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A242604

a(n) = (n - 1)*(n^3 + 1) = n^4 - n^3 + n - 1, for n >= 1.

0, 9, 56, 195, 504, 1085, 2064, 3591, 5840, 9009, 13320, 19019, 26376, 35685, 47264, 61455, 78624, 99161, 123480, 152019, 185240, 223629, 267696, 317975, 375024, 439425, 511784, 592731, 682920, 783029, 893760, 1015839, 1150016, 1297065

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OFFSET

1,2

COMMENTS

1/a(n), for n >= 2, is in base n represented by 0.repeat(0,0,0,1,1,1). This is instance p = 3 of the general formula for 0.repeat(0^(q),1^(q)) (meaning here q zeros followed by q 1's) in base b >= 2 which is 1/a(q,b) with a(q,b) = (b - 1)*(b^p + 1), for p >= 1.

LINKS

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

FORMULA

a(n) = (n - 1)*(n^3 + 1) = n^4 - n^3 + n - 1, n >= 1.

O.g.f.: x^2*(9 + 11*x + 5*x^2 - x^3)/(1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 23 2025

MAPLE

A242604:=n->(n-1)*(n^3+1); seq(A242604(n), n=1..50); # Wesley Ivan Hurt, Jun 20 2014

MATHEMATICA

Table[n^4-n^3+n-1, {n, 40}] (* Harvey P. Dale, Jun 20 2019 *)

PROG