A116103 Numbers k such that k concatenated with k-8 gives the product of two numbers which differ by 3.

58, 1552909498221642960203840517299022287877411846, 1723404686015903064905313664707396376746205762, 2153694542826970975147295558223788704098018912, 2872111611393408709716094936495178449746422388, 3420622776280099183231228659181978800586743708, 3671514956082179783197584591035293487267343366

Offset

1,1

Comments

A necessary condition for there to be terms where k-8 has d digits is that -23 is a quadratic residue mod 10^d + 1. The corresponding solutions are k = (y^2 + 3*y + 8)/(10^d+1) where y^2 + 3*y + 8 is divisible by 10^d + 1 and 10^(d-1) <= k - 8 < 10^d. The only d < 358 that work are 2, 46, and 178. - Robert Israel, Sep 30 2025

Links

Robert Israel, Table of n, a(n) for n = 1..24

John Brillhart et al., Cunningham Project [Factorizations of b^n +- 1, b = 2, 3, 5, 6, 7, 10, 11, 12 up to high powers]

Example

58//50 = 75 * 78, where // denotes concatenation.

Maple

Res:= NULL: count:= 0:
for d from 1 to 100 do
  F:= select(type, ifactors(10^d+1, easy)[2][.., 1], integer);
  if not andmap(t -> numtheory:-quadres(-23, t) = 1, F) then next fi;
  M:= map(t -> rhs(op(t)), [msolve(y^2 + 3*y + 8, 10^d + 1)]);
  for yy in M do
    kk:= (yy^2 + 3*yy + 8)/(10^d + 1);
    if kk >= 8 + 10^(d-1) and kk < 8 + 10^d then
       Res:= Res, kk; count:= count+1;
    fi
  od;
od:
sort([Res]); # Robert Israel, Sep 30 2025

Crossrefs

Cf. A116102, A116104, A116110, A116234.

Sequence in context: A243466 A201988 A116115 * A230852 A104380 A051321

Adjacent sequences: A116100 A116101 A116102 * A116104 A116105 A116106

Keyword

nonn,base

Author

Giovanni Resta, Feb 06 2006

Status

approved