A217414 2-adic valuation of A217413.
71, 79, 44, 70, 28, 53, 67, 80, 29, 21, 1653
Offset: 1
Examples
A217413(1)=(2^71)*(11^5), so this sequence's first term is 71.
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A217413(1)=(2^71)*(11^5), so this sequence's first term is 71.
A217413(1) = (2^71) * (11^5), so this sequence's first term is 5.
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
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.
A217426(1)=13 and A217427(1)=20, so this sequence's first term is 5^13 * 7^20. It has two copies each of the digits 1, 3, 4, 5, 7 and 9; three each of 0's, 6's and 8's; and five 2's.
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 *)
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 *)
3^29 * 11^22 has two each of 2's, 7's and 9's; three each of 3's and 6's; and five each of 0's, 1's, 4's, 5's and 8's. No smaller number with only 3 and 11 as prime factors has a prime number of each digit 0-9, so a(1) is this value.
{
k=33;a=[99];t=1;while(1,
v=vectorsmall(10);m=k;
while(m,d=m%10;m\=10;v[d+1];next());
f=1;for(i=1,10,if(isprime(v[i])==0,f=0;break()));
if(f,F=factor(k);print1(3"^"F[1,2]"*"11"^"F[2,2]"="k"\n"));
if(11^(t+1)
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