cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 10 results.

A100037 Positions of occurrences of the natural numbers as a second subsequence in A100035.

Original entry on oeis.org

4, 9, 18, 31, 48, 69, 94, 123, 156, 193, 234, 279, 328, 381, 438, 499, 564, 633, 706, 783, 864, 949, 1038, 1131, 1228, 1329, 1434, 1543, 1656, 1773, 1894, 2019, 2148, 2281, 2418, 2559, 2704, 2853, 3006, 3163, 3324, 3489, 3658, 3831, 4008, 4189, 4374, 4563
Offset: 1

Views

Author

Reinhard Zumkeller, Oct 31 2004

Keywords

Comments

For n > 1, A100035(a(n)) = n and A100035(m) != n for a(n-1) <= m < a(n);
A100036(n) < a(n) < A100038(n) < A100039(n).

Examples

			First terms (10 = A, 11 = B, 12 = C) of A100035(a(n)):
...1....2........3............4................5......
1231435425165764736271879869584938291A9BA8B7A6B5A4B3A2B;
a(1) = A084849(2) = 4, A100035(4) = 1;
a(2) = A014107(2) = 9, A100035(9) = 2;
a(3) = A033537(3) = 18, A100035(18) = 3;
a(4) = A100040(4) = 31, A100035(31) = 4;
a(5) = A100041(5) = 48, A100035(48) = 5.
		

Crossrefs

Formula

a(n) = 2*n^2 - n + 3 (conjectured). - Ralf Stephan, May 15 2007

A100038 Positions of occurrences of the natural numbers as third subsequence in A100035.

Original entry on oeis.org

11, 20, 33, 50, 71, 96, 125, 158, 195, 236, 281, 330, 383, 440, 501, 566, 635, 708, 785, 866, 951, 1040, 1133, 1230, 1331, 1436, 1545, 1658, 1775, 1896, 2021, 2150, 2283, 2420, 2561, 2706, 2855, 3008, 3165, 3326, 3491, 3660, 3833, 4010, 4191, 4376, 4565
Offset: 1

Views

Author

Reinhard Zumkeller, Oct 31 2004

Keywords

Comments

n>1: A100035(a(n))=n and A100035(m)<>n for a(n-1)<=m
A100036(n) < A100037(n) < a(n) < A100039(n).

Examples

			First terms (10=A,11=B,12=C) of A100035(a(n)):
..........1........2............3................4...
1231435425165764736271879869584938291A9BA8B7A6B5A4B3A2B1;
a(1) = A084849(3) = 11, A100035(11) = 1;
a(2) = A014107(3) = 20, A100035(20) = 2;
a(3) = A033537(4) = 33, A100035(33) = 3;
a(4) = A100040(5) = 50, A100035(50) = 4;
a(5) = A100041(6) = 71, A100035(71) = 5.
		

Crossrefs

Cf. A100037.

Formula

a(n) = 2*n^2 + 3*n + 6 (conjectured). - Ralf Stephan, May 15 2007

A100039 Positions of occurrences of the natural numbers as fourth subsequence in A100035.

Original entry on oeis.org

22, 35, 52, 73, 98, 127, 160, 197, 238, 283, 332, 385, 442, 503, 568, 637, 710, 787, 868, 953, 1042, 1135, 1232, 1333, 1438, 1547, 1660, 1777, 1898, 2023, 2152, 2285, 2422, 2563, 2708, 2857, 3010, 3167, 3328, 3493, 3662, 3835, 4012, 4193, 4378, 4567, 4760
Offset: 1

Author

Reinhard Zumkeller, Oct 31 2004

Keywords

Comments

n>1: A100035(a(n))=n and A100035(m)<>n for a(n-1)<=m
A100036(n) < A100037(n) < A100038(n) < a(n).

Examples

			First terms (10=A,11=B,12=C) of A100035(a(n)):
.....................1............2................3....
1231435425165764736271879869584938291A9BA8B7A6B5A4B3A2B1;
a(1) = A084849(4) = 22, A100035(22) = 1;
a(2) = A014107(4) = 35, A100035(35) = 2;
a(3) = A033537(5) = 52, A100035(52) = 3;
a(4) = A100040(6) = 73, A100035(73) = 4;
a(5) = A100041(7) = 98, A100035(98) = 5.
		

