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 A264450 a(n) = n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120. 7
 0, 3519, 8372, 14805, 23088, 33516, 46410, 62118, 81016, 103509, 130032, 161051, 197064, 238602, 286230, 340548, 402192, 471835, 550188, 638001, 736064, 845208, 966306, 1100274, 1248072, 1410705, 1589224, 1784727, 1998360, 2231318, 2484846, 2760240, 3058848, 3382071 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS It is well-known, and easy to prove, that the product of 5 consecutive integers n*(n + 1)*(n + 2)*(n + 3)*(n + 4) is divisible by 5!. It can be shown that the product of 5 integers in arithmetic progression n*(n + r)*(n + 2*r)*(n + 3*r)*(n + 4*r) is divisible by 5! if and only if r is not divisible by 2, 3 or 5 (see A007775 for these numbers). This is the case r = 11. LINKS Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1). FORMULA O.g.f.: x*(2408*x^4 - 10542*x^3 + 17358*x^2 - 12742*x + 3519)/(1 - x)^6. a(n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6) for n>5. - Vincenzo Librandi, Nov 16 2015 MAPLE seq( n*(n + 11)*(n + 22)*(n + 33)*(n + 44)/120, n = 0..33 ); MATHEMATICA Table[n (n + 11) (n + 22) (n + 33) (n + 44)/120, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 3519, 8372, 14805, 23088, 33516}, 40] (* Harvey P. Dale, Nov 27 2015 *) PROG (PARI) vector(100, n, n--; n*(n+11)*(n+22)*(n+33)*(n+44)/120) \\ Altug Alkan, Nov 15 2015 (MAGMA) [n*(n+11)*(n+22)*(n+33)*(n+44)/120: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015 CROSSREFS Cf. A007775, A264443, A264444, A264445, A264446, A264447, A264448, A264449. Sequence in context: A263302 A205278 A185920 * A187726 A267125 A241240 Adjacent sequences:  A264447 A264448 A264449 * A264451 A264452 A264453 KEYWORD nonn,easy AUTHOR Peter Bala, Nov 13 2015 STATUS approved

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Last modified January 20 05:37 EST 2020. Contains 331067 sequences. (Running on oeis4.)