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A225135
Squares that are a concatenation of primes.
2
25, 225, 289, 361, 529, 729, 2025, 2401, 2601, 2809, 3025, 4761, 5041, 5329, 5929, 7225, 7569, 11025, 11449, 11881, 13225, 15129, 19881, 20449, 21609, 22801, 23409, 24649, 25281, 26569, 27225, 29241, 29929, 31329, 32041, 32761, 34969, 36481, 39601, 47089
OFFSET
1,1
COMMENTS
Lim inf a(n)/n^2 >= 2. Is it finite? - Charles R Greathouse IV, Apr 30 2013
EXAMPLE
25 = 5^2 and can be separated into two prime numbers: 2|5.
231361 = 481^2 and can be separated into prime numbers in six ways: 2|3|1361, 2|3|13|61, 2|31|3|61, 2|313|61, 23|1361, and 23|13|61.
Leading zeros are allowed: 2025 = 2|02|5.
MATHEMATICA
r[d_] := Catch@ Block[{z = Length@d}, z<1 || Do[ If[ PrimeQ@ FromDigits@ Take[d, i] && r@ Take[d, i-z], Throw@ True], {i, z}]]; Select[ Range[1000]^2, r@ IntegerDigits@ # &] (* Giovanni Resta, Apr 30 2013 *)
PROG
(R) library(gmp); isprime2=function(x) isprime(x)>0
splithasproperty<-function(n, FUN, curdig=1, res=list(), curspl=c()) {
no0<-function(s){ while(substr(s, 1, 1)=="0" & nchar(s)>1) s=substr(s, 2, nchar(s)); s}
s=as.character(n)
if(curdig>nchar(s)) return(res)
if(length(curspl)>0) if(FUN(as.bigz(no0(substr(s, curdig, nchar(s)))))) res[[length(res)+1]]=curspl
for(i in curdig:nchar(s))
if(FUN(as.bigz(no0(substr(s, curdig, i)))))
res=splithasproperty(n, FUN, i+1, res, c(curspl, i))
res
}
which(sapply(1:100, function(x) length(splithasproperty(x^2, isprime2)))>0)^2
(PARI) has(n)=if(isprime(n), return(1)); if(n<202, return(isprime(n%10) && isprime(n\10))); my(k=n%10, v); if(k==5||k==2, return(if(n<6, 1, n\=10; has(n/10^valuation(n, 10))))); if(k%2==0, return(0)); v=digits(n); for(i=1, #v, if(isprime(n%10^i) && has(n\10^i), return(1))); 0
forstep(n=5, 1e3, 2, if(has(n^2), print1(n^2", ")))
\\ Charles R Greathouse IV, Apr 30 2013
CROSSREFS
KEYWORD
base,nonn
STATUS
approved