A146567 -id:A146567 - cp's OEIS frontend

A152492 a(n) = number of integers of the form (n*k)^2/(k^2 - n^2).

0, 0, 1, 1, 0, 2, 0, 1, 1, 1, 0, 8, 0, 0, 4, 1, 0, 2, 0, 4, 3, 0, 0, 9, 0, 0, 1, 2, 0, 7, 0, 1, 2, 0, 1, 8, 0, 0, 1, 4, 0, 5, 0, 1, 5, 0, 0, 9, 0, 1, 1, 1, 0, 2, 1, 4, 1, 0, 0, 23, 0, 0, 3, 1, 1, 4, 0, 1, 1, 2, 0, 10, 0, 0, 4, 1, 0, 4, 0, 4, 1, 0, 0, 17, 0, 0, 1, 1, 0, 8

Offset: 1

Ctibor O. Zizka, Dec 06 2008

k needs to be checked only up through n^2+1 since beyond this n^2 < (n*k)^2/(k^2 - n^2) < n^2 + 1 and thus can't be an integer. - Micah Manary, Aug 27 2022

Micah Manary, Table of n, a(n) for n = 1..1000

Cf. A071086, A146567, A146566, A146564.

a(n) = sum(k=1, n^2+1, if (k!=n, denominator((n*k)^2/(k^2 - n^2))==1)); \\ Michel Marcus, Oct 28 2022

More terms from Micah Manary, Aug 07 2022

A152491 Numbers n such that 1/c = 1/n + 1/S(n). c, n positive integers (A000027(n)), S(n) sum of digits of n (A007953(n)).

Original entry on oeis.org

2, 4, 6, 8, 18, 72

Offset: 1

Ctibor O. Zizka, Dec 06 2008

A000027(n)*A007953(n)/(A000027(n)+A007953(n))= c, c positive integer.
No further term < 10000000. - Michel Marcus, Jun 02 2013
For a given n let x be the minimal natural number such that n*x/(n+x)=c. I conjecture: from a certain n onward, x>S(n) for all n. Thus there is no other solution bigger than 72, and this sequence is finite. - Ctibor O. Zizka, Sep 13 2015
Sequence is complete. To prove it, let s denote the sum of digits of n and observe that 1/c - 1/s = 1/n is equivalent to (s-c)/(c*s) = 1/n. Hence we must have s > c and c*s >= n, otherwise the denominators cannot match. But if n is greater than, say, 1000, it is easy to see that s^2 < n and this implies c*s < n, since c < s. - Giovanni Resta, Sep 13 2015

Cf. A007953, A000027, A146567, A146564, A146747,

lista(nn) = {for (n=1, nn, digs = Vec(Str(n)); sn = sum(i=1, #digs, eval(digs[i])); if (type(1/(1/n+1/sn)) == "t_INT", print1(n, ", ")););} \\ Michel Marcus, Jun 02 2013

cp's OEIS Frontend

A152492 a(n) = number of integers of the form (n*k)^2/(k^2 - n^2).

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