A260143 Runs of consecutive integers with same prime signature.

2, 3, 14, 15, 21, 22, 33, 34, 35, 38, 39, 44, 45, 57, 58, 75, 76, 85, 86, 87, 93, 94, 95, 98, 99, 116, 117, 118, 119, 122, 123, 133, 134, 135, 136, 141, 142, 143, 145, 146, 147, 148, 158, 159, 171, 172, 177, 178, 201, 202, 203, 205, 206, 213, 214, 215, 217, 218, 219, 230, 231, 244, 245

Offset

1,1

Comments

This sequence is infinite, see A189982 and Theorem 4 in Goldston-Graham-Pintz-Yıldırım. - Charles R Greathouse IV, Jul 17 2015

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

D. A. Goldston, S. W. Graham, J. Pintz, and C. Y. Yıldırım, Small gaps between almost primes, the parity problem, and some conjectures of Erdos on consecutive integers, arXiv:0803.2636 [math.NT], 2008.

MathOverflow, Question on consecutive integers with similar prime factorizations

Eric Weisstein's World of Mathematics, Prime Signature

OEIS Wiki, Prime signatures

Example

Runs begin:
(terms)         (prime signature)
{2, 3},         [1]
{14, 15},       [1,1]
{21, 22},       [1,1]
{33, 34, 35},   [1,1]
{38, 39},       [1,1]
{44, 45},       [1,2]
{57, 58},       [1,1]
{75, 76},       [1,2]
{85, 86, 87},   [1,1]
{93, 94, 95},   [1,1]
{98, 99},       [1,2]
...

Mathematica

Split[Range[2, 250], Sort[FactorInteger[#1][[All, 2]]] === Sort[FactorInteger[#2][[All, 2]]]&] // Select[#, Length[#] > 1&]& // Flatten

Prog

(PARI) is(n)=my(f=vecsort(factor(n)[, 2])); f==vecsort(factor(n-1)[, 2]) || f==vecsort(factor(n+1)[, 2]) \\ Charles R Greathouse IV, Jul 17 2015
(Python)
from sympy import factorint
def aupto(limit):
    aset, prevsig = {2}, [1]
    for k in range(3, limit+2):
        sig = sorted(factorint(k).values())
        if sig == prevsig: aset.update([k - 1, k])
        prevsig = sig
    return sorted(aset)
print(aupto(250)) # Michael S. Branicky, Sep 20 2021

Crossrefs

Main sequence is A052213.

Sequence in context: A032806 A353870 A225756 * A353871 A275303 A281500

Adjacent sequences: A260140 A260141 A260142 * A260144 A260145 A260146

Keyword

nonn,easy,tabf

Author

Jean-François Alcover, Jul 17 2015

Status

approved