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A281661 The least common multiple of 1 + n^2 and 1 + n^3.

%I A281661 #27 Jun 10 2023 20:26:24
%S A281661 1,2,45,140,1105,1638,8029,8600,33345,29930,101101,81252,250705,
%T A281661 186830,540765,381488,1052929,712530,1895725,1241660,3208401,2046902,
%U A281661 5164765,3224520,7977025,4890938,11899629,7184660,17233105,10268190,24327901,14329952,33588225,19586210
%N A281661 The least common multiple of 1 + n^2 and 1 + n^3.
%C A281661 If d|(1 + n^2) and d|(1 + n^3), then d|((1 + n^2) - (n*(1 + n^2) - (1 + n^3))^2) = 2*n. If k|n and k|(1 + n^2), then k = 1 is only option since k|n^2 and k|(1 + n^2). So d must be 1 or 2, exactly. Obviously if n is odd, then the greatest d must be 2 since 1 + n^2 and 1 + n^3 are even. If n is even, then d must be 1 since 1 + n^2 and 1 + n^3 are odd.
%H A281661 Colin Barker, <a href="/A281661/b281661.txt">Table of n, a(n) for n = 0..1000</a>
%H A281661 <a href="/index/Rec#order_12">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,0,-15,0,20,0,-15,0,6,0,-1).
%F A281661 a(n) = lcm(1+n^2, 1+n^3) = (1+n^2)*(1+n^3)/gcd(1+n^2, 1+n^3).
%F A281661 a(n) = (1+n^2)*(1+n^3)/ A000034(n) with g.f. ( 1 +2*x +39*x^2 +128*x^3 +850*x^4 +828*x^5 +2054*x^6 +832*x^7 +861*x^8 +130*x^9 +35*x^10 ) / ( (x-1)^6 *(1+x)^6 ).
%F A281661 A006530(a(n)) = max( A081256(n), A014442(n)). - _R. J. Mathar_, Jan 28 2017
%F A281661 a(n) = (3 + (-1)^n)*(1 + n^2 + n^3 + n^5)/4. - _Colin Barker_, Feb 07 2017
%p A281661 A281661 := proc(n)
%p A281661         ilcm(1+n^2,1+n^3);
%p A281661 end proc:
%t A281661 Table[LCM[n^2+1,n^3+1],{n,0,50}] (* _Harvey P. Dale_, Jun 10 2023 *)
%o A281661 (PARI) a(n) = lcm(n^2+1, n^3+1); \\ _Michel Marcus_, Jan 29 2017
%o A281661 (PARI) a(n) = (n^2 + 1)*(n^3 + 1)/(1 + n%2); \\ _Altug Alkan_, Jan 29 2017
%Y A281661 Cf. A001093, A002522, A281660.
%K A281661 nonn,easy
%O A281661 0,2
%A A281661 _R. J. Mathar_, Jan 26 2017