A080075 Proth numbers: of the form k*2^m + 1 for k odd, m >= 1 and 2^m > k.

3, 5, 9, 13, 17, 25, 33, 41, 49, 57, 65, 81, 97, 113, 129, 145, 161, 177, 193, 209, 225, 241, 257, 289, 321, 353, 385, 417, 449, 481, 513, 545, 577, 609, 641, 673, 705, 737, 769, 801, 833, 865, 897, 929, 961, 993, 1025, 1089, 1153, 1217, 1281, 1345, 1409

Offset: 1

Author

Eric W. Weisstein, Jan 24 2003

Keywords

• nonn

Comments

A Proth number is a square iff it is of the form (2^(m-1)+-1)*2^(m+1)+1 = 4^m+-2^(m+1)+1 = (2^m+-1)^2 for m > 1. See A086341. - Thomas Ordowski, Apr 22 2019

Links

• Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
• Bertalan Borsos, Attila Kovács and Norbert Tihanyi, Tight upper and lower bounds for the reciprocal sum of Proth primes, The Ramanujan Journal (2022).
• Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 38.
• Amelia Carolina Sparavigna, Discussion of the groupoid of Proth numbers (OEIS A080075), Politecnico di Torino, Italy (2019).
• Eric Weisstein's World of Mathematics, Proth Number.
• Wikipedia, Proth number.

Crossrefs

Cf. A080076, A086341, A112714, A116882, A157892, A157893.

Programs

Mathematica

Select[Range[3, 1500, 2], And[OddQ[#[[1]] ], #[[-1]] >= 1, 2^#[[-1]] > #[[1]] ] &@ Append[QuotientRemainder[#1, 2^#2], #2] & @@ {#, IntegerExponent[#, 2]} &[# - 1] &] (* Michael De Vlieger, Nov 04 2019 *)

PARI

is_A080075 = isproth(x)={!bittest(x--,0) && (x>>valuation(x+!x,2))^2 < x } \\ M. F. Hasler, Aug 16 2010; edited by Michel Marcus, Apr 23 2019, M. F. Hasler, Jul 07 2022

PARI

next_A080075(N)=N+2^(exponent(N)\2+1)
A080075_first(N)=vector(N,i,if(i>1,next_A080075(N),3)) \\ M. F. Hasler, Jul 07 2022

Python

from itertools import count, islice
def A080075_gen(startvalue=3): # generator of terms >= startvalue
    return filter(lambda n:(n-1&-n+1)**2+1>=n,count(max(startvalue,3)))
A080075_list = list(islice(A080075_gen(),30)) # Chai Wah Wu, Oct 06 2024

Formula

a(n) = A116882(n+1)+1. - Klaus Brockhaus, Georgi Guninski and M. F. Hasler, Aug 16 2010

a(n) = A157892(n)*2^A157893(n) + 1. - M. F. Hasler, Aug 16 2010

a(n) ~ n^2/2. - Thomas Ordowski, Oct 19 2014

Sum_{n>=1} 1/a(n) = 1.09332245643583252894473574405304699874426408312553... (Borsos et al., 2022). - Amiram Eldar, Jan 29 2022

a(n+1) = a(n) + 2^round(L(n)/2), where L(n) is the number of binary digits of a(n); equivalently, floor(log_2(a(n))/2 + 1) in the exponent. [Lemma 2.2 in Borsos et al.] - M. F. Hasler, Jul 07 2022