A324298 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 6 (A324297).
3, 9, 15, 21, 25, 27, 33, 39, 41, 45, 51, 57, 63, 67, 69, 73, 75, 81, 87, 89, 93, 99, 105, 111, 117, 119, 121, 123, 129, 135, 137, 141, 145, 147, 153, 159, 165, 169, 171, 177, 183, 185, 189, 195, 197, 201, 207, 211, 213, 217, 219, 223, 225, 231, 233, 237, 243, 249
Offset: 1
Examples
145 is a term because 26*56 = 1456 = 145*10 + 6. - _Bernard Schott_, May 13 2019
Links
- Stefano Spezia, Table of n, a(n) for n = 1..10000
Programs
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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[10*a+6],AppendTo[a,n]]]]; a
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PARI
isok6(n) = (n%10) == 6; \\ A017341 isok(k) = {my(n=10*k+6, d=divisors(n)); fordiv(n, d, if (isok6(d) && isok6(n/d), return(1))); return (0);} \\ Michel Marcus, Apr 14 2019
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Python
def aupto(lim): return sorted(set(a*b//10 for a in range(6, 10*lim//6+2, 10) for b in range(a, 10*lim//a+2, 10) if a*b//10 <= lim)) print(aupto(249)) # Michael S. Branicky, Aug 21 2021
Formula
a(n) = (A324297(n) - 6)/10.
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 1.
The conjecture is true since a(n) = (A324297(n) - 6)/10 and lim_{n->infinity} A324297(n)/A324297(n-1) = 1. - Stefano Spezia, Aug 21 2021
Comments