%I #28 Oct 19 2024 15:47:05
%S 0,119,3104,29319,162104,643535,2040744,5502959,13129424,28468359,
%T 57167120,107793719,192849864,329995679,543506264,865980255,
%U 1340320544,2022007319,2981683584,4308073319,6111252440,8526292719,11717298824
%N Sum of consecutive non-fourth-powers.
%C Relationship with tetrahedral numbers: a(4) = (first term + last term)*(6*Tetra_n + n^3) = (82+255)*(6*10+27) = (337)*(87) = 29319.
%H Vincenzo Librandi, <a href="/A048397/b048397.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8, -28, 56, -70, 56, -28, 8, -1).
%F a(n) = 4*n^7 + 14*n^6 + 28*n^5 + 34*n^4 + 26*n^3 + 11*n^2 + 2*n.
%F G.f.: (119*x +2152*x^2 +7819*x^3 +7800*x^4 +2141*x^5 +128*x^6 +x^7)/(x-1)^8. - _Harvey P. Dale_, Apr 23 2011
%e Between 3^4 and 4^4 we have 82+83+...+254+255 which is 29319 or a(4).
%p A048397:=n->4*n^7 + 14*n^6 + 28*n^5 + 34*n^4 + 26*n^3 + 11*n^2 + 2*n; seq(A048397(n), n=0..40); # _Wesley Ivan Hurt_, Feb 10 2014
%t Table[Total[Range[n^4+1,(n+1)^4-1]],{n,0,40}] (* or *) Table[4n^7+ 14n^6+28n^5+34n^4+26n^3+11n^2+2n,{n,0,40}] (* _Harvey P. Dale_, Apr 23 2011 *)
%t LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{0,119,3104,29319,162104,643535,2040744,5502959},40] (* _Harvey P. Dale_, Jul 31 2021 *)
%o (PARI) a(n)=n*(2*n^2 + 3*n + 2)*(2*n^4 + 4*n^3 + 6*n^2 + 4*n + 1) \\ _Charles R Greathouse IV_, Jan 24 2022
%o (Python)
%o def A048397(n): return n*(n*((n<<1)+3)+2)*(n*(n*(n*((n+2)<<1)+6)+4)+1) # _Chai Wah Wu_, Oct 19 2024
%Y Cf. A048395, A048396, A000292.
%K nonn,nice,easy
%O 0,2
%A _Patrick De Geest_, Mar 15 1999