25

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25 is an integer, the smallest square that can be written as a sum of 2 (nonzero) squares (A134422). It is also the smallest composite number that is both a lucky number (A000959) and ludic number (A003309). Also, the smallest composite number, whose binary expansion is 11001, such that when interpreted as a polynomial in ring $GF(2)[X],(x^{4}+x^{3}+1)$ , is irreducible in that ring (A091214).

1 Membership in core sequences
2 Sequences pertaining to 25
3 Partitions of 25
4 Roots and powers of 25
5 Logarithms and 25th powers
6 Values for number theoretic functions with 25 as an argument
7 Factorization of some small integers in a quadratic integer ring adjoining the cubic root of 5
8 Factorization of 25 in some quadratic integer rings
9 Representation of 25 in various bases
10 See also

Membership in core sequences

Odd numbers	..., 19, 21, 23, 25, 27, 29, 31, ...	A005408
Semiprimes	..., 15, 21, 22, 25, 26, 33, 34, ...	A001358
Perfect squares	..., 4, 9, 16, 25, 36, 49, 64, 81, ...	A000290
Powers of primes	..., 17, 19, 23, 25, 27, 29, 31, ...	A000961
Composite numbers	..., 21, 22, 24, 25, 26, 27, 28, ...	A002808
Lucky numbers	..., 13, 15, 21, 25, 31, 33, 37, ...	A000959

In Pascal's triangle, 25 occurs twice.

Sequences pertaining to 25

Multiples of 25	0, 25, 50, 75, 100, 125, 150, 175, 200, 225, 250, 275, ...	A008607
$3x+1$ sequence beginning at 33	33, 100, 50, 25, 76, 38, 19, 58, 29, 88, 44, 22, 11, 34, ...	A008880
$3x-1$ sequence beginning at 17	17, 50, 25, 74, 37, 110, 55, 164, 82, 41, 122, 61, 182, 91, ...	A003124

Partitions of 25

There are 1958 partitions of 25.

The Goldbach represenations of 25 using distinct primes are: 2 + 23 = 3 + 5 + 17 = 5 + 7 + 13 = 25.

Roots and powers of 25

In the table below, irrational numbers are given truncated to eight decimal places.

${\sqrt {25}}$	5.00000000		25 2	625
${\sqrt[{3}]{25}}$	2.92401773	A010597	25 3	15625
${\sqrt[{4}]{25}}$	2.23606797	A002163	25 4	390625
${\sqrt[{5}]{25}}$	1.90365393	A011110	25 5	9765625
${\sqrt[{6}]{25}}$	1.70997594	A005481	25 6	244140625
${\sqrt[{7}]{25}}$	1.58381960		25 7	6103515625
${\sqrt[{8}]{25}}$	1.49534878	A011003	25 8	152587890625
${\sqrt[{9}]{25}}$	1.42996914		25 9	3814697265625
${\sqrt[{10}]{25}}$	1.37972966	A005534	25 5	95367431640625
				A009969

Logarithms and 25th powers

In the OEIS specifically and mathematics in general, $\log x$ refers to the natural logarithm of $x$ , whereas all other bases are specified with a subscript.

As above, irrational numbers in the following table are truncated to eight decimal places.

$\log _{25}2$	0.21533827		$\log _{2}25$	4.64385618		2 25	33554432
$\log _{25}e$	0.62133493		$\log 25$	1.60943791	A016628	$e^{25}$	72004899337.38587252
$\log _{25}3$	0.34130309		$\log _{3}25$	2.92994704	A228375	3 25	847288609443
$\log _{25}\pi$			$\log _{\pi }25$			$\pi ^{25}$
$\log _{25}4$	0.43067655	A152675	$\log _{4}25$	2.32192809	A020858	4 25	1125899906842624
$\log _{25}5$	0.50000000	A020761	$\log _{5}25$	2.00000000	A000038	5 25	298023223876953125
$\log _{25}6$	0.55664137		$\log _{6}25$	1.79648880		6 25	28430288029929701376
$\log _{25}7$	0.60453097		$\log _{7}25$	1.65417495		7 25	1341068619663964900807
$\log _{25}8$	0.64601483		$\log _{8}25$	1.54795206		8 25	37778931862957161709568
$\log _{25}9$	0.68260619	A152914	$\log _{9}25$	1.46497352	A113209	9 25	717897987691852588770249
$\log _{25}10$	0.71533827		$\log _{10}25$	1.39794000		10 25	10000000000000000000000000

(See A010813 for the 25th powers of integers).