Formula

2n^2 + 7n + 13 (conjectured). - Ralf Stephan, May 15 2007

A100036 a(n) = smallest m such that A100035(m) = n.

Original entry on oeis.org

1, 2, 3, 5, 7, 12, 14, 23, 25, 38, 40, 57, 59, 80, 82, 107, 109, 138, 140, 173, 175, 212, 214, 255, 257, 302, 304, 353, 355, 408, 410, 467, 469, 530, 532, 597, 599, 668, 670, 743, 745, 822, 824, 905, 907, 992, 994, 1083, 1085, 1178, 1180, 1277, 1279, 1380
Offset: 1

Author

Reinhard Zumkeller, Oct 31 2004

Keywords

Comments

Smallest positions of occurrences of the natural numbers as subsequence in A100035;
A100035(a(n)) = n and A100035(m) <> n for m < a(n);
a(n) < A100037(n) < A100038(n) < A100039(n).

Examples

			First terms (10=A,11=B,12=C) of A100035(a(n)):
123.4.5....6.7........8.9............A.B................C.
1231435425165764736271879869584938291A9BA8B7A6B5A4B3A2B1CBD;
a(1) = A084849(1) = 1, A100035(1) = 1;
a(2) = A014107(1) = 2, A100035(2) = 2;
a(3) = A033537(1) = 3, A100035(3) = 3;
a(4) = A100040(1) = 5, A100035(5) = 4;
a(5) = A100041(1) = 7, A100035(7) = 5.
		

Formula

Conjecture: a(n) = partial sums of sequence [1,1,1,2,2,5,2,9,2,13,2,17,2,21,2,25,2,29,2,33,...2,n/2-7,2,...]. In other words, a(n) consists of the numbers 1,2,3 and the sequences A096376 and A096376+2 interspersed. - Ralf Stephan, May 15 2007

A014105 Second hexagonal numbers: a(n) = n*(2*n + 1).

Original entry on oeis.org

0, 3, 10, 21, 36, 55, 78, 105, 136, 171, 210, 253, 300, 351, 406, 465, 528, 595, 666, 741, 820, 903, 990, 1081, 1176, 1275, 1378, 1485, 1596, 1711, 1830, 1953, 2080, 2211, 2346, 2485, 2628, 2775, 2926, 3081, 3240, 3403, 3570, 3741, 3916, 4095, 4278
Offset: 0

