cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A274968 Even numbers n >= 4 which are not m-gonal number for 3 <= m < n.

Original entry on oeis.org

4, 8, 14, 20, 26, 32, 38, 44, 50, 56, 62, 68, 74, 80, 86, 98, 104, 110, 116, 122, 128, 134, 140, 146, 152, 158, 164, 170, 182, 188, 194, 200, 206, 212, 218, 224, 230, 236, 242, 248, 254, 266, 272, 278, 284, 290, 296, 302
Offset: 1

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Author

Daniel Forgues, Jul 12 2016

Keywords

Comments

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 of order k >= 3.
Except for a(1) = 4, all a(n) are congruent to 2 (mod 6), although from 8 to 302, the numbers
92: 5-gonal of order 8,
176: 5-gonal of order 11, 8-gonal of order 8,
260: 11-gonal of order 8,
are not in this sequence.
Even numbers n for which A176948(n) = n.
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 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.
a(1) = 4 is the only square number: 4-gonal with order k = 2.
All integers of the form n = 6j + 4, with j >= 1, are m-gonal numbers of order k = 4, with m = j + 2, which means that none are in this sequence. - Daniel Forgues, Aug 01 2016

Examples

			20 is in this sequence because 20 is trivially a 20-gonal number of order k = 2 (element of A051872) but not an m-gonal number for 3 <= k <= (1/2)*{-1 + sqrt[1 + 8*20]}.
		

Crossrefs

Programs

  • PARI
    lista(nn) = {forstep(n=4, nn, 2, sp = n; forstep(k=n, 3, -1, if (ispolygonal(n, k), sp=k);); if (sp == n, print1(n, ", ")););} \\ Michel Marcus, Sep 06 2016
  • Python
    A274968_list = []
    for n in range(4,10**6,2):
        k = 3
        while k*(k+1) <= 2*n:
            if not (2*(k*(k-2)+n)) % (k*(k - 1)):
                break
            k += 1
        else:
            A274968_list.append(n) # Chai Wah Wu, Jul 28 2016
    
  • Sage
    def is_A274968(n):
        if is_odd(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_A274968(n)]) # Peter Luschny, Jul 28 2016