A227073 -id:A227073 - cp's OEIS frontend

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

Balarka Sen, Jul 26 2013

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

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

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

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

Cf. A227072, A227073.

(* 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 *)

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", ")))

Corrected by T. D. Noe, Jul 26 2013

A227072 Positive numbers with nondecreasing digits such that sum of cubes of the digits equals the square of the sum of the digits.

Original entry on oeis.org

1, 12, 22, 123, 333, 1224, 1234, 2244, 4444, 12235, 12345, 33336, 33346, 55555, 111225, 111445, 112455, 114555, 122346, 122446, 123456, 144466, 222226, 224466, 244557, 244666, 333357, 333666, 345567, 355567, 455667, 666666, 1122556, 1134457, 1145557, 1155666

Offset: 1

T. D. Noe, Jul 27 2013

Because the digits are nondecreasing, the search to 10^20 is fairly rapid.

			1234 is here because 1^3 + 2^3 + 3^3 + 4^3 = (1 + 2 + 3 + 4)^2 and its digits are nondecreasing..

T. D. Noe, Table of n, a(n) for n = 1..660

Cf. A225567 (primes in a related sequence), A227073.

(* complete sequence *) tx = {}; Do[d = {i1, i2, i3, i4, i5, i6, i7, i8, i9, i10, i11, i12, i13, i14, i15, i16, i17, i18, i19, i20}; If[Total[d^3] == Total[d]^2, n = FromDigits[d]; AppendTo[tx, n]], {i1, 0, 9}, {i2, i1, 9}, {i3, i2, 9}, {i4, i3, 9}, {i5, i4, 9}, {i6, i5, 9}, {i7, i6, 9}, {i8, i7, 9}, {i9, i8, 9}, {i10, i9, 9}, {i11, i10, 9}, {i12, i11, 9}, {i13, i12, 9}, {i14, i13, 9}, {i15, i14, 9}, {i16, i15, 9}, {i17, i16, 9}, {i18, i17, 9}, {i19, i18, 9}, {i20, i19, 9}]; tx = Rest[tx]
(* partial sequence *) nddQ[n_] := Module[{idn=IntegerDigits[n]}, Min[Differences[idn]] >= 0 && Total[idn^3] == Total[idn]^2]; Select[Range[2000000], nddQ] (* Harvey P. Dale, Sep 01 2013 *)

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

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