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A177025
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Number of ways to represent n as a polygonal number.
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13
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1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 1, 1, 3, 2, 1, 2, 1, 1, 3, 2, 1, 2, 2, 1, 2, 3, 1, 2, 1, 1, 2, 2, 2, 4, 1, 1, 2, 2, 1, 2, 1, 1, 4, 2, 1, 2, 2, 1, 3, 2, 1, 2, 3, 1, 2, 2, 1, 2, 1, 1, 2, 3, 2, 4, 1, 1, 2, 3, 1, 2, 1, 1, 3, 2, 1, 3, 1, 1, 4, 2, 1, 2, 2, 1, 2, 2, 1, 2, 3, 2, 2, 2, 2, 3, 1, 1, 2, 3
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OFFSET
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3,4
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COMMENTS
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Since n is always n-gonal number, a(n) >= 1.
Conjecture: Every positive integer appears in the sequence.
Records of 2, 3, 4, 5, ... are reached at n = 6, 15, 36, 225, 561, 1225, ... see A063778. [R. J. Mathar, Aug 15 2010]
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REFERENCES
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J. J. Tattersall, Elementary Number Theory in Nine chapters, 2nd ed (2005), Cambridge Univ. Press, page 22 Problem 26, citing Wertheim (1897)
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LINKS
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FORMULA
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G.f.: x * Sum_{k>=2} x^k / (1 - x^(k*(k + 1)/2)) (conjecture). - Ilya Gutkovskiy, Apr 09 2020
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MAPLE
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local ii, a, n, s, m ;
ii := 2*p ;
a := 0 ;
for n in numtheory[divisors](ii) do
if n > 2 then
s := ii/n ;
if (s-2) mod (n-1) = 0 then
a := a+1 ;
end if;
end if;
end do:
return a;
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MATHEMATICA
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nn = 100; t = Table[0, {nn}]; Do[k = 2; While[p = k*((n - 2) k - (n - 4))/2; p <= nn, t[[p]]++; k++], {n, 3, nn}]; t (* T. D. Noe, Apr 13 2011 *)
Table[Length[Intersection[Divisors[2 n - 2] + 1, Divisors[2 n]]] - 1, {n, 3, 100}] (* Jonathan Sondow, May 09 2014 *)
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PROG
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(PARI) a(n) = sum(i=3, n, ispolygonal(n, i)); \\ Michel Marcus, Jul 08 2014
(Python)
from sympy import divisors
def a(n):
i=2*n
x=0
for d in divisors(i):
if d>2:
s=i/d
if (s - 2)%(d - 1)==0: x+=1
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CROSSREFS
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Cf. A129654, A139601, A090428, A176949, A176948, A176774, A176744, A176747, A176775, A175873, A176874.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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