A069753 Smallest k>n such that the tetrahedral number n*(n+1)*(n+2)/6 divides the tetrahedral number k*(k+1)*(k+2)/6.

2, 4, 4, 8, 13, 14, 26, 54, 43, 20, 64, 26, 63, 48, 118, 270, 151, 170, 55, 54, 229, 207, 274, 350, 323, 350, 376, 174, 433, 928, 494, 1054, 119, 440, 259, 664, 701, 208, 778, 328, 859, 516, 944, 504, 1033, 2160, 1126, 2350, 1223, 1274, 1324, 1376, 1429, 350

Offset

1,1

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Tsz Ho Chan, The Diophantine equation b (b+1) (b+2) = t a (a + 1) (a + 2) and gap principle, arXiv preprint (2024). arXiv:2408.01306 [math.NT]

Formula

Chan proves that a(n) >> n (log n)^(1/6)/(log log n)^(1/3). - Charles R Greathouse IV, Aug 13 2024

Note that a(n) <= n^2 + n - 2 and that if n is odd, a(n) <= (n^2 + n - 4)/2. - Charles R Greathouse IV, Aug 13 2024

Mathematica

stn[n_]:=Module[{i=n+1, tn=(n(n+1)(n+2))/6}, While[!Divisible[(i(i+1) (i+2))/6, tn], i++]; i]; stn/@Range[70]  (* Harvey P. Dale, Apr 11 2011 *)

Prog

(PARI) for(s=1, 80, n=s+1; while(frac(n*(n+1)*(n+2)/(s*(s+1)*(s+2)))>0, n++); print1(n, ", "); )
(PARI) a(n)=my(t=n*(n+1)*(n+2)); for(k=n+1, oo, if(k*(k+1)*(k+2)%t==0, return(k))) \\ Charles R Greathouse IV, Aug 13 2024
(PARI) a(n)=if(n==1, return(2)); my(t=n*(n+1)*(n+2), p=vecmax(apply(k->vecmax(factor(k)[, 1]), [n..n+2])), start=((n+2)\p+1)*p-2); for(k=max(n+1, start-p+1), start-p+2, if(k*(k+1)*(k+2)%t==0, return(k))); forstep(k=start, oo, [1, 1, p-2], if(k*(k+1)*(k+2)%t==0, return(k))) \\ Charles R Greathouse IV, Aug 13 2024

Crossrefs

Sequence in context: A335974 A047073 A308528 * A181245 A216950 A240364

Adjacent sequences: A069750 A069751 A069752 * A069754 A069755 A069756

Keyword

nonn

Author

Benoit Cloitre, May 01 2002

Status

approved