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

Author

Balarka Sen, Jul 26 2013

Keywords

• nonn
• base
• fini
• easy

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

Examples

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

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

Crossrefs

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

Cf. A227072, A227073.

Programs

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 *)

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", ")))

Extensions

Corrected by T. D. Noe, Jul 26 2013