A348511 Numbers k for which 2k+1 can be obtained with successive prime shifts towards larger primes (by iterating A003961, starting from k).

2, 3, 4, 5, 10, 11, 23, 29, 41, 53, 57, 83, 89, 113, 131, 173, 179, 191, 233, 239, 251, 281, 293, 359, 419, 431, 443, 491, 509, 593, 641, 653, 659, 683, 719, 743, 761, 809, 911, 953, 1013, 1019, 1031, 1049, 1054, 1103, 1223, 1229, 1289, 1409, 1439, 1451, 1481, 1499, 1511, 1559, 1583, 1601, 1733, 1811, 1889, 1901, 1931

Offset: 1

Author

Antti Karttunen, Oct 30 2021

Keywords

• nonn

Comments

Numbers k for which A246277(2k+1) = A246277(k). This in turn implies a looser condition A046523(2k+1) = A046523(k).

Conjecture: Apart from 4 all other terms in this sequence are squarefree. No counterexamples <= 2^22.

Examples

			4 is present, because by applying prime shift once to it, we get A003961(4) = 9 = 2*4 + 1.
5 is present, because by applying prime shift twice to it, we get 5 -> 7 -> 11 = 2*5 + 1.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Index entries for sequences computed from indices in prime factorization

Crossrefs

Cf. A003961, A046523, A246277.

Subsequences: A005384 (primes present), A348514 (terms requiring only one iteration to reach 2k+1).

Cf. also A285706.

Programs

Maple

filter:= proc(n) local F,F1,F2,G,G1,G2;
  F:= sort(ifactors(n)[2],(a,b) -> a[1] a[1] nops(G) then return false fi;
  F2:= map(t -> t[2],F);
  G2:= map(t -> t[2],G);
  if F2 <> G2 then return false fi;
  if nops(F) = 1 then return true fi;
  F1:= map(t -> numtheory:-pi(t[1]),F);
  G1:= map(t -> numtheory:-pi(t[1]),G);
  nops(convert(G1-F1,set))=1;
end proc:
select(filter, [$2..10000]);# Robert Israel, Feb 18 2022

Mathematica

f[p_, e_] := NextPrime[p]^e; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; q[n_] := NestWhileList[s, n, # < 2*n + 1 &][[-1]] == 2*n + 1; Select[Range[2, 2000], q] (* Amiram Eldar, Oct 30 2021 *)

PARI

A246277(n) = if(1==n, 0, my(f = factor(n), k = primepi(f[1,1])-1); for (i=1, #f~, f[i,1] = prime(primepi(f[i,1])-k)); factorback(f)/2);
isA348511(n) = (A246277(n)==A246277(1+n+n));