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A339578 Numbers k such that k^3 can be obtained by taking a number m that is the product of exactly three primes (cf. A014612) and deleting the central digit if m has an odd number of digits, or by deleting the central pair of digits if m has an even number of digits. 2
3, 4, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 47, 49, 51, 53, 57, 59, 61, 63, 67, 69, 71, 73, 77, 79, 81, 83, 87, 89, 91, 93, 97, 99, 217, 219, 221, 223, 227, 229, 231, 233, 237, 239, 241, 243, 247, 249, 251, 253, 257, 259, 261, 263, 267, 269, 271, 273, 277, 279, 281, 283, 287, 289 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

This is a kind of inverse to A217297.

From David A. Corneth, Dec 20 2020: (Start)

If a(n) > 100 then gcd(a(n), 10) = 1. Proof: Suppose gcd(a(n), 10)) > 1 then a(n) is divisible by 2 or 5. As the last three digits of m (as defined in name) will have the last three digits such that m is divisible by 2^3 = 8 or 5^3 = 125 we have a number that is a product of strictly more than three primes. Contradiction.

Furthermore, a(n)^3 has an even number of digits as after deletion of the middle digits there always is an even number of digits left. (End)

LINKS

David A. Corneth, Table of n, a(n) for n = 1..10000

EXAMPLE

3 is a term because 3^3 = 27 can be obtained by deleting the central digit of 3*3*23 = 207.

4 is a term because 4^3 = 64 can be obtained by deleting the central pair of digits of 2*31*97 = 6014.

20 is not a term because 20^3 = 8000, and any candidate for m will end in 00, and therefore will have at least four prime factors.

MATHEMATICA

p3oQ[x_]:=AnyTrue[Table[With[{td=TakeDrop[IntegerDigits[x], IntegerLength[ x]/2]}, FromDigits[ Flatten[ Join[ {td[[1]], {v}, td[[2]]}]]]], {v, 0, 9}], PrimeOmega[#] == 3&]; p3eQ[x_]:= AnyTrue[ Table[With[ {td=TakeDrop[ IntegerDigits[x], IntegerLength[ x]/2]}, FromDigits[Flatten[ Join[ {td[[1]], PadLeft[{w}, 2], td[[2]]}]]]], {w, 0, 99}], PrimeOmega[#]==3&]; p3Q[m_]:=p3oQ[m] || p3eQ[m]; Module[{rng={Ceiling[Reduce[k^3>#, k][[-1]]], Floor[Reduce[k^3<10#, k][[-1]]] }&/@ Table[10^(2n-1), {n, 4}]}, Surd[#, 3]&/@Select[Flatten[Table[w^3, {w, #[[1]], #[[2]]}]&/@ rng], p3Q]] // Quiet (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 20 2020 *)

PROG

(PARI) is(n) = { my(d = digits(n^3), half); if(#d % 2 == 1, return(0)); half = #d \ 2; left = vector(half, i, d[i]); right = vector(half, i, d[i + half]); for(j = 1, 2, c = fromdigits(left) * 10^(half + j) + fromdigits(right); for(i = 0, 10^j - 1, if(bigomega(c + i*10^half) == 3, print(c + i*10^half); return(1) ) ) ); 0 } \\ David A. Corneth, Dec 20 2020

CROSSREFS

Cf. A014612, A217297.

Sequence in context: A232862 A344460 A195589 * A244005 A228236 A344346

Adjacent sequences:  A339575 A339576 A339577 * A339579 A339580 A339581

KEYWORD

nonn,base

AUTHOR

Harvey P. Dale and N. J. A. Sloane, Dec 20 2020

STATUS

approved

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Last modified August 7 23:50 EDT 2022. Contains 355995 sequences. (Running on oeis4.)