

A093497


Beginning with 6, numbers such that every partial concatenation has the same prime signature p*q.


0



6, 2, 2, 7, 7, 1, 1, 8, 7, 3, 3, 9, 1, 3, 13, 3, 7, 3, 11, 3, 7, 3, 3, 15, 9, 9, 17, 3, 102, 3, 22, 7, 7, 51, 3, 7, 26, 43, 9, 39, 8, 1, 29, 7, 39, 53, 7, 59, 81, 39, 41, 1, 11, 47, 23, 39, 13, 13, 89, 15, 101, 7, 17, 21, 7, 3, 81, 51
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

a(69) through a(85) are conjectured to be 23, 31, 23, 1, 3, 3, 39, 26, 3, 1, 53, 1, 1, 17, 51, 53, 7 but this is unproved. The concatenation of a(1) through a(69) is a 102digit composite number. I put this number in the factorization tool at the Alpertron link and found no factors in the first 20000 curves, so it is unlikely to have more than two. All the other concatenations have been factored.  David Wasserman, Jun 07 2007


LINKS

Table of n, a(n) for n=1..68.
Dario Alpern, Factorization using the Elliptic Curve Method


EXAMPLE

62, 622, 6227, 62277, etc. have the same prime signature p*q.


PROG

(PARI) digs(k) = if (k, digits(k), [0]);
isok(vd, k) = {newv = concat(vd, digs(k)); val = fromdigits(newv); (bigomega(val) == 2) && (omega(val) == 2); }
fnew(vd) = {k = 0; while (! isok(vd, k), k++); k; }
lista(nn) = {k = 6; v = [k]; vd = digs(k); print1(v[1], ", "); for (n=2, nn, k = fnew(vd); print1(k, ", "); v = concat(v, k); vd = concat(vd, digs(k)); ); } \\ Michel Marcus, Jan 21 2017


CROSSREFS

Cf. A006881.
Sequence in context: A259838 A256576 A201674 * A092138 A138995 A010133
Adjacent sequences: A093494 A093495 A093496 * A093498 A093499 A093500


KEYWORD

base,nonn,more


AUTHOR

Amarnath Murthy, Apr 17 2004


EXTENSIONS

More terms from David Wasserman, Jun 07 2007
Edited and a(69)a(85) moved to comments by Michel Marcus, Jan 21 2017


STATUS

approved



