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 A333261 #45 Jul 03 2025 17:04:41 %S A333261 1,5,51,68,87,116,171,176,591,2108,2403,7143,8787,9308,18548,19371, %T A333261 27387,32127,34887,37928,40131,140667,180548,192428,200996,433311, %U A333261 521727,934449,2476671,2617563,3960896,8198156,9670748,11892512,16585748,19113651,25367396,25643012 %N A333261 Numbers k such that A071324(k) = A071324(k+1). %C A333261 From _Shreyansh Jaiswal_, Jun 14 2025: (Start) %C A333261 If the density of the set containing all even terms exists, then it is less than 0.15. (Proposition 3 in Jaiswal.) %C A333261 Let k denote any even term. Then, the least prime factor of k+1 is either 3 or 5. (Theorem 11 in Jaiswal.) %C A333261 Each even term satisfies at least one of three specific congruences. (Theorem 2 in Jaiswal.) %C A333261 10519952096 and 16159802432 are also terms of this sequence. %C A333261 Conjecture: There are infinitely many terms of this sequence. (Conjecture 15 in Jaiswal.) (End) %H A333261 Amiram Eldar, <a href="/A333261/b333261.txt">Table of n, a(n) for n = 1..89</a> (terms below 10^10) %H A333261 Shreyansh Jaiswal, <a href="/A333261/a333261.pdf">On the Even Solutions of A071324(n) = A071324(n+1)</a>, Jun 14, 2025. %e A333261 1 is a term since A071324(1) = A071324(2) = 1. %e A333261 5 is a term since A071324(5) = A071324(6) = 4. %t A333261 f[n_] := Plus @@ (-(d = Divisors[n])*(-1)^(Range[Length[d],1,-1])); seq = {}; f1 = f[1]; Do[f2 = f[n]; If[f1 == f2, AppendTo[seq, n-1]]; f1 = f2, {n, 2, 50000}]; seq %t A333261 SequencePosition[Table[Total[Times@@@Partition[Riffle[Reverse[Divisors[n]],{1,-1},{2,-1,2}],2]],{n,2565*10^4}],{x_,x_}][[All,1]] (* _Harvey P. Dale_, Nov 06 2022 *) %o A333261 (Python) %o A333261 from sympy import divisors; from functools import lru_cache %o A333261 cached_divisors = lru_cache()(divisors) %o A333261 def c(n): return sum(d if i%2==0 else -d for i, d in enumerate(reversed(cached_divisors(n)))) %o A333261 for n in range(1,2201): %o A333261 if c(n) == c(n+1): %o A333261 print(n, end=", ") # _Shreyansh Jaiswal_, Apr 14 2025 %Y A333261 Cf. A002961, A071324, A206368. %K A333261 nonn,hard %O A333261 1,2 %A A333261 _Amiram Eldar_, Mar 13 2020