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

1, 2, 45, 140, 1105, 1638, 8029, 8600, 33345, 29930, 101101, 81252, 250705, 186830, 540765, 381488, 1052929, 712530, 1895725, 1241660, 3208401, 2046902, 5164765, 3224520, 7977025, 4890938, 11899629, 7184660, 17233105, 10268190, 24327901, 14329952, 33588225, 19586210

Offset

0,2

Comments

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.

Links

Colin Barker, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (0,6,0,-15,0,20,0,-15,0,6,0,-1).

Formula

a(n) = lcm(1+n^2, 1+n^3) = (1+n^2)*(1+n^3)/gcd(1+n^2, 1+n^3).

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 ).

A006530(a(n)) = max( A081256(n), A014442(n)). - R. J. Mathar, Jan 28 2017

a(n) = (3 + (-1)^n)*(1 + n^2 + n^3 + n^5)/4. - Colin Barker, Feb 07 2017

Maple

A281661 := proc(n)
        ilcm(1+n^2, 1+n^3);
end proc:

Mathematica

Table[LCM[n^2+1, n^3+1], {n, 0, 50}] (* Harvey P. Dale, Jun 10 2023 *)

Prog

(PARI) a(n) = lcm(n^2+1, n^3+1); \\ Michel Marcus, Jan 29 2017
(PARI) a(n) = (n^2 + 1)*(n^3 + 1)/(1 + n%2); \\ Altug Alkan, Jan 29 2017

Crossrefs

Cf. A001093, A002522, A281660.

Sequence in context: A305363 A304955 A316637 * A019579 A302220 A302666

Adjacent sequences: A281658 A281659 A281660 * A281662 A281663 A281664

Keyword

nonn,easy

Author

R. J. Mathar, Jan 26 2017

Status

approved