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COMMENTS
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n^3 divides a(n) for n = {35, 55, 77, 95, 115, 119, 143, 155, 161,...} = A121707.
It appears that p^(3k-1) divides a(p^k) for all integer k > 1 and prime p > 2:
for prime p > 2, p^2 divides a(p), p^5 divides a(p^2) and p^8 divides a(p^3).
Additionally, p^3 divides a(3p) for prime p > 2.
For prime p > 3, p divides a(p+1) and p^3 divides a(2p+1);
for prime p > 5, p divides a(3p+1) and p^3 divides a(4p+1);
for prime p > 7, p divides a(5p+1) and p^3 divides a(6p+1):
It appears that p divides a((2k+1)p+1) for integer k >= 0 and prime p > 2k+3, and p^3 divides a(2kp+1) for integer k > 0 and prime p > 2k+2.
p divides a((p+1)/2) for prime p = {7, 11, 19, 23, 31, 43, 47, 59, 67, 71,...} = A002145: primes of the form 4n+3, n >= 1.
p^2 divides a((p+1)/2) for prime p = {7, 23, 31, 47, 71, 79, 103, 127,...} = A007522: primes of the form 8n+7, n >= 0.
n*(2*n+1) divides a(2*n+1) for n >= 1. - Franz Vrabec, Dec 20 2020
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