This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A373763 #7 Jun 22 2024 22:41:18 %S A373763 1,27,63,81,238,243,247,279,322,580,671,729,1222,2074,2187,3172,3550, %T A373763 3577,4185,5589,6561,7805,7957,8239,8701,8890,9040,9064,9523,9730, %U A373763 9898,10087,10138,10549,11074,11176,11440,11473,11920,12232,12430,12604,13900,14287,14410 %N A373763 Numbers k that are neither primes nor squares of primes such that A006134(k) - A102283(k) is divisible by k. %C A373763 The congruence A006134(k) == A102283(k) (mod k) holds for all values of k that are primes or squares of primes. %H A373763 Amiram Eldar, <a href="/A373763/b373763.txt">Table of n, a(n) for n = 1..1000</a> %H A373763 Moa Apagodu and Doron Zeilberger, <a href="https://doi.org/10.4169/amer.math.monthly.124.7.597">Using the "Freshman's Dream" to Prove Combinatorial Congruences</a>, The American Mathematical Monthly, Vol. 124, No. 7 (2017), pp. 597-608; <a href="https://arxiv.org/abs/1606.03351">arXiv preprint</a>, arXiv:1606.03351 [math.CO], 2016. %H A373763 Ji-Cai Liu, <a href="https://arxiv.org/abs/2208.10275">Supercongruences involving Motzkin numbers and central trinomial coefficients</a>, arXiv:2208.10275 [math.NT], 2022. %H A373763 Zhi-Wei Sun and Roberto Tauraso, <a href="https://doi.org/10.1142/S1793042111004393">On some new congruences for binomial coefficients</a>, International Journal of Number Theory, Vol. 7, No. 3 (2011), pp. 645-662; <a href="https://arxiv.org/abs/0709.1665">arXiv preprint</a>, arXiv:0709.1665 [math.NT], 2007-2011. %t A373763 q[n_] := !PrimeQ[n] && !PrimeQ[Sqrt[n]] && Divisible[Sum[Binomial[2*k, k], {k, 0, n - 1}] - JacobiSymbol[n, 3], n]; Select[Range[1000], q] %o A373763 (PARI) is1(k) = !isprime(k) && !(issquare(k) && isprime(sqrtint(k))); %o A373763 lista(kmax) = {my(s0 = 1, s1 = 3); print1(1, ", "); for(k = 2, kmax, s2 = ((5*k - 2) * s1 - 2 * (2*k - 1) * s0 )/k; if(is1(k + 1) && !((s2 - [1, -1, 0][k % 3 + 1]) % (k + 1)), print1(k + 1, ", ")); s0 = s1; s1 = s2);} %Y A373763 Cf. A006134, A102283. %K A373763 nonn %O A373763 1,2 %A A373763 _Amiram Eldar_, Jun 18 2024