMod[Denominator[m],-1,n],n,-n/2]
n=0;Do[If[rMod[EulerE[Prime[k]-3,1/4],Prime[k]]==0,n=n+1;Print[n," ",Prime[k]]],{k,2,200}]

A245329 a(n) = sum_{k=0..n}C(n,k)^3*(-8)^k with C(n,k) = n!/(k!(n-k)!).

Original entry on oeis.org

1, -7, 1, 1001, -15359, 30233, 3126529, -61392247, 259448833, 11970181433, -287815672319, 1854020654633, 48800262650881, -1443188813338279, 12410505050039041, 198977188596692681, -7472188661349285887, 80331498114096555641

Offset: 0

Zhi-Wei Sun, Jul 18 2014

sign

Conjecture: (i) For any prime p > 3, we have sum_{k=0}^{p-1}(-1)^k*a(k) == (p/3) - p^2/12*B_{p-2}(1/3) (mod p^3) and sum_{k=0}^{p-1}(-1)^k*a(k)*H(k,2) == B_{p-2}(1/3) (mod p), where B_m(x) denotes the Bernoulli polynomial of degree m and H(k,2) stands for sum_{j=1..k} 1/j^2. Also, sum_{k=1}^{p-1}(-1)^k*a(k)/k == - 3*(2^p-2)/p (mod p) for any prime p.

(ii) For any positive integer n, the arithmetic mean (sum_{k=0}^{n-1}(6*k+5)(-1)^k*a(k))/n is an odd integer. Moreover, if p is a prime then sum_{k=0}^{p-1}(6*k+5)(-1)^k*a(k) == 3*p (mod p^2).

			a(2) = 1 since sum_{k=0}^2 C(2,k)^3*(-8)^k = 1 + 2^3*(-8) + (-8)^2 = 1.

Zhi-Wei Sun, Table of n, a(n) for n = 0..150
Zhi-Wei Sun, Congruences involving g_n(x) = sum_{k=0}^n C(n,k)^2*C(2k,k)*x^k, preprint, arXiv:1407.0967 [math.NT], 2014-2016.
Zhi-Wei Sun, Congruences for Franel numbers, preprint, arXiv:1112.1034 [math.NT], 2011-2013.
Z.-W. Sun, Congruences for Franel numbers, Adv. in Appl. Math. 51(2013), 524-535.

Cf. A000172, A245089.

a[n_]:=Sum[Binomial[n,k]^3*(-8)^k,{k,0,n}]
Table[a[n],{n,0,17}]

Recurrence (obtained via the Zeilberger algorithm):

343*(3n+7)*(n+1)^2*a(n) + (3n+5)*(363n^2+1331n+1113)*a(n+1) + 7*(9n^3+57n^2+116n+74)*a(n+2) + (3n+4)*(n+3)^2*a(n+3) = 0.

cp's OEIS Frontend

A277640 a(n) is the integer r with |r| < prime(n)/2 such that (T(prime(n)^2)-T(prime(n)))/prime(n)^4 == r (mod prime(n)), where T(k) denotes the central trinomial coefficient A002426(k).

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A245204 The unique integer r with |r| < prime(n)/2 such that E_{prime(n)-3}(1/4) == r (mod prime(n)), where E_m(x) denotes the Euler polynomial of degree m.

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A277860 a(n) = Sum_{k=0..n-1} binomial(4k, 2k+1)*binomial(2k, k)*48^(n-1-k).

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A245206 Odd primes p with E_{p-3}(1/4) == 0 (mod p), where E_n(x) denotes the Euler polynomial of degree n.

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A245329 a(n) = sum_{k=0..n}C(n,k)^3*(-8)^k with C(n,k) = n!/(k!(n-k)!).

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Formula

A277860 a(n) = Sum_{k=0..n-1} binomial(4k, 2k+1)binomial(2k, k)48^(n-1-k).