A202318 Let (n)_p denote the exponent of prime p in the prime power factorization of n. Then a(n) is defined by the formulas a(1)=1; for n >= 2, (a(n))_2 = (n)_2, (a(n))_3 = (n)_3 and, for p >= 5, (a(n))_p = 1 + ((2n)/(p-1))_p if p-1|2*n, and (a(n))_p = 0 otherwise.

1, 10, 21, 20, 11, 2730, 1, 680, 1197, 550, 23, 5460, 1, 290, 7161, 1360, 1, 5757570, 1, 45100, 6321, 230, 47, 185640, 11, 530, 3591, 580, 59, 283933650, 1, 2720, 32361, 10, 781, 840605220, 1, 10, 1659, 1533400, 83, 23830170, 1, 40940, 408177, 470, 1, 36014160, 1, 277750, 2163, 1060, 107, 1882725390

Offset: 1

Author

Vladimir Shevelev and Peter J. C. Moses, Dec 16 2011

Keywords

• nonn

Comments

a(n)=1 iff n has form 6n+-1 and, if d >= 5 is a divisor of n, then 2*d+1 is not prime. The places of 1's form sequence A045979.

If p is an odd prime and p^n is the side length of the odd leg of a primitive Pythagorean triangle (PPT) it constrains the other leg and hypotenuse to be (p^(2n)-1)/2 and (p^(2n)+1)/2 and the area to be (p^n-1)p^n(p^n+1)/4. Now consider the term (p^n-1)p^n(p^n+1): it must at least be divisible by 24 for all odd primes p because the area of a PPT is divisible by 6 (see A127922 for n=1). a(n) equals the common divisor of the term (p^n-1)p^n(p^n+1)/24 for all odd primes p. - Frank M Jackson, Dec 09 2017

Examples

			Let n=6. Since 2*6+1=13 is prime, the max p that should be considered is 13. We have
  (a(6))_2  = (a(6))_3 = 1,
  (a(6))_5  = (12/4)_5 + 1 = 1,
  (a(6))_7  = (12/6)_7 + 1 = 1,
  (a(6))_13 = (12/12)_13 + 1 = 1.
Thus a(6) = 2*3*5*7*13 = 2730.
		

Links

• Frank M Jackson, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A053657, A045979, A002445, A127922.

Programs

Mathematica

Table[Numerator[Exp[Re[Limit[Zeta[s] (Zeta[-1]^(s - 1) - Zeta[-(2*n - 1)]^(s - 1)), s -> 1]]]], {n, 1, 54}] (* Mats Granvik, Feb 05 2016 *)
Table[(lst=Table[p=Prime[m+1]; (p^n-1)p^n(p^n+1), {m, 1, 10}]; GCD@@lst/24), {n, 1, 100}] (* Frank M Jackson, Dec 09 2017 *)
a[n_] := Product[p^Sum[Floor[(n-1)/((p-1) p^k)], {k, 0, n}], {p, Prime[Range[n]]}]; Array[a[2#+1]/(24 a[2#-1]) &, 100] (* using Jean-François Alcover's program A053657 *)(* Frank M Jackson, Dec 16 2017 *)

PARI

a(n) = {my(r = 1); forprime(p=2, 2*n+1, if (p<=3, r *= p^valuation(n, p), if (! (2*n % (p-1)), r *= p^(1+valuation((2*n)/(p-1), p))););); r;} \\ Michel Marcus, Feb 06 2016

Formula

a(n) = (1/24)*b(2n+1)/b(2n-1), where b(n) = A053657(n).

a(p) = A002445(p)/6, for prime p >= 5.

a(n) = numerator of e^(real(lim_{s -> 1} (zeta(s)*(zeta(-1)^(s-1) - zeta(-(2*n-1))^(s-1))))). - Mats Granvik, Feb 05 2016

a(n) = A036283(n)/6. - Hugo Pfoertner, Dec 18 2022