

A217772


a(n) = ((p+1)*(3p)!/((2p1)!*(p+1)!*2p)  3)/(3p^3), where p is the nth prime.


2



1, 8, 113, 48469, 1232351, 1002175798, 30956114561, 32956274508457, 1386101220044940571, 50017672586399947073, 2548160990547719392420658, 3694160975065765801289789930, 142486973648670437070915061157
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OFFSET

2,2


COMMENTS

This sequence is based on Gary Detlefs's conjecture, which he posted as a comment to A005809. His conjecture is equivalent to the conjecture that the Diophantine equation ((n+1)*(3*n)!/((2*n1)!*(n+1)!*2*n)3)/n^3 = m has integer solutions m for all odd primes n.
Additionally I conjecture that all m are divisible by 3, therefore terms of this sequence a(n) = m/3.
It is also notable that for quite a few values of n (2, 3, 4, 5, 6, 7, 17, 19, 21, 22, 23, 24, 25, 26, 35, 39, 43, ...) a(n+1) = a(n) mod 7.
The values of this sequence's terms are replicated by conjectured general formula, given in A223886 (and also added to the formula section here) for k=3, j=1 and n>=2.  Alexander R. Povolotsky, Apr 18 2013
For n>=3 and k>=2 ((binomial(k*p,p)k)/p^3)/k is an integer. For k=2 this is the Wolstenholme quotient (A034602) and for k=3 the current sequence.  Peter Luschny, Feb 09 2016


LINKS

Table of n, a(n) for n=2..14.


FORMULA

a(n) = ((binomial (j*k*prime(n), j*prime(n))  binomial(k*j, j))/(k*prime(n)^3) for k=3, j=1 and n>=2 (conjectured).  Alexander R. Povolotsky, Apr 18 2013


MAPLE

WQ := proc(n, k) local p; p := ithprime(n); ((binomial(k*p, p)k)/p^3)/k end:
seq(WQ(n, 3), n=2..14); # Peter Luschny, Feb 09 2016


PROG

(PARI) a(n)=my(p=prime(n)); (binomial(3*p, p+1)*(p+1)/(2*p)3)/(3*p^3) \\ Charles R Greathouse IV, Mar 26 2013


CROSSREFS

Cf. A005809, A034602, A223886.
Sequence in context: A092084 A099715 A023814 * A062126 A222520 A220635
Adjacent sequences: A217769 A217770 A217771 * A217773 A217774 A217775


KEYWORD

nonn


AUTHOR

Alexander R. Povolotsky, Mar 24 2013


STATUS

approved



