%I #30 Sep 17 2022 03:45:25
%S 0,15,90,273,612,1155,1950,3045,4488,6327,8610,11385,14700,18603,
%T 23142,28365,34320,41055,48618,57057,66420,76755,88110,100533,114072,
%U 128775,144690,161865,180348,200187,221430,244125,268320,294063,321402,350385,381060,413475,447678,483717
%N a(n) = n*(2*n + 1)*(4*n + 1).
%C Sums of the consecutive integers from A000384(n) to A000384(n+1)-1. This is the case s=6 of the formula n*(n*(s-2) + 1)*(n*(s-2) + 2)/2 related to s-gonal numbers.
%C The inverse binomial transform is 0, 15, 60, 48, 0, ... (0 continued).
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F O.g.f.: 3*x*(5 + 10*x + x^2)/(1 - x)^4.
%F E.g.f.: x*(15 + 30*x + 8*x^2)*exp(x).
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
%F a(n) = 3*A258582(n).
%F a(n) = -3*A100157(-n).
%F Sum_{n>0} 1/a(n) = 2*(3 - log(4)) - Pi.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = log(2) + 2*sqrt(2)*log(1+sqrt(2)) + (sqrt(2)-1/2)*Pi - 6. - _Amiram Eldar_, Sep 17 2022
%e Row sums of the triangle:
%e | 0 | ................................................................. 0
%e | 1 | 2 3 4 5 .................................................... 15
%e | 6 | 7 8 9 10 11 12 13 14 ........................................ 90
%e | 15 | 16 17 18 19 20 21 22 23 24 25 26 27 ........................... 273
%e | 28 | 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 ............... 612
%e | 45 | 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 .. 1155
%e ...
%e where:
%e . first column is A000384,
%e . second column is A130883 (without 1),
%e . third column is A033816,
%e . diagonal is A014106,
%e . 0, 2, 8, 18, 32, 50, ... are in A001105.
%p seq(n*(2*n+1)*(4*n+1),n=0..40); # _Muniru A Asiru_, Jun 27 2018
%t Table[n (2 n + 1) (4 n + 1), {n, 0, 40}]
%o (PARI) vector(40, n, n--; n*(2*n+1)*(4*n+1))
%o (Sage) [n*(2*n+1)*(4*n+1) for n in (0..40)]
%o (Maxima) makelist(n*(2*n+1)*(4*n+1), n, 0, 40);
%o (GAP) List([0..40], n -> n*(2*n+1)*(4*n+1));
%o (Magma) [n*(2*n+1)*(4*n+1): n in [0..40]];
%o (Python) [n*(2*n+1)*(4*n+1) for n in range(40)]
%o (Julia) [n*(2*n+1)*(4*n+1) for n in 0:40] |> println
%Y First bisection of A059270 and subsequence of A034828, A047866, A109900, A290168.
%Y Sums of the consecutive integers from P(s,n) to P(s,n+1)-1, where P(s,k) is the k-th s-gonal number: A027480 (s=3), A055112 (s=4), A228888 (s=5).
%K nonn,easy
%O 0,2
%A _Bruno Berselli_, Jun 27 2018