A393327 - OEIS

%I #29 Mar 01 2026 07:39:54

%S 0,10,30,63,112,180,270,385,528,702,910,1155,1440,1768,2142,2565,3040,

%T 3570,4158,4807,5520,6300,7150,8073,9072,10150,11310,12555,13888,

%U 15312,16830,18445,20160,21978,23902,25935,28080,30340,32718,35217,37840,40590,43470,46483,49632,52920,56350

%N a(n) = (1/2)*n*(n+3)*(n+4).

%C a(n) is the sum of the smallest n+3 multiples of n. For example, a(4) = 112 since 112 = 4 + 8 + 12 + 16 + 20 + 24 + 28.

%C a(n) = row sum of the n+4 row in the triangular view of A001477 (excluding the last 4 entries) - T(n-1), where T(n) is the n-th triangular number. For example, a(2) = 15 + 16 - T(1) = 31 - 1 = 30;  a(4) = 28 + 29 + 30 + 31 - T(3) = 118 - 6 = 112.

%C a(n)*binomial(n+2,2) is the number of strings of length n+4 defined on {0,1,2,3} that contain exactly two 0's, exactly two 1's, exactly one 2 and any number of 3's (see Examples).

%H Paolo Xausa, <a href="/A393327/b393327.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).

%F a(n) = n*A000217(n+3).

%F a(n) - a(n-1) = A140229(n+2), n >= 1.

%F G.f.: x*(10 - 10*x + 3*x^2)/(1-x)^4.

%F E.g.f.: exp(x)*x*(20 + 10*x + x^2)/2. - _Stefano Spezia_, Feb 12 2026

%F From _Amiram Eldar_, Mar 01 2026: (Start)

%F Sum_{n>=1} 1/a(n) = 13/72.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/3 - 61/72. (End)

%e For n=1, a(1)*binomial(3,2) = 10*3 is the number of strings of length 5 described in the comment since there are 30 permutations of the string 00112.

%e For n=4, a(4)*binomial(6,2) = 112*15 is the number of strings of length 8 since there are 1680 permutations of the string 00112333.

%t A393327[n_] := n*(n+3)*(n+4)/2; Array[A393327, 50, 0] (* or *)

%t LinearRecurrence[{4, -6, 4, -1}, {0, 10, 30, 63}, 50] (* _Paolo Xausa_, Feb 23 2026 *)

%Y Cf. A000217, A001477, A027480, A005586, A140229.

%K nonn,easy

%O 0,2

%A _Enrique Navarrete_, Feb 11 2026