OFFSET
1,1
COMMENTS
If p >= 3 is prime, then A176948(p) = p. The sequence lists composite numbers with this property.
It is interesting that there is a large overlap with terms in A140164 (but there are exceptions, e.g., 77).
From Daniel Forgues, Jul 15 2016: (Start)
Composite numbers n which are not of form (k/2)*[(m-2)*k - (m-4)] for any m >= 3 and k >= 3, thus not m-gonal numbers for any order k >= 3.
An m-gonal number, m >= 3, i.e., of the form n = (k/2)*[(m-2)*k - (m-4)], yields a nontrivial factorization of n if and only if k >= 3. (End)
Since we are looking for solutions of (m-2)*k^2 - (m-4)*k - 2*n = 0,
with m >= 3 and k >= 3, the largest order k we need to consider is
k = {(m-4) + sqrt[(m-4)^2 + 8*(m-2)*n]}/[2*(m-2)] with m = 3, thus
k <= (1/2)*{-1 + sqrt[1 + 8*n]}.
Or, since we are looking for solutions of 2n = m*k*(k-1) - 2*k*(k-2),
with m >= 3 and k >= 3, the largest m we need to consider is
m = [2n + 2*k*(k-2)]/[k*(k-1)] with order k = 3, thus m <= (n+3)/3.
Composite numbers n which are divisible by 3 are m-gonal numbers of order 3, with m = (n + 3)/3. Thus all a(n) are coprime to 3.
Odd composite numbers n which are divisible by 5 are m-gonal numbers of order 5, with m = (n + 15)/10. Thus all odd a(n) are coprime to 5.
a(1) = 4 is the only square number: 4-gonal with order k = 2. (End)
An integer n which is congruent to k (mod t_{k-1}) with 3 <= t_{k-1} < n, i.e. n = j * t_{k-1} + k with k >= 3 and j >= 1, is an m-gonal number of order k, with m = j + 2, where t_{k-1} is a triangular number. If all the congruence tests fail, a composite n belongs to this sequence. - Daniel Forgues, Aug 02 2016
From Jonathan Dushoff, Apr 05 2022: (Start)
All numbers n>2 are trivially n-gonal numbers, and will thus have A176948(n)=n unless they have a nontrivial polygonal decomposition. Thus this is just the sequence of non-polygonal composite numbers.
Note that the 2nd through 13th terms are in arithmetic progression.
Some reasons: many of the smaller odd numbers are prime (and thus don't appear); numbers of the form 6x (or 6x+3) are always order-3 numbers; numbers of the form 6x+4 are always order-4 numbers; small odd composites not divisible by 3 are usually divisible by 5, and are thus order-5 numbers.
In fact, the first number to break the arithmetic progression is the first product of distinct primes > 5.
Conversely, 6x+2 numbers cannot be order-3 or -6 numbers (those are divisible by 3); order-4 numbers (2 is not a square (mod 6)); order-5 numbers (all odd); or order-7 numbers (all == 1 (mod 3)).
The first 6x+2 composite not in the list is order-8 pentagonal number 92.
(End)
LINKS
Giovanni Resta, Table of n, a(n) for n = 1..10000
OEIS Wiki, Polygonal numbers
EXAMPLE
8 is in the sequence since it is composite and is an octagonal number, but not a heptagonal number, hexagonal number, pentagonal number, etc. 10 is not in the sequence because even though it is composite and a decagonal number, it is also a triangular number: 10 = 1 + 2 + 3 + 4. - Michael B. Porter, Jul 16 2016
MATHEMATICA
Select[Range[302], CompositeQ@ # && FindInstance[n*(4 + n*(s-2) - s)/2 == # && s >= 3 && n >= 3, {s, n}, Integers] == {} &] (* Giovanni Resta, Jul 13 2016 *)
PROG
(Sage)
def is_a(n):
if is_prime(n): return False
for m in (3..(n+3)//3):
if pari('ispolygonal')(n, m):
return False
return True
print([n for n in (3..302) if is_a(n)]) # Peter Luschny, Jul 28 2016
(PARI) listc(nn) = {forcomposite(c=1, nn, sp = c; forstep(k=c, 3, -1, if (ispolygonal(c, k), sp=k); ); if (sp == c, print1(c, ", ")); ); } \\ Michel Marcus, Sep 06 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Apr 29 2010, Apr 30 2010
EXTENSIONS
Offset corrected and sequence extended by R. J. Mathar, May 03 2010
STATUS
approved