A035232 Number of substrings of n which are primes (counted with multiplicity).

0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 2, 0, 1, 0, 2, 0, 1, 1, 1, 2, 3, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 3, 1, 1, 0, 1, 1, 2, 0, 1, 0, 2, 0, 0, 1, 1, 2, 3, 1, 2, 1, 2, 1, 2, 0, 1, 1, 1, 0, 1, 0, 2, 0, 0, 1, 2, 2, 3, 1, 2, 1, 2, 1, 2, 0, 0, 1, 2, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1, 0, 2, 0, 0, 0, 1, 1, 2, 0, 1

Offset: 1

Author

Erich Friedman

Keywords

• base
• easy
• nonn

Comments

No leading 0's allowed in substrings.

Examples

			The primes occurring as substrings of 37 are 3, 7, 37, so a(37) = 3.
a(22) = 2, since the prime 2 occurs twice as a substring.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A039997, A039995.

Programs

Maple

a:= n-> (s-> nops(select(t -> t[1]<>"0" and isprime(parse(t)),
        [seq(seq(s[i..j], i=1..j), j=1..length(s))])))(""||n):
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 07 2022

Mathematica

a[n_] := Block[{s = IntegerDigits[n], c = 0, d = {}}, l = Length[s]; t = Flatten[ Table[ Take[s, {i, j}], {i, 1, l}, {j, i, l}], 1]; k = l(l + 1)/2; While[k > 0, If[ t[[k]][[1]] != 0, d = Append[d, FromDigits[ t[[k]] ]]]; k-- ]; Count[ PrimeQ[d], True]]; Table[ a[n], {n, 1, 105}]

Python

from sympy import isprime
def a(n):
    s = str(n)
    ss = (s[i:j] for i in range(len(s)) for j in range(i+1, len(s)+1))
    return sum(1 for w in ss if w[0] != "0" and isprime(int(w)))
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Aug 07 2022

Formula

Trivial upper bound: a(n) <= binomial(floor(log(n)/log(10)+2), 2) ~ k*log^2 n with k = 0.09430584850580... = 1/log(10)^2/2. - Charles R Greathouse IV, Nov 15 2022

Extensions

Edited by Robert G. Wilson v, Feb 24 2003