A324297 Positive integers k that are the product of two integers ending with 6.

36, 96, 156, 216, 256, 276, 336, 396, 416, 456, 516, 576, 636, 676, 696, 736, 756, 816, 876, 896, 936, 996, 1056, 1116, 1176, 1196, 1216, 1236, 1296, 1356, 1376, 1416, 1456, 1476, 1536, 1596, 1656, 1696, 1716, 1776, 1836, 1856, 1896, 1956, 1976, 2016, 2076, 2116

Offset: 1

Author

Stefano Spezia, Mar 16 2019

Keywords

• nonn
• base

Comments

All the terms end with 6 (A017341).

Examples

			36 = 6*6, 96 = 6*16, 216 = 6*36, 256 = 16*16, 276 = 6*46, ...
		

Links

• Stefano Spezia, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000400, A017341 (supersequence), A324298, A053742 (ending with 5).

Programs

Mathematica

a={}; For[n=0,n<=250,n++,For[k=0,k<=n,k++,If[Mod[10*n+6,10*k+6]==0 && Mod[(10*n+6)/(10*k+6),10]==6 && 10*n+6>Max[a],AppendTo[a,10*n+6]]]]; a

PARI

isok6(n) = (n%10) == 6; \\ A017341
isok(n) = {if (isok6(n), my(d=divisors(n)); fordiv(n, d, if (isok6(d) && isok6(n/d), return(1)));); return (0);} \\ Michel Marcus, Apr 14 2019

Python

def aupto(lim): return sorted(set(a*b for a in range(6, lim//6+1, 10) for b in range(a, lim//a+1, 10)))
print(aupto(2117)) # Michael S. Branicky, Aug 18 2021

Formula

Conjecture: Lim_{n->infinity} a(n)/a(n-1) = 1.

The conjecture is true since it can be proved that a(n) = (sqrt(a(n-1)) + g(n-1))^2 where [g(n): n > 1] is a bounded sequence of positive real numbers. - Stefano Spezia, Aug 18 2021