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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 102-digit 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

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Last modified December 12 22:06 EST 2019. Contains 329963 sequences. (Running on oeis4.)