

A175280


Base9 pandigital primes: primes having at least one of each digit 0,...,8 when written in base 9.


10



393474749, 393474821, 393475373, 393481069, 393486901, 393488437, 393492797, 393494477, 393499429, 393499517, 393500741, 393528029, 393528517, 393538157, 393541693, 393544709, 393545861, 393546149, 393551189, 393551357, 393552629
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OFFSET

1,1


COMMENTS

Terms in this sequence have at least 10 digits in base 9, i.e., are larger than 9^9, since sum(d_i 9^i) = sum(d_i) (mod 8), and 0+1+2+3+4+5+6+7+8 is divisible by 4. So there must be at least one repeated digit, which may not be even, else the resulting number is even. The smallest terms are therefore of the form "10123...." in base 9, where "...." is a permutation of "45678", cf. examples.


LINKS



EXAMPLE

The first terms of this sequence, i.e., smallest base9 pandigital primes, are "1012346785", "1012346875", "1012347658", "1012356487", "1012365487", "1012367584", "1012374568", "1012376845", "1012384657", ... (written in base 9).


MATHEMATICA

Select[Range[4*10^8], Min @ DigitCount[#, 9] > 0 && PrimeQ[#] &] (* Amiram Eldar, Apr 13 2021 *)


PROG

(PARI) pdp( b=9/*base*/, c=99/* # of terms to produce */) = { my(t, a=[], bp=vector(b, i, b^(bi))~, offset=b*(b^b1)/(b1)); for( i=1, b1, offset+=b^b; for( j=0, b!1, isprime(t=offsetnumtoperm(b, j)*bp)  next; #(a=concat(a, t))<c  return(vecsort(a))))} /* NOTE: Due to the implementation of numtoperm, the returned list may be incomplete towards its end. Thus computation of more than the required # of terms is recommended. [The initial digits of the base9 expansion of the terms allow one to know up to where it is complete.] One may use a construct of the form: vecextract(pdp(9, 199), "1..20")) */


CROSSREFS

Cf. A050288, A138837, A175271, A175272, A175273, A175274, A175275, A175276, A175277, A175278, A175279.


KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