Values for number theoretic functions with 25 as an argument

$\mu (25)$	0
$M(25)$	–2
$\pi (25)$	9
$\sigma _{1}(25)$	31
$\sigma _{0}(25)$	3
$\phi (25)$	20
$\Omega (25)$	2
$\omega (25)$	1
$\lambda (25)$	20	This is the Carmichael lambda function.
$\lambda (25)$	1	This is the Liouville lambda function.
$\zeta (25)$
25!	15511210043330985984000000
$\Gamma (25)$	620448401733239439360000

Factorization of some small integers in a quadratic integer ring adjoining the cubic root of 5

Since 25 is a perfect square, adjoining its square root to the field of rational numbers does not generate a field extension.

However, adjoining the cubic root of 5 to $\mathbb {Q}$ does generate a field extension, one in which the algebraic integers are of the form $a+b{\sqrt[{3}]{5}}+c{\sqrt[{3}]{25}}$ , with $a,b,c$ being integers from $\mathbb {Z}$ . The norm function is then $a^{3}+5b^{3}+25c^{3}-15abc$ .

These algebraic integers form a ring that is a unique factorization domain. The following table shows a few small primes factorized in this ring.

TABLE GOES HERE

Factorization of 25 in some quadratic integer rings

In $\mathbb {Z}$ , 25 is the square of 5. But it has different factorizations in some quadratic integer rings, and in a few cases it's not a simple matter of taking the factorization of 5 and adding in some exponent 2s.

$\mathbb {Z} [i]$	$(2-i)^{2}(2+i)^{2}$
$\mathbb {Z} [{\sqrt {-2}}]$	5 2	$\mathbb {Z} [{\sqrt {2}}]$	5 2
$\mathbb {Z} [\omega ]$	5 2	$\mathbb {Z} [{\sqrt {3}}]$	5 2
$\mathbb {Z} [{\sqrt {-5}}]$	$({\sqrt {-5}})^{4}$	$\mathbb {Z} [\phi ]$	$(-1+2\phi )^{4}$
$\mathbb {Z} [{\sqrt {-6}}]$	5 2 OR $(1-2{\sqrt {-6}})(1+2{\sqrt {-6}})$	$\mathbb {Z} [{\sqrt {6}}]$	$(1-{\sqrt {6}})^{2}(1+{\sqrt {6}})^{2}$
${\mathcal {O}}_{\mathbb {Q} ({\sqrt {-7}})}$	5 2	$\mathbb {Z} [{\sqrt {7}}]$	5 2
$\mathbb {Z} [{\sqrt {-10}}]$	5 2	$\mathbb {Z} [{\sqrt {10}}]$	5 2 OR $(-1)(15-5{\sqrt {10}})(15+5{\sqrt {10}})$
${\mathcal {O}}_{\mathbb {Q} ({\sqrt {-11}})}$	$\left({\frac {3}{2}}-{\frac {\sqrt {-11}}{2}}\right)^{2}\left({\frac {3}{2}}+{\frac {\sqrt {-11}}{2}}\right)^{2}$	$\mathbb {Z} [{\sqrt {11}}]$	$(4-{\sqrt {11}})^{2}(4+{\sqrt {11}})^{2}$
$\mathbb {Z} [{\sqrt {-13}}]$	5 2	${\mathcal {O}}_{\mathbb {Q} ({\sqrt {13}})}$	5 2
$\mathbb {Z} [{\sqrt {-14}}]$		$\mathbb {Z} [{\sqrt {14}}]$	$(3-{\sqrt {14}})^{2}(3+{\sqrt {14}})^{2}$
${\mathcal {O}}_{\mathbb {Q} ({\sqrt {-15}})}$		$\mathbb {Z} [{\sqrt {15}}]$	5 2 OR $(20-5{\sqrt {15}})(20+5{\sqrt {15}})$
$\mathbb {Z} [{\sqrt {-17}}]$		${\mathcal {O}}_{\mathbb {Q} ({\sqrt {17}})}$	5 2
${\mathcal {O}}_{\mathbb {Q} ({\sqrt {-19}})}$	$\left({\frac {1}{2}}-{\frac {\sqrt {-19}}{2}}\right)^{2}\left({\frac {1}{2}}+{\frac {\sqrt {-19}}{2}}\right)^{2}$	$\mathbb {Z} [{\sqrt {19}}]$	$(9-2{\sqrt {19}})^{2}(9+2{\sqrt {19}})^{2}$ .