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A346376
a(n) = n^4 + 14*n^3 + 63*n^2 + 98*n + 28.
2
28, 204, 604, 1348, 2580, 4468, 7204, 11004, 16108, 22780, 31308, 42004, 55204, 71268, 90580, 113548, 140604, 172204, 208828, 250980, 299188, 354004, 416004, 485788, 563980, 651228, 748204, 855604, 974148, 1104580, 1247668, 1404204, 1575004, 1760908, 1962780
OFFSET
0,1
COMMENTS
The product of eight consecutive positive integers can always be expressed as the difference of two squares: x^2 - y^2.
This sequence gives the x-values for each product. The y-values are A017113(n+4).
a(n) is always divisible by 4. In addition, we have (a(n)+16)/4 belongs to A028387.
Are 4 and 8 the unique values of k such that the product of k consecutive integers is always distant to upper square by a square?
FORMULA
a(n) = A239035(n)^2 - A017113(n+4)^2.
a(n) = 4*(A028387(A046691(n+2)) - 4).
G.f.: 4*(7 + 16*x - 34*x^2 + 22*x^3 - 5*x^4)/(1 - x)^5. - Stefano Spezia, Jul 14 2021
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Lamine Ngom, Jul 14 2021
STATUS
approved