A217426 5-adic valuation of A217425.
13, 4, 3, 3, 19, 29, 773
Offset: 1
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A217405(1)=36 and A217406(1)=15, giving this sequence's first value as (2^36)*(7^15). Its decimal representation can be seen to have two each of 0's, 1's, 3's, 5's, 6's, 7's, 8's and 9's; and three each of 2's and 3's (prime number counts of each digit).
N:= 10^100: # to get all terms <= N filter:= proc(n) local L,P,d; L:= convert(n,base,10); P:= Vector(10); for d in L do P[d+1]:= P[d+1]+1 od: andmap(isprime,P); end proc: sort(select(filter, [seq(seq(2^r*7^s, r=0..floor(log[2](N/7^s))),s=0..floor(log[7](N)))])); # Robert Israel, May 08 2017
prDigits(n)=my(d=digits(n),v=vector(10));for(i=1,#d,v[d[i]+1]++);for(i=1,10,if(!isprime(v[i]),return(0))); 1 list(lim)=my(v=List(),t); for(a=0,log(lim+.5)\log(7), t=7^a; while(t<=lim, if(prDigits(t), listput(v,t)); t<<=1)); vecsort(Vec(v)) \\ Charles R Greathouse IV, Sep 19 2013
The first term here is (3^35)*(5^17), corresponding to A217408(1)=35 and A217409(1)=17. Its decimal representation has two each of 0's, 2's, 7's, 8's and 9's; three each of 4's, 5's and 6's; and 5 each of 1's and 3's.
N:= 10^100: # to get all terms <= N filter:= proc(n) local L,P,d; L:= convert(n,base,10); P:= Vector(10); for d in L do P[d+1]:= P[d+1]+1 od: andmap(isprime,P); end proc: sort(select(filter, [seq(seq(3^r*5^s, r=0..floor(log[3](N/5^s))),s=0..floor(log[5](N)))])); # Robert Israel, May 08 2017
prDigits(n)=my(d=digits(n), v=vector(10)); for(i=1, #d, v[d[i]+1]++); for(i=1, 10, if(!isprime(v[i]), return(0))); 1 list(lim)=my(v=List(), t); for(a=0, log(lim+.5)\log(5), t=5^a; while(t<=lim, if(prDigits(t), listput(v, t)); t*=3)); vecsort(Vec(v)) \\ Charles R Greathouse IV, Sep 19 2013
A217411(1)=37 and A217412(1)=10, so this sequence's first term is (3^37)*(7^10). It is the smallest number with exactly 3 and 7 as its prime factors to have decimal representation with each digit 0-9 counted a prime number of times. The digits 0, 2, 3, 5, 6 and 9 occur two times each; 1, 4 and 8 occur three times each; and 7 occurs five times.
A217423(1)=47 and A217424(1)=8, so this sequence's first term is 2^47 * 17^8. It has in its decimal representation two copies each of the digits 0, 2, 3, 4, 6 and 7; and three copies each of 1, 5, 8 and 9.
a(1) = 3^25 * 13^8 (so A217432(1)=25 and A217432(1)=8). Indeed, it contains two copies of each digit other than 9 and three copies of 9. No smaller 21-digit number with this general character -- two copies of all but one digit -- and no 20-digit number with two copies of each digit has form 3^a*13^b with a,b > 0.
nd = 50; mx = 10^nd; pr = Prime@ Range@ PrimePi@ nd; pQ[n_] := Union[DigitCount@n, pr] == pr; Sort@ Select[ Flatten@ Table[3^p*13^q, {p, Log[3, mx/13]}, {q, Log[13, mx/3^p]}], pQ] (* terms < 10^50, Giovanni Resta, Jan 16 2014 *)
A217414(1)=71 and A217415(1)=5, so this sequence's first term is 2^71 * 11^5. It has two each of 0's, 3's, 4's, 6's, 8's and 9's; three each of 1's, 5's and 7's; and five 2's.
A217417(1)=10 and A217418(1)=20, so this sequence's first term can be simply written as 338^10 (338=2*13^2). It has two each of 0's, 1's, 5's, 6's, 7's and 8's; three each of 3's, 4's and 9's; and five 2's.
2^55 * 19^8 has two copies of each of the digits 0, 1, 2, 7 and 9; three copies of each of 3, 4, 5 and 8; and five copies of the digit 6. All smaller numbers of the required type have at least a digit counted 0, 1, or a composite number of times, so this is a(1).
nd = 50; mx = 10^nd; pr = Prime@Range@PrimePi@nd; pQ[n_] := Union[DigitCount@n, pr] == pr; Sort@ Select[Flatten@ Table[2^p 19^q, {p, Log[2, mx/19]}, {q, Log[19, mx/2^p]}], pQ] (* Giovanni Resta, Jan 16 2014 *)
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