Author

N. J. A. Sloane, Jun 14 1998

Keywords

Comments

Note that when starting from a(n)^2, equality holds between series of first n+1 and next n consecutive squares: a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2*n)^2; e.g., 10^2 + 11^2 + 12^2 = 13^2 + 14^2. - Henry Bottomley, Jan 22 2001; with typos fixed by Zak Seidov, Sep 10 2015
a(n) = sum of second set of n consecutive even numbers - sum of the first set of n consecutive odd numbers: a(1) = 4-1, a(3) = (8+10+12) - (1+3+5) = 21. - Amarnath Murthy, Nov 07 2002
Partial sums of odd numbers 3 mod 4, that is, 3, 3+7, 3+7+11, ... See A001107. - Jon Perry, Dec 18 2004
If Y is a fixed 3-subset of a (2n+1)-set X then a(n) is the number of (2n-1)-subsets of X intersecting Y. - Milan Janjic, Oct 28 2007
More generally (see the first comment), for n > 0, let b(n,k) = a(n) + k*(4*n + 1). Then b(n,k)^2 + (b(n,k) + 1)^2 + ... + (b(n,k) + n)^2 = (b(n,k) + n + 1 + 2*k)^2 + ... + (b(n,k) + 2*n + 2*k)^2 + k^2; e.g., if n = 3 and k = 2, then b(n,k) = 47 and 47^2 + ... + 50^2 = 55^2 + ... + 57^2 + 2^2. - Charlie Marion, Jan 01 2011
Sequence found by reading the line from 0, in the direction 0, 10, ..., and the line from 3, in the direction 3, 21, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Nov 09 2011
a(n) is the number of positions of a domino in a pyramidal board with base 2n+1. - César Eliud Lozada, Sep 26 2012
Differences of row sums of two consecutive rows of triangle A120070, i.e., first differences of A016061. - J. M. Bergot, Jun 14 2013 [In other words, the partial sums of this sequence give A016061. - Leo Tavares, Nov 23 2021]
a(n)*Pi is the total length of half circle spiral after n rotations. See illustration in links. - Kival Ngaokrajang, Nov 05 2013
For corresponding sums in first comment by Henry Bottomley, see A059255. - Zak Seidov, Sep 10 2015
a(n) also gives the dimension of the simple Lie algebras B_n (n >= 2) and C_n (n >= 3). - Wolfdieter Lang, Oct 21 2015
With T_(i+1,i)=a(i+1) and all other elements of the lower triangular matrix T zero, T is the infinitesimal generator for unsigned A130757, analogous to A132440 for the Pascal matrix. - Tom Copeland, Dec 13 2015
Partial sums of squares with alternating signs, ending in an even term: a(n) = 0^2 - 1^2 +- ... + (2*n)^2, cf. Example & Formula from Berselli, 2013. - M. F. Hasler, Jul 03 2018
Also numbers k with the property that in the symmetric representation of sigma(k) the smallest Dyck path has a central peak and the largest Dyck path has a central valley, n > 0. (Cf. A237593.) - Omar E. Pol, Aug 28 2018
a(n) is the area of a triangle with vertices at (0,0), (2*n+1, 2*n), and ((2*n+1)^2, 4*n^2). - Art Baker, Dec 12 2018
This sequence is the largest subsequence of A000217 such that gcd(a(n), 2*n) = a(n) mod (2*n) = n, n > 0 up to a given value of n. It is the interleave of A033585 (a(n) is even) and A033567 (a(n) is odd). - Torlach Rush, Sep 09 2019
A generalization of Hasler's Comment (Jul 03 2018) follows. Let P(k,n) be the n-th k-gonal number. Then for k > 1, partial sums of {P(k,n)} with alternating signs, ending in an even term, = n*((k-2)*n + 1). - Charlie Marion, Mar 02 2021
Let U_n(H) = {A in M_n(H): A*A^H = I_n} be the group of n X n unitary matrices over the quaternions (A^H is the conjugate transpose of A. Note that over the quaternions we still have A*A^H = I_n <=> A^H*A = I_n by mapping A and A^H to (2n) X (2n) complex matrices), then a(n) is the dimension of its Lie algebra u_n(H) = {A in M_n(H): A + A^H = 0} as a real vector space. A basis is given by {(E_{st}-E_{ts}), i*(E_{st}+E_{ts}), j*(E_{st}+E_{ts}), k*(E_{st}+E_{ts}): 1 <= s < t <= n} U {i*E_{tt}, j*E_{tt}, k*E_{tt}: t = 1..n}, where E_{st} is the matrix with all entries zero except that its (st)-entry is 1. - Jianing Song, Apr 05 2021

Examples

			For n=6, a(6) = 0^2 - 1^2 + 2^2 - 3^2 + 4^2 - 5^2 + 6^2 - 7^2 + 8^2 - 9^2 + 10^2 - 11^2 + 12^2 = 78. - _Bruno Berselli_, Aug 29 2013
		

References

  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, pp. 77-78. (In the integral formula on p. 77 a left bracket is missing for the cosine argument.)

Crossrefs

Second column of array A094416.
Equals A033586(n) divided by 4.
See Comments of A132124.
Second n-gonal numbers: A005449, A147875, A045944, A179986, A033954, A062728, A135705.
Row sums in triangle A253580.

