cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A116852 Number of partitions of n-th semiprime into 2 squares.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 0, 0, 2, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 1, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 2, 0, 1, 0, 2, 0, 1, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 1, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 1
Offset: 1

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Author

Jonathan Vos Post, Mar 15 2006

Keywords

Comments

See also A000161 Number of partitions of n into 2 squares (when order does not matter and zero is allowed).
From Robert Israel, Jun 10 2020: (Start)
a(1)=1 if A001358(n) = p^2 where p is not in A002144.
a(n)=1 if A001358(n) = 2*p where p is in A002144.
a(n)=2 if A001358(n) = p*q where p and q are in A002144 (not necessarily distinct).
a(n)=0 otherwise. (End)

Examples

			a(1) = 1 because semiprime(1) = 4 = 0^2 + 2^2, the unique sum of squares.
a(2) = 0 because semiprime(2) = 6 has no decomposition into sum of 2 squares because it has a prime factor p == 3 (mod 4) with an odd power.
a(3) = 1 because semiprime(3) = 9 = 0^2 + 3^2, the unique sum of squares.
a(4) = 1 because semiprime(4) = 10 = 2*5 = 1^2 + 3^2.
a(9) = 2 because semiprime(9) = 25 = 0^2 + 5^2 = 3^2 + 4^2, two distinct ways.
a(23) = 2 because semiprime(23) = 65 = 5*13 = 1^2 + 8^2 = 4^2 + 7^2.
a(28) = 2 because semiprime(28) = 85 = 5*17 = 2^2 + 9^2 = 6^2 + 7^2.
a(49) = 2 because semiprime(49) = 145 = 5*29 = 1^2 + 12^2 = 8^2 + 9^2.
a(56) = 2 because semiprime(56) = 169 = 0^2 + 13^2 = 5^2 + 12^2.
a(60) = 2 because semiprime(60) = 185 = 5*37 = 4^2 + 13^2 = 8^2 + 11^2.
		

Crossrefs

Programs

  • Maple
    R:= NULL: count:= 0:
    for n from 4 while count < 100 do
      if numtheory:-bigomega(n) = 2 then
        count:= count+1;
        F:= ifactors(n)[2];
        if nops(F) = 1 then
          if F[1][1] mod 4 = 1 then v:= 2
          else v:= 1
          fi
        elif F[1][1]=2 and F[2][1] mod 4 = 1 then v:= 1
        elif F[1][1] mod 4 = 1 and F[2][1] mod 4 = 1 then v:= 2
        else v:= 0
        fi;
        R:= R, v;
      fi
    od:
    R; # Robert Israel, Jun 10 2020

Formula

a(n) = A000161(A001358(n)).

Extensions

More terms from Giovanni Resta, Jun 15 2016