OFFSET
1,2
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
LINKS
Frank M Jackson, Table of n, a(n) for n = 1..10000
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
EXAMPLE
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.
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 *)
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev and Peter J. C. Moses, Dec 16 2011
STATUS
approved