login
A328146
Numbers k such that binomial(d(i),d(j)) = k for some proper divisors d(i) > d(j) of k.
1
10, 36, 78, 120, 136, 165, 210, 300, 406, 462, 528, 666, 820, 924, 969, 990, 1140, 1176, 1378, 1596, 1716, 1830, 2080, 2346, 2380, 2628, 2926, 3060, 3240, 3276, 3570, 3654, 3876, 3916, 4278, 4656, 5050, 5460, 5886, 6328, 6786, 7260, 7750, 7770, 8256, 8436
OFFSET
1,1
COMMENTS
The corresponding pairs of divisors (d(i), d(j)) are (5, 2), (9, 2), (13, 2), (10, 3), (17, 2), (11, 3), {(21, 2) and (10, 6)}, (25, 2), (29, 2), (11, 6), ...
We observe that the term 210 of the sequence generates two pairs of divisors having the property that binomial(d(i),d(j)) = 210 (see the example).
LINKS
EXAMPLE
10 is in the sequence because the proper divisors of 10 are {2, 5} and binomial(5, 2) = 10.
210 is in the sequence because the proper divisors of 210 are { 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105} with binomial(21, 2) = 210 and binomial(10, 6) = 210 where 2, 6, 10, 21 are divisors of 210.
MAPLE
with(numtheory):
for n from 3 to 10000 do:
ii:=0:d:=divisors(n):n0:=nops(d):
for i from n0-1 by -1 to 2 while (ii=0) do:
for j from i-1 by -1 to 2 while(ii=0) do:
f:=binomial(d[i], d[j]):
if f=n
then
ii:=1:printf(`%d, `, n):
else
fi:
od:
od:
od:
# Alternative:
N:= 10^6: # for all terms <= N
S:= {}:
for n from 2 while (n+1)*(n+2)/2 <= N do
for m from n+2 do
t:= binomial(m, n);
if t > N then break fi;
if t mod n = 0 and t mod m = 0 and t > m then S:= S union {t} fi
od;
od:
sort(convert(S, list)); # Robert Israel, Oct 06 2019
PROG
(PARI) isok(m) = {my(d=divisors(m)); for (i=1, #d-1, for (j=1, i-1, if (binomial(d[i], d[j]) == m, return(1)); ); ); } \\ Michel Marcus, Oct 05 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Oct 05 2019
STATUS
approved