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A391246
Number of natural number pairs (x, y) such that x >= y, x*y = n and n = A048720(x, A065621(y)).
7
1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 3, 1, 3, 1, 3, 2, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 2, 5, 1, 2, 1, 3, 1, 4, 1, 3, 2, 2, 1, 3, 2, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 3, 1, 2, 1, 4, 1, 2, 1, 3, 1, 3, 1, 6, 1, 2, 2, 3, 1, 2, 1, 4, 1, 2, 1, 5, 1, 2, 1, 4, 1, 4, 1, 3, 2, 2, 1, 4, 1, 3, 1, 3, 1, 2, 1, 4, 2
OFFSET
1,4
COMMENTS
a(n) = number of times n occurs in such a position of A075362 that also A391244 has n in the same position.
FORMULA
a(n) = Sum_{d|n, d <= (n/d)} [n = A277199(n/d, d)], where [ ] is the Iverson bracket.
For all n >= 1, a(n) <= A038548(n).
EXAMPLE
Let A(n, k) = A277199(n, k), then A(36,1) = A(18,2) = A(12,3) = A(9,4) = A(6,6) = 36, therefore a(36) = 5.
Similarly, A(45,1) = A(15,3) = 45, but on the other hand A(9,5) = 101 <> 45, thus a(45) = 2, different from A038548(45) = 3.
PROG
(PARI)
A048720(b, c) = fromdigits(Vec(Pol(binary(b))*Pol(binary(c)))%2, 2);
A065621(n) = bitxor(n-1, n+n-1);
A391245_tr(x, y) = if(x*y == A048720(x, A065621(y)), x*y, 0);
A391246(n) = sumdiv(n, d, ((d <= (n/d)) && n==A391245_tr(n/d, d)));
CROSSREFS
Number of times n occurs in triangular table A391245.
Cf. A038548, A075362, A391244, A391247 (positions of 1's), A391248.
Cf. also A358235, A391239 for sequences with a similar theme.
Sequence in context: A183025 A072410 A342824 * A344713 A072491 A051034
KEYWORD
nonn
AUTHOR
Antti Karttunen, Dec 06 2025
STATUS
approved