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A349696
Smallest number in a set of three consecutive triangular numbers each with three prime factors (counted with multiplicity).
2
153, 406, 861, 39621, 2166321, 3924201, 11146281, 14804961, 19198306, 73951041, 83417986, 97951006, 209643526, 310415986, 522339681, 526225461, 583333246, 611153241, 801460666, 1601581906, 2520251506, 2690954841, 4455349606, 6681853401, 9895642221, 13878029901
OFFSET
1,1
COMMENTS
153 is the only known number in the sequence which is not squarefree.
From Robert Israel, Mar 11 2025: (Start)
Terms are of the form A000217(k) with either
k prime == 5 (mod 12), k + 1 = 6 * prime, k + 2 prime, k + 3 = 4 * prime
or
k = 4 * prime == 4 (mod 12), k + 1 prime, k + 2 = 6 * prime, k + 3 prime.
In particular, for k > 17 (where k = 17 corresponds to a(1) = 153), the primes mentioned above are > 3, and so the terms except for 153 are squarefree. (End)
LINKS
Shyam Sunder Gupta, Beauty of Number 153, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 15, 399-410.
FORMULA
a(n) = A000217(A255200(n)). - Michel Marcus, Dec 25 2021
EXAMPLE
a(1) = 153 because 153 is the smallest number in the first set of three consecutive triangular numbers with three prime factors (counted with multiplicity), i.e., (153 = 3*3*17, 171 = 3*3*19, 190 = 2*5*19).
MAPLE
R:= NULL: count:= 0:
for i from 1 while count < 100 do
k:= 12*i+4;
if isprime(k+1) and isprime((k+2)/6) and isprime(k+3) then
if isprime(k/4) then R:= R, k*(k+1)/2; count:= count+1; fi;
if isprime((k+4)/4) then R:= R, (k+1)*(k+2)/2; count:= count+1; fi;
fi;
od:
R; # Robert Israel, Mar 11 2025
MATHEMATICA
t[n_] := n*(n + 1)/2; q[n_] := PrimeOmega[n] == 3; Select[Partition[t /@ Range[10^5], 3, 1], AllTrue[#, q] &][[;; , 1]] (* Amiram Eldar, Nov 25 2021 *)
(#(#+1))/2&/@SequencePosition[PrimeOmega[Accumulate[Range[170000]]], {3, 3, 3}][[;; , 1]] (* Harvey P. Dale, Oct 20 2023 *)
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Shyam Sunder Gupta, Nov 25 2021
EXTENSIONS
Definition clarified by Harvey P. Dale, Oct 20 2023
STATUS
approved