Programs

Formula

a(n) = 3*Sum_{k=1..n} tan^2(k*Pi/(2*(n + 1))). - Ignacio Larrosa Cañestro, Apr 17 2001
a(n)^2 = n*(a(n) + 1 + a(n) + 2 + ... + a(n) + 2*n); e.g., 10^2 = 2*(11 + 12 + 13 + 14). - Charlie Marion, Jun 15 2003
From N. J. A. Sloane, Sep 13 2003: (Start)
G.f.: x*(3 + x)/(1 - x)^3.
E.g.f.: exp(x)*(3*x + 2*x^2).
a(n) = A000217(2*n) = A000384(-n). (End)
a(n) = A084849(n) - 1; A100035(a(n) + 1) = 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A126890(n, k) + A126890(n, n-k), 0 <= k <= n. - Reinhard Zumkeller, Dec 30 2006
a(2*n) = A033585(n); a(3*n) = A144314(n). - Reinhard Zumkeller, Sep 17 2008
a(n) = a(n-1) + 4*n - 1 (with a(0) = 0). - Vincenzo Librandi, Dec 24 2010
a(n) = Sum_{k=0.2*n} (-1)^k*k^2. - Bruno Berselli, Aug 29 2013
a(n) = A242342(2*n + 1). - Reinhard Zumkeller, May 11 2014
a(n) = Sum_{k=0..2} C(n-2+k, n-2) * C(n+2-k, n), for n > 1. - J. M. Bergot, Jun 14 2014
a(n) = floor(Sum_{j=(n^2 + 1)..((n+1)^2 - 1)} sqrt(j)). Fractional portion of each sum converges to 1/6 as n -> infinity. See A247112 for a similar summation sequence on j^(3/2) and references to other such sequences. - Richard R. Forberg, Dec 02 2014
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3, with a(0) = 0, a(1) = 3, and a(2) = 10. - Harvey P. Dale, Feb 10 2015
Sum_{n >= 1} 1/a(n) = 2*(1 - log(2)) = 0.61370563888010938... (A188859). - Vaclav Kotesovec, Apr 27 2016
From Wolfdieter Lang, Apr 27 2018: (Start)
a(n) = trinomial(2*n, 2) = trinomial(2*n, 2*(2*n-1)), for n >= 1, with the trinomial irregular triangle A027907; i.e., trinomial(n,k) = A027907(n,k).
a(n) = (1/Pi) * Integral_{x=0..2} (1/sqrt(4 - x^2)) * (x^2 - 1)^(2*n) * R(4*(n-1), x), for n >= 0, with the R polynomial coefficients given in A127672, and R(-m, x) = R(m, x). [See Comtet, p. 77, the integral formula for q = 3, n -> 2*n, k = 2, rewritten with x = 2*cos(phi).] (End)
a(n) = A002943(n)/2. - Ralf Steiner, Jul 23 2019
a(n) = A000290(n) + A002378(n). - Torlach Rush, Nov 02 2020
a(n) = A003215(n) - A000290(n+1). See Squared Hexagons illustration. Leo Tavares, Nov 23 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/2 + log(2) - 2. - Amiram Eldar, Nov 28 2021

Extensions

Link added and minor errors corrected by Johannes W. Meijer, Feb 04 2010

A084849 a(n) = 1 + n + 2*n^2.

Original entry on oeis.org

1, 4, 11, 22, 37, 56, 79, 106, 137, 172, 211, 254, 301, 352, 407, 466, 529, 596, 667, 742, 821, 904, 991, 1082, 1177, 1276, 1379, 1486, 1597, 1712, 1831, 1954, 2081, 2212, 2347, 2486, 2629, 2776, 2927, 3082, 3241, 3404, 3571, 3742, 3917, 4096, 4279, 4466
Offset: 0

Author

Paul Barry, Jun 09 2003

Keywords

Comments

Equals (1, 2, 3, ...) convolved with (1, 2, 4, 4, 4, ...). a(3) = 22 = (1, 2, 3, 4) dot (4, 4, 2, 1) = (4 + 8 + 6 + 4). - Gary W. Adamson, May 01 2009
a(n) is also the number of ways to place 2 nonattacking bishops on a 2 X (n+1) board. - Vaclav Kotesovec, Jan 29 2010
Partial sums are A174723. - Wesley Ivan Hurt, Apr 16 2016
Also the number of irredundant sets in the n-cocktail party graph. - Eric W. Weisstein, Aug 09 2017

