Problems & Puzzles: Puzzles

Puzzle 53.- Sequences of consecutive economical numbers.

A number E is called (after B. Recam) an economical number, if the expression of E as a product of powered primes uses fewer digits than the digits of E.

The first 8 economical numbers are:

125, 128, 243, 256, 343, 512, 625, 729, …. (See Neils sequence A046759)

125 =5^3, 128 =2^7, etceteras.

Do exist consecutive economical numbers? The answer is yes. Here we are interested only in the earliest sequences of K=>2 consecutive economical numbers. I have found the earliest sequences for K = 2, 3 and 4:

K=2
4374 = 2* 3^ 7
4375 = 5^ 4* 7

K=3
1097873 = 7* 47^ 2* 71
1097874 = 2* 3^ 7* 251
1097875 = 5^ 3* 8783

179210312 = 2^ 3* 4733^ 2
179210313 = 3^ 5* 97* 7603
179210314 = 2* 29* 37^ 3* 61
179210315 = 5* 79^ 2* 5743

Q. Find the earliest sequence of 5<=K<=10 consecutive economical numbers.

Around ten years later, J. K. Andersen sent more solutions to this old -but not dead- and beloved puzzle of mine:

The earliest for K=5:
10546514090622 = 2*3^6*7233548759
10546514090623 = 7*67^2*335630401
10546514090624 = 2^7*82394641333
10546514090625 = 3*5^5*6079*185057
10546514090626 = 2*73^2*989539697

The earliest for K>5 looks hard. It is easier to find large solutions.
In http://emis.impa.br/EMIS/journals/INTEGERS/papers/a5int2003/a5int2003.pdf
by Jean-Marie De Koninck and Florian Luca there are 10 numbers starting at
1187615078125922863258960810793892104104920690716348822

I found 15 starting at 1105970172560682507458205276016\
3613095334199423826725902703901860203178192999295637491

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