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A347254 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 4 (A347253). 1
1, 5, 9, 13, 17, 19, 21, 25, 29, 33, 37, 41, 45, 47, 49, 53, 57, 61, 65, 69, 73, 75, 77, 81, 85, 89, 93, 97, 101, 103, 105, 109, 113, 115, 117, 121, 125, 129, 131, 133, 137, 141, 145, 149, 153, 157, 159, 161, 165, 169, 173, 177, 181, 183, 185, 187, 189, 193, 197 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,2
COMMENTS
Since an integer 10*k + 6 = (10*a + 4)*(10*b + 4) implies that k = 10*a*b + 4*(a + b) + 1, all the terms of this sequence are odd.
LINKS
FORMULA
a(n) = (A347253(n) - 6)/10.
Lim_{n->infinity} a(n)/a(n-1) = 1.
EXAMPLE
13 is a term because 4*34 = 136 = 13*10 + 6.
MATHEMATICA
a={}; For[n=0, n<=200, n++, For[k=0, k<=n, k++, If[Mod[10*n+6, 10*k+4]==0 && Mod[(10*n+6)/(10*k+4), 10]==4 && n>Max[a], AppendTo[a, n]]]]; a
PROG
(Python)
def aupto(lim): return sorted(set(a*b//10 for a in range(4, 10*lim//4+3, 10) for b in range(a, 10*lim//a+3, 10) if a*b//10 <= lim))
print(aupto(197)) # Michael S. Branicky, Aug 24 2021
(PARI) isok(k) = my(x=10*k+6); sumdiv(x, d, (Mod(d, 10)==4) && Mod(x/d, 10)==4); \\ Michel Marcus, Oct 04 2021
CROSSREFS
Cf. A016873 (ending with 5), A017341, A324298 (ending with 6), A346951 (ending with 3), A347253.
Sequence in context: A051541 A188261 A190813 * A314665 A314666 A314667
KEYWORD
nonn,base
AUTHOR
Stefano Spezia, Aug 24 2021
STATUS
approved

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Last modified June 25 04:03 EDT 2024. Contains 373697 sequences. (Running on oeis4.)