Programs

Formula

a(n) = A058331(n) + A000027(n).
G.f.: (1 + x + 2*x^2)/(1 - x)^3.
a(n) = A014105(n) + 1; A100035(a(n)) = 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = ceiling((2*n + 1)^2/2) - n = A001844(n) - n. - Paul Barry, Jul 16 2006
From Gary W. Adamson, Oct 07 2007: (Start)
Row sums of triangle A131901.
(a(n): n >= 0) is the binomial transform of (1, 3, 4, 0, 0, 0, ...). (End)
Equals A134082 * [1,2,3,...]. -
a(n) = (1 + A000217(2*n-1) + A000217(2*n+1))/2. - Enrique Pérez Herrero, Apr 02 2010
a(n) = (A177342(n+1) - A177342(n))/2, with n > 0. - Bruno Berselli, May 19 2010
a(n) - 3*a(n-1) + 3*a(n-2) - a(n-3) = 0, with n > 2. - Bruno Berselli, May 24 2010
a(n) = 4*n + a(n-1) - 1 (with a(0) = 1). - Vincenzo Librandi, Aug 08 2010
With an offset of 1, the polynomial a(t-1) = 2*t^2 - 3*t + 2 is the Alexander polynomial (with negative powers cleared) of the 3-twist knot. The associated Seifert matrix S is [[-1,-1], [0,-2]]. a(n-1) = det(transpose(S) - n*S). Cf. A060884. - Peter Bala, Mar 14 2012
E.g.f.: (1 + 3*x + 2*x^2)*exp(x). - Ilya Gutkovskiy, Apr 16 2016

A014107 a(n) = n*(2*n-3).

Original entry on oeis.org

0, -1, 2, 9, 20, 35, 54, 77, 104, 135, 170, 209, 252, 299, 350, 405, 464, 527, 594, 665, 740, 819, 902, 989, 1080, 1175, 1274, 1377, 1484, 1595, 1710, 1829, 1952, 2079, 2210, 2345, 2484, 2627, 2774, 2925, 3080, 3239, 3402, 3569, 3740, 3915, 4094, 4277
Offset: 0

Keywords

Comments

Positive terms give a bisection of A000096. - Omar E. Pol, Dec 16 2016

Programs

Formula

a(n) = A100345(n, n - 3) for n > 2.
a(n) = A033537(n) - 8*n^2; A100035(a(n)) = 2 for n > 1. - Reinhard Zumkeller, Oct 31 2004
a(n) = A014106(-n) for all n in Z. - Michael Somos, Nov 06 2005
From Michael Somos, Nov 06 2005: (Start)
G.f.: x*(-1 + 5*x)/(1 - x)^3.
E.g.f: x*(-1 + 2*x)*exp(x). (End)
a(n) = A097070(n)/A000108(n - 2), n >= 2. - Philippe Deléham, Apr 12 2007
a(n) = 2*a(n-1) - a(n-2) + 4, n > 1; a(0) = 0, a(1) = -1, a(2) = 2. - Zerinvary Lajos, Feb 18 2008
a(n) = a(n-1) + 4*n - 5 with a(0) = 0. - Vincenzo Librandi, Nov 20 2010
a(n) = (2*n-1)*(n-1) - 1. Also, with an initial offset of -1, a(n) = (2*n-1)*(n+1) = 2*n^2 + n - 1. - Alonso del Arte, Dec 15 2012
(a(n) + 1)^2 + (a(n) + 2)^2 + ... + (a(n) + n)^2 = (a(n) + n + 1)^2 + (a(n) + n + 2)^2 + ... + (a(n) + 2n - 1)^2 starting with a(1) = -1. - Jeffreylee R. Snow, Sep 17 2013
a(n) = A014105(n-1) - 1 for all n in Z. - Michael Somos, Nov 23 2021
From Amiram Eldar, Feb 20 2022: (Start)
Sum_{n>=1} 1/a(n) = -2*(1 - log(2))/3.
Sum_{n>=1} (-1)^n/a(n) = Pi/6 + log(2)/3 + 2/3. (End)
For n > 0, A002378(a(n)) = A000384(n-1)*A000384(n). - Charlie Marion, May 21 2023

A033537 a(n) = n*(2*n+5).

