

A242462


a(1) = 10; for n>1, a(n) = (a(n1)1) * (smallest odd prime factor of a(n1)) + 1.


0



10, 46, 1036, 7246, 26248636, 11628145306, 461742021916246, 7849614372576166, 44750651538056716666, 17139499539075722482696, 188534494929832947309646, 69192159639248691662639716, 2144956948816709441541831166, 13721289601580490297543093962506
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OFFSET

1,1


COMMENTS

Note: this sequence will terminate if a power of 2 occurs.
Conjecture: this sequence is infinite (so it contains no powers of 2). The similarly defined sequence starting with a(1) = 6 terminates after 2 terms: 6 and 16.
Conjecture is true if it turns out that one number in this sequence is one more than a multiple of 41. Any number of the form 2^n1 that is divisible by 41 is also divisible by 25, which is the square of a prime number greater than 3. Because numbers one less than this sequence's terms are always 9 times a squarefree number, this proves that if a number one less than this sequence is divisible by 41, then this sequence is infinite.  J. Lowell, Jul 17 2017
Conjecture is also true if it turns out that one number in this sequence is a multiple of either 19 or 73. Any number of the form 2^n1 that is divisible by 19 is divisible by 27, and any number of the form 2^n1 that is divisible by both 73 and 9 is divisible by 27. Numbers one less than this sequence are always divisible by 9 but not by 27.  J. Lowell, Feb 27 2022


LINKS



EXAMPLE

a(1) = 10; 101 = 9; 9*5 (smallest odd prime factor of 10) is 45; 45+1=46, so a(2) = 46.


MATHEMATICA

NestList[DeleteCases[FactorInteger[#], w_ /; First@ w == 2][[1, 1]] (#  1) + 1 &, 10, 13] (* Michael De Vlieger, Jul 18 2017 *)


PROG

(PARI) opf(n) = n = n>>valuation(n, 2); if (n==1, 1, factor(n)[1, 1]);
lista(nn) = {a = 10; for (n=2, nn, olda = a; print1(a, ", "); a = 1 + (olda1)*opf(olda); ); } \\ Michel Marcus, May 17 2014


CROSSREFS



KEYWORD

nonn


AUTHOR



EXTENSIONS



STATUS

approved



