A225567 Primes with nonzero digits such that sum of cubes of digits equal to square of sums.

1423, 2143, 2341, 4231, 12253, 21523, 22153, 22531, 23251, 25321, 32251, 35221, 36343, 36433, 43633, 52321, 64333, 114451, 144511, 224461, 244261, 246241, 365557, 415141, 424261, 426421, 446221, 446461, 451411, 462421, 466441, 541141, 555637, 556537, 556573

Offset

1,1

Comments

Largest term of this sequence is the 20-digit prime 99151111111111111111.

The Pagni article mentioned below has no bearing on this problem because it deals with the well-known identity sum_{i=1..n} i^3 = (sum_{i=1..n} i)^2. However, the article is interesting. - T. D. Noe, Jul 26 2013

This sequence has exactly 14068465 provable primes. This result required about one hour of Mathematica on fairly fast computer having 16 GB of memory. - T. D. Noe, Jul 30 2013

Links

T. D. Noe, Table of n, a(n) for n = 1..1201 (terms < 10^7)

David Pagni, 82.27 An interesting number fact, The Mathematical Gazette 82:494 (1998), pp. 271-273.

C. Rivera, PP&P Puzzle 158: Sum of Cubes equal to Square of Sum

Example

a(5) = 12253 since 1^3 + 2^3 + 2^3 + 5^3 + 3^3 = (1 + 2 + 2 + 5 + 3)^2.

Mathematica

(* let tz[[i]] be numbers computed in A227073 *) Select[tz, PrimeQ] (* T. D. Noe, Jul 30 2013 *)
pQ[n_]:=Module[{idn=IntegerDigits[n]}, FreeQ[idn, 0]&&Total[idn^3] == Total[ idn]^2]; Select[Prime[Range[50000]], pQ] (* Harvey P. Dale, Sep 17 2013 *)

Prog

(PARI)forprime(n=1, 10^7, v=digits(n); if(sum(i=1, length(v), v[i]^3)==sum(i=1, length(v), v[i])^2 & setsearch(Set(v), 0)!=1, print1(n", ")))

Crossrefs

Cf. A055012 (sum of cubes of digits), A118881 (square of sum of the digits).

Cf. A227072, A227073.

Sequence in context: A337671 A237960 A023937 * A216444 A083428 A376793

Adjacent sequences: A225564 A225565 A225566 * A225568 A225569 A225570

Keyword

nonn,base,fini,easy

Author

Balarka Sen, Jul 26 2013

Extensions

Corrected by T. D. Noe, Jul 26 2013

Status

approved