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Partial sums of (base 10) Pandigital primes. Note that almost all primes are pandigital. a(59) is (after the first value) the first prime in this sequence. What is the smallest pandigital prime partial sum of (base 10) pandigital primes? In other bases?

This is a finite sequence with 9*9! = 3265920 terms: a(9*9!) = 9876543210.

A171102 is the infinite version, where each digit must appear at least once.

More precisely, this is exactly the subset of the first 9*9! terms of A171102. - M. F. Hasler, Jan 05 2020

Subsequence of A134336 and of A178403; A178401(a(n)) = 1. - Reinhard Zumkeller, May 27 2010

Smallest prime factors: A178775(n) = A020639(a(n)). - Reinhard Zumkeller, Jun 11 2010

A178788(a(n)) = 1. - Reinhard Zumkeller, Jun 30 2010

All these numbers are composite because the sum of the digits, 45, is divisible by 9. - T. D. Noe, Nov 09 2011

This is the 10th row of the array T(k,n) = n-th number in which the number of distinct base-10 digits is k. A031969 is the 4th row. A220063 is the 5th row. A220076 is the 6th row. A218019 is the 7th row. A219743 is the 8th row. - Jonathan Vos Post, Dec 05 2012

From Hieronymus Fischer, Feb 13 2013: (Start)

The sum of all terms is 9!*49444444440 = 17942399998387200.

General formula for the sum of all terms of the finite sequence of the corresponding base-p pandigital numbers with p places: sum = ((p^2 - p - 1)*(p^p - 1) + p - 1)*(p-2)!/2.

General formula for the sum of all terms (interpreted as decimal permutational numbers with exactly d+1 different digits from the range 0..d < 10): sum = (d+1)!*((10d - 1)*10^d - d + 1)/18, d > 1.

(End)

This is the infinite version. See A050278 for the finite version.

The first 9*9!=3265920 terms of this sequence are permutations of the digits 0-9 with a(9*9!)=9876543210 (see Version 1, A050278). - Jeremy Gardiner, May 29 2010

Subsequence of A134336 and of A178403; A178401(a(n))>0. - Reinhard Zumkeller, May 27 2010

Smallest prime factors: A178775(n) = A020639(a(n)). - Reinhard Zumkeller, Jun 11 2010

A178788(a(n)) = 1, for n <= 9*9!, else A178788(a(n)) = 0. - Reinhard Zumkeller, Jun 30 2010 [corrected by Hieronymus Fischer, Feb 02 2013]

A230959(a(n)) = 0. - Reinhard Zumkeller, Nov 02 2013

The first term of the sequence absent in A050278 is a(3265921) = 10123456789. Also, the first prime is a(3306373) = 10123457689 = A050288(1). - Zak Seidov, Sep 23 2015

Almost all numbers are in this sequence, in the sense that it has asymptotic density equal to 1. Indeed, the fraction of n-digit numbers which don't have a given digit d is roughly 0.9^n (not exactly because the first digit is chosen among {1..9}) which tends to zero as n -> oo. - M. F. Hasler, Jan 05 2020

Base-8 pandigital primes must have at least 9 octal digits, since sum(d_i 8^i) = sum(d_i) (mod 7), and 0+1+...+6+7 is divisible by 7. So the smallest ones should be of the form "10123...." in base 8, where "...." is a permutation of "4567". By chance, the identical permutation already yields a prime: a(1)="101234567" in base-8.

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.

Terms in this sequence have at least 8 digits in base 7, i.e., are larger than 7^7, since sum(d_i 7^i) = sum(d_i) (mod 6), and 0+1+2+3+4+5+6 is divisible by 3. So there must be at least one repeated digit, which may not be 0 nor 6 neither odd (else the resulting number is even). The smallest terms are therefore of the form "1022...." in base 7, where "...." is a permutation of "3456", cf. examples.

These numbers need to have at least 13 digits in base 12 since any permutation of the digits 0,...,9,A,B will result in a number divisible by 11. For the same reason, it must be digit different from 0 which is repeated. Thus the smallest terms in this sequence are written "10123456....." in base 12, where ..... is a permutation of {7,8,9,A,B}.

Note: Due to the implementation of numtoperm(), the PARI script will not necessarily print the terms in the correct order. In some cases, more than the desired number of terms have to be calculated, and vecsort() to be used to get the correct sequence. - M. F. Hasler, May 27 2010

Base-16 (a.k.a. hexadecimal, sexadecimal, senidenary or hexadecadic) pandigital primes must have at least 17 hexadecimal digits (i.e. they are larger than 16^16 = 2^64 > 10^19), since sum(d_i 16^i) = sum(d_i) (mod 15), and 0+1+...+14+15 is divisible by 15. So the smallest ones should be of the form "101234567...." in base 16, where "...." is a permutation of "89ABCDEF".

The same reasoning shows that numbers of this form ("1012...") are congruent to 1 modulo 15 and thus modulo 30 (since also = 1 [mod 2]). This explains that all terms < 2*16^16 end in the (decimal!) digit 1.

a(n) == 1 (mod 30) for a(n) < 2^65 = 3.69*10^19.

Base-20 pandigital primes must have at least 21 base-20 digits (i.e. they are larger than 20^20 > 10^26), since sum(d_i 20^i) = sum(d_i) (mod 19), and 0+1+...+18+19 is divisible by 19. So the smallest ones should be of the form "10123456789ABCD..." in base 20, where "..." is a permutation of "EFHGIJ" (with A..J representing digits 10..19).