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A264449 a(n) = n*(n + 7)*(n + 14)*(n + 21)*(n + 28)/120. 7
0, 638, 1656, 3162, 5280, 8151, 11934, 16807, 22968, 30636, 40052, 51480, 65208, 81549, 100842, 123453, 149776, 180234, 215280, 255398, 301104, 352947, 411510, 477411, 551304, 633880, 725868, 828036, 941192, 1066185, 1203906, 1355289, 1521312, 1702998, 1901416, 2117682 (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 = 7.

LINKS

Table of n, a(n) for n=0..35.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

FORMULA

O.g.f.: x*(351*x^4 - 1612*x^3 + 2796*x^2 - 2172*x + 638)/(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 + 7)*(n + 14)*(n + 21)*(n + 28)/120, n = 0..35 );

MATHEMATICA

Table[n (n + 7) (n + 14) (n + 21) (n + 28)/120, {n, 0, 40}] (* Vincenzo Librandi, Nov 16 2015 *)

PROG

(PARI) vector(100, n, n--; n*(n+7)*(n+14)*(n+21)*(n+28)/120) \\ Altug Alkan, Nov 15 2015

(Magma) [n*(n+7)*(n+14)*(n+21)*(n+28)/120: n in [0..40]]; // Vincenzo Librandi, Nov 16 2015

CROSSREFS

Cf. A007775, A264443, A264444, A264445, A264446, A264447, A264448, A264450.

Sequence in context: A203879 A317395 A212399 * A210883 A281884 A200710

Adjacent sequences:  A264446 A264447 A264448 * A264450 A264451 A264452

KEYWORD

nonn,easy

AUTHOR

Peter Bala, Nov 13 2015

STATUS

approved

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Last modified October 6 12:35 EDT 2022. Contains 357264 sequences. (Running on oeis4.)