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| A '''sum-product number''' is an integer that in a given base is equal to the sum of its digits times the product of its digits. Or, to put it algebraically, given an integer ''n'' that is ''l'' digits long in base ''b'' (with ''d''<sub>''x''</sub> representing the ''x''th digit), if
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| <math>n = (\sum_{i = 1}^l d_i)(\prod_{j = 1}^l d_j)</math>
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| then ''n'' is a sum-product number in base ''b''. In base 10, the only sum-product numbers are [[0 (number)|0]], [[1 (number)|1]], [[135 (number)|135]], [[144 (number)|144]] {{OEIS|id=A038369}}. Thus, for example, 144 is a sum-product number because 1 + 4 + 4 = 9, and 1 × 4 × 4 = 16, and 9 × 16 = 144.
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| 1 is a sum-product number in any base, because of the multiplicative identity. 0 is also a sum-product number in any base, but no other integer with significant zeroes in the given base can be a sum-product number. 0 and 1 are also unique in being the only single-digit sum-product numbers in any given base; for any other single-digit number, the sum of the digits times the product of the digits works out to the number itself squared.
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| Any integer shown to be a sum-product number in a given base must, by definition, also be a [[Harshad number]] in that base.
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| In [[binary numeral system|binary]], 0 and 1 are the only sum-product numbers. The following table lists some sum-product numbers in a few selected bases:
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| {| class="wikitable"
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| !Base
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| !Sum-product numbers
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| !Values in base 10
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| |-
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| |4
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| |0, 1, 12
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| |0, 1, 6
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| |-
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| |5
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| |0, 1, 341
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| |0, 1, 96
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| |-
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| |7
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| |0, 1, 22, 242, 1254, 2343
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| |0, 1, 16, 128, 480, 864
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| |-
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| |9
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| |0, 1, 13
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| |0, 1, 12
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| |-
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| |10
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| |0, 1, 135, 144
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| |0, 1, 135, 144
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| |-
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| |12
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| |0, 1, 128, 173, 353
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| |0, 1, 176, 231, 495
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| |-
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| |36
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| |0, 1, 16, 22O<!--Note this is O, not zero. This is base 36 so the digits run from zero through nine plus A through Z-->
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| |0, 1, 42, 2688
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| |}
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| The finiteness of the list for [[decimal|base 10]] was proven by David Wilson. First he proved that a base 10 sum-product number will not have more than 84 digits. Next, he ruled out numbers with significant zeroes. Thereafter he concentrated on digit products of the forms <math>2^i3^j7^k</math> or <math>3^i5^j7^k</math>, which the previous constraints reduce to a set small enough to be testable by brute force in a reasonable period of time.
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| From Wilson's proof, Raymond Puzio developed the proof that in any positional base system there is only a finite set of sum-product numbers. First he observed that any number ''n'' of length ''l'' must satisfy <math>n \ge b^{l - 1}</math>. Second, since the largest digit in the base represents ''b'' - 1, the maximum possible value of the sum of digits of ''n'' is <math>l(b - 1)</math> and the maximum possible value of the product of digits is <math>(b - 1)^l</math>. Multiplying the maximum possible sum by the maximum possible product gives <math>l(b - 1)^{l + 1}</math>, which is an upper bound of the value of any sum-product number of length ''l''. This suggests that <math>l(b - 1)^{l + 1} \ge n \ge b^{l - 1}</math>, or dividing both sides, <math>l(b - 1)^2 \ge (b / (b - 1))^{l - 1}</math>. Puzio then deduced that, because of the growth of exponential function, this inequality can only be true for values of ''l'' less than some limit, and thus that there can only be finitely many sum-product numbers ''n''.
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| In Roman numerals, the only sum-product numbers are 1, 2, 3, and possibly 4 (if written IIII).
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| ==External links==
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| * PlanetMath, [http://planetmath.org/?op=getobj&from=objects&id=7743 proof that number of sum-product numbers in any base is finite]
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| {{Classes of natural numbers}}
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| [[Category:Base-dependent integer sequences]]
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Oscar is how he's called and he completely loves this title. North Dakota is our birth location. One of the very very best things in the globe for me is to do aerobics and I've been doing it for quite a while. For years I've been working as a payroll clerk.
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