Original entry on oeis.org

0, 7, 18, 33, 52, 75, 102, 133, 168, 207, 250, 297, 348, 403, 462, 525, 592, 663, 738, 817, 900, 987, 1078, 1173, 1272, 1375, 1482, 1593, 1708, 1827, 1950, 2077, 2208, 2343, 2482, 2625, 2772, 2923, 3078, 3237, 3400, 3567, 3738, 3913, 4092, 4275, 4462, 4653, 4848, 5047, 5250, 5457, 5668
Offset: 0

Keywords

Comments

Permutations avoiding 12-3 that contain the pattern 32-1 exactly once.
a(n) = A014107(n) + 8*n^2; A100035(a(n)) = 3 for n>1. - Reinhard Zumkeller, Oct 31 2004
If Y is a 3-subset of an (2n+1)-set X then, for n>=1, a(n-1) is the number of (2n-1)-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 16 2007

Crossrefs

Programs

Formula

a(n) = a(n-1) + 4*n + 3 (with a(0)=0). - Vincenzo Librandi, Nov 17 2010
From L. Edson Jeffery, Oct 14 2012: (Start)
G.f.: x*(7-3*x)/(1-x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n>=3, a(0)=0, a(1)=7, a(2)=18. (End)
E.g.f.: x*(7 + 2*x)*exp(x). - G. C. Greubel, Jul 15 2017
From Amiram Eldar, Feb 06 2022: (Start)
Sum_{n>=1} 1/a(n) = 46/75 - 2*log(2)/5.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/10 + log(2)/5 - 26/75. (End)

A100040 a(n) = 2*n^2 + n - 5.

Original entry on oeis.org

-5, -2, 5, 16, 31, 50, 73, 100, 131, 166, 205, 248, 295, 346, 401, 460, 523, 590, 661, 736, 815, 898, 985, 1076, 1171, 1270, 1373, 1480, 1591, 1706, 1825, 1948, 2075, 2206, 2341, 2480, 2623, 2770, 2921, 3076, 3235, 3398, 3565, 3736, 3911, 4090, 4273
Offset: 0

Author

Reinhard Zumkeller, Oct 31 2004

Keywords

Comments

a(n) is the result of taking five consecutive numbers starting at n-2, then adding the products of the first and the last and of the second with the fourth and finally adding the middle term. That is, a(n) = (n^2-4) + (n^2-1) + n. - J. M. Bergot, Mar 06 2018

Crossrefs

Programs

Formula

A100035(a(n)) = 4 for n>3;
a(n) = A014105(n) - 5 = A084849(n) - 6 = A100041(n) + 2.
a(n) = 2*a(n-1)-a(n-2)+4; a(0)=-5, a(1)=-2. - Vincenzo Librandi, Dec 26 2010
G.f.: (-5 + 13*x - 4*x^2)/(1 - x)^3. - Arkadiusz Wesolowski, Dec 25 2011
E.g.f.: (2*x^2 + 3*x - 5)*exp(x). - G. C. Greubel, Jul 15 2017

A100041 a(n) = 2*n^2 + n - 7.

Original entry on oeis.org

-7, -4, 3, 14, 29, 48, 71, 98, 129, 164, 203, 246, 293, 344, 399, 458, 521, 588, 659, 734, 813, 896, 983, 1074, 1169, 1268, 1371, 1478, 1589, 1704, 1823, 1946, 2073, 2204, 2339, 2478, 2621, 2768, 2919, 3074, 3233, 3396, 3563, 3734, 3909, 4088, 4271, 4458, 4649
Offset: 0

Author

Reinhard Zumkeller, Oct 31 2004

Keywords

Crossrefs

Programs

Formula

A100035(a(n)) = 5 for n>3.
a(n) = A014105(n) - 7 = A084849(n) - 8 = A100040(n) - 2.
From G. C. Greubel, Jul 15 2017: (Start)
G.f.: (7 - 17 x + 6 x^2)/(-1 + x)^3.
E.g.f.: (2*x^2 + 3*x - 7)*exp(x). (End)
Showing 1-10 of 10 results.