A286290 A bisection of A064736.
1, 6, 12, 20, 35, 56, 72, 90, 110, 143, 182, 210, 240, 272, 306, 342, 399, 462, 506, 552, 600, 650, 702, 756, 812, 870, 930, 992, 1056, 1122, 1224, 1332, 1406, 1482, 1560, 1640, 1722, 1806, 1892, 1980, 2070, 2162, 2256, 2352, 2450, 2550, 2652, 2756, 2862, 2970, 3135, 3306, 3422, 3540
Offset: 1
Links
- Ray Chandler, Table of n, a(n) for n = 1..10000
- Ray Chandler, Table of n, a(n) for n = 1..1000000 (large gzipped file)
Programs
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PARI
A286290_list(Nmax,a=List(1),c=2)={while(#a
M. F. Hasler, May 23 2017 -
PARI
a(n) = my(r = 1); for(i = 2, n, r = nxt(r)); r is(n) = if(n < 6, return(n==1)); if(issquare(n+1, &n), is(n), if(sqrtint(4*n+1)^2 == 4*n+1, s = sqrtint(4*n+1); !(is(s\2) || is(s\2+1)), return(0))) nxt(n) = n==1&&return(6); if(issquare(n+1, &n), (n+1) * (n+2), my(m = sqrtint(n)); if(is(m + 2), (m + 1) * (m + 3), (m + 1) * (m + 2))) lista(n) = my(c = 1, l = List([1])); for(i=2, n, c = nxt(c); listput(l, c)); l \\ David A. Corneth, May 25 2017
Formula
a(n) ~ n^2*(1 + 1.5/n^c) with c=1/2. (Conjectured, although for small n around 10^5 a smaller c ~ 0.478 is a better fit to the data.) - M. F. Hasler, May 23 2017
For n around 10^8, c ~ 0.4848 is a better fit. - David A. Corneth, May 25 2017
Comments