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A336535
a(n) = (m(n)^2 + 3)*(m(n)^2 + 7)/32, where m(n) = 2*n - 1.
1
1, 6, 28, 91, 231, 496, 946, 1653, 2701, 4186, 6216, 8911, 12403, 16836, 22366, 29161, 37401, 47278, 58996, 72771, 88831, 107416, 128778, 153181, 180901, 212226, 247456, 286903, 330891, 379756, 433846, 493521, 559153, 631126, 709836, 795691, 889111, 990528, 1100386, 1219141, 1347261, 1485226, 1633528, 1792671
OFFSET
1,2
COMMENTS
For m(n) = 3,5,11, and 181, the perfect numbers (A000396), 6, 28, 496, and 33550336 are produced, respectively. 3,5,11, and 181 are the numbers m(n) such that (m(n)^2+7) is a power of 2. cf A038198.
REFERENCES
David M. Burton, Elementary Number Theory, McGraw-Hill (2011), 25.
FORMULA
From Stefano Spezia, Jul 25 2020: (Start)
O.g.f.: x*(1 + x + 8*x^2 + x^3 + x^4)/(1 - x)^5.
a(n) = (1 - n + n^2)*(2 - n + n^2)/2.
a(n) = A002061(n)*A014206(n-1)/2.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5. (End)
EXAMPLE
m(2) = 2*2-1 = 3 and (3^2+3)*(3^2+7)/32 = 6, so 6 is in the sequence.
MATHEMATICA
Table[((n^2+3)*(n^2+7))/32, {n, 1, 100, 2}]
PROG
(PARI) a(n)=(1-n+n^2)*(2-n+n^2)/2 \\ Charles R Greathouse IV, Oct 21 2022
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Jeff Brown, Jul 24 2020
STATUS
approved