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

1,1

Comments

All the terms of this sequence are odd.

Why? If an integer 10*k+6 = (10*a+6) * (10*b+6), then k = 10*a*b + 6*(a+b) + 3, so k is odd. - Bernard Schott, May 13 2019

Links

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

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

Example

145 is a term because 26*56 = 1456 = 145*10 + 6. - Bernard Schott, May 13 2019

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

Prog

(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
(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

Crossrefs

Cf. A017341, A053742 (ending with 5), A324297, A337856, A346389.

Sequence in context: A282031 A307202 A162488 * A194041 A029506 A299790

Adjacent sequences: A324295 A324296 A324297 * A324299 A324300 A324301

Keyword

nonn,base

Author

Stefano Spezia, Mar 16 2019

Status

approved